Organic Chemistry I

🧭 Overview

🧠 One-sentence thesis

Chemical bonding occurs through either electron transfer (ionic) or electron sharing (covalent), with covalent bond polarity determined by electronegativity differences between atoms.

📌 Key points (3–5)

• Two bonding types: ionic bonds form through electron transfer; covalent bonds form through electron pair sharing, with both atoms achieving filled outer shells.
• Polar vs non-polar covalent bonds: identical atoms share electrons equally (non-polar); different atoms share unequally based on electronegativity differences (polar).
• Electronegativity trend: EN increases left to right across periods and decreases top to bottom down groups; ΔEN (difference) measures bond polarity.
• Common confusion: C-H bonds are technically non-polar because carbon (2.5) and hydrogen (2.1) have close EN values, even though they are different atoms.
• Lewis structures: represent bonding electrons as lines and non-bonding valence electrons as dots, showing how atoms share electrons to achieve stable configurations.

⚛️ Ionic vs covalent bonding

⚛️ Ionic bonds through electron transfer

• Atoms transfer electrons completely from one to another.
• The atom losing electrons becomes a positive ion; the atom gaining electrons becomes a negative ion.
• Examples from the excerpt:
  • Calcium atom → Ca²⁺ + 2 electrons (loses 2)
  • Chlorine atom + 1 electron → Cl⁻ (gains 1)
  • Oxygen atom + 2 electrons → O²⁻ (gains 2)

🔗 Covalent bonds through electron sharing

A covalent bond is a bond formed through the sharing of electron pairs between the two bonding atoms.

• Shared electron pairs are attracted by both nuclei simultaneously.
• Both atoms gain a filled outer shell (octet) through sharing.
• Almost all bonds in organic compounds are covalent bonds.
• Why sharing works: mutual attraction to both nuclei holds the atoms together while satisfying both atoms' electron needs.

🎯 Bond polarity and electronegativity

🎯 Non-polar covalent bonds

• Formed between two identical atoms (homonuclear molecules).
• Electron pairs are shared equally between the two nuclei.
• Electron density is distributed evenly through the bond.
• ΔEN equals zero.
• Examples: H-H, Cl-Cl, O=O, N≡N (all bonds between the same element).

⚡ Polar covalent bonds

• Formed between two different atoms (heteronuclear bonds).
• Electron pairs are not shared evenly.
• The atom with stronger electron-pulling ability attracts the shared pairs more.
• ΔEN is not zero; greater ΔEN means more polar bond.

📏 Electronegativity (EN) concept

Electronegativity: the ability of an atom to pull electron pairs towards itself in a bond.

• Measured using the Pauling scale.
• Trend (Main Group elements):
  • Increases left to right across periods
  • Decreases top to bottom down groups
• Key values to know:
  • F: 4.0 (highest)
  • O: 3.5
  • N: 3.0
  • C: 2.5
  • H: 2.1
• Important note: knowing the trend is more important than memorizing exact values.

🔍 Special case: C-H bonds

• Carbon EN = 2.5, Hydrogen EN = 2.1
• The values are close, making C-H bonds technically non-polar.
• Don't confuse: even though C and H are different atoms, the small ΔEN means the bond behaves as non-polar.
• This matters because C-H bonds are involved in all organic compounds.

🎨 Representing bond polarity

🎨 Partial charges

• The atom with higher EN bears a slightly negative charge (δ⁻).
• The atom with lower EN bears a slightly positive charge (δ⁺).
• Example: in H-Cl, chlorine has higher EN, so Cl is δ⁻ and H is δ⁺.

➡️ Polarity arrow notation

• Arrow head points to the negative end (higher EN atom).
• Short perpendicular line near the tail marks the positive end (lower EN atom).
• Example: H-Cl molecule shows the arrow pointing from H toward Cl.

📐 Lewis structures

📐 What Lewis structures show

A Lewis structure shows the bonding between atoms as short lines (some books use pairs of dots) and non-bonding valence electrons as dots.

• Bonding electrons: represented as short lines (or pairs of dots).
• Non-bonding valence electrons: represented as dots.
• Purpose: visualize how atoms share electrons to achieve stable configurations.

🔤 Lewis symbols

• The chemical symbol of an element with valence electrons shown as dots.
• Building block for constructing Lewis structures.
• Examples mentioned: aluminum, tin, nitrogen, chlorine, bromine.

🔬 Diatomic molecule examples

Hydrogen (H₂)

• Each H needs only two electrons (called a duet, not an octet).
• Two H atoms share one pair of electrons.

Fluorine (F₂)

• Each F atom shares one pair of electrons with the other.
• Both atoms achieve filled outer shells through sharing.

🛠️ How to build simple Lewis structures

• Start with the Lewis symbol of each element.
• Combine symbols by pairing up unpaired electrons to form shared pairs (bonds).
• Remaining unpaired electrons stay as non-bonding dots.
• Check that each atom achieves a stable configuration (duet for H, octet for most others).

🧭 Overview

🧠 One-sentence thesis

Lewis structures systematically represent bonding and non-bonding electrons in molecules and ions, with formal charge guidelines and exceptions to the octet rule determining the most plausible structures.

📌 Key points (3–5)

• What Lewis structures show: bonding between atoms (as lines) and non-bonding valence electrons (as dots), built from Lewis symbols of individual elements.
• Systematic drawing procedure: for polyatomic species, follow a step-by-step method—count total valence electrons, build skeleton, complete terminal octets first, then form multiple bonds if needed, and check formal charges.
• Formal charges guide plausibility: smaller absolute values are better; negative formal charges appear on high-EN atoms, positive on low-EN atoms; structures with same-sign charges on adjacent atoms are unlikely.
• Common confusion—Kekulé vs Lewis: Kekulé structures omit lone pairs (except when highlighting special properties), while complete Lewis structures show all electrons; remember N has 1 lone pair, O has 2, halogens have 3.
• Exceptions to the octet rule: odd-electron radicals, incomplete octets (e.g., BH₃, carbocations), and expanded valence shells (Period 3+ elements can exceed 8 electrons using d orbitals).

🔤 Fundamental concepts

🔤 Lewis symbol

The Lewis symbol is the chemical symbol of an element with valence electrons represented as dots.

• Shows only the valence electrons of a free atom.
• Example: the excerpt shows symbols for aluminum, tin, nitrogen, chlorine, and bromine with their valence electrons as dots.

🔗 Lewis structure

A Lewis structure shows the bonding between atoms as short lines (some books use pairs of dots) and non-bonding valence electrons as dots.

• Bonding pairs: pairs of valence electrons involved in a covalent bond; drawn as short lines; one line = one pair = 2 electrons.
• Lone pairs: pairs of valence electrons not involved in a covalent bond; also called non-bonding electron pairs.
• Special case: unpaired (single) electrons exist in radicals (free radicals).

🧪 Simple diatomic molecules

For simple diatomic molecules, combine the Lewis symbols of each element:

• H₂: each H needs only two electrons (a duet, not an octet).
• F₂: each F shares one pair; each F has three lone pairs.
• HCl: H and Cl share one pair; Cl has three lone pairs.
• O₂: the two O atoms share two pairs (a double bond); each O has two lone pairs.

Don't confuse: bonding pairs (shared, shown as lines) vs lone pairs (not shared, shown as dots).

📝 Drawing procedure for polyatomic species

📝 Step-by-step method

The excerpt emphasizes: "It is very important that you use the following procedure to get the correct Lewis structures for polyatomic molecules and ions."

1. Count total valence electrons: for ions, adjust for charge (subtract for cations, add for anions).
   • Example: NH₄⁺ has 5 (N) + 4×1 (H) − 1 (charge) = 8 valence electrons.
2. Write skeletal structure:
   • Hydrogen is always terminal.
   • Central atoms are generally those with lowest EN; carbon is always central.
   • Connect central atom to each terminal atom with a single bond.
3. Subtract bonding electrons: for each single bond, subtract 2 from the total.
4. Complete octets: use remaining electrons to complete terminal atoms first, then central atoms as much as possible.
5. Check completion: if all atoms have octets, done.
6. Form multiple bonds if needed: move lone pairs from terminal atoms to central atom to complete central octets.
7. Calculate formal charges: label non-zero formal charges.

🧮 Formal charge calculation

Formal Charge on an atom = No. of valence electrons in free atoms − No. of lone pair electrons − ½ (No. of bonding electrons)

• Alternative (derived) formula: Formal Charge = No. of valence electrons in free atoms − No. of lone pair electrons − No. of covalent bonds around the atom.
• Double bonds count as 2, triple bonds as 3 in the derived formula.
• The excerpt notes Formula 1.2 (derived) is "easier to use and can be regarded as the most practical one based on experience."

🔬 CO₂ example walkthrough

1. Total valence electrons: 4 (C) + 2×6 (O) = 16.
2. Skeletal structure: O—C—O (carbon is always central).
3. Four electrons used; 12 remain.
4. Complete terminal O octets first: uses all 12 remaining electrons; central C does not have octet yet.
5. Not done; proceed to next step.
6. Move one lone pair from each terminal O to form double bonds: O=C=O.
7. Formal charges: FC(C) = 4 − ½×(4×2) = 0; FC(O) = 6 − 4 − ½×(2×2) = 0; all zero, so structure is correct.

Key rule: "the remaining electrons should be used to give the octet of terminal atoms first!"

Key rule: "the lone-pairs can only be moved from terminal atoms to the central atom to form multiple bonds, not the other way around."

⚖️ Formal charge guidelines

⚖️ Purpose and meaning

The purpose of formal charges is to compare the difference between the number of valence electrons in the free atom and the number of electrons the atom "owns" when it is bonded.

• Smaller difference → "happier" (more stable) atom.
• The atom owns all lone pair electrons and half of the bonding electrons.

⚖️ Rules for plausibility

Rule	Explanation
Sum equals total charge	The sum of all formal charges must equal the molecule or ion's total charge
Minimize absolute values	Formal charges should be as small as possible
Negative on high-EN atoms	"−" FC appears on most electronegative atoms (they "win" electrons in sharing)
Positive on low-EN atoms	"+" FC appears on least electronegative atoms (they "lose" electrons in sharing)
Avoid same-sign adjacency	Structures with same-sign formal charges on adjacent atoms are unlikely

⚖️ Example—why one CO₂ structure is better

The excerpt asks: "Why is the following structure not the best way to show the Lewis structure of CO₂?"

• The better structure (O=C=O) has all formal charges equal to zero.
• A structure with non-zero formal charges would be less plausible even if it satisfies the octet rule.

🚫 Special case—fluorine

Key rule: "F is the atom with the highest electronegativity; therefore, F atom NEVER has the '+' formal charge in any plausible Lewis structure!"

🎨 Kekulé structures

🎨 What they omit

Organic species are usually shown as Kekulé structures with all the lone pair electrons completely omitted (with exception to the lone pairs shown to highlight special properties).

• Complete Lewis structures show all bonding and lone pair electrons.
• Kekulé structures show only bonds (lines); lone pairs are understood but not drawn.
• Examples given: ethanol, acetic acid, ethyl amine, ethyl bromide.

🎨 Counting invisible lone pairs

"To count how many lone pairs should be involved on a certain atom, apply the octet rule."

Atom	Typical lone pairs
N	1 lone pair
O	2 lone pairs
Halogens	3 lone pairs

• All atoms (except H) should have 8 electrons around them.
• Don't confuse: just because lone pairs aren't drawn in Kekulé structures doesn't mean they aren't there.

🔓 Exceptions to the octet rule

🔓 Odd number of electrons

If total valence electrons is odd, the octet rule cannot be applied to all atoms.

The neutral species that contain an unpaired electron are called radicals (or free radicals).

• NO molecule: two possible structures; formal charge guideline selects the one with zero formal charges as better.
• NO₂ molecule: contains an unpaired electron.
• Alkyl radicals: e.g., •CH₃ with 7 total valence electrons; the carbon has an unpaired electron.

Example: •CH₃ structure shows three C–H bonds and one unpaired electron on carbon.

🔓 Incomplete octet

An incomplete octet means the atom has less than 8 electrons involved.

This occurs when:

• Total valence electrons < 8, or
• Formal charge concerns favor incomplete octet.

BH₃ molecule: total valence electrons = 6; central boron does not get an octet.

BF₃ molecule: even though all atoms could get octets, the actual structure keeps boron with an incomplete octet because formal charge guidelines favor it. Similar examples: BeF₂, AlCl₃.

CH₃⁺ (carbocation): a reactive intermediate in organic reactions; formal charge calculations show the "+" charge lies on C; carbon has an incomplete octet.

🔓 Expanded valence shell

For elements in Period 3 or higher, they can have more than 8 electrons if it helps to lower the formal charges.

• Common examples: central atoms P, S, Cl, etc.
• Sometimes multiple double bonds are necessary to minimize formal charge.
• Example: phosphate anion (PO₄³⁻) structure shows expanded valence on P.

Why this works: "Elements in Period 3 (or higher) have 3 (or more than 3) principal shells, so the d orbital is available in the valence shell. That is why they can accommodate more than 8 electrons."

🔓 Key limitation

Key Takeaway: "For elements in 2nd period, C, N, O, F and Ne, the maximum number of electrons involved in Lewis structure is eight!!!"

• Second-period elements cannot expand their valence shell.
• Don't confuse: Period 3+ elements can exceed 8 electrons; Period 2 elements cannot.

🔄 Resonance structures (preview)

🔄 What resonance means

In cases in which more than one reasonable (plausible) Lewis structure can be drawn for a species, these structures are called resonance structures or resonance contributors.

• Resonance structures can be equivalent or non-equivalent.
• The actual structure is a hybrid of all resonance contributors.

🔄 Equivalent resonance—carbonate anion

CO₃²⁻ example: following Step 6, the double bond can be built between central C and any of the three terminal O atoms, generating three structures.

• They are not identical; they are equivalent.
• Connected by double-headed resonance arrows.
• All three are at the same energy level and have the same stability, so they make equal contributions to the actual structure.
• Experimental evidence: all carbon-oxygen bonds in CO₃²⁻ are the same length, longer than a regular double bond.

Don't confuse: the resonance arrow (double-headed) is different from reaction arrows; "Resonance structures have to be connected using resonance arrows."

🧭 Overview

🧠 One-sentence thesis

Resonance structures show that electrons and charges can be distributed across multiple atoms in a molecule, stabilizing the species through charge delocalization, with more resonance contributors leading to greater stability.

📌 Key points (3–5)

• Charge delocalization: negative or positive charges spread evenly among atoms rather than being localized on one atom, which stabilizes the molecule.
• Resonance stabilization effect: the stability gained from having multiple resonance contributors; more contributors = more stable species.
• Equivalent vs non-equivalent structures: equivalent resonance structures have the same energy and bonding patterns; non-equivalent structures differ in energy and stability.
• Common confusion: the actual structure is not any single resonance structure but a hybrid of all contributors; conventional Lewis structures cannot show partial charges.
• Stability guidelines for non-equivalent structures: complete octets, smaller formal charges, negative charges on more electronegative atoms, and less charge separation all increase stability.

🔄 Charge delocalization and resonance stabilization

⚡ What charge delocalization means

Charge delocalization: negative or positive charges are not localized on any single atom but are spread evenly among multiple atoms.

• In the carbonate anion (CO₃²⁻), the two negative charges are not stuck on one or two oxygen atoms; instead, they are distributed equally across all three oxygen atoms.
• Each oxygen atom carries two-thirds of a full negative charge.
• This spreading of charge stabilizes the molecule.

🛡️ Resonance stabilization effect

Resonance stabilization effect: the stability a species gains from having charge delocalization through resonance contributors.

• The more resonance structures (contributors) a species has, the greater the resonance stabilization effect.
• Greater resonance stabilization = more stable species.
• Example: a molecule with three equivalent resonance structures is more stable than one with only two.

🖼️ Representing the actual structure

🎨 The hybrid structure

• The actual structure of a molecule with resonance is a hybrid (combination) of all resonance contributors.
• It cannot be accurately shown with a single conventional Lewis structure because Lewis structures do not include partial charges.
• Don't confuse: the actual structure is not one resonance structure "flipping" to another; it is all of them at once.

🖍️ Dashed lines and partial charges

• To approximate the hybrid, dashed lines can be drawn to show bonds that are between single and double bonds.
• Partial charges are shown on atoms (e.g., each oxygen in CO₃²⁻ has a partial negative charge).
• Example: in CO₃²⁻, the carbon-oxygen bond is somewhere between a single and double bond, not purely one or the other.

🌈 Electrostatic potential maps

• An electrostatic potential map uses colors to show charge distribution.
• Red regions = higher negative charge; blue regions = more positive charge.
• In the carbonate anion map, all three oxygen atoms show the same shade of red, indicating equal charge distribution.

⚖️ Equivalent vs non-equivalent resonance structures

🟰 Equivalent resonance structures

• Equivalent resonance structures have the same bonding patterns and charge distributions.
• They are at the same energy level.
• Example: the three resonance structures of CO₃²⁻ are equivalent because each has the same arrangement of bonds and charges, just rotated.

🔀 Non-equivalent resonance structures

• Non-equivalent resonance structures have different bonding and charge distributions.
• They exist at different energy levels, so some are more stable (lower energy) than others.
• Example: the three resonance structures of OCN⁻ are non-equivalent because the placement of multiple bonds and formal charges differs.

📏 Stability guidelines for non-equivalent structures

📐 Four key rules

The following guidelines help predict which non-equivalent resonance structure is most stable (lower energy = more stable):

Guideline	What it means	Example
Complete octets	Structures with complete octets are usually more stable (except known exceptions)	A structure where all atoms have eight electrons is preferred
Smaller formal charges	Lower absolute values of formal charges increase stability	A structure with -1 charge is more stable than one with -2 and +1
Charge placement	Negative charges on more electronegative atoms; positive charges on less electronegative atoms	Negative charge on oxygen (more EN) is better than on carbon
Less charge separation	Structures with charges closer together or fewer separated charges are more stable	A structure with one -1 charge is better than one with -2 and +1 far apart

🧪 Applying the rules

• For OCN⁻, structure 3 is least stable because it has the highest formal charges.
• Structures 1 and 2 both have a formal charge of -1.
• Structure 2 is most stable because the negative charge is on oxygen (more electronegative) rather than nitrogen.
• Don't confuse: "most stable" means lowest energy, not "most common to draw first."

🧭 Overview

🧠 One-sentence thesis

The resonance effect stabilizes organic molecules through multiple valid electron arrangements, and drawing resonance structures correctly—by moving only π electrons and lone pairs—allows chemists to predict which structures contribute most to a molecule's actual form.

📌 Key points (3–5)

• Resonance stabilization effect: more resonance contributors → greater stabilization → more stable species.
• What moves, what doesn't: only π electrons and lone pair electrons can move; σ bonds and atoms NEVER move.
• Three allowed electron transformations: π bond → π bond, π bond → lone pair, or lone pair → π bond.
• Common confusion: resonance structures differ only in electron arrangement, not atom positions; atoms stay fixed while electrons relocate.
• Stability ranking: structures with complete octets, smaller formal charges, and negative charges on more electronegative atoms are more stable.

🎯 Core rules for drawing resonance structures

🎯 What must stay the same

All resonance structures must maintain:

• Valid Lewis structure format (all Lewis rules still apply)
• Same atom connectivity (atoms never move)
• Same total number of electrons and same net charge

Key principle: Atoms NEVER move; only electrons move.

⚡ What can move

Only two types of electrons are allowed to move:

• π electrons (from multiple bonds)
• Lone pair electrons

Never move σ bonds!

Movement direction: from higher electron density areas to lower electron density areas.

🔄 Three allowed transformations

Starting electron type	Becomes	Result
π bond	another π bond	Multiple bond shifts position
π bond	lone pair electrons	Bond breaks, lone pair forms
lone pair electrons	π bond	Lone pair becomes bonding electrons

🏹 Arrow pushing technique

🏹 How to use curved arrows

Arrow pushing: using curved arrows to show electron transfer between resonance structures.

• Arrows must be shown in the "original" structure
• The "new" structure is automatically obtained by following the arrows
• This is a fundamental skill in organic chemistry

🏹 Step-by-step approach

1. Identify high electron density areas (e.g., atoms with negative formal charges, lone pairs)
2. Identify low electron density areas (e.g., positive charges, electron-deficient atoms)
3. Move electrons from high to low density
4. Ensure no atom exceeds its octet (for C, N, O)
5. Calculate and label formal charges in the new structure

Example from excerpt: When moving a lone pair from nitrogen (with "-" charge) to form a π bond, if that creates 5 bonds on carbon (not allowed), the π electrons in an adjacent C=O bond must simultaneously move to become a lone pair on oxygen.

⚖️ Comparing stability of resonance contributors

⚖️ Equivalent vs non-equivalent structures

• Equivalent structures: same bonding pattern and charge distribution, same energy level
• Non-equivalent structures: different bonding and charge distributions, different energy levels

⚖️ Stability guidelines (most to least stable)

1. Complete octets: structures with maximum octets are most important (except known exceptions)
2. Smaller formal charges: structures with lower formal charge magnitudes are more stable
3. Charge placement:
   • Negative charges preferentially on more electronegative atoms
   • Positive charges preferentially on less electronegative atoms
4. Charge separation: less separation → more stable (separation increases energy)

Major resonance contributor: the most stable structure that makes the greatest contribution to the actual molecular structure.

⚖️ Applying the rules

Example from excerpt: For a structure with "-" formal charge on nitrogen vs oxygen, the structure with negative charge on oxygen is more stable because oxygen is more electronegative than nitrogen.

⚠️ Common errors to avoid

⚠️ Five critical mistakes

1. Moving σ bonds (only π electrons and lone pairs can move)
2. Moving atoms (atom connectivity must remain unchanged)
3. Exceeding octets (more than 8 electrons around C, N, or O)
4. Incorrect arrow notation (arrows not shown properly)
5. Moving electrons too far (electrons should only move to the next position/atom, not skip positions)

⚠️ Don't confuse

• Resonance structures are NOT different molecules; they are different electron arrangements of the same molecule
• The actual structure is a hybrid of all contributors, not a mixture flipping between forms
• More contributors means more stabilization, but not all contributors are equally important—the most stable one dominates

🧭 Overview

🧠 One-sentence thesis

VSEPR theory predicts molecular shapes by considering all electron groups around a central atom, but the actual molecular shape is determined only by the positions of atoms, making lone pairs "invisible" in the final geometry.

📌 Key points (3–5)

• Core principle: Electron groups (bonding pairs and lone pairs) arrange themselves to minimize repulsion, creating a geometry for all electron groups.
• Key distinction: The geometry of all electron groups differs from the molecular shape when lone pairs are present—lone pairs occupy space but are not counted in the molecular shape.
• Common confusion: Tetrahedral electron-group geometry vs bent molecular shape—water has four electron groups (tetrahedral arrangement) but only two atoms visible, making it bent.
• Bond angles: Lone pairs cause deviations; angles become smaller than ideal values (e.g., bent from trigonal planar is less than 120°, trigonal pyramidal is less than 109.5°).
• Systematic patterns: Total electron groups (2–6) determine the base geometry; the number of bonding pairs vs lone pairs determines the specific molecular shape.

🔍 Core distinction: electron-group geometry vs molecular shape

🔍 What electron groups include

• Electron groups = bonding pairs (BP) + lone pairs (LP) on the central atom.
• All electron groups arrange themselves in space to minimize repulsion.
• This arrangement creates the geometry of all electron groups (e.g., tetrahedral, trigonal bipyramidal, octahedral).

👁️ Why lone pairs are "invisible"

Even though lone pairs occupy space, there are no terminal atoms connected with the lone pairs, so the lone pairs become "invisible" for the shape of the species.

• The molecular shape counts only the positions of atoms, not lone pairs.
• Example: Water (H₂O) has a central oxygen with 2 BP and 2 LP. The electron-group geometry is tetrahedral (four groups total), but the molecular shape is bent because only the two hydrogen atoms are visible.
• Don't confuse: the electron groups still repel each other (lone pairs take up space), but the shape name describes only where the atoms are.

📐 How total electron groups determine base geometry

📐 The five main base geometries

The geometry of all electron groups depends on the total number of electron groups around the central atom:

Total e-groups	Base geometry of all electron groups	Example molecule (all BP)	Ideal angles
2	Linear	—	180°
3	Trigonal planar	—	120°
4	Tetrahedral	—	109.5°
5	Trigonal bipyramidal	PCl₅	120°, 90°, 180°
6	Octahedral	—	90°, 180°

• Example: PCl₅ has five bonding pairs (no lone pairs), so both the electron-group geometry and the molecular shape are trigonal bipyramidal.

🔢 Why the number matters

• The total number of electron groups sets the framework.
• Once you know the base geometry, the number of lone pairs determines the actual molecular shape.

🧩 How lone pairs change molecular shapes

🧩 Lone pairs reduce visible atoms

• When lone pairs are present, fewer atoms are visible, so the molecular shape name changes.
• The base geometry remains the same (electron groups still repel), but the shape name describes only the atom positions.

🔻 Bond angle deviations

• Lone pairs occupy more space than bonding pairs, so they push bonding pairs closer together.
• Result: actual bond angles become smaller than ideal values.
• Example: Bent shape from trigonal planar (2 BP, 1 LP) has angles less than 120°; bent shape from tetrahedral (2 BP, 2 LP) has angles less than 109.5°.

📋 Common shape patterns

The excerpt provides a systematic table. Key examples:

Total e-groups	BP / LP	Molecular shape	Angles
3	3BP, 0LP	Trigonal planar	120°
3	2BP, 1LP	Bent	<120°
4	4BP, 0LP	Tetrahedral	109.5°
4	3BP, 1LP	Trigonal pyramidal	<109.5°
4	2BP, 2LP	Bent	<109.5°
5	5BP, 0LP	Trigonal bipyramidal	120°, 90°, 180°
5	4BP, 1LP	See-saw	<120°, 90°, 180°
5	3BP, 2LP	T-shape	90°, 180°
5	2BP, 3LP	Linear	180°
6	6BP, 0LP	Octahedral	90°, 180°
6	5BP, 1LP	Square pyramidal	90°, 180°
6	4BP, 2LP	Square planar	90°, 180°

• Notice: with 5 total electron groups and 2 BP + 3 LP, the molecular shape is linear (only two atoms visible), even though the electron-group geometry is trigonal bipyramidal.

🎯 Applying VSEPR: the water example

💧 Step-by-step for H₂O

1. Count electron groups on central atom (oxygen): 2 bonding pairs (two H atoms) + 2 lone pairs = 4 total electron groups.
2. Determine electron-group geometry: 4 groups → tetrahedral arrangement.
3. Determine molecular shape: Only 2 atoms visible → bent shape.
4. Bond angles: Less than 109.5° (ideal tetrahedral angle) because lone pairs push the bonding pairs closer.

🧪 Why this matters

• The same base geometry (tetrahedral) can produce different molecular shapes depending on lone pairs.
• To describe a shape correctly, use the specific name (e.g., "bent," not "tetrahedral") and note the bond angle information.

🛠️ Practical tips

🛠️ How to determine any molecular shape

1. Draw the Lewis structure to find bonding pairs and lone pairs on the central atom.
2. Count total electron groups (BP + LP).
3. Look up the base geometry for that total number.
4. Use the number of bonding pairs to find the specific molecular shape in the table.
5. Note that bond angles will be smaller than ideal if lone pairs are present.

🛠️ Common confusion checklist

• Don't confuse electron-group geometry with molecular shape: Water is not "tetrahedral" in shape; it is "bent" with a tetrahedral electron-group arrangement.
• Don't ignore lone pairs when counting total groups: They affect the base geometry and push atoms closer.
• Don't forget angle deviations: Lone pairs make actual angles smaller than the ideal values listed for the base geometry.

🧭 Overview

🧠 One-sentence thesis

Valence bond theory explains covalent bonding through orbital overlap, and hybridization extends this theory to account for the bonding patterns in organic molecules like methane that cannot be explained by simple atomic orbitals alone.

📌 Key points (3–5)

• Valence bond theory: covalent bonds form when half-filled atomic orbitals on different atoms overlap and share electrons.
• Sigma (σ) bonds: formed by head-to-head orbital overlap; cylindrically symmetrical around the bond axis.
• Hybridization: mathematical combination of atomic orbitals to create new hybrid orbitals that explain bonding in molecules like CH₄.
• Common confusion: valence bond theory alone predicts carbon should form only two bonds (two half-filled p orbitals), but carbon always forms four bonds in stable compounds—hybridization resolves this discrepancy.
• Multiple bonds: double and triple bonds contain one sigma bond plus one or two pi (π) bonds formed by side-by-side p orbital overlap.

🔬 Valence bond theory fundamentals

🔬 How orbital overlap creates bonds

Valence bond theory: describes the covalent bond formed from the overlap of two half-filled atomic orbitals on different atoms.

• Bonding occurs when orbitals from separate atoms overlap, allowing electrons to pair up.
• The shared electron pair is attracted to both nuclei simultaneously, acting as "glue" holding the nuclei together.
• Example: In H₂, two 1s orbitals (each with one electron) overlap to form the H-H bond.

📉 Energy changes during bond formation

• When atoms are far apart: no overlap, no interaction.
• As atoms approach: orbitals overlap, attraction between nucleus and electron lowers total energy.
• At optimal distance: energy reaches minimum—this distance is the bond length (74 pm for H₂).
• Bond dissociation energy: energy difference between the most stable state and complete separation (435 kJ/mol for H-H).
• If atoms get too close: nuclear repulsion dominates and energy increases sharply.

🔗 Sigma (σ) bonds

σ (sigma) bond: formed by head-to-head overlap of orbitals.

• Cylindrically symmetrical: a cross-section at any point forms a circle.
• Can form from:
  • s-s overlap (H₂)
  • s-p overlap (H-F: hydrogen 1s overlaps with fluorine half-filled 2p)
  • p-p overlap (F₂: two half-filled 2p orbitals overlap head-to-head)

🧩 Hybridization concept

🧩 Why hybridization is needed

• Carbon's valence configuration is 2s² 2p², showing only two half-filled orbitals.
• Valence bond theory predicts carbon should form only two bonds.
• Reality: carbon always forms four bonds in stable organic compounds.
• Don't confuse: simple valence bond theory works for H₂ and HF but fails for carbon-containing molecules.

🔄 What hybridization means

Hybridization: the mathematical combination of several orbitals to generate a set of new hybrid orbitals.

• Key rule: number of hybrid orbitals = total number of atomic orbitals combined.
• Example: combining one 2s and three 2p orbitals produces four sp³ hybrid orbitals.
• All hybrid orbitals in a set have the same energy level (between the original s and p levels).

🔺 sp³ hybridization in methane (CH₄)

• Carbon combines one 2s and three 2p orbitals → four sp³ hybrid orbitals.
• The "sp³" label identifies: one s + three p orbitals (1 + 3 = 4 total).
• Geometry: four sp³ orbitals point toward corners of a tetrahedron with 109.5° angles.
• Each sp³ orbital has two lobes (one large, one small); the larger lobe participates in bonding.
• Each of carbon's four valence electrons occupies one sp³ orbital → four half-filled orbitals available for bonding.
• C-H bonds form by overlap of carbon sp³ and hydrogen 1s orbitals.

📐 3D representation

• Tetrahedral shape matches VSEPR prediction.
• Drawing convention:
  • Two bonds in the paper plane: ordinary lines
  • One bond pointing out: solid wedge
  • One bond pointing behind: dashed wedge

🔗 Connecting hybridization and VSEPR

🔗 Determining hybridization from electron groups

• Total number of electron groups (bonding pairs + lone pairs) = total number of orbitals in hybridization.
• Example: CH₄ has four bonding pairs around carbon → sp³ (1 s + 3 p = 4 orbitals).
• Example: five electron groups → sp³d (1 s + 3 p + 1 d = 5 orbitals).

📊 Hybridization-VSEPR correlation table

Hybridization	Total electron pairs	Geometry of electron groups
sp	2	Linear
sp²	3	Trigonal planar
sp³	4	Tetrahedral
sp³d	5	Trigonal bipyramidal
sp³d²	6	Octahedral

🧪 Applying to organic molecules

• Organic molecules have multiple central atoms—analyze each atom individually.
• Must include lone pairs (often omitted in structural formulas).
• Example: oxygen in ethanol's OH group has two bonding pairs + two lone pairs = four electron groups → sp³ hybridization, tetrahedral shape.

🔀 Multiple bonds and pi (π) bonds

🔀 Ethene (C₂H₄) and double bonds

• Each carbon has three electron groups → sp² hybridization.
• sp² means: one 2s + two 2p orbitals hybridize; one 2p remains unhybridized.
• The three sp² orbitals lie in a trigonal planar arrangement (120° angles).
• The unhybridized 2p orbital is perpendicular to the sp² plane.

🔗 Formation of C=C double bond

• One sp² orbital from each carbon overlaps head-to-head → σ (sigma) bond.
• The two unhybridized 2p orbitals overlap side-by-side → π (pi) bond.
• Double bond = one σ bond + one π bond.
• The π bond makes the entire C₂H₄ molecule co-planar.
• Other sp² orbitals overlap with hydrogen 1s orbitals → four C-H σ bonds.

🔗 Ethyne (C₂H₂) and triple bonds

• Each carbon has two electron groups → sp hybridization (linear, 180°).
• sp means: one 2s + one 2p hybridize; two 2p orbitals remain unhybridized.
• One sp orbital from each carbon overlaps head-to-head → C-C σ bond.
• Two pairs of unhybridized 2p orbitals overlap side-by-side → two π bonds.
• Triple bond = one σ bond + two π bonds.
• Other sp orbitals overlap with hydrogen 1s → two C-H σ bonds.

🔍 Sigma vs pi bonds comparison

Bond type	Overlap type	Symmetry	Found in
σ (sigma)	Head-to-head	Cylindrical	Single, double, triple bonds
π (pi)	Side-by-side	Not cylindrical	Double and triple bonds only

Don't confuse: every bond has one σ bond; additional bonds in double/triple bonds are π bonds formed by unhybridized p orbitals.

🧭 Overview

🧠 One-sentence thesis

Alkanes are the simplest hydrocarbons with only single bonds, and their structures can be represented in multiple ways while exhibiting constitutional isomers that dramatically increase in number as carbon count grows.

📌 Key points (3–5)

• What alkanes are: hydrocarbons containing only C-C single bonds, fitting the formula C_n H_(2n+2) for chain alkanes.
• Multiple structure formats: Kekulé, condensed, short-line, and perspective formulas each show the same molecule differently with varying levels of detail.
• Constitutional isomers: different compounds with the same molecular formula but atoms connected in different orders, leading to different physical properties.
• Common confusion: short-line structures omit carbon and hydrogen atoms—each bend or terminus represents a carbon unless another atom is shown explicitly.
• Carbon classification: carbons are categorized as 1°, 2°, 3°, or 4° based on how many other carbons they connect to, affecting structural and reactive properties.

🔗 What alkanes are and their sources

🔗 Basic definition and formula

Alkane: the simplest hydrocarbon with only C-C single bonds.

• Chain alkanes fit the general formula C_n H_(2n+2) (n is a positive integer).
• The number of hydrogen atoms reaches the maximum level in chain alkanes.
• The excerpt provides a table listing straight-chain alkanes from methane (1 carbon) to decane (10 carbons).

⛽ Natural sources and uses

• Natural gas: contains mainly methane (70–90%) and some ethane.
• Petroleum refining: separates crude oil into fractions based on carbon number.
• Practical applications: propane and butane are common fuels; alkanes with 5–8 carbons form gasoline; diesel contains alkanes with 9–16 carbons.
• Physical trend: as carbon number increases, boiling point and viscosity increase.

📐 Structure representation formats

📐 Kekulé structures

• Similar to Lewis structures with all bonding electrons shown as short lines.
• All atoms included as element symbols.
• Key difference: lone pair electrons are left out (unlike Lewis structures).

📐 Condensed structure formulas

• C-H bonds are omitted.
• Hydrogen atoms attached to a carbon are shown as groups (CH3, CH2, NH2, OH).
• C-C bonds sometimes omitted as well.
• When branches exist: bonding between parent structure and branch shown with a short line.
• Advantage: faster to draw and less bulky than Kekulé structures.

📐 Short-line (skeletal) structures

This is a very common format in organic chemistry due to its simplicity.

Conventions to understand:

• Each short line represents a bond.
• Carbon chains shown in zig-zag way.
• No carbon atoms shown (exception: CH3 groups at chain ends or branches may optionally be shown).
• Each bend or terminus represents a carbon unless another atom is shown explicitly.
• Hydrogen atoms bonded to carbons are not shown; hydrogens bonded to other atoms are shown explicitly.
• Heteroatoms (N, O, Cl, etc.) must be shown explicitly.
• How to find hydrogens: calculate the number of hydrogen atoms on each carbon by applying the octet rule and checking formal charges.

📐 Perspective formulas (3D)

Used to highlight spatial arrangement around tetrahedral sp3 carbons for conformation or stereochemistry purposes.

Convention:

• Two bonds lie within the paper plane (ordinary lines).
• Solid wedge: bond pointing out of the paper plane.
• Dashed wedge: bond pointing behind the paper plane.

🧬 Constitutional isomers

🧬 What they are

Constitutional (structural) isomers: different compounds with the same molecular formula, but their atoms are arranged in a different order (i.e., the atoms are bonded in different ways).

• They are different compounds with different names and different physical properties.
• Example from the excerpt: butane and isobutane both have formula C4H10 but different structures, boiling points, and densities.

🧬 How isomer count grows

• For C4H10: 2 constitutional isomers.
• For C5H12: 3 constitutional isomers.
• For C6H14: 5 constitutional isomers.
• For C7H16: the excerpt asks students to draw all isomers (answer not provided in excerpt).
• For C20H42: 366,319 constitutional isomers.
• No simple formula exists to predict the total number for a given carbon count.
• This phenomenon partially explains the high diversity of organic structures.

🧬 Types of constitutional isomers

The excerpt mentions three types (though only one is detailed here):

• Skeletal constitutional isomers: different lengths of carbon "backbones" (the examples shown so far).
• Positional and functional constitutional isomers: mentioned as topics to be encountered later.

🧬 Strategy for building isomers

The excerpt provides a figure showing strategies for building constitutional isomers of C5H12, though the specific strategies are in notes beside structures not fully reproduced here.

🏷️ Carbon classification (1°, 2°, 3°, 4°)

🏷️ How carbons are categorized

Classification depends on how many other carbons a given carbon connects with:

Type	Definition	Connects to
Primary (1°)	Attached directly to only one other C	1 carbon
Secondary (2°)	Attached directly to two other C atoms	2 carbons
Tertiary (3°)	Attached directly to three other C atoms	3 carbons
Quaternary (4°)	Attached to four other C atoms	4 carbons

🏷️ Hydrogen classification

• Hydrogen atoms attached to 1°, 2°, and 3° carbons are labeled as 1°, 2°, and 3° hydrogens respectively.
• (Note: 4° carbons have no hydrogens attached.)

🏷️ Why it matters

• In one compound, carbons (or hydrogens) belonging to different categories show different structural and reactive properties.
• The excerpt states this concept has many applications in later sections.

🧭 Overview

🧠 One-sentence thesis

IUPAC nomenclature provides a systematic, rule-based method for naming alkanes and their derivatives, ensuring that each unique structure receives a consistent and unambiguous name.

📌 Key points (3–5)

• Why systematic naming matters: As the number of constitutional isomers grows dramatically with more carbons, common names become impractical, requiring a rule-based system.
• Core IUPAC principle: Identify the longest continuous carbon chain as the parent, number it to give substituents the lowest positions, and name substituents by changing "-ane" to "-yl".
• Cycloalkanes follow similar rules: The ring becomes the parent structure (named "cycloalkane"), numbered to give substituents the lowest sequence, unless a longer chain or higher-priority functional group is present.
• Common confusion: Branched alkyl groups (like isopropyl, isobutyl, tert-butyl) have both common names (widely used) and systematic IUPAC names; understanding their origin from parent alkanes (propane, butane, isobutane) helps distinguish them.
• Ring vs chain priority: When both a ring and a chain are present, the one with more carbons becomes the parent structure; the other is treated as a substituent.

🔤 IUPAC naming rules for alkanes

🔗 Identify the parent chain

• Longest continuous carbon chain: This chain determines the parent name (e.g., hexane for 6 carbons, octane for 8 carbons).
• Tie-breaker: If two chains have the same length, choose the one with the greatest number of branches (substituents).
• Why: The parent chain forms the "last name" of the compound; all other groups are named relative to it.

🔢 Number the chain strategically

• Lowest-number rule: Start numbering from the end that gives substituents the lowest possible position numbers.
• Example: If a substituent could be at position 3 or position 5, number so it becomes position 3.
• Why: This ensures a unique, standardized name for each structure.

🏷️ Name and locate substituents

• Substituent naming: Replace "-ane" with "-yl" (e.g., methane → methyl, ethane → ethyl).
• Position numbers: Use numbers separated by commas; separate numbers from letters with hyphens.
• Multiple identical groups: Use prefixes di (2), tri (3), tetra (4), penta (5), hexa (6).
• Alphabetical order: List substituents alphabetically by their base name (ignore di-, tri-, etc., but include "iso" and "cyclo").
• Example: "4-ethyl-3-methyloctane" (ethyl comes before methyl alphabetically).

🌿 Branched alkyl groups

🌿 Three-carbon groups from propane

• Propane has two types of hydrogens: primary (1°) on end carbons, secondary (2°) on the middle carbon.
• Removing a 1° H gives propyl (n-propyl).
• Removing a 2° H gives isopropyl (1-methylethyl in systematic naming).
• Don't confuse: The attachment point determines which type of alkyl group forms.

🌿 Four-carbon groups from butane and isobutane

• From butane (straight chain):
  • Removing a 1° H → butyl (n-butyl)
  • Removing a 2° H → sec-butyl (1-methylpropyl)
• From isobutane (2-methylpropane):
  • Removing a 1° H → isobutyl (2-methylpropyl)
  • Removing the 3° H → tert-butyl (1,1-dimethylethyl)
• Understanding origin: Knowing which parent alkane (butane vs isobutane) and which hydrogen (1°, 2°, or 3°) is removed helps memorize these common names.

🔧 Systematic naming of branched substituents

• Treat the substituent as if it were a compound: number from the attachment point to the parent chain.
• Enclose the systematic name in parentheses when writing the full compound name.
• Example: Isobutyl can be named "2-methylpropyl" systematically—the propyl part attaches to the parent chain, with a methyl on position 2 of that propyl group.

🔄 Cycloalkanes

🔄 Basic cycloalkane naming

Cycloalkane: An alkane containing a ring structure; general formula for one ring is CₙH₂ₙ (two fewer hydrogens than the non-cyclic alkane).

• Parent name: "cycloalkane" (e.g., cyclopentane, cyclohexane).
• Numbering the ring: Assign positions to give the lowest possible numbers to substituents.
• Alphabetical priority: When two numbering sequences are equally low, cite substituents alphabetically and give position 1 to the first-cited substituent.
• Example: "1-ethyl-3-methylcyclohexane" (ethyl at position 1, methyl at position 3).

⚖️ Ring vs chain: which is the parent?

• Compare carbon counts: The structure with more carbons becomes the parent; the other is a substituent.
• Example: A four-carbon ring with a three-carbon chain attached → "propylcyclobutane" (ring has more carbons, so it's the parent).
• Functional group priority: If a higher-priority functional group is present (e.g., aldehyde, carboxylic acid), the parent structure must contain that group, regardless of ring or chain size.
• Example: A cyclobutane ring attached to a three-carbon chain ending in an aldehyde → "3-cyclobutylpropanal" (the aldehyde-containing chain is the parent).

📝 Writing the complete name

📝 Assembly rules

• Single word: Write the entire name as one word.
• Punctuation:
  • Hyphens separate numbers from letters (e.g., "3-methyl")
  • Commas separate numbers from numbers (e.g., "2,3-dimethyl")
• Order: Position numbers + substituent names (alphabetically) + parent name with appropriate suffix.
• Example: "5,7-diethyl-3,4,7-trimethyl-5-propyldecane" shows multiple substituents listed alphabetically (diethyl, trimethyl, propyl) with their positions, all attached to a decane parent chain.

🎯 Common pitfalls

• Don't forget: Ignore numerical prefixes (di-, tri-) when alphabetizing, but do include "iso" and "cyclo".
• Don't confuse: The parent chain must be continuous; a longer non-continuous sequence doesn't count.
• Check: After naming, verify that the numbering gives the lowest possible set of position numbers for all substituents combined.

🧭 Overview

🧠 One-sentence thesis

Functional groups are structural units within organic molecules that determine chemical behavior, and recognizing their patterns—especially distinguishing similar groups like aldehydes vs. ketones or nitrile vs. nitro—is essential for understanding organic chemistry.

📌 Key points (3–5)

• Functional groups define chemistry: groups like alcohols, ethers, amines, and carbonyl-containing groups determine how molecules behave.
• Classification by substitution: alkyl halides, alcohols, and amines are categorized as primary (1°), secondary (2°), or tertiary (3°) based on the carbon or nitrogen substitution pattern.
• Carbonyl group variations: aldehyde, ketone, carboxylic acid, ester, anhydride, and amide all share a C=O unit, but differ in what is attached to the carbonyl carbon.
• Common confusion: nitrile (C≡N) vs. nitro (NO₂); aldehyde (C=O with H) vs. ketone (C=O between two R groups); ester (COOR) is not "ketone + ether."
• Position matters: some groups (aldehyde, nitrile) can only be at the end of a structure, while others (ketone, nitro) can be in the middle or on a ring.

🔗 Halides, Alcohols, and Ethers

🧪 Alkyl halides (haloalkanes)

Alkyl halide: a functional group where a halogen is connected to carbon.

• Categorized as primary (1°), secondary (2°), or tertiary (3°) depending on the carbon atom to which the halogen is attached.
• Example: a chlorine attached to a secondary carbon is a 2° chloride.

🍷 Alcohols

Alcohol: a compound containing the OH (hydroxy) group.

• Also classified as primary (1°), secondary (2°), or tertiary (3°) based on the position of the OH group.
• Example: an OH group on a primary carbon is a 1° alcohol.
• Don't confuse: the term "alcohol" in organic chemistry specifically refers to the OH functional group, not the beverage.

🔗 Ethers

Ether: a functional group where an oxygen atom connects two carbon-containing R groups through two C–O single bonds.

• Common naming: "alkyl alkyl ether"; when both alkyl groups are the same, use "dialkyl."
• Example: diethyl ether has two ethyl groups attached to oxygen.
• Ethers can also be cyclic; labeling carbon atoms helps identify the ether oxygen in ring structures.

🔵 Nitrogen-containing groups

⚠️ Nitrile vs. nitro

Group	Structure	Position	Key distinction
Nitrile	C≡N triple bond	Only at the end of a structure	Triple bond to nitrogen
Nitro	NO₂	Any position on chain or ring	Two oxygens attached to nitrogen

• Common confusion: both contain nitrogen, but nitrile has a triple bond and must be terminal, while nitro is NO₂ and can be anywhere.

🧬 Amines

Amine: the organic derivative of ammonia (NH₃), where hydrogen atom(s) are replaced with R groups.

• Classified as primary (1°), secondary (2°), or tertiary (3°) depending on how many R groups are connected to nitrogen.
• Example: one R group = primary amine; two R groups = secondary amine; three R groups = tertiary amine.
• Amines can also be referred to with common names.

🎯 Carbonyl-containing functional groups

🧱 General carbonyl structure

Carbonyl group: a C=O double bond; the different structure of "W" attached to the carbonyl carbon determines the nature of the functional group.

• All groups in this category share the C=O unit but differ in what is bonded to the carbonyl carbon.
• More challenging to identify and draw correctly because they are similar; practice is needed.

🔶 Aldehyde and ketone

Aldehyde: a carbonyl group where one side is H (can be regarded as a special case of ketone where "H" is R with zero carbon).

Ketone: a carbonyl group where both sides are connected with R groups.

Feature	Aldehyde	Ketone
Structure	C=O with H on one side	C=O with R groups on both sides
Position	Only at the end of a structure	Must be in the middle position
Cyclic form	Not applicable	Can be in a cyclic structure

• Example: because H must be on one side of C=O in aldehyde, it can only be terminal.
• Ketone can also be in a ring.

🧪 Carboxylic acid and its derivatives

Carboxylic acid: a functional group where an OH group is connected with C=O, forming COOH.

• The other three groups—ester, anhydride, and amide—are all derivatives of carboxylic acid, meaning they can be prepared starting from carboxylic acid.

Derivative	Structure	Key point
Ester	COOR (or CO₂R)	The "W" part must be considered together with C=O
Anhydride	Two C=O groups linked by oxygen	Formed from two carboxylic acid molecules
Amide	C=O connected to nitrogen	Nitrogen replaces the OH of carboxylic acid

• Important: for ester, anhydride, and amide, the "W" part has to be considered together with the C=O, since overall they define the functional group.
• Common confusion: COOR is ester; it cannot be recognized as a "ketone" plus an "ether."
• Example: in ester, the entire COOR unit is the functional group, not separate C=O and OR parts.

🔍 Special structures

🔷 Benzene

• Benzene is a conjugation system; its structure has nothing to do with alkenes (despite drawings that show alternating double and single bonds).
• The structure of benzene and the definitions of aromatic/aromaticity are discussed in detail in Organic Chemistry II.
• Benzene rings can be shown with various structure drawings.

🧭 Overview

🧠 One-sentence thesis

IUPAC naming of organic compounds follows a priority-based system where the highest-priority functional group determines the parent name and suffix, while other groups become substituents with prefixes.

📌 Key points (3–5)

• Priority hierarchy: functional groups are ranked by priority (carboxylic acid highest); the highest-priority group determines the parent name and gets a suffix, while lower-priority groups become substituents with prefixes.
• Core naming steps: find the longest chain containing the highest-priority group, change the parent alkane/alkene/alkyne ending to the appropriate suffix, number from the end closest to that group, and name other groups as substituents.
• Common confusion: for carbonyl-containing groups (ester, amide, anhydride), the entire unit (e.g., COOR for ester) defines the functional group—do not split it into separate parts like "ketone + ether."
• Benzene naming: when benzene has 6+ carbons attached, treat the carbon chain as parent and benzene as the "phenyl" substituent; for smaller substituents, benzene is the parent.
• Special benzene derivatives: phenol, benzoic acid, and benzaldehyde use their common names as IUPAC parent names, with the base functional group always at position #1.

🏗️ The five-step naming procedure

🔍 Step 1: Identify the longest chain with the highest-priority group

• Locate the longest continuous carbon chain that contains the functional group with the highest priority (refer to Table 2.3 in the excerpt).
• This chain becomes the parent structure and determines the base name.
• Example: if a molecule has both a carboxylic acid and a ketone, the carboxylic acid (higher priority) determines the parent chain.

🔤 Step 2: Change the ending to the appropriate suffix

• Take the parent alkane/alkene/alkyne name and replace its ending with the suffix for the highest-priority group.
• Usually drop the final "e" before adding the suffix (exception: nitrile keeps the "e").
• Example: a 6-carbon chain with a carboxylic acid and a double bond → "hexene" becomes "hexenoic acid."

🔢 Step 3: Number the chain from the end closest to the highest-priority group

• Start numbering from whichever end gives the highest-priority functional group the lowest possible number.
• The carboxylic acid group is always at position #1, so its number is not written.
• Example: in "4-hexenoic acid," the double bond is at position 4; the carboxylic acid is implicitly at position 1.

🏷️ Step 4: Name other groups as substituents with prefixes

• All functional groups not chosen as the parent become substituents.
• Use the appropriate prefix from the priority table (e.g., "hydroxy" for OH when it is a substituent, not the parent alcohol).
• List substituents in alphabetical order (or as specified in Table 2.4 for subordinate groups).
• Example: "5-bromo-7-chloro-6-hydroxy-2,2,5-trimethyl-7-octen-4-one" lists bromo, chloro, hydroxy, and methyl as substituents around the ketone parent.

🧬 Step 5: Assign stereochemistry (E/Z or R/S) as necessary

• The excerpt notes this step but defers details to Chapter 5.
• Include stereochemical descriptors when relevant to fully specify the compound.

🎯 Functional group priority and common pitfalls

📊 Priority ranking (Table 2.3)

Priority rank	Functional group	Suffix (as parent)	Prefix (as substituent)
Highest	Carboxylic acid (COOH)	-oic acid	carboxy-
↓	Ester (COOR)	-oate	(R named first)
↓	Amide	-amide	amido-
↓	Aldehyde (CHO)	-al	formyl- or oxo-
↓	Ketone (C=O)	-one	oxo-
↓	Alcohol (OH)	-ol	hydroxy-
↓	Amine (NH₂, NHR, NR₂)	-amine	amino-
Lower	Alkene (C=C)	-ene	alkenyl-
Lower	Alkyne (C≡C)	-yne	alkynyl-

• The order is decreasing priority from top to bottom.
• Subordinate groups (Table 2.4) have no priority differences and are listed alphabetically.

⚠️ Don't split carbonyl-containing groups

• The excerpt emphasizes: for ester (COOR), anhydride, and amide, the "W" part (the atom/group attached to C=O) must be considered together with the carbonyl.
• Common mistake: recognizing COOR as "ketone + ether" is incorrect; the entire COOR unit is the ester functional group.
• Why: these groups are derivatives of carboxylic acid and have distinct chemical properties as a whole unit.

🧪 Aldehyde vs ketone

• Aldehyde: one side of C=O must be H, so aldehydes are always at the end of a structure.
• Ketone: both sides of C=O are R groups (carbon-containing), so ketones are in the middle or in a cyclic structure.
• Think of aldehyde as a special case of ketone where one R is replaced by H (zero carbons).
• Example: a cyclic ketone (cyclohexanone) has the C=O within the ring; an aldehyde cannot be fully within a ring without an H on one side.

🧪 Detailed naming examples

🧪 Example 1: 4-hexenoic acid

• Parent structure: 6-carbon chain with a carboxylic acid and a double bond.
• Base name: "hexene" (6 carbons, double bond).
• Drop the "e" and add "-oic acid" → "hexenoic acid."
• Number the double bond position: 4-hexenoic acid.
• The carboxylic acid is always at #1, so no need to write "1-."

🧪 Example 2: 3-ethylcyclohexanone

• Parent structure: cycloalkane with a ketone.
• Base name: "cyclohexane."
• Add suffix "-one" → "cyclohexanone."
• Substituent: ethyl group at position 3.
• Complete name: 3-ethylcyclohexanone.

🧪 Example 3: Complex multi-group compound

• Full name: "5-bromo-7-chloro-6-hydroxy-2,2,5-trimethyl-7-octen-4-one."
• Highest priority: ketone (decides the parent name "octenone").
• 8-carbon chain with a double bond and ketone → "octenone."
• Number from the left to give the ketone the lowest number (position 4).
• Substituents: bromo (position 5), chloro (position 7), hydroxy (position 6, not "alcohol" because ketone is parent), and multiple methyl groups (positions 2, 2, 5).
• The double bond is at position 7.

🧪 Example 4: Cyclopropanol with a phenyl substituent

• Parent structure: cyclic alcohol → "cyclopropanol."
• Substituent: a benzene ring (called "phenyl" when it is a substituent) with an isopropyl group on it.
• The benzene substituent is named "3-isopropylphenyl" (the isopropyl is at position 3 on the phenyl ring).
• Complete name: "2,2-dimethyl-3-(3-isopropylphenyl)cyclopropanol."
• Don't confuse: benzene as a substituent is "phenyl," not "benzene."

🧪 Example 5: Ester naming (tert-butyl propanoate)

• Ester structure: an OR group replaces the OH of a carboxylic acid.
• Naming rule: state the R group name first, then the acid name with "-oic acid" replaced by "-oate."
• Here: the R is tert-butyl, the acid part is propanoic acid.
• Complete name: tert-butyl propanoate.
• The R in OR is treated like a "substituent" even though it is part of the ester functional group.

🔷 Benzene and benzene derivatives

🔷 Substituted benzene basics

• Benzene ring is the parent structure.
• Positions and names of substituents are added to the front.
• Examples:
  • Methylbenzene (one methyl group on benzene).
  • Chlorobenzene (one chlorine on benzene).
  • 1,3-dinitrobenzene (two nitro groups at positions 1 and 3).
  • 1,2,4-trimethylbenzene (three methyl groups at positions 1, 2, and 4).

🔷 Ortho-, meta-, para- system for di-substituted benzene

• An alternative common naming system for indicating the relative positions of two substituents on a benzene ring.
• ortho- (o-): 1,2- (next to each other).
• meta- (m-): 1,3- (separated by one carbon).
• para- (p-): 1,4- (across from each other).
• This system is widely used in books and academic literature, even though it is a common (not IUPAC) naming convention.

🔷 Special benzene derivatives: phenol, benzoic acid, benzaldehyde

• These three compounds have common names adopted in the IUPAC system:
  • Phenol: benzene with OH.
  • Benzoic acid: benzene with COOH.
  • Benzaldehyde: benzene with CHO.
• When other substituents are added, the common name becomes the parent name, and the base functional group is always at position #1.
• Examples:
  • 2,4-dichlorophenol: phenol with chlorine at positions 2 and 4 (OH is at #1).
  • 2-bromo-4-methylbenzoic acid: benzoic acid with bromine at position 2 and methyl at position 4 (COOH is at #1).

🔷 When benzene becomes a substituent: "phenyl"

• If the carbon chain has six or more carbons, treat the carbon chain as the parent structure.
• The benzene ring becomes a substituent and is called phenyl.
• Example: 2-phenylheptane has a 7-carbon chain (heptane) as the parent, with a phenyl group attached at position 2.
• Don't confuse: benzene is "phenyl" when it is a substituent, but "benzene" when it is the parent.

🧩 Related concepts from the excerpt

🧩 Carbonyl-containing functional groups

• The excerpt groups aldehyde, ketone, carboxylic acid, ester, anhydride, and amide together because they all contain a carbonyl group (C=O).
• The nature of "W" (the atom or group attached to the carbonyl) determines which functional group it is.
• Carboxylic acid derivatives: ester, anhydride, and amide can all be prepared from carboxylic acid as the starting material.
• Remember: the entire unit (e.g., COOH, COOR, amide group) defines the functional group, not just the C=O.

🧩 Amines

• Amines have nitrogen connected to R groups.
• They can be primary (one R), secondary (two R), or tertiary (three R).
• Amines can also be referred to with common names (details not fully covered in this excerpt).

🧩 Degree of unsaturation preview

• The excerpt briefly introduces the concept at the end: comparing pentane (C₅H₁₂, saturated), 1-pentene (C₅H₁₀, two fewer H), and cyclopentane (C₅H₁₀, two fewer H).
• Pentane has zero degrees of unsaturation (maximum hydrogen, fully saturated).
• 1-pentene and cyclopentane each have one degree of unsaturation (two fewer H than the saturated level).
• This concept (also called Index of Hydrogen Deficiency) helps explore constitutional isomers, but full details are deferred to section 2.5.

🧭 Overview

🧠 One-sentence thesis

The physical properties of organic compounds—such as boiling point and solubility—are determined by the types and strengths of intermolecular forces acting between molecules, which in turn depend on molecular polarity and structure.

📌 Key points (3–5)

• Three main intermolecular forces: dispersion forces (weakest, apply to all molecules), dipole-dipole forces (polar molecules), and hydrogen bonds (strongest, require H bonded to O/N/F).
• Molecular polarity determines force types: nonpolar molecules have only dispersion forces; polar molecules have dipole-dipole forces; molecules with O-H, N-H, or F-H bonds can form hydrogen bonds.
• Boiling point correlates with force strength: stronger total intermolecular forces require more energy to overcome, leading to higher boiling points.
• "Like dissolves like" rule for solubility: nonpolar substances dissolve in nonpolar solvents; polar/ionic substances dissolve in polar solvents; mismatched polarities are insoluble.
• Common confusion—hydrophobic vs hydrophilic parts: large organic molecules can have both nonpolar (hydrophobic) and polar (hydrophilic) regions; the dominant part determines overall solubility in water.

🔗 Types of Intermolecular Forces

🌀 Dispersion forces (London forces)

Dispersion forces result from the instantaneous dipole and induced dipole of molecules.

• How they work: Electrons constantly shift, creating temporary (instantaneous) dipoles that induce corresponding dipoles in neighboring molecules.
• Applies to: All molecules, including nonpolar ones (the only force for nonpolar molecules).
• Strength: Weakest intermolecular force (0.1–5 kJ/mol), but most fundamental.
• What affects magnitude:
  • Polarizability: Larger atoms/molecules with more loosely held electrons → stronger dispersion forces; generally, higher molar mass → stronger force.
  • Surface area: Longer, flatter, or cylindrical shapes have greater surface area than bulky/branched shapes → stronger dispersion forces.
• Example: Butane (linear) has stronger dispersion forces than isobutane (branched) despite identical molecular formulas.

⚡ Dipole-dipole forces

Dipole-dipole force is the attraction between the positive end of one polar molecule and the negative end of a neighboring polar molecule.

• Applies to: Polar molecules only (molecules with permanent dipoles).
• Strength: Intermediate (5–20 kJ/mol).
• How it works: One end of the molecule has a partial negative charge, the other a partial positive charge; opposite charges attract.
• Example: Acetone molecules attract each other through dipole-dipole interactions between the partially negative oxygen and partially positive carbon regions.

🔗 Hydrogen bonds

A hydrogen bond is the force between an H atom bonded to O, N, or F and a neighboring electronegative atom.

• Not a covalent bond: Despite the name, it is an intermolecular force.
• Applies to: Polar molecules with N-H, O-H, or F-H bonds.
• Strength: Strongest intermolecular force (5–50 kJ/mol).
• Why so strong: High electronegativity of O, N, F creates strong partial charges.
• Example: Water has an extraordinarily high boiling point (100°C) for its small molar mass (18 g/mol) because each water molecule can form multiple hydrogen bonds; methane (similar size) boils at -167.7°C.
• Important for organic compounds: Determines properties of alcohols (R-OH), carboxylic acids (R-COOH), amines (R-NH₂), and amides (RCONH₂).

⚛️ Ion-dipole force (not an intermolecular force)

Ion-dipole force is responsible for the interaction between ions and polar substances.

• Key application: Dissolving ionic solids (salts) in water.
• Solvation process: Water molecules surround and separate cations and anions through strong ion-dipole interactions, overcoming the ionic lattice.
• Example: Table salt (NaCl) dissolves in water because ion-dipole forces between Na⁺/Cl⁻ and water molecules are strong enough to break the crystal structure.

🎯 Determining Molecular Polarity

🔍 Diatomic molecules

• Homonuclear (same atoms): Always nonpolar (e.g., H₂, N₂, O₂, F₂).
• Heteronuclear (different atoms): Always polar (e.g., HF, HCl).

🔍 Polyatomic molecules

• Depends on both bonds and shape: Even if polar bonds are present, molecular geometry determines whether bond polarities cancel out.
• Polar example—water (H₂O): Bent shape means two O-H bond polarities add up → polar molecule.
• Nonpolar example—carbon dioxide (CO₂): Linear shape means two C=O bond polarities cancel out → nonpolar molecule.
• Other nonpolar molecules with polar bonds: BF₃, CCl₄, PCl₅, XeO₄ (symmetrical shapes cause cancellation).

🔍 Organic compounds

• Hydrocarbons (CₓHᵧ): Always nonpolar (small electronegativity difference between C and H).
• Compounds with heteroatoms: Functional groups like R-O-R, C=O, OH, NH make molecules polar.

🌡️ Boiling Point and Intermolecular Forces

📈 General trend

Stronger total intermolecular forces → higher boiling point.

• Why: Boiling requires overcoming intermolecular forces to convert liquid to gas; stronger forces need more energy (higher temperature).
• Comparison table:

Compound	Molar Mass	Intermolecular Forces	Boiling Point
Butane	Similar	Dispersion only (nonpolar)	Lowest
Propanal	Similar	Dispersion + dipole-dipole (polar)	Middle
Propanol	Similar	Dispersion + dipole-dipole + hydrogen bonds (O-H)	Highest

• All three have similar molar masses (similar dispersion forces), but different polarities lead to different total forces and boiling points.

💧 Solubility and the "Like Dissolves Like" Rule

🧪 Basic solubility rules

• Nonpolar dissolves in nonpolar: Nonpolar substances are soluble in nonpolar solvents (e.g., hexane, benzene, toluene).
• Polar/ionic dissolves in polar: Polar and ionic substances are soluble in polar solvents (e.g., water, methanol, ethanol).
• Mismatch is insoluble: Polar and nonpolar substances do not dissolve in each other.

🧪 Polar solvent examples

• Very polar (hydrogen bonding): Water, methanol, ethanol.
• Moderately polar: Ether, ketone, halides, esters.

🔄 Hydrophilic vs hydrophobic regions

Hydrophilic (water-loving): Polar parts that dissolve in water.
Hydrophobic (water-fearing): Nonpolar parts that do not dissolve in water.

• Hydrocarbon chains: Hydrophobic (nonpolar).
• Functional groups (OH, COOH, NH₂): Hydrophilic (polar).
• Overall polarity depends on the dominant part:
  • Short carbon chains (1–3 carbons): Hydrophilic effect dominates → water-soluble.
  • Longer chains (4–5+ carbons): Hydrophobic effect dominates → water-insoluble.

📊 Alcohol solubility trend

Alcohol	Carbon Chain Length	Solubility in Water
Methanol, ethanol, propanol	1–3	Miscible (all proportions)
1-Butanol	4	9 g/100mL
1-Pentanol	5	2.7 g/100mL
1-Octanol	8	0.06 g/100mL

• Don't confuse: The same functional group (OH) behaves differently depending on the length of the carbon chain.

🧂 Converting insoluble compounds to soluble salts

• Strategy: Use acid-base reactions to convert water-insoluble organic compounds into water-soluble salt derivatives.
• Example: Benzoic acid (water-insoluble) + strong base → sodium benzoate (water-soluble salt).
• Why it works: The salt form can interact with water through ion-dipole forces.
• Application: Commonly used in labs to separate organic compounds.

🧭 Overview

🧠 One-sentence thesis

The Brønsted-Lowry definition frames acid-base reactions as proton transfers, and the acid dissociation constant Ka quantifies how strongly an acid donates protons, which applies equally to organic and inorganic acids.

📌 Key points (3–5)

• Brønsted-Lowry definition: acids donate protons (H⁺), bases accept protons; the reaction is a proton transfer process.
• Conjugate pairs: when an acid loses a proton it forms its conjugate base; when a base accepts a proton it forms its conjugate acid.
• Ka measures acid strength: larger Ka means stronger acid (more complete proton donation); Ka > 10 indicates a strong acid.
• Common confusion: strong vs weak acids—strong acids donate protons completely (→), weak acids reach equilibrium (⇌).
• Inverse relationship: the stronger the acid, the weaker its conjugate base, and vice versa.

🔬 Brønsted-Lowry acid-base framework

🔬 Core definitions

Brønsted-Lowry Acid: a substance that can donate a proton (H⁺).

Brønsted-Lowry Base: a substance that can accept a proton (H⁺).

• An acid-base reaction is a proton transfer process.
• The acid gives away a proton; the base accepts it.
• General equation: HA + B ⇌ A⁻ + HB⁺ (where HA is the acid and B is the base).

🔄 Conjugate acid-base pairs

Conjugate base: the species that forms when an acid loses its proton.

Conjugate acid: the species that forms when a base accepts a proton.

• In the general equation HA + B ⇌ A⁻ + HB⁺:
  • HA is the conjugate acid of A⁻.
  • A⁻ is the conjugate base of HA.
  • HA and A⁻ form a conjugate acid-base pair.
  • HB⁺ and B form another conjugate acid-base pair.
• Example: when HCl donates a proton, Cl⁻ is its conjugate base.

⚖️ Strong vs weak acids and equilibrium

⚖️ Strong acids

• A strong acid donates the proton completely.
• The reaction goes to completion; use the arrow "→" (not an equilibrium arrow).
• Example: HCl (g) + H₂O (l) → H₃O⁺ (aq) + Cl⁻ (aq).
• Don't confuse: strong acids do not establish equilibrium; the forward reaction is essentially complete.

⚖️ Weak acids

• A weak acid (general formula HA) donates the proton only partially.
• The reaction stays at equilibrium; use the equilibrium arrow "⇌".
• Example: HA + H₂O ⇌ H₃O⁺ + A⁻.
• The system contains both reactants and products at equilibrium.

📏 Acid dissociation constant Ka

📏 What Ka measures

Acid dissociation constant (Ka): the equilibrium constant for the dissociation of a weak acid; it measures the relative strength of an acid.

• For the equilibrium HA + H₂O ⇌ H₃O⁺ + A⁻, the expression for Ka is:
  • Ka = [H₃O⁺][A⁻] / [HA]
• The larger the Ka value, the stronger the ability of the acid to donate protons, and the stronger the acid is.
• Technically, when Ka is larger than 10, the acid can be regarded as a strong acid.

📏 Inverse relationship in conjugate pairs

• For a conjugate acid-base pair, the stronger the acid, the weaker the conjugate base, and vice versa.
• This is a key principle: if an acid readily donates its proton (high Ka), the resulting conjugate base has little tendency to accept a proton back.
• Don't confuse: a strong acid and its conjugate base are not both strong; they are inversely related.

🧪 Organic acids

🧪 What makes an acid "organic"

• Inorganic acids (e.g., HCl, H₂SO₄, HF) are discussed in general chemistry.
• If the structure of the acid contains a "carbon" part, it is an organic acid.
• Organic acids donate protons in the same way as inorganic acids, but their structure may be more complicated due to the nature of organic structures.

🧪 Common organic acid types

Type	General formula	Example	Ka range	Notes
Carboxylic acid	R-COOH	Acetic acid (CH₃COOH)	1.8×10⁻⁵	Most common organic acid; ingredient in vinegar
Sulfonic acid	RSO₃H or ArSO₃H	p-toluenesulfonic acid	~10⁶	Strong organic acid; derived from H₂SO₄ by replacing one OH with R or Ar

• Carboxylic acid: the most common organic acid; general formula R-COOH.
• Sulfonic acid: formed by replacing one OH group in H₂SO₄ with a carbon-containing R (alkyl) or Ar (aromatic) group.
• Example: p-toluenesulfonic acid (also called tosylic acid) has the structure CH₃C₆H₄SO₃H.

🧪 Broader view of organic acids

• Technically, any organic compound could be an acid because organic compounds always have hydrogen atoms that could potentially be donated as H⁺.
• The excerpt shows examples with hydrogen atoms highlighted, indicating that many positions in organic molecules can act as acidic sites.
• Don't confuse: not all organic hydrogens are equally acidic; the structure and functional groups determine acidity.

🧭 Overview

🧠 One-sentence thesis

In organic chemistry, the scope of acids and bases extends far beyond simple inorganic compounds—any organic molecule with hydrogen can act as an acid, and any with electron pairs (lone pairs or π electrons) can act as a base, with their relative strengths measured by pKa values.

📌 Key points (3–5)

• Organic acids include more than carboxylic acids: any organic compound with hydrogen atoms can potentially donate H⁺, though most are weak acids.
• Organic bases share a common feature: they all have electron pairs (lone pairs on neutral/charged species or π electrons) that can accept protons.
• Reaction mechanisms use curved arrows: arrows start at electrons and end where electrons are received, showing step-by-step electron transfer between species (not just within one molecule like resonance).
• Common confusion—identifying bases: functional groups like C=O, alcohols, and ethers are bases even though they don't "look like" bases, because the oxygen lone pairs can accept protons.
• pKa measures relative acidity: smaller pKa means larger Ka and stronger acid; most organic acids have pKa between 5–60.

🧪 Types of organic acids

🧪 Carboxylic acids (R-COOH)

Carboxylic acid: organic acid with the general formula R-COOH, where R is a carbon-containing group.

• Most common and familiar organic acid.
• Example: acetic acid (CH₃COOH), found in vinegar, has Ka = 1.8×10⁻⁵.
• Donates protons in the same way as inorganic acids, but structure may be more complicated due to organic nature.

🧪 Sulfonic acids (RSO₃H or ArSO₃H)

• Organic derivative of sulfuric acid H₂SO₄.
• Formed by replacing one OH group in H₂SO₄ with a carbon-containing R (alkyl) or Ar (aromatic) group.
• Strong organic acid with Ka in the range of 10⁶.
• Example: p-toluenesulfonic acid (tosylic acid), CH₃C₆H₄SO₃H.

🧪 Broader definition—any organic compound

• Technically, any organic compound can be an acid because all organic compounds have hydrogen atoms that could potentially be donated as H⁺.
• The excerpt shows examples with hydrogen atoms highlighted, emphasizing that the scope of acids is broadly extended in organic chemistry.
• Most of these are weak acids, which makes it difficult to recognize them as acids at first.

🔋 Types of organic bases

🔋 What makes an organic base

Base: a species able to accept protons; organic bases must have electron pairs (lone pairs or π electrons) that can accept protons.

• It is not always easy to identify organic bases because they involve a variety of structures.
• The common feature: electron pairs available to accept protons.
• Electron pairs can be:
  • Lone pair electrons on neutral or negatively charged species
  • π electron pairs in double bonds

⚡ Negatively charged organic bases

• Examples: RO⁻ (alkoxide), RNH⁻ (amide), R⁻ (alkide, conjugate base of alkane).
• High electron density makes them usually stronger bases than neutral ones.
• Important note: Lone pairs are usually omitted in organic structures; for example, CH₃NH⁻ actually has two pairs of lone pair electrons on N.

🧲 Neutral organic bases—amines

• Examples: RNH₂, R₂NH, R₃N, ArNH₂.
• Organic derivatives of NH₃ (an inorganic weak base).
• Weak bases with lone pair electrons on nitrogen that can accept protons.

🧲 Neutral organic bases—oxygen-containing groups

Functional groups with oxygen atoms can act as bases:

Functional group	Formula	Why it's a base
Carbonyl	C=O	Lone pairs on O accept protons
Alcohol	R-OH	Lone pairs on O accept protons
Ether	R-O-R	Lone pairs on O accept protons

• Aldehyde, ketone, alcohol, and ether are all organic bases.
• Don't confuse: These groups may not "look like" bases, but they are bases according to the definition because the oxygen lone pairs can accept protons.
• The excerpt emphasizes: "Adjust your thinking here to embrace the broader scope of acids and bases in an organic chemistry context."

🧲 Neutral organic bases—alkenes (C=C)

• Although there are no lone pair electrons, the π electrons of the C=C double bond can accept protons.
• The double bond acts as a base by using its π electrons.

🔄 Reaction mechanisms and curved arrows

🔄 What is a reaction mechanism

Reaction mechanism: a step-by-step electron transfer process that converts reactants to products.

Curved arrows: used to illustrate the reaction mechanism; they always start at the electrons and end in the spot that is receiving the electrons.

• Curved arrows show how electrons move during a reaction.
• Key distinction: Curved arrows for mechanisms connect between species, whereas curved arrows for resonance structures show electron transfer within the same molecule.

🔄 Example—protonation of acetone (ketone)

The excerpt provides a detailed example of acetone reacting with H⁺:

• The curved arrow starts from the electron pair on oxygen (in C=O) and points to H⁺.
• A new O-H bond forms as a result of this electron pair movement.
• This reaction is called "protonation of ketone."
• The product is an "oxonium ion," which is stabilized by resonance with a carbocation structure.

🔄 Determining acid and base in a reaction

When there is no obvious acid or base:

• Compare electron density: the species with higher electron density acts as the base.
• Example from the excerpt: methanol (CH₃OH) is neutral, and amide (NH₂⁻) is negatively charged.
• The negatively charged amide has higher electron density, so NH₂⁻ is the base and CH₃OH is the acid that donates H⁺.

📏 pKa and relative acidity

📏 Definition and relationship to Ka

pKa = -log Ka

• The smaller the pKa value, the larger the Ka, and the stronger the acidity.
• pKa is used more often in organic chemistry than Ka because it is more convenient than remembering values like 1.8×10⁻⁵.

📏 Range for organic acids

• Most organic acids are weak acids with small Ka.
• The pKa of most organic acids ranges between 5 and 60.
• Example: acetic acid CH₃COOH has Ka = 1.8×10⁻⁵.
• Many other organic acids are even weaker than acetic acid, making it difficult to realize they are acids.

📏 Importance in organic chemistry

• Weak acidity is very important in organic chemistry.
• Understanding pKa (and acidity) based on functional groups is very useful, even though it is impossible to know the pKa of every organic compound.
• The excerpt notes that pKa can be used to predict acid-base reaction outcomes (though this section is cut off in the provided text).

🧭 Overview

🧠 One-sentence thesis

The pKa value allows chemists to predict which direction an acid-base reaction will proceed by comparing acid strengths, with reactions always favoring formation of the weaker acid and weaker base.

📌 Key points (3–5)

• What pKa measures: pKa = -log Ka; smaller pKa means stronger acid, and most organic acids have pKa between 5 and 60.
• Prediction rule: acid-base reactions favor the side with the weaker acid (larger pKa) and weaker base because equilibrium favors more stable products.
• Functional group patterns: different functional groups have characteristic pKa ranges (carboxylic acids ~5, alkanes >50), making it possible to estimate acidity without memorizing exact values.
• Common confusion: OH⁻ is considered a "strong base" in general chemistry, but it's not strong enough to deprotonate very weak acids like HC≡CH (pKa ~25); stronger bases like NH₂⁻ are needed.
• Practical application: knowing pKa values helps choose the right reagent in the lab—for example, selecting NH₂⁻ over OH⁻ to deprotonate ethyne.

📊 Understanding pKa as a measure of acidity

📐 Definition and relationship to Ka

pKa = -log Ka

• Ka is the acid dissociation constant; pKa converts it to a more convenient scale.
• The smaller the pKa, the larger the Ka, and the stronger the acid.
• Example: acetic acid has Ka = 1.8×10⁻⁵, which is inconvenient to remember; pKa makes comparisons easier.

🔢 Typical pKa ranges for organic compounds

• Most organic acids are weak acids with pKa values ranging from 5 to 60.
• The excerpt emphasizes that all organic compounds can theoretically act as acids because they all contain hydrogen atoms that could be donated.
• Carboxylic acids are the strongest organic acids listed (pKa ~5).
• Alkanes are the weakest organic acids (pKa >50).

🧪 Functional group approach

• It's impossible to memorize the pKa of every organic compound.
• Instead, understand pKa based on functional groups: the same functional groups usually have similar pKa values.
• The approximate range is usually sufficient for most organic chemistry purposes; exact values are rarely necessary.

⚠️ Important note about specific hydrogen atoms

• When discussing acidity of an organic compound, it refers to a specific H atom, not the molecule as a whole.
• Different H atoms in the same compound have different acidity and pKa values.
• Example: in methanol, the H on the oxygen (O-H) has different acidity than the H atoms on carbon (C-H).

🎯 Predicting reaction outcomes using pKa

🧲 The fundamental rule

Acid-base reactions always favor the formation of the weaker acid and the weaker base.

• This is because equilibrium favors more stable products.
• Weaker acids and bases are more stable than stronger ones.
• The reaction proceeds toward the side with the larger pKa (weaker acid).

🔄 How to apply the rule

Step-by-step approach:

1. Identify the acids on both sides of the reaction (reactant side and product side).
2. Compare their pKa values.
3. The side with the larger pKa (weaker acid) will be favored.

Why comparing acids alone is sufficient:

• If one acid is stronger than another, its conjugate base must be weaker than the other base.
• You only need to compare the acids to determine the equilibrium position.

📝 Example comparison

The excerpt shows methanol (CH₃OH) reacting with amide (NH₂⁻):

• Reactant acid: CH₃OH
• Product acid: NH₃ (ammonia)
• NH₃ has a larger pKa than CH₃OH, meaning NH₃ is weaker.
• Therefore, the reaction proceeds to the right, forming CH₃O⁻ and NH₃ as major products.

🔬 Practical laboratory applications

🧪 Choosing the right reagent

Reaction 1: Ethyne with amide

• HC≡CH (pKa ~25) + NH₂⁻ → reaction proceeds
• NH₂⁻ is strong enough to deprotonate ethyne.

Reaction 2: Ethyne with hydroxide

• HC≡CH (pKa ~25) + OH⁻ → no reaction
• OH⁻ is not strong enough to deprotonate ethyne.

💡 Resolving the "strong base" confusion

Common misconception:

• In general chemistry, OH⁻ is taught as a strong base.
• This conflicts with the finding that OH⁻ cannot deprotonate HC≡CH.

Resolution:

• OH⁻ is indeed a pretty strong base in general terms.
• However, it's just barely not strong enough to deprotonate HC≡CH.
• HC≡CH is a very weak acid (pKa ~25), much weaker than the "weak acids" from general chemistry.
• Very weak acids require much stronger bases like NH₂⁻ for deprotonation.

🎓 Lab decision-making

• If you need to deprotonate HC≡CH in the lab, you now know to choose NH₂⁻ over OH⁻.
• This demonstrates how pKa predictions have direct practical value in experimental design.

📋 Functional group acidity hierarchy

📊 Acidity trends in the table

Direction	Acidity	Conjugate base basicity
Top to bottom	Decreases	Increases

• Carboxylic acids (top, pKa ~5) are most acidic.
• Alkanes (bottom, pKa >50) are least acidic.
• Inverse relationship: the stronger the acid, the weaker its conjugate base.

🧠 Memorization strategy

• The excerpt recommends memorizing the approximate pKa ranges for major functional groups.
• This provides a framework for estimating acidity without needing exact values.
• Seven major functional groups are listed in the referenced table, spanning the range from ~5 to >50.

🧭 Overview

🧠 One-sentence thesis

The acidity of organic compounds is determined by four structural factors—element effect, resonance effect, inductive effect, and hybridization effect—all of which work by stabilizing the conjugate base to varying degrees.

📌 Key points (3–5)

• Core principle: A more stable (weaker) conjugate base corresponds to a stronger acid; all four structural effects are explained through conjugate base stability.
• Element effect: Acidity increases with electronegativity across a period (left to right) and with atomic size down a group (top to bottom).
• Resonance and inductive effects: Resonance delocalizes charge over multiple atoms (e.g., acetate vs. ethoxide); inductive effect spreads charge through σ bonds via electronegative atoms (e.g., chlorinated acetic acids).
• Hybridization effect: More s character (sp > sp² > sp³) means electrons are closer to the nucleus, stabilizing the anion and increasing acidity.
• Common confusion: Electronegativity dominates within a period, but atomic size dominates within a group—don't apply the same reasoning to both directions on the periodic table.

🔬 Element Effect

🔬 Periodic (horizontal) trend: Electronegativity

For elements in the same period, the more electronegative an atom, the stronger the acid is; the acidity increases from left to right across the period.

• Why it works: Higher electronegativity means the atom can better accommodate negative charge on the conjugate base, stabilizing it.
• Example comparison: Ethane (C–H), methylamine (N–H), and ethanol (O–H) show increasing acidity from C to N to O along the second row.
• The more stable the conjugate base, the weaker it is as a base, and therefore the stronger the original acid is.

🔬 Group (vertical) trend: Atomic size

For elements in the same group, the larger the size of the atom, the stronger the acid is; the acidity increases from top to bottom along the group.

• Why size matters: A larger atom spreads the negative charge over a bigger volume, lowering charge density and stabilizing the anion.
• Example: Haloacids HX increase in acidity from HF to HI; iodide ion is much larger than fluoride, so the negative charge is spread out and I⁻ is more stable (less basic), making HI more acidic.
• Another example: Thiol (CH₃CH₂SH, pKₐ ~10) is much more acidic than the corresponding alcohol (CH₃CH₂OH, pKₐ ~16) because sulfur is larger than oxygen.
• Don't confuse: Within a period, electronegativity dominates; within a group, size dominates—the two trends use different reasoning.

🌀 Resonance Effect

🌀 Charge delocalization stabilizes the conjugate base

• What it is: When multiple resonance contributors can be drawn, the negative charge is delocalized (shared) over multiple atoms, increasing stability.
• Key comparison: Acetic acid vs. ethanol
  • Both have H bonded to oxygen (no element effect difference).
  • Acetate (conjugate base of acetic acid) has two resonance contributors, so the negative charge is delocalized over two oxygens.
  • Ethoxide (conjugate base of ethanol) has no additional resonance contributors; the negative charge is localized on one oxygen.
  • Result: Acetate is much more stable and weaker as a base, so acetic acid is much stronger (pKₐ difference of over 10 units, meaning a factor of about 10¹⁰ in Kₐ).

🌀 Phenol as a special case

• Phenol (OH on a benzene ring) has a pKₐ of about 10, much more acidic than typical alcohols (pKₐ ~15).
• Why: The conjugate base of phenol is stabilized by resonance with the benzene ring—four additional resonance contributors delocalize the negative charge onto the ring.
• Practical implication: NaOH can deprotonate phenol effectively but not normal alcohols like ethanol, because phenol is acidic enough for the reaction to proceed.

⚡ Inductive Effect

⚡ Electron-withdrawing groups through σ bonds

The inductive effect is the charge dispersal effect of electronegative atoms through σ bonds.

• How it works: Electronegative atoms (like chlorine or fluorine) pull electron density toward themselves through covalent bonds, spreading out the negative charge on the conjugate base and stabilizing it.
• Example: Chlorinated acetic acids
  • Acetic acid vs. mono-, di-, and tri-chlorinated derivatives show increasing acidity.
  • Chlorine atoms are more electronegative than hydrogen, so they induce electron withdrawal and stabilize the carboxylate anion.
  • The effect is additive: more chlorine atoms → stronger overall effect → higher acidity.

⚡ Factors that influence inductive effect strength

Factor	Effect on acidity	Example
Electronegativity	Higher EN → stronger inductive effect	Trifluoroacetic acid (TFA, pKₐ = -0.25) is stronger than trichloroacetic acid (pKₐ = 0.64) because F is more electronegative than Cl
Distance	Effect decreases significantly with distance through bonds	A chlorine two carbons away from the carboxylic acid has a weaker effect than one carbon away

• Don't confuse with resonance: Inductive effect works through σ bonds and depends on distance; resonance involves π systems and delocalization over multiple atoms.

🔗 Hybridization Effect

🔗 s character and orbital energy

• Key comparison: Alkane (C–H), alkene (C=C–H), and alkyne (C≡C–H) show increasing acidity.
• No element effect (all H bonded to C) and no extra resonance contributors in the conjugate bases.
• The difference: Hybridization of the carbon atom.

Hybridization	s character	Stability of anion	Acidity
sp³ (alkane)	25% (1 s, 3 p)	Least stable	Weakest acid
sp² (alkene)	33.3% (1 s, 2 p)	Intermediate	Intermediate
sp (alkyne)	50% (1 s, 1 p)	Most stable	Strongest acid

🔗 Why s character matters

• Electrons in s orbitals are closer to the nucleus than those in p orbitals (lower energy level).
• More s character → electrons experience stronger nuclear attraction → anion is lower in energy and more stable.
• Electrostatic potential maps show this: lighter color (less red) in sp anion indicates less electron density and higher stability.

🔗 General principle

For the same atom, an sp hybridized atom is more electronegative than an sp² hybridized atom, which is more electronegative than an sp³ hybridized atom.

• This means hybridization changes the effective electronegativity of the atom.
• Example: The conjugate base of alkyne (sp carbon) is much more stable than that of alkane (sp³ carbon), making alkyne significantly more acidic.

🧪 Practical Applications

🧪 Choosing the right base in the lab

• Scenario: Deprotonating ethyne (HC≡CH, pKₐ ~25).
• Amide (NH₂⁻) can deprotonate ethyne (reaction proceeds to products).
• Hydroxide (OH⁻) cannot deprotonate ethyne effectively (no reaction).
• Why: Ethyne is a very weak acid (pKₐ ~25), much weaker than typical "weak acids" from general chemistry, so a much stronger base like NH₂⁻ is required.
• Don't confuse: OH⁻ is generally a strong base, but it is just barely not strong enough for this specific very weak acid—context matters.

🧪 Predicting reaction outcomes with pKₐ

• By comparing pKₐ values, you can predict whether an acid-base reaction will proceed.
• If you have a choice of bases in the lab, knowing these structural effects helps you select the appropriate one for your target molecule.

🧭 Overview

🧠 One-sentence thesis

The Lewis definition of acids and bases expands beyond proton transfer to include any species that can accept or donate electron pairs, covering a broader range of chemical reactions.

📌 Key points (3–5)

• Lewis definition is broader: includes all Brønsted-Lowry acids/bases but also covers reactions without proton transfer.
• Lewis acid: a species that can accept an electron pair (e.g., electron-deficient species with incomplete octets).
• Lewis base: a species that can donate an electron pair (e.g., species with lone pair electrons).
• Common confusion: all Brønsted-Lowry acids/bases are Lewis acids/bases, but not all Lewis acids/bases involve protons—the Lewis definition is more inclusive.
• LA-LB adduct: the product formed when a Lewis acid and Lewis base join together.

🔄 From Brønsted-Lowry to Lewis definitions

📖 Brønsted-Lowry limitation

• The Brønsted-Lowry definition works well for many reactions but limits acid-base reactions to those where the proton H⁺ must be involved.
• This restriction excludes important reactions that don't involve proton transfer.

🌐 Lewis definition (1923)

Lewis Acid: a species that can accept an electron pair
Lewis Base: a species that can donate an electron pair

• First introduced by G.N. Lewis in 1923.
• More inclusive way to define acids and bases.
• The key shift: focus on electron pair movement rather than proton movement.

🔗 How Brønsted-Lowry fits into Lewis

• All Brønsted-Lowry acids and bases fit into the Lewis definition.
• Why: proton transfer is essentially the base using its electron pair to accept a proton.
• The mechanism arrow shows this electron pair donation.
• Example: In the reaction with NH₃ and H⁺:
  • H⁺ is both the Brønsted-Lowry acid and the Lewis acid
  • NH₃ is both the Brønsted-Lowry base and the Lewis base
  • NH₃ donates its electron pair to accept the proton

🧪 Lewis acids and bases without protons

⚛️ Electron-deficient Lewis acids

• The Lewis definition covers situations beyond proton transfer.
• Example: B(CH₃)₃ (trimethylborane) as a Lewis acid:
  • Boron has an incomplete octet
  • The empty 2p orbital on boron can accept electrons
  • No proton is involved in this reaction

🎯 Electron-donating Lewis bases

• Example: (CH₃)₃N (trimethylamine) as a Lewis base:
  • Has a lone pair of electrons on nitrogen
  • The lone pair can be donated to the Lewis acid
  • Forms a bond by sharing the electron pair

🔗 LA-LB adduct formation

LA-LB adduct: the product between Lewis acids and Lewis bases, usually a species that has the acid and base joined together.

• The Lewis acid and Lewis base combine to form a single product.
• The bond forms through the electron pair donation from base to acid.
• Example: B(CH₃)₃ + (CH₃)₃N → adduct with B-N bond

📋 Common Lewis acids and bases

🧲 Examples of Lewis acids

Lewis acids include electron-deficient species:

• H⁺ (proton)
• M⁺ (metal cations)
• M²⁺ (metal dications)
• BH₃ (borane)
• BF₃ (boron trifluoride)
• AlCl₃ (aluminum chloride)

Common feature: all have the ability to accept electron pairs, either through positive charge or incomplete octet.

💧 Examples of Lewis bases

Lewis bases include species with lone pair electrons:

• Amines (nitrogen with lone pair)
• Ethers (oxygen with lone pairs)
• Other species that have lone pair electrons to donate

Common feature: all possess at least one lone pair of electrons available for donation.

🔍 Don't confuse

• Not all Lewis acid-base reactions involve protons, even though all proton-transfer reactions are Lewis acid-base reactions.
• The presence of an incomplete octet or positive charge suggests a Lewis acid; the presence of lone pairs suggests a Lewis base.

🧭 Overview

🧠 One-sentence thesis

The rotation of single (sigma) bonds in alkanes produces different spatial arrangements called conformations, which differ in energy due to torsional and steric strain, with staggered conformations being more stable than eclipsed ones.

📌 Key points (3–5)

• What conformations are: different spatial arrangements of atoms resulting from single bond rotation; molecules with different conformations are called conformers.
• Two extreme types: staggered conformations (atoms spread out, more stable) vs. eclipsed conformations (atoms overlapped, less stable due to torsional strain).
• Two types of strain: torsional strain (electron repulsion when bonds eclipse) and steric strain (repulsion when large groups are close together, ≤60°).
• Common confusion: anti vs. gauche conformations—both are staggered, but anti has largest groups 180° apart (most stable), while gauche has them 60° apart (less stable due to steric strain).
• Why it matters: energy differences determine which conformations predominate at room temperature (e.g., 99% of ethane molecules are staggered at any moment).

🔄 Nature of conformations and bond rotation

🔄 What causes conformations

• Sigma bonds in alkanes rotate continuously at the molecular level.
• The C-C bond is formed by sp³-sp³ orbital overlap and is cylindrically symmetrical, allowing easy rotation.
• Example: in ethane (CH₃CH₃), one methyl group can rotate around the C-C bond freely.
• The molecule appears unchanged during rotation, but a closer look reveals different spatial arrangements of hydrogen atoms.

Conformations: the different spatial arrangements of atoms/groups that result from single bond rotation.

Conformational isomers (conformers): molecules with different conformations.

📐 How to visualize conformations

Three types of structural formulas are used:

Formula type	Description	View angle
Perspective formula	Uses solid/dashed wedges to show spatial arrangement	Side view
Sawhorse formula	Shows tilted arrangement	Tilted top view
Newman projection	Most commonly used; views directly along C-C bond	End-on view

Newman projection rules:

• Front carbon: shown as a point with three bonds radiating out
• Rear carbon: shown as a circle with three bonds
• In eclipsed conformers, rear bonds are drawn slightly tilted to remain visible (though they should be completely overlapped)

⚡ Energy differences and strain

⚡ Torsional strain (eclipsing strain)

Torsional strain: repulsion between electrons of C-H bonds when they overlap in eclipsed conformations.

• Occurs when bonds on adjacent carbons eclipse (overlap) each other.
• Makes eclipsed conformations higher in energy than staggered ones.
• Example: in ethane, the eclipsed conformer is 12 kJ/mol (about 2.9 kcal/mol) higher in energy than the staggered conformer.

🔨 Steric strain

Steric strain: strain caused when atoms or groups are close enough together (≤60°) that their electron clouds repel each other.

• Only matters when groups are close to each other (60° or less).
• Does NOT apply in anti-conformations (180° apart).
• Magnitude depends on group size: larger groups → higher steric strain.
• No steric strain between two small hydrogen atoms, even when close.

🔄 Energy barriers and interconversion

• Energy barriers exist between conformations but are small enough to be overcome at room temperature.
• Conformers interconvert millions of times per second at room temperature.
• Cannot be separated from each other due to continuous interconversion.
• At any moment, molecules predominantly exist in the most stable conformation (e.g., 99% of ethane is staggered).

🧪 Ethane conformation analysis

🧪 Two conformations of ethane

Staggered conformation:

• All H atoms spread out as far apart as possible
• Lower energy, more stable
• Predominant form (99% at room temperature)

Eclipsed conformation:

• All H atoms overlapped
• Higher energy due to torsional strain
• 12 kJ/mol (2.9 kcal/mol) higher than staggered

📊 Potential energy diagram

• Shows energy vs. angle of rotation about the C-C bond
• Energy varies periodically as rotation occurs
• Three staggered conformations (energy minima) at 0°, 120°, 240°
• Three eclipsed conformations (energy maxima) at 60°, 180°, 300°

🔬 Propane conformation analysis

🔬 Similar pattern to ethane

• Still two types of conformations: staggered and eclipsed
• Difference: one methyl (CH₃) group on the rear carbon instead of hydrogen
• This does not affect relative stability: staggered is still more stable
• Same principle applies: torsional strain makes eclipsed conformations higher in energy

🧬 Butane conformation analysis

🧬 Complexity with larger groups

• Butane has three C-C bonds; rotation can occur about each
• Most complex case: rotation along the C2-C3 bond (both carbons have two H atoms and one CH₃ group)
• Produces six different conformers: three staggered and three eclipsed

🎯 Four distinct conformations

Two staggered conformations:

1. Anti-conformation (most stable):

   • Two CH₃ groups 180° opposite to each other
   • Largest groups as far apart as possible
   • No steric strain

2. Gauche conformation (less stable):

   • Two CH₃ groups 60° adjacent to each other
   • Experiences steric strain due to large groups being close
   • Higher energy than anti

Two eclipsed conformations:

1. Eclipsed with CH₃ groups overlapping (least stable):

   • Two CH₃ groups eclipsing each other
   • Experiences both torsional strain AND steric strain
   • Highest energy

2. Eclipsed with CH₃ and H overlapping (intermediate):

   • Less steric strain than when CH₃ groups eclipse
   • Still has torsional strain
   • Higher energy than staggered, lower than the other eclipsed

📈 Complex energy diagram

• Four different energy levels corresponding to four distinct conformers
• More complex curve than ethane
• Energy barriers larger than ethane but still not high enough to prevent rotation at room temperature

Don't confuse: All staggered conformations are lower in energy than eclipsed ones, but not all staggered conformations have the same energy—anti is more stable than gauche due to steric strain in gauche.

🛠️ Tips for drawing all conformers

1. View along the C-C bond of interest; identify what atoms/groups are on each carbon
2. Start with the anti-conformation (largest groups opposite, 180° apart)
3. Fix groups on one carbon; rotate the other carbon in 60° increments
4. Repeat rotation five times for a total of six conformers

🧭 Overview

🧠 One-sentence thesis

Cyclohexane is the most stable cycloalkane because it adopts a chair conformation that eliminates both angle strain and torsional strain, unlike smaller rings which are forced into strained geometries.

📌 Key points (3–5)

• Stability varies by ring size: 5- and 6-membered rings are common in nature; 3- and 4-membered rings are rare due to high strain.
• Baeyer Strain Theory limitation: the theory assumed all rings are flat, predicting cyclopentane as most stable, but experiments show cyclohexane is actually strain-free.
• Strain energy measurement: heat of combustion reveals how much strain energy a cycloalkane contains—higher heat release means more strain.
• Common confusion: flat vs non-flat rings—most cycloalkanes are not planar; they pucker or fold to reduce strain.
• Why cyclohexane is special: its chair conformation achieves perfect 109.5° bond angles and staggered hydrogens, eliminating all strain.

🔬 Understanding cycloalkane stability

🔬 What determines stability

Angle strain: the difference between the ideal bond angle (109.5° for sp³ carbon) and the actual angle in the cycloalkane ring.

• Cycloalkanes have different stabilities depending on ring size.
• Observation from nature: 5- and 6-membered rings are common; 3- and 4-membered rings are rare.
• Stability is not just about size—it depends on how much strain the ring geometry creates.

📐 Baeyer Strain Theory and its flaw

• The theory's assumption: all rings are flat (planar).
• The prediction: cyclopentane would be most stable (108° vs ideal 109.5° = only 1.5° deviation); cyclohexane would be less stable (10.5° deviation).
• The problem: experimental results contradicted this—cyclohexane is actually the most stable and strain-free.
• Why it failed: the assumption that rings must be flat was wrong; rings can pucker or adopt 3D shapes.

🔥 Measuring strain with heat of combustion

• Method: measure heat released when cycloalkanes burn completely with oxygen.
• Logic: strained cycloalkanes are at higher energy levels, so they release more heat when burned.
• Calculation: strain energy = difference in heat of combustion between cycloalkane and a "strainless" chain alkane.

Cycloalkane	Strain Energy (kJ/mol)
Cyclopropane	114
Cyclobutane	110
Cyclopentane	25
Cyclohexane	0

• Larger difference = higher strain energy = less stable.

🔺 Small ring cycloalkanes (3- and 4-membered)

🔺 Cyclopropane—extremely strained

• Must be planar: with only three carbons, no other shape is possible.
• Angle strain: bond angle is 60°, a huge 49.5° deviation from ideal 109.5°.
• Weak bonds: sp³-sp³ orbitals overlap only partially due to angle deviation, creating "banana bonds" (bonds that resemble a banana shape).
• Torsional strain: all adjacent C-H bonds are eclipsed (overlapping), adding more strain.
• Result: 3-membered rings are very unstable, rarely exist in nature, and undergo ring-opening reactions easily.

🔷 Cyclobutane—still highly strained

• Not planar: the ring puckers (folds) slightly to release some torsional strain.
• Angle strain remains: internal angles are about 88°, still far from ideal 109.5°.
• Overall: still an unstable structure with high strain, though less than cyclopropane.

🔶 Medium and large rings (5- and 6-membered)

🔶 Cyclopentane—moderately stable

• Not planar: puckers into a bent "envelope" conformation.
• Why it puckers: one carbon atom sticks out of the plane, allowing some hydrogens to become almost staggered, reducing torsional strain.
• Strain level: significantly lower than 3- or 4-membered rings (25 kJ/mol).
• Ring flipping: the envelope conformation can undergo ring flipping due to C-C bond rotation.

⭐ Cyclohexane—the most stable

• Strain-free: zero strain energy, as stable as chain alkanes.
• Chair conformation: the most stable shape, resembling a chair.
• Perfect geometry: all carbons at 109.5° bond angles (no angle strain); all adjacent hydrogens perfectly staggered (no torsional strain).

🪑 Chair conformation properties

• Two types of C-H bonds: axial ("a") and equatorial ("e").
• Axial bonds: vertical, perpendicular to the average plane of the ring.
• Equatorial bonds: more "flat," extending from the perimeter of the ring.
• Pattern: each carbon has one axial and one equatorial bond; if one points up ↑, the other points down ↓.
• Alternating orientation: for the same bond type, up ↑ and down ↓ alternate from carbon to carbon.

Don't confuse: axial vs equatorial is about orientation relative to the ring, not about which is "better"—both are present in the chair conformation.

🧭 Overview

🧠 One-sentence thesis

Cyclohexane achieves strain-free stability through its chair conformation, where all bonds adopt staggered positions and substituents prefer equatorial positions to minimize repulsion.

📌 Key points (3–5)

• Why cyclohexane is special: it is the most stable cycloalkane because the chair conformation eliminates both angle strain (all 109.5° bond angles) and torsional strain (all staggered hydrogens).
• Two bond types in the chair: axial bonds are vertical and perpendicular to the ring plane; equatorial bonds are "flat" and extend from the ring perimeter.
• Ring flipping: cyclohexane rapidly interconverts between two chair conformations (~1 million times per second at room temperature), converting all axial bonds to equatorial and vice versa while keeping up/down positions unchanged.
• Common confusion: for unsubstituted cyclohexane the two chair conformers are equivalent, but for substituted cyclohexane (e.g., methylcyclohexane) the equatorial-substituent conformer is more stable due to 1,3-diaxal repulsion in the axial-substituent conformer.
• Boat conformation exists but is unstable: it has no angle strain but suffers from torsional strain (eclipsed hydrogens) and flagpole interaction, making it ~30 kJ/mol higher in energy than the chair.

🪑 The chair conformation and its properties

🪑 Why the chair is strain-free

Chair conformation: the most stable conformation of cyclohexane, resembling a chair shape.

• No angle strain: all carbons maintain 109.5° bond angles, matching the ideal tetrahedral geometry.
• No torsional strain: hydrogens on adjacent carbons are perfectly staggered, avoiding eclipsing interactions.
• This combination makes cyclohexane as stable as chain alkanes, unlike smaller cycloalkanes that suffer from angle or torsional strain.

🔵🔴 Axial and equatorial bonds

Axial bonds ("a"): vertical bonds perpendicular to the average plane of the ring.
Equatorial bonds ("e"): "flat" bonds extending from the perimeter of the ring.

• Each carbon has one axial and one equatorial bond.
• If one bond points up ↑ (above the ring), the other must point down ↓ (below the ring).
• For the same bond type, orientation alternates on adjacent carbons: if one carbon has a↑, the next has a↓.
• Total distribution: 3 a↑, 3 a↓, 3 e↑, and 3 e↓ (12 C-H bonds total).

Example: On carbon #1, if the axial bond points down (a↓), the equatorial bond must point up (e↑); on the adjacent carbon #2, the axial bond points up (a↑) and the equatorial bond points down (e↓).

✏️ How to draw the chair conformation

The excerpt provides a step-by-step procedure:

1. Draw two parallel lines of equal length, both pointing slightly down (forming a parallelogram with ~60°/120° internal angles).
2. Connect the right endpoints with a "V" shape, vertex pointing upper right.
3. Connect the left starting points with another "V" shape, vertex pointing bottom left.
4. Add all axial bonds as vertical lines, following the alternating up/down trend on adjacent carbons.
5. Add equatorial bonds following two rules:
   • On each carbon, if axial points up, equatorial points down (and vice versa).
   • Each equatorial bond is parallel to the C-C bond one bond away.

Tip from the excerpt: Use a molecular model set to practice; "practice makes perfect" for mastering chair drawing.

🔄 Ring flipping

🔄 What happens during ring flipping

Ring flipping: the chair-chair conformation conversion that cyclohexane undergoes.

• Not rotation: ring flipping comes from partial C-C bond rotation limited within the ring structure.
• Rapid interconversion: energy barrier is ~45 kJ/mol; at room temperature, thermal energy causes ~1 million flips per second.
• Bond transformation: all axial bonds become equatorial, and all equatorial bonds become axial.
• Position preservation: the up/down orientation relative to the ring remains the same.

Example: A bond that was a↓ (axial pointing down) becomes e↓ (equatorial pointing down) after flipping; a bond that was e↑ becomes a↑.

⚖️ Equivalence for unsubstituted cyclohexane

• The two chair conformers are conformation isomers (conformers).
• For cyclohexane itself (no substituents), the two conformers have the same energy level and are equivalent.
• This changes when substituents are present (discussed later).

Don't confuse: ring flipping is not a rotation of the entire molecule; it is a conformational change where the ring "flips" between two chair shapes.

🔍 Newman projection of the chair

🔍 Visualizing the staggered conformation

The chair conformation's staggered arrangement is not obvious in standard drawings; Newman projection clarifies this.

How to construct the Newman projection:

• Pick two parallel C-C bonds to view along (e.g., C1-C2 and C5-C4).
• Each bond is represented by a Newman projection circle (front and rear carbons).
• The two projections are connected through the remaining carbons (C3 and C6).

🔍 Key features in the Newman projection

• Staggered conformation is clearly visible: hydrogens on adjacent carbons do not eclipse each other.
• Axial bonds appear as vertical red lines; equatorial bonds appear as "flat" blue lines.
• A dashed line represents the average plane of the ring: bonds above are ↑, bonds below are ↓.
• Four carbons (C1, C2, C4, C5) show explicit "a" or "e" bonds; C3 and C6 are shown as CH₂ groups.

🚤 Other conformations: the boat

🚤 Boat conformation structure

Boat conformation: a higher-energy conformation of cyclohexane that resembles a boat shape.

• Formed by partial C-C bond rotation (flipping one carbon up from the chair).
• No angle strain: all carbons still have 109.5° bond angles.
• Has torsional strain: hydrogens on the base of the boat are in eclipsed positions, visible in Newman projection along C6-C5 and C2-C3 bonds.

🚤 Why the boat is unstable

• Torsional strain: eclipsed hydrogens on the boat base create repulsion.
• Flagpole interaction: the two hydrogens on C1 and C4 are very close to each other, causing steric strain.
• Combined strains make the boat conformation ~30 kJ/mol higher in energy than the chair.

Conformation	Angle strain	Torsional strain	Steric strain	Relative energy
Chair	None	None	None	Lowest (0)
Boat	None	Yes (eclipsed H)	Yes (flagpole)	+30 kJ/mol

🔬 Substituted cyclohexanes

🔬 Monosubstituted cyclohexane: methylcyclohexane example

For substituted cyclohexane, the two chair conformers from ring flipping are no longer equivalent.

Methylcyclohexane has two conformers:

• Conformation I: methyl group in axial position (less stable).
• Conformation II: methyl group in equatorial position (more stable, ~7.6 kJ/mol lower energy).

🔬 Why equatorial is more stable: 1,3-diaxal interaction

1,3-diaxal interaction: repulsion between an axial substituent (position #1) and axial hydrogens one carbon away (position #3).

• In the axial-methyl conformer, the CH₃ group is very close to axial hydrogens on carbons one bond away.
• This repulsion is essentially the same as gauche steric strain (the CH₃ and CH are in a gauche position).
• In the equatorial-methyl conformer, no such strains occur because the methyl group extends away from the ring.

Example: When the methyl group occupies an axial position, it "bumps into" nearby axial hydrogens, creating unfavorable steric repulsion; when it occupies an equatorial position, it has more space and avoids this repulsion.

Don't confuse: for unsubstituted cyclohexane, axial and equatorial positions are energetically equivalent; for substituted cyclohexane, equatorial substituents are generally more stable.

🧭 Overview

🧠 One-sentence thesis

Substituents on cyclohexane rings prefer equatorial positions over axial positions because axial placement causes repulsive 1,3-diaxial interactions, and this preference determines the most stable conformations of mono- and disubstituted cyclohexanes.

📌 Key points (3–5)

• Monosubstituted cyclohexanes: the equatorial conformer is more stable than the axial conformer due to 1,3-diaxial interactions.
• 1,3-diaxial interaction: repulsion between an axial substituent and axial hydrogens one carbon away; essentially the same as gauche steric strain.
• Size matters: larger substituents cause greater 1,3-diaxial strain, so the energy difference between axial and equatorial conformers increases with substituent size.
• Disubstituted cyclohexanes: geometric isomers (cis/trans) exist; the most stable conformer maximizes equatorial substituents, with larger groups preferring equatorial positions.
• Common confusion: ring flipping interconverts chair conformations (axial ↔ equatorial) but does NOT interconvert geometric isomers (cis vs trans).

🪑 Monosubstituted cyclohexane conformations

🔄 Ring flipping creates two non-equivalent conformers

• Unlike unsubstituted cyclohexane (where ring flipping produces equivalent conformers), substituted cyclohexane has two distinct chair conformations.
• One conformer places the substituent in an axial position; the other places it in an equatorial position.
• Example: methylcyclohexane has an axial-methyl conformer (I) and an equatorial-methyl conformer (II).

⚖️ Equatorial is more stable

For mono-substituted cyclohexane, the equatorial-conformer is more stable than the axial-conformer because of the 1,3-diaxial interaction.

• The equatorial-methyl conformer of methylcyclohexane is about 7.6 kJ/mol lower in energy than the axial-methyl conformer.
• At equilibrium, about 95% of methylcyclohexane molecules are in the equatorial conformation.

🔍 The 1,3-diaxial interaction

🧲 What causes the interaction

• When a substituent occupies an axial position (regarded as position #1), it is very close to the axial hydrogens one carbon away (regarded as position #3).
• This proximity causes repulsion between the substituent and those hydrogens.
• The excerpt notes this is "essentially the same as the gauche steric strain" because the CH₃ group and the CH are in a gauche position.

📐 Newman projection view

• The 1,3-diaxial interaction can be visualized more clearly using Newman projections.
• In the axial conformer, the substituent and nearby groups are gauche (close together).
• In the equatorial conformer, the substituent and nearby groups are anti (farther apart), eliminating the strain.

📏 Size effect

• Since 1,3-diaxial interaction is a type of steric strain, larger substituents cause greater interaction.
• Example: t-butylcyclohexane with the t-butyl group equatorial is about 21 kJ/mol more stable than the axial conformation.
• For t-butylcyclohexane, about 99.9% of molecules are in the equatorial conformation at equilibrium.
• Don't confuse: the energy difference is not fixed—it scales with substituent size.

🔀 Disubstituted cyclohexane

🧬 Geometric isomers: cis and trans

Geometric isomers are a type of stereoisomer where two substituents on different carbons can be either on the same side or the opposite side of the ring.

• Cis isomer: both groups on the same side of the ring.
• Trans isomer: groups on opposite sides of the ring.
• These isomers cannot interconvert because the C-C bonds in the ring cannot rotate freely.
• How to identify: check whether bonds point up (↑) or down (↓) relative to the ring; same direction = cis, opposite = trans.

📊 Stability guidelines for disubstituted conformers

Guideline	Explanation
Maximize equatorial positions	The more substituents in equatorial positions, the more stable the conformer
Larger group equatorial	When substituents differ in size, the conformer with the larger substituent equatorial is more stable

🔬 Cis-1,2-dimethylcyclohexane

• Both chair conformations have one methyl equatorial and one methyl axial.
• The two conformers are equivalent in energy and stability.
• Ring flipping does not change the relative stability.

🔬 Trans-1,2-dimethylcyclohexane

• One chair conformation has both methyls axial; the other has both methyls equatorial.
• These conformers are not equivalent.
• The di-equatorial conformer is much more stable (less 1,3-diaxial strain).

🔬 Different substituents: cis-1-fluoro-4-isopropylcyclohexane

• Both chair conformations have one axial and one equatorial substituent.
• The conformer with the larger substituent (isopropyl) equatorial is more stable.
• If the large group is axial, stronger steric strain is generated, making it less stable.

🎯 Determining the most stable isomer

🧪 Strategy for comparing isomers

1. Draw all chair conformers for each geometric isomer (cis and trans).
2. Identify which conformer of each isomer is most stable (maximize equatorial, prioritize larger groups equatorial).
3. Compare the most stable conformers across isomers to determine the overall most stable structure.

💡 Example approach

• For cis-1-ethyl-2-methylcyclohexane vs trans-1-ethyl-2-methylcyclohexane:
  • The trans isomer can have a conformer with both substituents equatorial.
  • The cis isomer must always have one axial and one equatorial.
  • Therefore, the trans isomer is more stable overall.

🧭 Overview

🧠 One-sentence thesis

Organic molecules can exist as constitutional isomers (different bonding) or stereoisomers (same bonding, different spatial arrangements), with stereoisomers including geometric isomers and chirality-based isomers that have distinct properties.

📌 Key points (3–5)

• Two major categories: constitutional isomers differ in bonding; stereoisomers have identical bonding but different spatial arrangements.
• Geometric isomers arise from restricted rotation: caused by cyclic structures or double bonds, preventing groups from freely rotating.
• E/Z naming system: IUPAC method for designating geometric isomers of alkenes based on priority rules and spatial arrangement.
• Common confusion: cis/trans vs E/Z—cis/trans is simpler but can be ambiguous; E/Z is systematic and applies priority rules.
• Chirality centers introduce new stereoisomers: molecules with chirality centers form a distinct category beyond geometric isomers.

🗂️ The isomer hierarchy

🗂️ Constitutional vs stereoisomers

Constitutional isomers: molecules with different bonding patterns.

Stereoisomers: molecules with the same bonding but groups in different spatial arrangements.

• The fundamental split divides all isomers into these two types.
• Constitutional isomers were covered in earlier chapters; this chapter focuses on stereoisomers.
• Example: two molecules with the same molecular formula but different connectivity are constitutional isomers; two molecules with identical connectivity but different 3D arrangements are stereoisomers.

🌳 Stereoisomer subcategories

The excerpt identifies two main types of stereoisomers:

• Geometric isomers: result from restricted rotation (cyclic structures or double bonds).
• Isomers with chirality centers: a distinct category involving chiral molecules.

🔄 Geometric isomers

🔄 What causes geometric isomers

• Restricted rotation is the key requirement.
• Two sources of restriction:
  • Cyclic structures (e.g., disubstituted cycloalkanes)
  • Double bonds (e.g., alkenes with sp² carbons)
• Without restricted rotation, groups can freely rotate and no distinct geometric isomers exist.

🔗 Cycloalkane geometric isomers

• Example from the excerpt: 1,2-dimethylcyclohexane.
• The ring prevents C-C bond rotation.
• Two arrangements are possible:
  • cis: both substituents on the same side of the ring.
  • trans: substituents on opposite sides.
• Don't confuse: the ring itself doesn't create isomers; it's the restricted rotation combined with multiple substituents.

🔗 Alkene geometric isomers

• Double bonds also restrict rotation because sp² carbons form trigonal planar shapes.
• Example: 2-butene (CH₃-CH=CH-CH₃).
• A simple condensed formula doesn't show the geometry; a Kekulé structure reveals two distinct shapes:
  • CH₃ groups on the same side of the double bond.
  • CH₃ groups on opposite sides of the double bond.
• These are geometric isomers, similar to cycloalkane cis/trans isomers.

🏷️ E/Z naming system

🏷️ Why E/Z is needed

• cis/trans is a common designation but can be ambiguous for some structures.
• E/Z is the IUPAC systematic naming method for geometric isomers of alkenes.
• It provides unambiguous designation by using priority rules.

📐 How E/Z assignment works

Step 1: Assign priority

• Groups connected to each sp² double bond carbon are assigned priority based on atomic number.
• Higher atomic number = higher priority.

Step 2: Determine E or Z

• Z (from German "Zusammen" = same side): the higher-priority groups on each carbon are on the same side of the double bond.
• E (from German "Entgegen" = opposite): the higher-priority groups are on opposite sides of the double bond.

Designation	Priority group positions	Meaning
Z	Same side of double bond	"Together"
E	Opposite sides of double bond	"Opposite"

🔍 E/Z vs cis/trans comparison

• Both describe geometric isomers, but E/Z is more systematic.
• cis/trans may be unclear when substituents are complex or when there are more than two different groups.
• E/Z uses priority rules to handle all cases unambiguously.
• Example: for simple cases like 2-butene, cis corresponds to Z and trans corresponds to E, but this equivalence doesn't always hold for more complex molecules.

🧭 Overview

🧠 One-sentence thesis

The E/Z naming system provides an unambiguous IUPAC method for designating geometric isomers of alkenes by assigning priority to groups based on atomic number and determining whether higher-priority groups are on the same side (Z) or opposite sides (E) of the double bond.

📌 Key points (3–5)

• What causes geometric isomers in alkenes: restricted rotation around double bonds creates different spatial arrangements, similar to how cyclic structures create cis/trans isomers.
• How E/Z designation works: assign priority to groups on each sp² carbon based on atomic number, then label Z ("Zusammen" = same side) if high-priority groups are together, or E ("Entgegen" = opposite) if apart.
• Priority assignment rules: compare atomic numbers of directly bonded atoms first; if tied, move outward to next atoms; treat multiple bonds as multiple single bonds.
• Common confusion: E/Z and cis/trans are different systems and don't always match—a compound can be cis but E, or trans but Z.
• Why it matters: E/Z eliminates ambiguity in naming geometric isomers, especially for complex structures where cis/trans designations become unclear.

🔄 What are geometric isomers of alkenes

🔄 Restricted rotation creates isomers

• Double bonds prevent free rotation around the C=C bond, just as cyclic structures restrict rotation.
• The sp² carbons in a double bond have trigonal planar geometry.
• Groups attached to the double bond can be arranged on the same side or opposite sides, creating different isomers.

🧪 Example: 2-butene

• The condensed formula CH₃-CH=CH-CH₃ doesn't show the spatial arrangement.
• Drawing the Kekulé structure reveals two possibilities: both CH₃ groups on the same side or on opposite sides.
• These are geometric isomers, traditionally labeled cis (same side) or trans (opposite side).

⚠️ Limitations of cis/trans naming

• The excerpt notes that cis/trans designation "might be ambiguous for some structures."
• This motivates the need for the more systematic E/Z naming system.

🏷️ The E/Z naming system

🏷️ Basic principle

The isomer with the same priority group on the same side of the double bond is assigned "Z", and the isomer with the same priority group on the opposite side of the double bond is called "E".

• Z comes from German "Zusammen" = "same side"
• E comes from German "Entgegen" = "opposite"
• Priority is assigned separately for groups on each sp² carbon of the double bond.

📋 How to apply E/Z

1. Identify the two groups attached to each sp² carbon of the double bond
2. Assign priority to groups on each carbon (higher priority = #1, lower = #2)
3. Compare the spatial relationship of the two #1 groups across the double bond
4. If both #1 groups are on the same side → Z isomer
5. If both #1 groups are on opposite sides → E isomer

🎯 Priority assignment rules

🎯 Rule 1: Compare atomic numbers directly

Priority is assigned based on the atomic number of the atoms bonded directly to the sp² double bond carbon – the larger the atomic number, the higher the priority.

• Order: S > O > N > C > H
• Isotopes with higher mass number have higher priority.
• Example: In 2-pentene, on the left sp² carbon, CH₃ (carbon) is higher priority than H; on the right sp² carbon, CH₂CH₃ (carbon) is higher than H.

🔗 Rule 2: When there's a tie, look at next atoms

• If both groups start with the same atom attached to the sp² carbon, move outward to the next set of atoms.
• Compare the atoms bonded to the "tied" atom.

Example from excerpt: Comparing CH₃ vs CH₂CH₃

• Both have carbon directly attached to sp² carbon (tie)
• CH₃: that carbon is bonded to H, H, H
• CH₂CH₃: that carbon is bonded to H, H, C
• Since C > H, the ethyl group (CH₂CH₃) has higher priority

Important note: Compare the single atom with the greatest number in one group versus the single atom with the greatest number in the other group. Do NOT add atomic numbers.

• Example: C, C, C vs C, O, H → the C, O, H side is higher because O > C

🔄 Rule 3: Repeat if necessary

• Continue moving outward through the structure until priority can be assigned.

🔗 Rule 4: Multiple bonds

When a multiple bond is part of the group, the multiple bond is treated as if it was singly bonded to several of those atoms.

• A double bond is treated as two single bonds to that atom
• A triple bond is treated as three single bonds to that atom
• This allows comparison when groups contain C=C, C=O, C≡C, etc.

⚠️ Common confusions and distinctions

⚠️ E/Z vs cis/trans mismatch

System	Basis	Can they differ?
cis/trans	Same or opposite sides (traditional)	Yes—they are different systems
E/Z	Priority-based (IUPAC)	that don't always match

Key example from excerpt: A compound can be a cis-isomer (both ethyl groups on same side) but an E-isomer in the E/Z system, because priority assignment may make the "higher priority" groups end up on opposite sides.

🔍 Don't confuse: atomic number comparison

• When comparing H, H, H vs H, H, C: compare the highest single atom in each group (H vs C), not the sum
• When comparing C, C, C vs C, O, H: the side with O wins because O > C, even though the other side has three carbons

📝 Working through examples

📝 Determining E or Z from a structure

Approach:

1. Identify groups on left sp² carbon and assign priority
2. Identify groups on right sp² carbon and assign priority
3. Check if #1 groups are on same side (Z) or opposite sides (E)

Example: 3-chloro-4-methyl-3-hexene

• Right sp² carbon: Cl > C (of CH₂CH₃), so Cl is #1
• Left sp² carbon: CH₂CH₃ vs CH₃ → tie on first carbon, so compare next atoms → CH₂CH₃ is #1
• The two #1 groups (Cl and CH₂CH₃) are on opposite sides
• Result: E isomer, named (E)-3-chloro-4-methyl-3-hexene

📝 Drawing a structure from E/Z name

Example: (E)-3-methyl-2-pentene

1. Draw the base chain with the double bond at position 2
2. Add the methyl substituent at position 3
3. Assign priorities on each sp² carbon
4. Arrange groups so higher-priority groups are on opposite sides (E designation)

🧭 Overview

🧠 One-sentence thesis

Chirality—the property of being non-superimposable on one's mirror image—creates pairs of enantiomers that are distinguished by the R/S naming system and differ in how they interact with plane-polarized light.

📌 Key points (3–5)

• What chirality means: the property of any object or molecule being non-superimposable on its mirror image, like left and right hands.
• Chirality center requirements: an sp³ carbon bonded to four different groups creates a chirality center, making the molecule chiral.
• Enantiomers: a pair of non-superimposable mirror images of each other; both are chiral and one has R configuration while the other has S configuration.
• Common confusion: not all carbons are chirality centers—CH₃ or CH₂ carbons and sp² double-bond carbons are NEVER chirality centers.
• Why it matters: enantiomers share most physical properties (boiling point, melting point, density) but differ in how they rotate plane-polarized light (optical activity).

🖐️ Understanding chirality

🖐️ What chirality is

Chirality: the property of any object (molecule) being non-superimposable on its mirror image.

• The term comes from the Greek word cheir, meaning "hand"—chirality means "handedness."
• Left and right hands are mirror images of each other but cannot be superimposed (overlaid) no matter how you try.
• Both left and right hands are chiral and show chirality.
• Example: A screw and a book also show chirality—their mirror images cannot be superimposed on the originals.

🔄 Chiral vs achiral

• Chiral: an object that is non-superimposable on its mirror image.
• Achiral: an object that is superimposable on its mirror image (the object and its mirror image are identical).
• Example: A cup and a Lego piece are achiral—their mirror images are identical to the originals.

🧪 Chirality in molecules

• When a central carbon is sp³ and bonded to four different groups (represented by four different colors in molecular models), the molecule is chiral.
• The two mirror-image structures cannot be superimposed on each other, so both are chiral.
• The chirality results from the structure of the central carbon, called the chirality center (or asymmetric center).
• A molecule with one chirality center must be chiral.

It is highly recommended to use a molecular model set to assemble the structures and understand the concept of chirality and for R/S assignment later.

🎯 Identifying chirality centers

🎯 Two requirements for a chirality center

A chirality (asymmetric) center must meet both:

1. sp³ carbon (not sp² double-bond carbon)
2. Bonded with four different groups

⚠️ What is NEVER a chirality center

• CH₃ or CH₂ carbons are NEVER chirality centers.
• sp² double-bond carbons are NEVER chirality centers.
• The chirality center must be a carbon bonded with a branch (or branches).
• Carbons in a ring can be chirality centers as long as they meet the two requirements.
• Not all compounds have a chirality center.

🔍 Approach to finding chirality centers

• Look for sp³ carbons bonded to branches.
• Check that all four groups attached to the carbon are different.
• Label each chirality center with a star.
• Example: In 2-pentanol, C2 is the chirality center because it is sp³ and bonded to four different groups (H, OH, CH₃, and CH₂CH₂CH₃).

🪞 Enantiomers: mirror-image stereoisomers

🪞 What enantiomers are

Enantiomers: molecules that are a pair of non-superimposable mirror images of each other.

• For 2-butanol, C2 is the chirality center.
• The two mirror images are different molecules with the same bonding but differ in how atoms are arranged in space—they are stereoisomers.
• This specific type of stereoisomer is defined as an enantiomer.

🔑 Important properties of enantiomers

Property	Description
Mirror images	Enantiomers are a pair of non-superimposable mirror images
Both chiral	Enantiomers are both chiral and show chirality (enantiomers must be chiral)
Always paired	For any chiral molecule, it must have its enantiomer (the mirror image)
Achiral molecules	Achiral molecules do not have enantiomers; their mirror image is identical to themselves

✏️ Drawing enantiomers

• Use a perspective formula with solid and dashed wedges to show tetrahedral arrangements around the sp³ carbon.
• Solid wedge: bond pointing out of the paper plane.
• Dashed wedge: bond pointing behind the paper plane.
• Ordinary lines: two bonds lying within the paper plane.
• For the first enantiomer, draw the four groups with any arrangement.
• Then draw the other enantiomer by drawing the mirror image of the first one.
• Although there seem to be different ways to show enantiomers, there are only two total enantiomers.

🏷️ The R/S naming system

🏷️ Why we need R/S designations

• The two enantiomers are different compounds (though very similar), so we need a nomenclature system to distinguish them.
• The R/S naming system (defined in IUPAC) gives each enantiomer a different designation.
• For a pair of enantiomers with one chirality center, one has the R configuration and the other has the S configuration.

📋 The Cahn-Ingold-Prelog rule

The R/S designation is determined by following the Cahn-Ingold-Prelog rule (devised by R. S. Cahn, C. Ingold, and V. Prelog):

Step 1: Assign priorities

• Assign priorities of the groups (or atoms) bonded to the chirality center by following the same priority rules as for the E/Z system.
• The highest priority group is labeled #1, and the lowest priority group is labeled #4.

Step 2: Orient the molecule

• Orient the molecule so the lowest priority group (#4) is pointing away from you.

Step 3: Determine direction

• Look at the direction in which priority decreases for the other three groups: 1 → 2 → 3.
• Clockwise direction: the designation is R (rectus, meaning "right" in Latin).
• Counterclockwise direction: the designation is S (sinister, meaning "left" in Latin).

🔧 Practical hints for R/S assignment

Challenge 1: Assigning priority

• Review and practice the priority guidelines from section 5.2 (E/Z system).

Challenge 2: Re-orienting the molecule

• A molecular model is very helpful.
• Assemble a model with four different colors connected to carbon.
• Match the assigned priority to each color (e.g., red is #1, blue is #2).
• Rotate the model to arrange the lowest (#4) group away from you.
• See how the other groups are located to get the answer.

🔄 Switching groups and configuration

Important properties when switching groups in perspective formulas:

• One (odd number of) switch for a pair of groups inverts the configuration of the chirality center.
• Two (even number of) switches get the original configuration back.

Example:

• One switch of structure A leads to B: if A is R, then B is S, so A and B are enantiomers.
• One switch of B leads to C: if B is S, then C is R, so B and C are enantiomers.
• Two switches of C lead to A: both C and A are R, so C and A are identical.

Caution: Switch with care—do not switch unless really necessary because it is easy to get lost. Doing R/S assignment is a safer (and easier for most cases) way to compare the relationship between two structures.

📝 Complete naming

• Once the configuration is determined, include it in the complete name.
• Example: (R)-1-chloroethanol or (S)-1-chloroethanol.

💡 Optical activity

💡 What optical activity is

Optical activity: the property of a compound being able to rotate the plane of polarization of plane-polarized light.

• A compound with such activity is labeled as optically active.
• A stereoisomer that is optically active is chiral.

🌈 How plane-polarized light is generated

• Normal light: the electric field oscillates in all directions.
• When normal light passes through a polarizing filter, only light oscillating in one single plane can go through.
• The resulting light that oscillates in one single direction is called plane-polarized light.

🔬 Discovery and interaction with chiral molecules

• First discovered by Jean-Baptiste Biot in 1815: some naturally occurring organic substances (like camphor) can rotate the plane of polarization of plane-polarized light.
• Some compounds rotated the plane clockwise and others counterclockwise.
• Further studies indicate that the rotation is caused by the chirality of the substances.

🔁 Enantiomers and physical properties

• The two enantiomers are mirror images and share many properties in common: same boiling point, melting point, density, color, and solubility.
• The pair of enantiomers have the same physical properties, except the way they interact with plane-polarized light.
• This difference in optical activity is what distinguishes enantiomers physically.

🧭 Overview

🧠 One-sentence thesis

Chiral compounds rotate plane-polarized light in characteristic directions and magnitudes, a property called optical activity that allows enantiomers to be distinguished experimentally and explains why enantiomers interact differently with biological systems.

📌 Key points (3–5)

• What optical activity is: the ability of chiral compounds to rotate the plane of polarization of plane-polarized light; achiral compounds cannot do this.
• How enantiomers differ: a pair of enantiomers rotate light by the same angle but in opposite directions (one clockwise/dextrorotatory, one counterclockwise/levorotatory).
• Common confusion: d/l (or +/–) labels indicate the direction of rotation (measured experimentally), while R/S labels indicate the spatial arrangement of groups (determined by structure)—an R compound can be either d or l.
• Racemic mixtures are optically inactive: equal amounts of two enantiomers cancel each other's rotation, resulting in zero observed rotation.
• Why chirality matters biologically: enzymes and receptors are chiral (made from chiral amino acids), so they bind selectively to one enantiomer over the other, leading to different biological effects.

🔬 What is optical activity and how it works

💡 Plane-polarized light

• Normal light oscillates in all directions.
• When normal light passes through a polarizing filter, only light oscillating in one single plane goes through.
• The resulting light is called plane-polarized light.

🌀 Rotation by chiral substances

Optical activity: the property of a compound being able to rotate the plane of polarization of plane-polarized light.

• Jean-Baptiste Biot discovered in 1815 that some naturally occurring organic substances (like camphor) rotate the plane of polarization.
• Some compounds rotate the plane clockwise, others counterclockwise.
• The rotation is caused by the chirality of the substances.
• Chiral compounds are optically active; achiral compounds are optically inactive.
• A stereoisomer that is optically active is also called an optical isomer.

🔄 How enantiomers rotate light differently

• For a pair of enantiomers with the same concentration under the same conditions:
  • They rotate the plane of polarization with the same angle but in opposite directions.
  • One rotates clockwise, the other counterclockwise.

🏷️ Labeling rotation direction: d/l and +/–

➡️ Dextrorotatory (d or +)

• The enantiomer that rotates the plane of polarization clockwise (to the right in Latin).
• Labeled with prefix (d) or (+).

⬅️ Levorotatory (l or –)

• The enantiomer that rotates the plane of polarization counterclockwise (to the left in Latin).
• Labeled with prefix (l) or (–).

⚠️ d/l vs R/S: don't confuse them

Label	What it tells you	How it is determined
d/l (or +/–)	Direction of rotation of plane-polarized light	Measured experimentally with a polarimeter
R/S	Spatial arrangement of groups around the chirality center	Determined by knowing the exact 3D structure

• Key point: d/l (or +/–) has nothing to do with R/S.
• A compound with an R configuration can be either d or l.
• A compound with an S configuration can also be either d or l.
• Example: both compounds in the excerpt are S-isomers, but one is d(+) and the other is l(–).
• The only certainty: for a pair of enantiomers, if one is d, the other must be l, and vice versa.

📐 Measuring optical rotation with a polarimeter

🔧 How a polarimeter works

• Plane-polarized light passes through a sample tube containing the solution.
• The angle of rotation is received and recorded by the analyzer.
• Most polarimeters use light from a sodium atomic spectrum with wavelength 589 nm (the sodium D-line).

📏 Observed rotation (α)

• The rotation degree measured by the polarimeter.
• Depends on:
  • Length of the sample tube
  • Concentration of the sample
  • Temperature

🧮 Specific rotation [α]

Specific rotation: the rotation caused by a solution with a concentration of 1.0 g/mL in a sample tube of 1.0 dm length, usually at 20°C.

• Used to compare optical rotation between different compounds under consistent conditions.
• Formula: specific rotation = (observed rotation) divided by (length in dm times concentration in g/mL)
• Note: the unit of concentration is g/mL (not the familiar units), and length is in dm (decimeters, not cm).
• The unit of specific rotation is degrees (°).

Example: 10.0 g of (R)-2-methyl-1-butanol in 50 mL solution in a 20-cm polarimeter tube has an observed rotation of +2.3° at 20°C.

• Concentration = 10.0 g / 50 mL = 0.2 g/mL
• Length = 20 cm = 2.0 dm
• Specific rotation = 2.3° / (2.0 × 0.2) = +5.75°

🔍 Using specific rotation to identify compounds

• Specific rotation is a characteristic property of an optically active compound.
• The specific rotation value of an authentic compound in literature can confirm the identity of an unknown compound.
• If (R)-2-methyl-1-butanol has a specific rotation of +5.75°, then (S)-2-methyl-1-butanol must have a specific rotation of –5.75° without further measurement.

🧪 Optical activity of mixtures

🧴 Enantiomerically pure sample

• Contains only one enantiomer.
• Shows the full rotation characteristic of that enantiomer.

⚖️ Racemic mixture (racemate)

Racemic mixture: a mixture with an equal amount (50%/50%) of two enantiomers.

• Racemic mixtures do not rotate the plane of polarization.
• Observed rotation = 0°.
• Racemic mixtures are optically inactive.
• Why: for every molecule rotating the plane in one direction, there is an enantiomer molecule rotating it in the opposite direction by the same angle → rotations cancel out.
• Symbol (±) sometimes indicates a racemic mixture.

📊 Mixtures with unequal amounts: enantiomeric excess (ee)

Enantiomeric excess (ee): tells how much of an excess of one enantiomer is in the mixture.

• Formula: ee = (percentage of one enantiomer minus percentage of the other enantiomer)
• Alternative formula: ee = (observed rotation of mixture divided by specific rotation of pure enantiomer) times 100%

Hypothetical examples (assume specific rotation of (+)-enantiomer is +100°, sample tube 1 dm, concentration 1.0 g/mL):

Sample	Composition	Observed rotation	Explanation
Pure (+)	100% (+)	+100°	Full rotation
Pure (–)	100% (–)	–100°	Full rotation, opposite direction
Racemic	50% (+), 50% (–)	0°	Rotations cancel
75% (+), 25% (–)	75% (+), 25% (–)	+50°	ee of (+) = 50%; net rotation from excess (+)
20% (+), 80% (–)	20% (+), 80% (–)	–60°	ee of (–) = 60%; net rotation from excess (–)

🧮 Understanding the net rotation

• In a mixture, the rotation of the minor enantiomer is canceled by part of the major enantiomer's rotation.
• The overall observed rotation depends on the "net amount" of the enantiomer in excess.
• Example: 75% (+) and 25% (–) → the 25% (–) cancels 25% of the (+), leaving 50% (+) to produce the net rotation.

🔢 Advanced calculation example

Problem: The (+)-enantiomer has a specific rotation of +100°. A mixture (1 g/mL in 1 dm cell) has an observed rotation of –45°. What is the percentage of (+)-enantiomer?

Solution:

• Observed rotation is negative → (–)-enantiomer is in excess.
• ee of (–)-enantiomer = (45° / 100°) × 100% = 45%
• Let x = % of (–)-enantiomer, y = % of (+)-enantiomer
• x + y = 100%
• x – y = 45%
• Solving: x = 72.5%, y = 27.5%
• Answer: 27.5% (+)-enantiomer, 72.5% (–)-enantiomer.

Note: To calculate ee, the sign of the rotation angle is not necessary in the formula; just remember that the sign (+/–) of the observed rotation indicates which enantiomer is in excess.

🧬 Chirality and biological properties

🤚 The "hand and glove" analogy

• Enantiomers of a chiral molecule usually show different properties when interacting with other chiral substances.
• Analogy: a right hand only fits into a right glove; wearing a left glove on the right hand feels weird and uncomfortable.
• A chiral object only fits into a specific chiral environment.

🧫 Enzymes and receptors are chiral

• In the human body, biological functions are modulated by enzymes and receptors.
• Enzymes and receptors are proteins, made up of amino acids.
• Amino acids are naturally occurring chiral substances.
• Most amino acids have a chirality (asymmetric) center at the carbon with an amino (NH₂) group.
• Only one enantiomer (usually an S-enantiomer) exists in nature.
• Because amino acids are chiral → proteins are chiral → enzymes and receptors are chiral.

🔐 Selective binding to enantiomers

• The binding site of an enzyme or receptor is chiral.
• It only binds with the enantiomer whose groups are in proper positions to fit into the binding site.
• Only one enantiomer binds with the site, not the other enantiomer.

🍊 Smell examples: limonene and carvone

Limonene:

• Two enantiomers smell completely different because they interact with different receptors in the nose.
• (R)-(+)-limonene: smell of oranges.
• (S)-(–)-limonene: smell of lemon.

Carvone:

• (S)-(+)-carvone: smell of caraway bread.
• (R)-(–)-carvone: smell of spearmint oil (much different odor).

💊 Medicine examples: ibuprofen and thalidomide

Ibuprofen (Advil):

• Only the (S)-enantiomer is the active anti-inflammatory agent.
• The (R)-enantiomer has no anti-inflammatory effects.
• Fortunately, the (R)-enantiomer has no harmful side effects and slowly converts to the (S)-enantiomer in the body.
• Ibuprofen is usually marketed as a racemate (50/50 mixture).

Thalidomide (tragic case):

• Sold in more than 40 countries (mainly Europe) in the early 1960s as a sleeping aid and antiemetic (prevents vomiting) for pregnant women with morning sickness.
• Only the R-enantiomer has the antiemetic property.
• The S-enantiomer was a teratogen (causes congenital deformations).
• The drug was marketed as a racemic mixture.
• Caused damage in about 10,000 children before withdrawal in November 1961.
• Not approved in the US thanks to Dr. Frances O. Kelsey (FDA physician) who insisted on additional tests; she was awarded the President's Medal in 1962.

📜 Historical impact

• The issue of chiral drugs (single enantiomer, not racemate) did not gain attention until 1960.
• Before then, drugs were approved in racemate form if a chirality center was involved, with no further study on biological differences of enantiomers.
• The thalidomide tragedy changed this approach.

🧭 Overview

🧠 One-sentence thesis

Fisher projections provide a systematic way to represent three-dimensional chirality centers on paper, with specific rules for assigning R/S configurations and manipulating the structure that preserve or invert stereochemistry.

📌 Key points (3–5)

• What a Fisher projection shows: chirality centers as intersections where horizontal lines = bonds pointing out of the plane, vertical lines = bonds pointing behind the plane.
• How to assign R/S: if the lowest-priority group (#4) is on a vertical bond, read #1→#2→#3 directly; if #4 is on a horizontal bond, read the direction then reverse the answer.
• Switching groups: one switch inverts configuration (enantiomer), two switches restore the original isomer.
• Common confusion—rotation: 180° rotation keeps the same structure; 90° rotation inverts configuration (enantiomer)—avoid 90° rotations unless necessary.
• Why it matters: Fisher projections encode spatial arrangement, not just connectivity; misreading the representation changes the stereochemistry.

🎨 What Fisher projections represent

🎨 The intersection convention

In a Fisher projection, the chirality center is shown as the intersection of two perpendicular lines.

• Horizontal lines: bonds pointing out of the plane (toward you).
• Vertical lines: bonds pointing behind the plane (away from you).
• The lines are not generic bonds—they encode specific spatial arrangements and stereochemistry.
• Example: if you see a horizontal bond to a chlorine atom, that chlorine is coming out toward you in 3D space.

⚠️ Don't confuse with flat structures

• Fisher projections look two-dimensional, but they represent three-dimensional geometry.
• The same connectivity drawn without the Fisher convention does not carry the same stereochemical information.

🔢 Assigning R/S configuration in Fisher projections

🔢 Step 1: Assign group priority

• Use the usual priority rules (not detailed in the excerpt, but referenced as "as we usually do").
• Identify groups #1 (highest), #2, #3, and #4 (lowest priority).

🔢 Step 2: When #4 is on a vertical bond

• Determine the priority decrease direction from #1 → #2 → #3.
• Clockwise = R; counterclockwise = S.
• Example: (R)-2-chlorobutane is shown with #4 on a vertical bond, and the #1→#2→#3 path is clockwise.

🔢 Step 3: When #4 is on a horizontal bond

• Determine the priority decrease direction #1 → #2 → #3 as in step 2.
• Reverse the answer to get the final configuration.
• Example: if the path looks clockwise, the actual configuration is S; the excerpt shows (S)-2-chlorobutane this way.

🤔 Why reverse when #4 is horizontal?

• The excerpt poses this as an exercise (5.6) but does not provide the answer in the given text.
• The key is that horizontal bonds point out of the plane, so the #4 group is in front rather than behind, which inverts the perspective.

🔄 Manipulating Fisher projections

🔄 Switching two groups

Operation	Effect	Relationship
One switch	Inverts configuration	Produces the enantiomer
Two switches	Restores original	Produces an identical structure

• Example from the excerpt:
  • One switch of A → B: A and B are enantiomers.
  • One switch of B → C: B and C are enantiomers.
  • Two switches of C → A: A and C are identical.

🔄 Rotating 180°

• Effect: produces the same structure; configuration is retained.
• Example: 180° rotation of A → B, and A and B are identical.
• This is a safe operation for checking equivalence.

🔄 Rotating 90°

• Effect: inverts the configuration; produces the enantiomer.
• Example: 90° rotation of A → B, and A and B are enantiomers.
• Warning from the excerpt: "Do NOT rotate the Fisher projection 90º, unless you have to. Keep in mind that the configuration gets inverted by a 90º rotation."
• Don't confuse: 90° looks like a simple rotation, but it flips the spatial meaning of horizontal and vertical bonds, changing the stereochemistry.

🧪 Compounds with more than one chirality center

🧪 Maximum number of stereoisomers

For a compound that has n chirality centers, the maximum number of stereoisomers for that compound is 2 to the power of n.

• Example: 2-bromo-3-chlorobutane has 2 chirality centers (C2 and C3).
• Each center can be R or S, so total stereoisomers = 2 squared = 4.
• The four configurations are RR, SS, RS, and SR.

🧪 Diastereomers

Diastereomers are stereoisomers that are not enantiomers.

• In the 2-bromo-3-chlorobutane example:
  • Isomers A (RR) and B (SS) are enantiomers (non-superimposable mirror images).
  • Isomers C (RS) and D (SR) are enantiomers.
  • Isomers A and C are not identical, not enantiomers, but are stereoisomers → they are diastereomers.
• The four stereoisomers form:
  • Two pairs of enantiomers: (A, B) and (C, D).
  • Four pairs of diastereomers: (A, C), (A, D), (B, C), (B, D).

🧪 Updated isomer classification

• The excerpt mentions that the introduction of diastereomers updates the isomer categorization (Fig. 5.6a replaces Fig. 5.1a).
• Stereoisomers now have two sub-types:
  • Enantiomers: non-superimposable mirror images.
  • Diastereomers: any stereoisomers that are not enantiomers.
• Don't confuse: all enantiomers are stereoisomers, but not all stereoisomers are enantiomers—the rest are diastereomers.

🧭 Overview

🧠 One-sentence thesis

When organic compounds contain multiple chirality centers, they can form several stereoisomers including enantiomer pairs and diastereomers, but the presence of symmetry can reduce the total count through meso compounds.

📌 Key points (3–5)

• Maximum stereoisomer count: A compound with n chirality centers has a maximum of 2^n stereoisomers (each center can be R or S).
• Diastereomers defined: Stereoisomers that are not enantiomers; they have different physical and chemical properties, unlike enantiomers which share most properties.
• Meso compounds: Achiral molecules that contain chirality centers but possess a plane of symmetry, reducing the actual stereoisomer count below the theoretical maximum.
• Common confusion: Not all compounds with chirality centers are chiral—meso compounds are achiral despite having multiple chirality centers.
• Geometric isomers are diastereomers: All cis/trans (geometric) isomers are diastereomers, but not all diastereomers are geometric isomers.

🔢 Counting stereoisomers

🔢 The 2^n rule

• For a compound with n chirality centers, the maximum number of stereoisomers is 2 to the power of n.
• Each chirality center has two possible configurations: R and S.
• Example: 2-bromo-3-chlorobutane has 2 chirality centers (C2 and C3), so the maximum is 2^2 = 4 stereoisomers (RR, SS, RS, SR).

⚠️ Why "maximum" matters

• The actual number may be less than 2^n due to meso compounds.
• Symmetry in the molecule can make some theoretical stereoisomers identical.
• Don't assume: always check for planes of symmetry before concluding the total count.

🔄 Diastereomers explained

🔄 What diastereomers are

Diastereomers: Stereoisomers that are not enantiomers.

• They are not identical, not mirror images, but still stereoisomers (same bonding, different spatial arrangement).
• Example: In 2-bromo-3-chlorobutane with four stereoisomers (A, B, C, D), if A and B are enantiomers and C and D are enantiomers, then A and C are diastereomers (also A and D, B and C, B and D).

🆚 Diastereomers vs enantiomers

Property	Enantiomers	Diastereomers
Mirror image relationship	Yes, non-superimposable	No
Physical properties	Same (except optical rotation)	Different (b.p., density, polarity, solubility, color)
Chemical properties	Generally same (except with chiral reagents)	Different

🔗 Geometric isomers connection

• All geometric isomers (cis/trans) are diastereomers.
• However, not all diastereomers are geometric isomers—diastereomers is the broader category.
• Example: cis and trans isomers of cyclic compounds are diastereomers to each other.

🪞 Meso compounds

🪞 What makes a compound meso

Meso compound: An achiral compound that contains chirality centers.

• A meso compound is superimposable on its mirror image despite having chirality centers.
• It is optically inactive (does not rotate plane-polarized light).
• Example: 2,3-dichlorobutane can form a meso compound when the two chirality centers have opposite configurations that create internal symmetry.

🔍 How to identify meso compounds

• Draw the mirror image and check if it can be superimposed on the original (often by 180° rotation for Fisher projections).
• If superimposable, the compound is achiral and therefore meso.
• Example: For 2,3-dichlorobutane, stereoisomers with RS and SR configurations are actually identical—they are the same meso compound.

📉 Impact on stereoisomer count

• Because meso compounds are identical to their "mirror images," they reduce the total stereoisomer count.
• Example: 2,3-dichlorobutane theoretically has 4 stereoisomers (2^2), but actually has only 3 because one is a meso compound.
• The 2^n rule gives the maximum; meso compounds bring the actual number below that maximum.

✂️ Plane of symmetry method

✂️ Quick chirality test

Plane of symmetry: A plane that cuts the molecule in half so each half is the mirror image of the other.

• If a plane of symmetry exists → the molecule is achiral.
• If no plane of symmetry exists in any conformation → the molecule is chiral.
• This method is faster than comparing a structure to its mirror image.

🔄 Checking conformations

• Sometimes you need to rotate groups to find the proper conformation that reveals the plane of symmetry.
• Example: A molecule might not show symmetry in one conformation but does in an eclipsed conformation after rotation around a bond.
• Don't confuse: Rotating around a single bond changes conformation but not configuration; the molecule remains the same compound.

🧪 Application to meso compounds

• Meso compounds always have a plane of symmetry.
• Example: The meso isomer of 2,3-dichlorobutane has a plane cutting between the two chirality centers.
• For cyclic compounds like 1,2-dibromocyclopentane, the cis isomer can be meso if it has a plane of symmetry through the ring.

🔬 Working with cyclic compounds

🔬 Cis/trans and enantiomers

• Cyclic compounds with two chirality centers can have both cis and trans isomers.
• Each geometric isomer (cis or trans) may have an enantiomer.
• Example: 1-bromo-2-chlorocyclobutane has two cis-isomers (enantiomers of each other) and two trans-isomers (also enantiomers of each other).

🔗 Relationships between isomers

• Any cis-isomer compared to any trans-isomer → diastereomers.
• Within cis or within trans → enantiomers (if no plane of symmetry).
• Example: For 1,2-dibromocyclopentane, there are 3 total stereoisomers: two trans-isomers (enantiomers) and one cis-isomer (meso compound with plane of symmetry).

🎯 Determining relationships

Method I: Assign R/S configuration to each chirality center and compare.

• If all configurations are opposite → enantiomers.
• If some are same and some opposite → diastereomers.
• If all are the same → identical.

Method II: Flip or rotate the structure (use molecular models) to see if it matches the other structure.

• If it matches after flipping → check if they are mirror images (enantiomers) or identical.
• This method can be confusing; Method I is safer if you're uncertain.

🧭 Overview

🧠 One-sentence thesis

Electromagnetic radiation spans a broad spectrum from gamma rays to radio waves, and molecular spectroscopy exploits how molecules absorb specific wavelengths to reveal structural information about organic compounds.

📌 Key points (3–5)

• What electromagnetic radiation is: oscillating electrical and magnetic fields traveling as waves across a spectrum from gamma rays to radio waves; visible light is only a narrow band.
• Wave-frequency-energy relationship: wavelength and frequency are inversely proportional (longer waves = lower frequency), and higher frequency means higher energy.
• How molecular spectroscopy works: molecules absorb energy at certain wavelengths, jumping from ground state to excited state; unabsorbed wavelengths pass through and a detector records the absorption pattern.
• Common confusion: wavelength vs frequency—they move in opposite directions (shorter wavelength = higher frequency = higher energy).
• Why it matters for structure: absorption patterns in different regions (IR for bond vibrations, NMR using radio waves) reveal the structure of organic molecules.

🌈 The electromagnetic spectrum

🌈 What electromagnetic radiation is

Electromagnetic radiation: radiation composed of oscillating electrical and magnetic fields.

• It covers a broad range: gamma rays, X-rays, ultraviolet (UV) light, visible light, microwaves, and radio-frequency waves.
• Each type has different uses:
  • Gamma rays: emitted by radioactive nuclei.
  • X-rays: medical examination of bones.
  • UV light: causes sunburns, used for disinfection.
  • Microwaves and radio waves: communication (radio, TV, cell phones).
• Visible light (what our eyes see, commonly called "light") is only a very narrow band of the full spectrum.

📏 Wavelength ranges and energy

• High-energy radiation (gamma, X-rays): very short waves, as short as 10^-16 meters.
• Visible light: 400–700 nanometers (1 nm = 10^-9 m).
• Radio waves: can be several hundred meters long.
• Longer wavelengths are much less energetic and less harmful to living things.

🔗 Wave properties and formulas

🔗 Wavelength and frequency relationship

• Electromagnetic radiation exhibits wave-like properties.
• The formula connecting wavelength (λ, lambda) and frequency (ν, nu, in Hz or s^-1, where 1 Hz = 1 s^-1):
  • c = λν (Formula 6.1)
  • c is the speed of light: 2.998 × 10^8 m/s in vacuum (speed in air is slightly slower but usually treated as the same).
• Because speed is constant, wavelength and frequency are inversely proportional:
  • Longer waves → lower frequencies.
  • Shorter waves → higher frequencies.
• Example: a wave with twice the wavelength has half the frequency.

⚡ Energy of electromagnetic radiation

• Energy formula: E = hν = hc/λ (Formula 6.2)
  • E: energy of each photon (in Joules, J).
  • h: Planck's constant, 6.626 × 10^-34 J·s.
• Higher frequency corresponds to higher energy.
• Don't confuse: wavelength and energy move in opposite directions—shorter wavelength means higher energy (because frequency is higher).

Property	Relationship	Implication
Wavelength vs frequency	Inversely proportional (c = λν)	Longer waves have lower frequency
Frequency vs energy	Directly proportional (E = hν)	Higher frequency = higher energy
Wavelength vs energy	Inversely proportional (E = hc/λ)	Shorter waves = higher energy

🔬 How molecular spectroscopy works

🔬 The absorption experiment

• In a molecular spectroscopy experiment:
  1. Electromagnetic radiation of a specified range of wavelengths passes through a sample containing a compound of interest.
  2. Sample molecules absorb energy from some wavelengths and jump from a lower energy "ground state" to a higher energy "excited state."
  3. Other wavelengths are not absorbed and pass through.
  4. A detector records which wavelengths were absorbed and how much was absorbed.

🧪 What the absorption pattern reveals

• By quantifying how a molecule absorbs (or does not absorb) different wavelengths, we can learn a lot about the structure of an organic molecule.
• Different spectroscopy techniques use different regions:
  • IR spectroscopy: involves absorption of radiations in the infrared region (studies bond vibrations).
  • NMR technique: applies radio waves.
• Example: if a molecule absorbs strongly at certain IR wavelengths, it indicates the presence of specific functional groups or bond types.

🔄 Ground state vs excited state

• Ground state: the lower energy state of a molecule before absorbing radiation.
• Excited state: the higher energy state after the molecule absorbs energy from radiation.
• The energy difference between these states determines which wavelengths are absorbed.
• Don't confuse: not all wavelengths are absorbed—only those matching the energy gap between ground and excited states.

🧭 Overview

🧠 One-sentence thesis

IR spectroscopy reveals molecular structure by detecting which infrared frequencies are absorbed when covalent bonds vibrate at their characteristic frequencies.

📌 Key points (3–5)

• What IR spectroscopy studies: how covalent bonds vibrate (stretch, bend, twist) when they absorb infrared radiation.
• Bonds as springs, not sticks: covalent bonds behave like vibrating springs at room temperature, not rigid structures.
• Absorption mechanism: molecules absorb only IR radiation that matches the frequency of one of their bond vibrations, increasing the amplitude (not frequency) of that vibration.
• Common confusion: absorption increases the amplitude of vibration (how much the bond moves), but the vibrational frequency itself stays the same.
• How the instrument works: an IR spectrophotometer compares radiation transmitted through the sample versus a blank to identify which frequencies were absorbed.

🌊 Bond vibration modes

🔗 Bonds behave like vibrating springs

Covalent bonds in organic molecules behave as if they were vibrating springs, not rigid sticks.

• At room temperature, organic molecules are always in motion.
• This motion involves several vibration modes: stretching, bending, and twisting.
• Each mode changes the bond in a different way.

📏 Stretching mode

Stretching: the vibration occurring along the line of the bond that changes the bond length.

• The bond lengthens and shortens along its axis.
• Example: a C-H bond can stretch, making the distance between C and H oscillate.

🪃 Bending mode

Bending: the vibration that, like a swing, does not occur along the line, but changes the bond angles.

• The bond angle changes, but the bond does not stretch along its length.
• Specific bending modes are described with terms like scissoring, twisting, etc.
• Don't confuse: bending changes angles; stretching changes bond length.

🔄 One bond, multiple modes

• A single covalent bond may vibrate in different vibrational modes simultaneously.
• Example: the C-H bond can be in both a stretching mode and a bending mode at the same time.
• Each mode has its own characteristic frequency.

🎯 How IR absorption works

🎵 Characteristic ground state frequency

• Each vibrational mode for a given bond occurs with a characteristic ground state frequency.
• These frequencies correspond to the IR region of the electromagnetic spectrum:
  • Frequency: 10¹³ to 10¹⁴ Hz
  • Wavelength: 2.5 to 17 micrometers

💡 Matching frequencies trigger absorption

• When a molecule is exposed to IR radiation, it will absorb only the radiation that matches the frequency of the vibration of one of its bonds.
• If the frequency does not match, the radiation passes through without being absorbed.
• Example: if a bond vibrates at frequency X, only IR radiation at frequency X will be absorbed by that bond.

📈 What happens after absorption

• The absorbed IR radiation allows the bond to vibrate a bit more.
• Specifically: the amplitude of vibration increases (the bond moves more).
• Important: the vibrational frequency itself remains the same.
• Don't confuse: absorption adds energy to make the vibration bigger, not faster.

🔬 The IR spectrophotometer

🛠️ How the instrument works

An infrared spectrophotometer passes a beam of IR radiation through the sample and compares it with a blank to detect absorption.

Two beams are used:

Beam	Path	Result
Sample beam	Passes through the sample	Some radiation is absorbed; remaining radiation goes through
Blank beam	Passes through a cell with no sample	All radiation goes through (no absorption)

📊 Detection and output

• The detector records and compares the radiation transmitted through the sample with that transmitted in the absence of the sample.
• Any frequencies absorbed by the sample will be apparent by the difference between the two beams.
• The computer plots the result as a graph showing transmittance versus frequency.
• Frequency is displayed in the format of wavenumber (explained in a later section).

🔍 Why comparison matters

• By comparing sample and blank, the instrument isolates which frequencies the molecule absorbed.
• This reveals which bond vibrations are present in the molecule, providing structural information.

🧭 Overview

🧠 One-sentence thesis

IR spectroscopy identifies functional groups in molecules by measuring characteristic absorption frequencies that correspond to specific bond vibrations, enabling structural analysis through pattern recognition of absorption bands.

📌 Key points (3–5)

• How IR spectra are read: vertical axis shows % transmittance (downward spikes = absorption bands); horizontal axis shows wavenumber (cm⁻¹) decreasing left to right (higher wavenumber = higher energy).
• Why IR works for identification: different functional groups have different characteristic absorption frequencies, so the presence or absence of specific bands reveals which groups are present.
• Key frequency regions: O-H (3200–3600 cm⁻¹, broad), C=O (1650–1750 cm⁻¹, strong), C-H (2800–3300 cm⁻¹), C=C (1620–1680 cm⁻¹), and fingerprint region (400–1400 cm⁻¹).
• Common confusion: wavenumber vs wavelength—wavenumber is the reciprocal of wavelength; larger wavenumbers mean shorter wavelengths, higher frequencies, and higher energy.
• Intensity clues: band intensity depends on bond polarity (more polar = stronger) and number of bonds involved; O-H bands are strong and broad due to hydrogen bonding.

📊 Reading an IR spectrum

📊 Vertical axis: % transmittance

• What it shows: how strongly light was absorbed at each frequency.
• 100% transmittance = no absorption at that frequency.
• Lower % transmittance = energy absorbed by the compound → downward spikes.
• These spikes are called absorption bands.
• A molecule has multiple covalent bonds with different vibration modes, so IR spectra usually show multiple absorption bands.

📏 Horizontal axis: wavenumber

Wavenumber: the reciprocal of wavelength, in units of cm⁻¹.

• Formula: wavenumber = 1 / wavelength.
• IR wavenumbers typically range from 4000 cm⁻¹ to 600 cm⁻¹ (corresponding to wavelengths of 2.5 μm to 17 μm).
• Direction: wavenumber decreases from left to right in IR spectra.
• Larger wavenumbers (left side) = shorter wavelengths = higher frequencies = higher energy.
• Example: in the 2-hexanone spectrum, the C=O stretching appears at 1716 cm⁻¹.

🔍 Don't confuse wavenumber with wavelength

• Wavenumber is the reciprocal of wavelength, not the same thing.
• Higher wavenumber means shorter wavelength and higher energy.
• The horizontal axis runs "backwards" compared to wavelength scales.

🎯 Characteristic absorption frequencies

🎯 Why functional groups have unique signatures

• Different functional groups have different characteristic absorption frequencies (in wavenumber).
• This makes IR spectroscopy a powerful tool for identifying which functional groups are present in a molecule.
• Example: a strong absorption band in the 1650–1750 cm⁻¹ region indicates a carbonyl group (C=O) is present.

🗂️ Major functional group regions

Functional Group	Bond	Frequency Range (cm⁻¹)	Intensity/Shape
Alcohol	O-H stretching	3200–3600	Strong, broad
Carboxylic acid	O-H stretching	2500–3300	Strong, broad
Carboxylic acid	C=O stretching	1700–1725	Strong
Carbonyl (general)	C=O stretching	1650–1750	Strong
Amine	N-H stretching	3300–3500	Medium
Alkane	C-H stretching	2850–2950	—
Alkene	=C-H stretching	3020–3080	—
Alkene	C=C stretching	1620–1680	Weak
Terminal alkyne	≡C-H stretching	3250–3350	—
Alkyne	C≡C stretching	2100–2250	Weak
Aldehyde	C-H stretching	~2800 and ~2700	Medium (two bands)
Benzene	C=C stretching	~1600 and 1500–1430	Strong to weak

📌 Within the carbonyl region

• Carboxylic acids, esters, ketones, and aldehydes absorb at the higher end (1700–1750 cm⁻¹).
• Conjugated unsaturated ketones and amides absorb at the lower end (1650–1700 cm⁻¹).
• Example: 2-hexanone shows C=O stretching at 1716 cm⁻¹, a very strong band.

🔬 Stretching vibrations and their characteristics

🔬 Why stretching appears at higher wavenumbers

• Stretching vibrations generally require more energy than bending.
• They show absorption bands in the higher wavenumber/frequency region.

💧 O-H bond (alcohol and carboxylic acid)

• Alcohol O-H: 3200–3600 cm⁻¹, strong and broad.
• Carboxylic acid O-H: 2500–3300 cm⁻¹, strong and broad.
• The broad shape results from hydrogen bonding between molecules.
• High polarity of O-H makes these bands easy to identify.
• Example: 1-hexanol shows a very broad peak centered at about 3400 cm⁻¹.

🧪 N-H bond (amine and amide)

• Absorption at 3300–3500 cm⁻¹, medium intensity.
• N-H polarity is weaker than O-H, so the band is not as intense or as broad.

⚛️ C-H bond stretching (hydrocarbons)

• All hydrocarbons show C-H stretching in the range 2800–3300 cm⁻¹.
• The exact location distinguishes between alkane, alkene, and alkyne:
  • Terminal alkyne (≡C-H, sp C-H): ~3300 cm⁻¹
  • Alkene (=C-H, sp² C-H): 3000–3100 cm⁻¹
  • Alkane (-C-H, sp³ C-H): ~2900 cm⁻¹
• Example: 2-hexanone shows C-H absorption at about 2900 cm⁻¹.

🏷️ Aldehyde C-H: a special case

• Aldehyde C-H stretching shows two absorption bands: one at ~2800 cm⁻¹ and another at ~2700 cm⁻¹.
• This makes aldehydes relatively easy to identify (together with C=O stretching at ~1700 cm⁻¹).
• Essentially no other absorptions occur at these wavenumbers.
• Example: butanal shows this characteristic two-band pattern.

🔗 Triple bonds (C≡C and C≡N)

• Absorption at 2100–2200 cm⁻¹, medium to weak intensity.
• Alkynes: weak but sharp bands in the range 2100–2250 cm⁻¹ (C≡C stretching).
• Terminal alkynes: additional absorption at ~3300 cm⁻¹ (sp C-H stretching).

🎭 C=O stretching (carbonyl)

• Strong absorption band in the 1650–1750 cm⁻¹ region.
• One of the most characteristic and easily identified bands in IR spectroscopy.

🌐 Other double bonds (C=C and C=N)

• Absorb at lower frequencies: ~1550–1650 cm⁻¹.
• Alkene C=C: one band at ~1600 cm⁻¹, weak intensity.
• Benzene ring: two sharp bands—one at ~1600 cm⁻¹ and one at 1500–1430 cm⁻¹.
• Example: ethyl benzene shows the characteristic two-band benzene pattern.

🔍 Band intensity and the fingerprint region

🔍 What determines band intensity

• Bond polarity: higher polarity → more intense absorption band.
• Number of bonds: more bonds involved → higher intensity.
• Intensity is usually described as strong (s), medium (m), weak (w), broad, or sharp.

🖐️ The fingerprint region (400–1400 cm⁻¹)

Fingerprint region: the lower-frequency region (400–1400 cm⁻¹) where the pattern of absorbance bands is characteristic of the compound as a whole.

• Similar to a human fingerprint, this pattern is unique to each compound.
• Even if two molecules have the same functional groups, their IR spectra will not be identical—the difference shows up in the fingerprint region.
• Use: the IR spectrum of an unknown sample can be compared to a database of known standards to confirm identification.

🧪 Practical interpretation strategy

🧪 What to look for (not everything)

• Do not try to identify all absorption bands in an IR spectrum.
• Instead, look for characteristic absorption bands to confirm the presence or absence of specific functional groups.
• IR alone usually does not provide enough information to determine the complete structure of a molecule.
• Other methods (e.g., NMR) must be used in conjunction for more specific structural information.

🔎 Example: 2-hexanone

• Most characteristic band: C=O stretching at 1716 cm⁻¹ (very strong).
• This strong band in the 1650–1750 cm⁻¹ region confirms the presence of a carbonyl group.
• Also shows sp³ C-H stretching at ~2900 cm⁻¹ (alkane portion).

🔎 Example: 1-hexanol

• sp³ C-H stretching at 2800–3000 cm⁻¹ (alkane portion).
• Very broad peak centered at ~3400 cm⁻¹: characteristic O-H stretching of alcohol.
• The broad shape indicates hydrogen bonding.

🧭 Overview

🧠 One-sentence thesis

IR spectrum interpretation focuses on identifying characteristic absorption bands to confirm the presence or absence of specific functional groups, rather than attempting to identify every band in the spectrum.

📌 Key points (3–5)

• Core strategy: look for characteristic absorption bands to confirm functional groups; do not try to identify all bands.
• IR limitations: an IR spectrum alone usually does not provide enough information to determine the complete molecular structure.
• Fingerprint region uniqueness: the 400–1400 cm⁻¹ region is characteristic of the compound as a whole, like a human fingerprint, and can be used to confirm identity by comparison to databases.
• Common confusion: distinguishing alkene vs benzene ring—alkene shows one band at ~1600 cm⁻¹, while benzene shows two sharp bands (one at ~1600 cm⁻¹ and one at 1500–1430 cm⁻¹).
• Complementary methods needed: other techniques like NMR must be used alongside IR to determine complete molecular structures.

🎯 Interpretation strategy

🎯 Focus on characteristic bands, not all bands

• The excerpt emphasizes that we do not try to identify all absorption bands in an IR spectrum.
• Instead, the goal is to look at characteristic absorption bands to confirm whether a functional group is present or absent.
• This approach simplifies interpretation and makes IR a practical screening tool.

🔬 IR as part of a toolkit

• IR alone usually does not provide enough information to determine the complete structure of a molecule.
• Other instrumental methods must be applied in conjunction, such as NMR.
• NMR is described as "a more powerful analytical method to give more specific information about molecular structures."

🔍 The fingerprint region

🔍 What it is and why it matters

Fingerprint region: the 400–1400 cm⁻¹ region in the IR spectrum, where the pattern of absorbance bands is characteristic of the compound as a whole.

• Similar to a human fingerprint, this region is unique to each compound.
• Even if two different molecules have the same functional groups, their IR spectra will not be identical, and the difference will be reflected in the fingerprint region.

🗂️ Using the fingerprint for identification

• The IR from an unknown sample can be compared to a database of IR spectra of known standards.
• This comparison confirms the identification of the unknown sample.
• Example: two compounds both contain an alcohol group, but their fingerprint regions will differ, allowing distinction.

📊 Worked examples from spectra

🍷 1-hexanol (Fig. 6.4a)

Band location	Assignment	Notes
~2800–3000 cm⁻¹	sp³ C-H stretching (alkane)	Expected for alkyl chains
~3400 cm⁻¹ (very broad)	O-H stretching (alcohol)	Characteristic of alcohols; broadness is key

• The very broad peak centered at about 3400 cm⁻¹ is the characteristic band of the O-H stretching mode of alcohols.
• Don't confuse: the broadness distinguishes O-H from sharper peaks like N-H or sp C-H.

🌿 1-octene (Fig. 6.4b)

Band location	Assignment	Notes
1642 cm⁻¹	C=C stretching (alkene)	Carbon-carbon double bond
3079 cm⁻¹	sp² C-H stretching	σ bond between sp²-hybridized alkene carbons and their attached hydrogens

• These two bands are characteristic of alkenes.
• The excerpt earlier noted that alkene C=C stretching shows one band at ~1600 cm⁻¹, distinguishing it from benzene rings.

🧪 Other spectra mentioned

The excerpt references additional IR spectra from a free organic compounds spectral database:

• Acetic acid (Fig. 6.4c): key bands labeled on the spectrum.
• Butanal (Fig. 6.4d): key bands labeled on the spectrum.
• Ethyl benzene (Fig. 6.4e): key bands labeled; earlier text noted benzene rings show two sharp absorption bands (one at ~1600 cm⁻¹ and one at 1500–1430 cm⁻¹).

🔄 Distinguishing similar functional groups

🔄 Alkene vs benzene ring

• Alkene: C=C stretching shows one band at ~1600 cm⁻¹.
• Benzene ring: shows two sharp absorption bands—one at ~1600 cm⁻¹ and one at 1500–1430 cm⁻¹.
• Example: ethyl benzene (Fig. 6.4e) demonstrates the two-band pattern characteristic of benzene.
• Don't confuse: the presence of a second band at 1500–1430 cm⁻¹ is the key to identifying a benzene ring versus a simple alkene.

🧭 Overview

🧠 One-sentence thesis

Nuclear magnetic resonance (NMR) spectroscopy exploits the magnetic properties of certain atomic nuclei to reveal the molecular structure and bonding framework of organic compounds atom by atom.

📌 Key points (3–5)

• What NMR detects: NMR-active nuclei (those with odd numbers of protons/neutrons) that spin and generate magnetic moments, including ¹H and ¹³C.
• How resonance works: nuclei in an external magnetic field occupy two spin states (α and β) with an energy gap; when radio frequency radiation matches this gap, nuclei absorb energy and flip between states.
• Why different signals appear: the electronic environment around each nucleus creates shielding or deshielding effects, changing the effective magnetic field and resonance frequency.
• Common confusion: shielding vs deshielding—more electron density shields protons (lower frequency), while electronegative groups withdraw electrons and deshield protons (higher frequency).
• Practical importance: NMR allows chemists to piece together molecular structures bond by bond, similar to how MRI scanners reveal internal body structures.

🧲 NMR-active nuclei and magnetic properties

🧲 Which nuclei are NMR-active

NMR-active nuclei: atomic nuclei that spin about their axes and generate their own magnetic field or magnetic moment.

• Only nuclei with an odd number of protons and/or neutrons have magnetic moments.
• Important for organic chemistry: ¹H (hydrogen), ¹³C (carbon), ¹⁴N (nitrogen), ¹⁹F (fluorine), and ³¹P (phosphorus) are all NMR-active.
• Common isotopes like ¹²C and ¹⁶O do not have magnetic moments and cannot be directly observed by NMR.
• In practice, ¹H and ¹³C are most commonly observed; ¹H NMR is called "proton NMR" because the ¹H nucleus is a single proton.

🔄 Spin states without and with magnetic field

• Without external field: in a flask on a bench, magnetic moments of all protons are oriented randomly.
• With external field (B₀): each proton assumes one of two possible orientations corresponding to two spin states labeled α and β.
  • α spin state: magnetic moment aligned with the direction of B₀ (lower energy).
  • β spin state: magnetic moment aligned opposed to the direction of B₀ (higher energy).
• The energy gap ΔE between the two states depends on the strength of B₀: stronger field → larger ΔE.

⚡ Resonance and energy absorption

⚡ How resonance occurs

Resonance: when a nucleus in an external magnetic field absorbs radio frequency (RF) radiation with energy matching the energy gap ΔE, causing the nucleus to flip from the lower energy state (α) to the higher energy state (β).

• Energy is required to excite the proton from α to β spin state.
• In an NMR spectrometer, energy is supplied by electromagnetic radiation in the radio frequency (RF) region.
• When the RF energy matches ΔE exactly, the proton absorbs the energy and flips its magnetic moment.

📻 Resonance frequency

Resonance frequency (ν): the frequency of radiation absorbed by a nucleus during a spin transition in an NMR experiment.

• Resonance frequency depends on B₀: larger external field → higher resonance frequency.
• The relationship follows a specific formula involving the magnetogyric ratio γ (different for different nuclei).
• Examples for protons:
  • B₀ ≅ 1.41 Tesla → RF frequency = 60 MHz
  • B₀ ≅ 7.04 Tesla → RF frequency = 300 MHz
• This frequency is the most important parameter for an NMR spectrometer; higher frequency → more sensitive instrument and higher resolution spectrum.

📊 Population difference between spin states

• In a large population of molecules in an external magnetic field:
  • Slightly more than half of protons occupy the lower energy α spin state.
  • Slightly less than half occupy the higher energy β spin state.
• This population difference is what NMR exploits, and it increases with the strength of B₀.

🔬 The NMR experiment and instrumentation

🔬 How the experiment works (simplified)

1. Sample is placed in strong external magnetic field B₀; protons begin to spin in one of two spin states.
2. Initially, slightly more protons are in α-spin states (aligned with B₀) than β-spin states (aligned against B₀).
3. Sample is exposed to a range of radio frequencies.
4. Only frequencies matching the resonance frequency are absorbed, causing protons aligned with B₀ to "spin flip" and align against B₀.
5. When flipped protons flip back to ground state, they emit energy as radio-frequency radiation.
6. The instrument detects and records the frequency and intensity using a Fourier transform (FT).
7. FT converts time versus amplitude signals to frequency versus amplitude signals—what we see in an NMR spectrum.

🧊 Modern NMR spectrometers

Type	Frequency	Features
High-field FT-NMR	100–800 MHz	Superconducting magnets; very high resolution; operate in liquid nitrogen or helium bath at very low temperature; costly to purchase and maintain
Benchtop NMR	60–90 MHz	Lower cost; good resolution for basic organic structures; suitable for teaching; allows hands-on student experience

🛡️ Shielding and deshielding effects

🛡️ The shielding effect

Shielding effect: the external magnetic field B₀ causes σ electrons to circulate and generate an induced local magnetic field (B_local) at the proton in the opposite direction to B₀, reducing the net magnetic field experienced by the proton.

• The proton experiences an effective magnetic field B_eff = B₀ – B_local (smaller than the applied field).
• As a result, the proton responds to a lower frequency (since resonance frequency is proportional to magnetic field).
• This local field shields the proton from experiencing the full force of B₀.
• Different hydrogen atoms in different electronic environments have different electron densities → different B_local → different B_eff → different resonance frequencies → different signals in the spectrum.

🔓 The deshielding effect

Deshielding: hydrogen atoms close to electronegative groups experience reduced shielding because electronegative groups withdraw electron density from nearby atoms, diminishing the shielding by circulating electrons.

• Deshielded protons experience more of the external magnetic field and have a higher resonance frequency than shielded protons.
• As the electronegativity (EN) of the substituent increases, the extent of deshielding increases.
• Example pattern: H atoms get more deshielded as:
  • The EN of the substituent increases (e.g., comparing different halogens).
  • More electronegative substituents are involved (e.g., one vs. multiple electronegative groups).

🔍 Don't confuse shielding and deshielding

• More electron density around a proton → more shielding → lower resonance frequency.
• Less electron density (due to nearby electronegative groups) → deshielding → higher resonance frequency.
• The electronic environment determines whether a proton is shielded or deshielded, which is why different hydrogens in a molecule show different signals.

🧭 Overview

🧠 One-sentence thesis

The ¹H NMR spectrum reveals molecular structure through four key features—number of signals, chemical shift positions, integration ratios, and signal splitting—with chemical equivalence and the anisotropy effect being critical for interpreting proton environments.

📌 Key points (3–5)

• Four aspects of ¹H NMR: chemical equivalent/non-equivalent protons (number of signals), chemical shift, integration, and signal splitting provide structural information.
• Chemical equivalence determines signal count: protons in the same chemical environment are equivalent and produce only one signal; symmetry helps identify equivalent protons.
• Chemical shift (δ) reflects electronic environment: shielded protons appear upfield (lower δ, lower frequency), deshielded protons appear downfield (higher δ, higher frequency).
• Common confusion—anisotropy vs simple deshielding: aromatic and vinylic protons resonate much further downfield than expected from electronegativity alone because circulating π electrons create an induced magnetic field that adds to the external field.
• TMS as zero reference: tetramethylsilane is defined as 0 ppm; all other signals are reported relative to TMS, and using ppm (parts per million) makes values instrument-independent.

🔢 Chemical Equivalence and Signal Count

🔢 What chemical equivalence means

Chemical equivalent protons: protons in the same chemical environment that have the same resonance frequency and show only one signal in the ¹H NMR spectrum.

• The number of signals in a ¹H NMR spectrum equals the number of distinct proton sets in the molecule.
• Example: methyl acetate has six hydrogens but only two signals—three Ha protons bonded to C=O are equivalent (one signal), and three Hb protons bonded to O are equivalent (another signal).

🪞 Symmetry and equivalence

• Protons that are symmetric to each other by a plane of symmetry are chemical equivalent.
• Benzene: all six protons are equivalent → one signal.
• Acetone: both methyl groups (six protons total) are in the same environment → one signal.

🧪 Predicting signal count in substituted benzenes

The excerpt provides detailed examples of dimethylbenzene isomers:

Compound	Aromatic proton sets	Methyl groups	Total signals
1,4-dimethylbenzene	1 set (all four H equivalent by symmetry)	1 set (both CH₃ equivalent)	2
1,2-dimethylbenzene	2 sets (Ha adjacent to CH₃; Hc two carbons away)	1 set	3
1,3-dimethylbenzene	3 sets (Hb between two CH₃; Hc one carbon away; Hd two carbons away)	1 set	4

• Don't confuse: protons on the same carbon are not automatically equivalent—their relationship to other substituents matters.

📏 Chemical Shift and the δ Scale

📏 What chemical shift measures

Chemical shift (δ): the position of a signal along the x-axis of an NMR spectrum, determined by the structural electronic environment of the nuclei producing that signal.

• The x-axis scale runs opposite to normal convention: smaller values (lower frequency) are on the right (upfield), larger values (higher frequency) are on the left (downfield).
• Shielded protons → lower resonance frequency → smaller δ values (upfield).
• Deshielded protons → higher resonance frequency → larger δ values (downfield).

🎯 TMS as the zero reference

• Tetramethylsilane (TMS) is defined as 0 ppm on the chemical shift scale.
• Why TMS was chosen:
  • Silicon is less electronegative than carbon, so TMS hydrogens are highly shielded with very low resonance frequency.
  • Twelve equivalent hydrogens produce a single strong signal.
  • Inert and easy to remove (boiling point 27°C).
  • Modern instruments calibrate electronically without actually adding TMS.

🧮 Why use ppm instead of Hz

• Formula: δ (ppm) = (resonance frequency of sample − resonance frequency of TMS) / operating frequency of instrument × 10⁶
• Different NMR instruments have different magnetic field strengths, so actual resonance frequencies in Hz vary.
• Chemical shift in ppm remains constant across all instruments (e.g., a signal at 2.0 ppm is always 2.0 ppm whether measured at 400 MHz or 60 MHz).
• Example: in methyl acetate measured at 400 MHz, signals at 2.0 and 3.6 ppm correspond to 800 Hz and 1440 Hz, respectively.

📊 Typical chemical shift ranges

Most protons in organic compounds fall between 0 and 12 ppm. Key ranges:

Proton type	Chemical shift (ppm)	Notes
C-H (no nearby functional groups)	1–2	Baseline alkyl protons
C-H beside C=C or C=O	2–2.5	Slightly deshielded
O-C-H	3–4	Deshielded by electronegative oxygen
Vinylic H (on C=C)	4.5–6	Anisotropy effect
Aromatic H (benzene ring)	~7	Strong anisotropy effect
OH (alcohol) or NH (amine)	1–5	Wide range, variable
Aldehyde H (-CHO)	9–10	Highly deshielded
Carboxylic acid H (COOH)	10–12	Most deshielded

• Exact values may vary by ±0.5 ppm depending on specific structure and solvent.
• Example application: in methyl acetate, the 2.0 ppm signal is assigned to Ha (CH₃ beside C=O), and the 3.6 ppm signal to Hb (CH₃ connected to O).

🌀 Anisotropy Effect and π Electrons

🌀 What anisotropy means

Anisotropy effect: the non-uniform shielding or deshielding that depends on the location of protons relative to an induced magnetic field generated by circulating π electrons.

• When π electrons are exposed to the external magnetic field B₀, they circulate and generate their own induced magnetic field (Binduced).
• Whether shielding or deshielding occurs depends on the proton's location in this induced field.

🔄 Aromatic protons and ring current

• In benzene, six π electrons form a delocalized system that circulates when exposed to B₀, creating a "ring current."
• At point A (above or below the ring plane): Binduced opposes B₀ → shielding effect.
• At point B (at the ring edge, where aromatic protons sit): Binduced is in the same direction as B₀ → deshielding effect.
• Result: benzene protons are highly deshielded and resonate far downfield (6.5–8.5 ppm), much more than electronegativity alone would predict.

🎭 Anisotropy in other functional groups

• Vinylic protons (on C=C double bonds): π electrons circulate to create an induced field that adds to B₀ at the proton locations → deshielded → 4–6.5 ppm.

• Aldehyde protons: anisotropy from the C=O π bond contributes to very high chemical shift (9.5–11 ppm).

• Carboxylic acid protons: even further downfield (9.5–12 ppm) due to combined influence of electronegative oxygen and nearby π bond anisotropy.

• Don't confuse: the high chemical shift of aromatic/vinylic protons is not just from nearby electronegative atoms—it's primarily the anisotropy effect from circulating π electrons that causes the large downfield shift.

🧭 Overview

🧠 One-sentence thesis

Integration and signal splitting in ¹H NMR spectra reveal both the relative number of protons in each set and the number of neighboring non-equivalent protons, enabling detailed structural assignments.

📌 Key points (3–5)

• Integration shows relative ratios: the area under each signal is proportional to the number of protons, but the integration number is a ratio, not the actual count.
• Splitting (coupling) reveals neighbors: a signal splits into n+1 peaks, where n is the number of vicinal (3-bond-away) non-equivalent hydrogens on adjacent carbons.
• Common confusion: equivalent protons do not couple with each other; only non-equivalent protons cause splitting.
• OH and NH protons are special: they do not couple with vicinal hydrogens because they exchange rapidly between molecules.
• Four features for structure determination: number of signals, chemical shift, integration, and splitting pattern together allow you to deduce molecular structure.

📏 Integration of signal areas

📏 What integration measures

Signal integration: the computer-calculated area under a signal, proportional to the number of protons that produce the signal.

• The integration line appears as a stepped curve generated by the instrument.
• Integration numbers are usually decimals but can be simplified to whole-number ratios.
• Key principle: integration shows the relative ratio of protons, not the absolute number.

🔢 How to use integration

• Compare the integration values across signals to find the ratio of protons in different sets.
• Example: In 1,4-dimethylbenzene, the peak at 2.6 ppm has an integration 1.5 times greater than the peak at 7.4 ppm, giving a ratio of 3:2.
  • This matches the actual ratio of 6 methyl protons (Hb) to 4 aromatic protons (Ha).
• Don't confuse: the integration number itself (e.g., 3) is not the actual number of protons; it is a ratio that must be scaled to match the molecular formula.

🧩 Combining integration with chemical shift

• Integration alone tells you how many protons are in each set.
• Chemical shift tells you what type of protons they are (aromatic, methyl, etc.).
• Together, these allow you to assign each peak to specific protons in the structure.
• Example: In 1,4-dimethylbenzene, the 7.4 ppm peak (smaller integration) corresponds to aromatic protons on the benzene ring, and the 2.6 ppm peak (larger integration) corresponds to the two methyl groups.

🔀 Signal splitting (coupling)

🔀 What causes splitting

Splitting (or coupling): the phenomenon where a signal splits into two or more peaks due to spin-spin coupling, the magnetic interactions between non-equivalent hydrogen atoms separated by 2 or 3 σ bonds.

• Nearby protons have magnetic moments that can align with or against the external field, splitting the energy levels of the observed protons.
• The terms "splitting" and "coupling" are used interchangeably.
• Most common type: vicinal coupling (or three-bond coupling), which occurs between non-equivalent hydrogens on adjacent carbons.

🧮 The n+1 rule

• Rule: number of peaks = n + 1, where n is the number of vicinal non-equivalent hydrogens.
• This rule predicts the splitting pattern for a given set of protons.
• Example: If a proton has two vicinal non-equivalent neighbors, its signal will split into 2+1 = 3 peaks (a triplet).

📊 Splitting patterns by n value

n	Pattern	Number of peaks	Peak area ratio	Example
0	Singlet	1	—	No vicinal neighbors
1	Doublet	2	1:1	Ha in 1,1,2-trichloroethane
2	Triplet	3	1:2:1	Hb in 1,1,2-trichloroethane
3	Quartet	4	1:3:3:1	Hb in ethyl acetate
≥4	Multiplet	n+1 (or merged)	Complex	Small side peaks may merge into noise

🔍 Doublet (n=1)

• Two peaks of equal height and area.
• The space between the two peaks is the coupling constant, Jab, measured in Hz.
• Example: In 1,1,2-trichloroethane, the Ha protons (at 3.96 ppm) have one vicinal neighbor (Hb), so they appear as a doublet.

🔍 Triplet (n=2)

• Three peaks with an area ratio of 1:2:1 (the middle peak is taller).
• Example: In 1,1,2-trichloroethane, the Hb proton (at 5.76 ppm) has two vicinal neighbors (2 Ha), so it appears as a triplet.

🔍 Quartet (n=3)

• Four peaks with an area ratio of 1:3:3:1.
• Example: In ethyl acetate, the Hb protons are adjacent to three vicinal protons (3 Hc), so they appear as a quartet.
• Note: The carbon with Hb is connected to oxygen on the other side, and oxygen has no hydrogen atoms, so only the three vicinal protons on the adjacent carbon cause splitting.

🔍 Multiplet (n≥4)

• Theoretically splits into n+1 peaks, but small side peaks may be too weak to observe and merge into noise.
• Signals with more than four peaks are generally called multiplets, and it is not critical to count the exact number of peaks.

⚠️ Special cases and exceptions

⚠️ Equivalent protons do not couple

• Key rule: Splitting only occurs between non-equivalent protons.
• Equivalent protons do not couple with each other, even if they are on adjacent carbons.
• Example: In succinic acid, the protons on the two middle carbons are equivalent (Ha), so there is no coupling between them and they show a singlet.
• Don't confuse: just because protons are on adjacent carbons does not mean they will couple; they must be non-equivalent.

⚠️ OH and NH protons do not couple

• Protons on OH or NH groups generally do not couple with vicinal hydrogens.
• Reason: OH and NH protons are acidic enough to rapidly exchange between different molecules, so neighboring protons never "feel" their influence.
• Example: In the 1-heptanol spectrum, the OH proton does not show coupling with the adjacent CH₂ protons.

🧪 Practice and structure determination

🧪 Signal assignment from a given structure

• Approach: Draw the structure and predict the expected signals based on:
  • Number of signals (sets of equivalent protons)
  • Chemical shift (type of protons)
  • Integration (relative number of protons)
  • Splitting pattern (number of vicinal neighbors)
• Example workflow: Match a spectrum to one of several candidate structures by eliminating mismatches.
  • If the spectrum has no signal at ~9 ppm, it is not an aldehyde.
  • If the spectrum has no signal at ~4–5 ppm, it is not an alkene.
  • If the splitting patterns and chemical shifts match, assign the structure.

🧪 Structure determination from a spectrum

• Four critical features to analyze:
  1. Number of signals: indicates how many different sets of protons are in the molecule.
  2. Chemical shift: tells you the electronic environment of each set (aromatic, alkyl, near oxygen, etc.).
  3. Integration: provides the ratio of protons in each set (not the absolute number).
  4. Splitting pattern: indicates the number of vicinal non-equivalent protons on adjacent carbons.
• By combining these four pieces of information, you can deduce the exact structure of an unknown compound.
• Note: The excerpt mentions that detailed examples of structure determination will appear in section 6.9.

🧭 Overview

🧠 One-sentence thesis

Carbon-13 NMR spectroscopy provides a separate window for observing carbon atoms in molecules, with signals spread over a wider chemical shift range than proton NMR and typically simplified by broadband decoupling to show all carbons as singlets.

📌 Key points (3–5)

• Chemical equivalence: equivalent carbons show only one signal, just like equivalent protons in ¹H NMR, making it essential to identify equivalent carbons for correct interpretation.
• Wider chemical shift range: ¹³C signals span about 220 ppm (much wider than ¹H NMR), with carbonyl carbons furthest downfield at 170–220 ppm.
• Different resonance frequency: ¹³C nuclei resonate at a completely different frequency than protons (e.g., 75 MHz vs 300 MHz in the same magnetic field), allowing separate observation.
• Integration is not useful: unlike ¹H NMR, peak areas in ¹³C NMR cannot reliably indicate the number of carbons because different carbon types produce inherently different signal strengths.
• Common confusion: broadband decoupling "turns off" C-H coupling, so all carbon signals appear as singlets rather than multiplets—this is different from the splitting patterns seen in ¹H NMR.

🔬 How ¹³C NMR differs from ¹H NMR

🔬 Separate observation window

• ¹³C nuclei have a different magnetogyric ratio (g) compared to ¹H nuclei.
• In a 7.05 Tesla magnet:
  • Protons resonate at about 300 MHz
  • Carbons resonate at about 75 MHz
• This frequency difference allows chemists to observe ¹³C signals using a completely separate "window" of radio frequencies.
• Why this matters: you can look at carbon signals independently without interference from proton signals.

📏 Reference standard

Tetramethylsilane (TMS) is used as the standard compound to define 0 ppm in ¹³C NMR.

• The same compound (TMS) is used as in ¹H NMR, but the signal from the four equivalent carbon atoms in TMS serves as the standard (not the hydrogen atoms).
• This provides a consistent reference point for measuring chemical shifts.

🎯 Chemical equivalence and signal counting

🎯 Identifying equivalent carbons

• Carbons that are chemical equivalent show only one signal in ¹³C NMR.
• This principle is similar to how equivalent protons behave in ¹H NMR.
• Example: Toluene has five sets of different carbon atoms, so its ¹³C NMR spectrum shows five signals.
• Key skill: correctly identifying equivalent carbons is very important for interpreting the spectrum correctly.

🔢 One signal per set

• Each set of equivalent carbons produces one peak.
• The number of signals tells you how many different carbon environments exist in the molecule.
• Don't confuse: the number of signals does NOT tell you how many total carbons are present (because integration is unreliable).

📊 Chemical shifts in ¹³C NMR

📊 Much wider range

• ¹³C chemical shifts are spread out over about 220 ppm.
• This is much wider than the typical range for ¹H NMR.
• Advantage: better separation of signals makes it easier to distinguish different carbon types.

⚗️ Factors affecting chemical shift

The chemical shift of a ¹³C nucleus is influenced by the same factors as proton shifts:

Factor	Effect	Result
Electronegative atoms	Deshielding effect	Shifts signals downfield (higher ppm)
Anisotropy effects	Deshielding	Shifts signals downfield
sp² hybridization	Large effect	Large downfield shift

🎯 Carbonyl carbons are furthest downfield

• ¹³C signals for carbonyl carbons generally appear at 170–220 ppm.
• This is the furthest downfield region.
• Two reasons: both sp² hybridization AND the double bond to oxygen contribute to this large shift.
• Example: In ethyl acetate, the carbonyl carbon signal appears in this characteristic downfield region.

⚠️ Integration and coupling limitations

⚠️ Why integration doesn't work

• Unlike ¹H NMR, the area under a ¹³C NMR signal cannot easily determine the number of carbons.
• The problem: signals for some carbon types are inherently weaker than others.
• Example: peaks for carbonyl carbons are much smaller than peaks for methyl or methylene (CH₂) carbons.
• Practical consequence: signal integration is generally not useful in ¹³C NMR spectroscopy.

🔇 Broadband decoupling technique

• ¹³C nuclei are coupled to nearby protons, which would create complicated spectra.
• Because of low natural abundance of ¹³C, spin-spin coupling between two nonequivalent ¹³C atoms is negligible.
• Solution: chemists use broadband decoupling, which essentially "turns off" C-H coupling.
• Result: all carbon signals appear as singlets (single peaks).
• Example: The proton-decoupled ¹³C NMR spectrum of ethyl acetate shows four singlets, one for each of the four carbons.

🔍 Clarity benefit

• Without decoupling, each carbon signal would be split by attached protons into multiplets.
• Broadband decoupling simplifies the spectrum dramatically.
• Don't confuse: this is a deliberate technique applied during data collection, not a natural property of ¹³C nuclei.

🛠️ Practical use

🛠️ Supporting information role

• For most purposes, ¹³C NMR spectra are used as supporting information to confirm the structure of a compound.
• They complement ¹H NMR and other spectroscopic data (IR, molecular formula).
• The number of signals helps confirm how many different carbon environments exist.
• Chemical shift values help identify functional groups (especially carbonyl groups in the 170–220 ppm region).

🧭 Overview

🧠 One-sentence thesis

¹H NMR spectroscopy, combined with molecular formula, IR, and ¹³C NMR data, provides a systematic approach to determining the structure of unknown organic compounds by identifying functional groups and assembling structural fragments.

📌 Key points (3–5)

• Systematic four-step approach: calculate degree of unsaturation, use IR to narrow functional groups, interpret NMR signals to identify structural fragments, then assemble the complete structure.
• Degree of unsaturation reveals structure type: zero degrees means no rings or double bonds, pointing to open-chain structures with single bonds only.
• Integration ratios are crucial: the relative integration areas in ¹H NMR directly indicate the number of hydrogens in each set, helping identify structural units like CH₃, CH₂, or CH groups.
• Common confusion—singlet signals: a singlet CH₃ means no adjacent hydrogens, so it must connect directly to the heteroatom (e.g., oxygen in ethers).
• Splitting patterns confirm connectivity: triplet-quartet pairs indicate ethyl groups (CH₂CH₃), doublet-multiplet pairs suggest isopropyl or similar branching.

🔢 The systematic four-step strategy

🔢 Step 1: Calculate degree of unsaturation

Degree of unsaturation (or IHD): a calculation based on molecular formula that reveals the presence of rings and/or double bonds in the structure.

• This is always the first step when solving structure determination problems.
• For the C₅H₁₂O example: degree of unsaturation = 0.
• Zero degrees means no rings and no double bonds → all structures are open-chain with single bonds only.
• With one oxygen atom, the possible functional groups narrow to open-chain alcohol or open-chain ether.

🔍 Step 2: Use IR to narrow functional groups

• IR spectroscopy eliminates or confirms specific functional groups before detailed NMR analysis.
• In the example: no strong bands above 3000 cm⁻¹ and no strong bands at 1700 cm⁻¹.
• No strong band above 3000 cm⁻¹ → excludes alcohol (which would show O-H stretch).
• Therefore, the only remaining option is open-chain ether.

🧩 Step 3: Identify structural fragments from NMR

• This step uses mainly ¹H NMR data (with ¹³C NMR as supporting evidence when available).
• Check all four aspects of ¹H NMR signals learned previously: number of signals, chemical shift, integration, and splitting pattern.
• Integration is key: relative integration areas indicate the actual number of hydrogens in each set.
• Example: integration ratio 3:1 with 12 total hydrogens → actual numbers are 9 and 3 hydrogens.

🧩 Step 4: Assemble the puzzle and verify

• Put the identified structural fragments together logically.
• Double-check that the proposed structure fits all available data.
• The excerpt notes: "There is no simple rule to follow...it takes practice to interpret ¹H NMR signals and translate them into structures."

🧪 Worked examples from C₅H₁₂O isomers

🧪 Compound 1: Simplest spectrum (two singlets)

Key observations:

• Only two signals, both singlets → two sets of non-equivalent hydrogens with no adjacent hydrogens.
• Integration ratio 3:1 → with 12 total hydrogens, this means 9 and 3 hydrogens.
• 9 equivalent hydrogens → three equivalent CH₃ groups → must be a t-butyl group.
• 3 hydrogens → one CH₃ group.

Structure: t-butyl methyl ether (oxygen connects a methyl group and a t-butyl group).

Why this works: The only way to have three equivalent CH₃ groups is in a t-butyl structure.

🧪 Compound 2: Isopropyl ethyl ether

Key observations from integration:

• One CH₃ as triplet
• Two equivalent CH₃ groups as doublet
• One CH as multiplet
• One CH₂ as quartet

Assembly logic:

• Triplet CH₃ + quartet CH₂ → ethyl group (CH₂CH₃), consistent with splitting pattern.
• Two equivalent CH₃ + one CH → isopropyl group, consistent with splitting pattern.

Structure: isopropyl ethyl ether.

🧪 Compound 3: sec-butyl methyl ether

Key observations:

• One CH₃ as singlet at ~3.2 ppm
• One CH₃ as triplet
• One CH₃ as doublet
• One CH₂ as multiplet
• One CH as quartet

Assembly logic:

• Singlet CH₃ at 3.2 ppm → no adjacent hydrogens, bonded to oxygen (chemical shift confirms this).
• Triplet CH₃ + quartet CH₂ → ethyl fragment.
• Doublet CH₃ + CH → these connect to each other.
• CH at ~3 ppm with quartet splitting → connects to oxygen.

Structure: sec-butyl methyl ether.

Don't confuse: A singlet doesn't mean the group is isolated; it means no hydrogens on adjacent carbons.

🧪 Compound 4: Ethyl propyl ether

Key observations:

• One CH₃ as triplet
• Another CH₃ as triplet
• One CH₂ as multiplet
• Two CH₂ groups with overlapping signals

Assembly logic:

• Two CH₃ groups both as triplets → both connect to CH₂ groups → two ethyl (CH₂CH₃) groups.
• The two ethyl groups are not equivalent (different signals).
• One more CH₂ group remains.
• Only one possible structure: ethyl propyl ether.

🎯 Pattern recognition tips

🎯 Common structural units and their signatures

Integration	Splitting	Likely fragment	Reasoning
3H	Triplet	CH₃ in ethyl group	Coupled to adjacent CH₂
2H	Quartet	CH₂ in ethyl group	Coupled to adjacent CH₃
6H	Doublet	Two CH₃ in isopropyl	Both coupled to same CH
1H	Multiplet (septet)	CH in isopropyl	Coupled to two CH₃ groups
3H	Singlet	CH₃ on heteroatom	No adjacent hydrogens

🎯 Chemical shift clues for ethers

• Hydrogens on carbons directly bonded to oxygen appear around 3-4 ppm.
• This helps confirm which fragments connect to the oxygen atom.
• Example: In compound 3, the CH quartet at ~3 ppm confirms it's bonded to oxygen.

🧭 Overview

🧠 One-sentence thesis

Nucleophilic substitution reactions occur when electron-rich nucleophiles displace leaving groups from electron-deficient electrophiles, and they proceed through two distinct mechanisms (SN2 and SN1) that differ in their kinetics and step-by-step processes.

📌 Key points (3–5)

• Three critical components: every nucleophilic substitution requires an electrophile (electron-deficient substrate), a nucleophile (electron-rich reagent), and a leaving group.
• Fundamental reaction rule: when electron-rich nucleophiles meet electron-deficient electrophiles, organic reactions can occur.
• Two distinct mechanisms: SN2 (second-order, bimolecular) and SN1 (first-order, unimolecular) differ in their kinetics and reaction pathways.
• Common confusion—kinetics vs mechanism: second-order means the rate depends on both substrate and nucleophile concentrations; first-order means the rate depends only on substrate concentration—the difference reveals different underlying mechanisms.
• Why it matters: understanding these mechanisms allows prediction of products, reaction rates, and the effects of changing substrates, nucleophiles, or reaction conditions.

🔑 The Three Essential Players

⚡ Electrophile (electron-deficient species)

An electrophile is an electron-deficient species that seeks an electron-rich reagent to connect with ("phile" = love, so "loves electrons").

• In alkyl halides like CH₃Br, the C-X bond is polar because halogen is more electronegative than carbon.
• Carbon carries a partial positive charge and becomes electron-deficient.
• This carbon atom acts as the electrophilic carbon in the reaction.
• Other examples: H⁺, CH₃⁺, BH₃, BeF₂, AlCl₃ (positively charged ions or atoms with incomplete octets).
• The compound undergoing substitution (e.g., CH₃Br) is called the substrate.

🎯 Nucleophile (electron-rich species)

A nucleophile is an electron-rich species that seeks positively charged or electron-poor species to react with ("nucleo" = nucleus/positive charge).

• OH⁻ is a nucleophile: oxygen has three lone pairs and a negative charge, giving high electron density.
• Generally, any species with an electron pair available for sharing can be a nucleophile.
• Nucleophiles can be:
  • Negatively charged: Nu:⁻ (e.g., OR⁻, RCOO⁻)
  • Neutral: Nu: (e.g., H₂O, ROH, NH₃, RNH₂)
• Key rule: when electron-rich nucleophiles meet electron-deficient electrophiles, organic reactions can occur.

🚪 Leaving group (departing species)

The leaving group (LG) leaves with the bonding pair of electrons and is replaced by the nucleophile in the substitution reaction.

• In CH₃Br + OH⁻ → CH₃OH + Br⁻, the Br leaves as Br⁻ (the leaving group).
• Without a proper leaving group, even if a nucleophile is attracted to an electrophile, the substitution cannot proceed.
• Leaving groups can be negatively charged or neutral.
• Don't confuse: the leaving group is not just "what gets replaced"—it must take the bonding electrons with it when it leaves.

🧪 Putting it together: nucleophilic substitution

The general pattern:

• Nucleophile displaces the leaving group in a substrate.
• Example: CH₃Br + OH⁻ → CH₃OH + Br⁻
  • Br is substituted by OH
  • CH₃OH is the major product
  • Br⁻ is the side product

⚙️ Two Mechanisms: SN2 vs SN1

📊 Kinetics reveals mechanism

Kinetics is the study of the rate of a chemical reaction or how fast the reaction occurs.

• Reaction rate data helps understand reaction mechanisms (the step-by-step electron transfer process).
• Kinetic studies show two different rate law expressions for nucleophilic substitution reactions.
• The only reason for different kinetic rates is that the reactions go through different reaction mechanisms.

🔄 SN2: Second-order, bimolecular mechanism

Feature	Description
Order	Second-order reaction
Rate depends on	Concentration of both substrate and nucleophile
Name meaning	Substitution, Nucleophilic, Bimolecular
Example	CH₃Br + OH⁻ → CH₃OH + Br⁻

How concentration affects rate:

• If [CH₃Br] is doubled → rate doubles
• If [OH⁻] is doubled → rate doubles
• If both [CH₃Br] and [OH⁻] are doubled → rate increases by a factor of four

Mechanism details (from SN2 section):

• Involves two electron pair transfers occurring at the same time: nucleophile attacking (one arrow) and leaving group leaving (another arrow).
• Nucleophile approaches the electrophilic carbon from the back side (opposite to the leaving group direction).
• New C—OH bond formation and old C—Br bond breaking occur simultaneously.
• Transition state: a very short transient moment where carbon is partially connected with both OH and Br (highest energy level).
  • Carbon is "pentacoordinated" (five groups around it).
• Eventually, the new bond is completely formed and the old bond is completely broken.
• Single-step process involving both nucleophile and substrate.

Example: In CH₃Br + OH⁻:

• OH⁻ attacks carbon from the back
• As OH⁻ gets closer, Br starts to leave
• At the transition state, carbon is bonded to five groups
• Br leaves with the bonding electron pair
• Product CH₃OH forms

🔀 SN1: First-order, unimolecular mechanism

Feature	Description
Order	First-order reaction
Rate depends on	Concentration of substrate only
Rate independent of	Nucleophile concentration
Name meaning	Substitution, Nucleophilic, Unimolecular
Example	(CH₃)₃CBr + H₂O → product

Substrate difference:

• The substrate here is a tertiary bromide (CH₃)₃CBr.
• The nucleophile is a neutral water molecule.

Don't confuse SN1 vs SN2:

• SN2 rate changes when you change nucleophile concentration; SN1 rate does not.
• SN2 involves both reactants in the rate-determining step; SN1 involves only the substrate.
• The substrate structure matters: the excerpt shows SN1 with a tertiary substrate, SN2 with a primary substrate (CH₃Br).

🎯 What You Need to Master

The excerpt lists learning objectives for Chapters 7 and 8:

🧠 Mechanism understanding

• Show the mechanism of nucleophilic substitution and elimination reactions, including:
  • Intermediates
  • Transition state
  • Reaction coordination diagram
  • Proper terms and curved arrows for SN1, SN2, E1, E2 mechanisms

🔮 Prediction skills

• Predict major/minor products given reactants, reagents, and reaction conditions.
• Compare relative reactivity of different substrates.
• Understand effects of different factors on each mechanism: nucleophiles, leaving group, and solvent.

🔬 Application abilities

• Apply and draw proper reaction mechanisms to explain or predict products, including stereochemistry.
• Provide proper reaction conditions to prepare a particular product.

🧭 Overview

🧠 One-sentence thesis

The SN2 mechanism proceeds through a single concerted step in which nucleophile attack and leaving group departure occur simultaneously from opposite sides, causing configuration inversion at the carbon center and making the reaction rate highly sensitive to steric hindrance.

📌 Key points (3–5)

• What SN2 means: Substitution, Nucleophilic, and Bimolecular—a second-order reaction involving both nucleophile and substrate in a single step.
• The concerted mechanism: nucleophile attacks from the backside while the leaving group departs at the same time, passing through a five-coordinate transition state.
• Steric effect on rate: methyl and primary halides react fastest; secondary halides are much slower; tertiary halides do not undergo SN2 at all because bulky groups block backside attack.
• Stereochemistry outcome: the configuration at the carbon inverts like an umbrella flipped inside out (Walden inversion).
• Common confusion: configuration inversion means the spatial arrangement flips, not that R always becomes S—the actual R/S label must be determined separately for the product.

⚙️ The SN2 mechanism and transition state

⚙️ How the mechanism works

An SN2 mechanism involves two electron pair transfers that occur at the same time: nucleophile attacking and leaving group leaving.

• The nucleophile approaches the electrophilic carbon from the backside—the side opposite to where the leaving group departs.
• As the nucleophile gets closer, the leaving group starts to leave.
• New bond formation and old bond breaking happen simultaneously—this is called a concerted mechanism.
• Example: In the reaction between CH₃Br and OH⁻, the OH⁻ attacks the carbon from the back while Br leaves from the front, both processes occurring at the same time.

🔺 The transition state

• At a very short transient moment, the carbon is partially connected to both the nucleophile and the leaving group.
• This highest-energy state is called the transition state.
• In the transition state, the carbon is pentacoordinated (five groups around it).
• The transition state has no appreciable lifetime and can never be isolated; it involves partial bonds (partially formed and partially broken).
• The structure is usually shown in square brackets with a double-dagger superscript.

📐 Drawing the mechanism correctly

When drawing an SN2 mechanism, you must show:

• Two arrows: one for nucleophile attack (red arrow) and one for leaving group departure (blue arrow).
• Proper direction: nucleophile attacks from the direction opposite to where the leaving group leaves (backside attack).
• Configuration change: the three hydrogens (or other groups) around the carbon are pushed to the other side, inverting the overall configuration.
• The transition state is optional depending on the question, but understanding that the reaction passes through it is important.

📊 Energy diagram and reaction order

📊 Single-step energy profile

• SN2 is a single-step reaction, so the energy diagram has only one curve.
• The top of the curve corresponds to the transition state, the highest-energy structure.
• In the example of CH₃Br + OH⁻ → CH₃OH + Br⁻, the products are at lower energy than the reactants, indicating the reaction is exothermic and the products are more stable.

🔢 Why it is second-order

• The reaction proceeds in a single step that involves both the nucleophile and the substrate.
• Increasing the concentration of either the nucleophile or the substrate increases the possibility of collision.
• This explains the second-order kinetics of an SN2 reaction.

🧱 Steric effect on reaction rate

🧱 How alkyl halide structure affects rate

The structure category of the electrophilic carbon dramatically affects the SN2 reaction rate:

Type of Alkyl Halide	Structure	Relative Rate
Methyl	CH₃X	30
Primary (1°)	RCH₂–X	1
Secondary (2°)	R₂CH–X	0.03
Tertiary (3°)	R₃C–X	negligible (no SN2)

Reactivity order: Methyl > Primary (1°) > Secondary (2°) >> Tertiary (3°) (no reaction)

🚧 Why the steric effect matters

• A key feature of SN2 is that the nucleophile attacks from the backside.
• It is easiest to get close to the methyl carbon because hydrogen atoms are small.
• As the size of groups connected to the carbon increases, it becomes more difficult to access the carbon.
• For tertiary carbons with three bulky alkyl groups, backside approach is completely blocked.
• This is called the steric effect: bulky groups create steric hindrance that prevents nucleophiles from doing backside attacks.
• Tertiary substrates never undergo SN2 because the reaction rate is too slow to be practical.

Don't confuse: The steric effect is about the volume and accessibility of the carbon, not about the electronic properties of the groups.

🔄 Stereochemistry and Walden inversion

🔄 Configuration inversion

• The overall configuration of the carbon in the product gets inverted compared to the reactant.
• This is like an umbrella flipped inside out in a windstorm.
• This inversion is called Walden inversion.

🧪 Stereochemical outcome

• For CH₃OH in the example, the inversion does not make a noticeable difference because the carbon is not a chirality center.
• When the carbon is a chirality center, the inversion has important consequences.
• Example: Starting with (R)-2-bromobutane, the SN2 reaction produces only one enantiomer of 2-butanol, and the configuration is predictable to be S because of the inversion.

⚠️ Important distinction

Inversion means the arrangement of the groups gets inverted, not necessarily means the absolute configuration, R/S, is inverted.

• The product does get an inverted R/S configuration compared to the reactant in many cases, but not guaranteed.
• The actual configuration of the product has to be determined accordingly using R/S rules.
• Don't confuse: spatial inversion always happens, but the R/S label change depends on the priority of the groups.

🚪 Leaving groups (introduction)

🚪 What makes a good leaving group

• When a leaving group departs, it takes the electron pair from the broken bond with it.
• A good leaving group should be one that can accommodate the electron pair very well.
• In other words, the good leaving group should be stable with the pair of electrons.
• Halogens are common leaving groups in alkyl halides, but other appropriate groups can serve as leaving groups as well.

🧭 Overview

🧠 One-sentence thesis

The success of an SN2 reaction depends critically on both the leaving group's ability to depart (weaker bases leave better) and the nucleophile's strength (stronger nucleophiles speed up the reaction).

📌 Key points

• Leaving group quality: weaker bases are better leaving groups because they are more stable and less reactive.
• Basicity vs leaving ability: strong bases hold electrons tightly and are poor leaving groups; weak bases release easily and are good leaving groups.
• Nucleophile strength: negatively charged nucleophiles are stronger than neutral ones; nucleophilicity also depends on periodic trends and size.
• Common confusion: strong bases like OH⁻ or RO⁻ cannot leave directly, but protonation converts them to neutral H₂O or ROH, which can leave.
• Expanded scope: SN2 reactions are not limited to alkyl halides—any compound with a good leaving group can undergo nucleophilic substitution.

🚪 Leaving groups

🔗 What makes a good leaving group

The weaker the basicity of a group, the better the leaving group is.

• Basicity measures a species' ability to share its electron pair.
• A strong base has high reactivity to share electrons → not stable → poor leaving group.
• A weak base has low tendency to share electrons → more stable → good leaving group.
• The key is stability: a group that can stabilize negative charge (or remain neutral) after departure is a better leaving group.

🧪 Halides as leaving groups

For alkyl halides, the relative reactivity order is:

Best leaving group	→	Weakest leaving group
I⁻	> Br⁻ > Cl⁻	> F⁻

• This order matches the relative basicity of halide anions.
• I⁻ is the weakest base → most stable after departure → best leaving group.
• F⁻ is the strongest base → least stable after departure → worst leaving group.

🧬 Other good leaving groups

Besides halides, conjugate bases of strong acids are good leaving groups because they are very weak bases.

• Examples include conjugate bases of tosylic acid (TsOH) and other strong organic acids.
• The excerpt highlights these groups in blue in Figure 7.3a (structure details not reproduced here).
• Principle: if the acid is strong, its conjugate base is weak and stable → good leaving group.

⚠️ Poor leaving groups and protonation

Strong bases are very poor leaving groups:

• OH⁻, RO⁻, NH₂⁻, and R⁻ cannot leave directly.

Don't confuse: OH⁻ or RO⁻ themselves are poor leaving groups, but upon protonation they become neutral molecules (H₂O or ROH), which are suitable leaving groups.

• This conversion will be covered in section 7.6.
• Example: OH⁻ + H⁺ → H₂O (now a good leaving group).

📦 Expanded scope of SN2 reactions

Note: With the scope of leaving groups expanded, substitution reactions are not limited to alkyl halides.

• Any compound with a good leaving group can undergo nucleophilic substitution.

⚡ Nucleophiles

🎯 What is nucleophilicity

Nucleophilicity: the relative strength of a nucleophile, measured in terms of the relative rate of its SN2 reaction with the same substrate.

• It is not an absolute property; it is a comparison of reaction rates.
• Stronger nucleophiles → faster SN2 reactions.
• The nucleophile is one of the rate-determining factors in SN2 reactions.

📊 Structural features that affect nucleophilicity

The excerpt lists four key trends:

Feature	Trend	Example
Charge	Negative charge > neutral	OH⁻ > H₂O; RO⁻ > ROH
Across a period	Nucleophilicity decreases left to right	NH₃ > H₂O; RNH₂ > ROH
Down a group	Nucleophilicity increases	RSH > ROH; RS⁻ > RO⁻; I⁻ > Br⁻ > Cl⁻ > F⁻ (in protic solvent)
Size/bulkiness	Smaller group > bulky group	t-BuO⁻ is a very poor nucleophile because of its bulky size

🔋 Charge effect

• A nucleophile with a negative charge is always stronger than the corresponding neutral one.
• Reason: negative charge means more electron density available to attack the electrophile.

➡️ Periodic trends

• Across a period (left to right): nucleophilicity decreases.
  • Example: nitrogen is more nucleophilic than oxygen (NH₃ > H₂O).
• Down a group (top to bottom): nucleophilicity increases.
  • Example: sulfur is more nucleophilic than oxygen (RSH > ROH).
  • For halides in protic solvent: I⁻ > Br⁻ > Cl⁻ > F⁻.

🧱 Size and steric hindrance

• A smaller group is a better nucleophile than a bulky group.
• Example: t-BuO⁻ (tert-butoxide) is very poor because its bulky size hinders approach to the substrate.

📋 Common strong and weak nucleophiles

Strong (good) nucleophiles:

• OH⁻, RO⁻ (small alkoxide), RS⁻ (thiolate), N₃⁻ (azide), CN⁻ (cyanide), Cl⁻, Br⁻, I⁻ (halide), RCO₂⁻ (carboxylate), RNH₂ (amine)

Weak (poor) nucleophiles:

• ROH, H₂O, t-BuO⁻

Don't confuse: t-BuO⁻ is negatively charged but still a poor nucleophile because of its bulky size—charge alone does not guarantee strong nucleophilicity.

🔬 Functional group interconversions via SN2

🔄 Diverse products from diverse nucleophiles

Because nucleophiles have such diverse structures, SN2 reactions can synthesize compounds with a variety of functional groups.

• The excerpt references Figure 7.3b showing functional group interconversions (structure details not reproduced here).
• Principle: by choosing different nucleophiles, you can introduce different functional groups into the substrate.

🧪 Examples and exercises

The excerpt includes examples and exercises (7.2) asking to show reaction mechanisms for the reactions discussed.

• Answers are referenced in "Answers to Chapter 7 Practice Questions."
• These exercises reinforce understanding of how leaving groups and nucleophiles work together in SN2 mechanisms.

🧭 Overview

🧠 One-sentence thesis

SN1 reactions proceed through a multi-step mechanism with a carbocation intermediate, producing racemic mixtures when the reaction occurs at a chiral center, and their rate depends primarily on substrate structure rather than nucleophile strength.

📌 Key points (3–5)

• Multi-step mechanism: SN1 involves three steps—leaving group departure (rate-determining), nucleophilic attack, and deprotonation—with a carbocation intermediate.
• Rate-determining step: only the substrate participates in the slowest step, so the rate law is first-order in substrate only.
• Substrate reactivity order: tertiary > secondary > primary/methyl—opposite to SN2—because more stable carbocations form faster.
• Stereochemistry outcome: reactions at chiral centers produce racemic mixtures (50:50 R and S) due to planar carbocation allowing attack from either side.
• Common confusion: SN1 vs SN2 substrate preferences are reversed; tertiary substrates favor SN1 but cannot undergo SN2, while primary/methyl favor SN2 but not SN1.

🔬 The SN1 mechanism step-by-step

🔬 Step 1: Leaving group departure (rate-determining)

• The C—Br bond breaks, and the leaving group (Br) departs with both bonding electrons, forming a carbocation and bromide anion.
• This is a highly endothermic bond-breaking process and the slowest step in the mechanism.
• Because this step determines the overall reaction rate, it is called the rate-determining step.
• Only the alkyl halide substrate is involved in this step—no nucleophile participates—which explains why the rate law is first-order: Rate = k[substrate].

Intermediate: an unstable, highly reactive species with a very short lifetime that is the product of one step and the reactant for the next step.

• The carbocation intermediate has a trigonal planar shape with an empty 2p orbital perpendicular to the plane.
• The central carbon is sp² hybridized with an incomplete octet, making it highly reactive and electrophilic.

🔬 Step 2: Nucleophilic attack

• The nucleophile (e.g., H₂O) uses its lone pair to attack the carbocation intermediate.
• Because the carbocation is planar, there is equal possibility for attack from either the front or back side of the plane.
• This produces a protonated alcohol (oxonium ion).
• Example: with tert-butyl carbocation, attack from either side gives the same product, but with a chiral center, each side produces a different stereoisomer in equal amounts.

🔬 Step 3: Deprotonation

• A base (often another solvent molecule like H₂O) accepts a proton from the oxonium ion.
• The final neutral product is formed.
• This step is very fast and sometimes combined with step 2 as a single step.

⚡ Energy diagram and reaction profile

⚡ Multiple curves for multiple steps

• SN1 has multiple curves, one for each step.
• Step 1 has the highest activation energy, confirming it is the rate-determining step.
• The connection between curves represents the carbocation intermediate—at lower energy than transition states but still highly reactive and unstable.

⚡ Energy levels

• Intermediate: relatively lower energy than transition states (peaks) but still unstable.
• Transition states: at the peak of each curve, representing the highest energy point for that step.

🏗️ Substrate structure and carbocation stability

🏗️ Reactivity trend for SN1

Tertiary (3°) > Secondary (2°) >> Primary (1°) and Methyl

• This trend is opposite to SN2, where primary and methyl are most reactive.
• Tertiary substrates are most reactive for SN1 but do not undergo SN2 at all.
• Primary and methyl substrates are unreactive for SN1 but best for SN2.

🏗️ Why substrate structure matters: carbocation stability

• A carbocation forms in the rate-determining step.
• More stable carbocation → forms more easily → faster reaction rate.
• Carbocation stability order: Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl

🏗️ Hyperconjugation effect

Hyperconjugation: the partial orbital overlap between a filled bonding orbital and an adjacent unfilled (or half-filled) orbital.

• Carbocations have an incomplete octet and empty 2p orbital (electron-deficient).
• Adjacent C—C or C—H sigma bonds have filled orbitals that can partially overlap with the empty 2p orbital.
• This shares electron density with the carbocation, stabilizing it.
• More R groups → stronger hyperconjugation → more stable carbocation.
• Tertiary carbocation has three R groups (most stable); methyl carbocation (CH₃⁺) has no R groups (least stable).

🔄 Stereochemistry of SN1 reactions

🔄 Racemization at the reaction center

• Starting with one pure enantiomer (e.g., (S)-3-bromo-3-methylhexane), SN1 produces a 50:50 mixture of both R and S enantiomers (racemic mixture).
• This occurs because the carbocation intermediate is trigonal planar.
• Nucleophile can attack from either the front side or back side with equal probability.
• Each side of attack produces one enantiomer in equal amounts.

Racemization: a reaction that converts an optically active compound into a racemic form.

🔄 When racemization occurs

• Racemization happens when:
  • The reactant is optically active (one enantiomer).
  • The chirality center is also the electrophilic carbon (reaction occurs at the chiral center).
• Don't confuse: If the chirality center is not the reaction center, or if there is more than one chiral center, SN1 does not necessarily produce a racemic mixture.

🧪 Reaction conditions for SN1

🧪 Leaving groups

• A good leaving group is required for SN1, just as for SN2.
• The same principles for leaving group quality apply to both mechanisms.

🧪 Nucleophiles in SN1

• The rate-determining step does not include nucleophiles, so nucleophile strength theoretically does not affect the SN1 rate.
• However, strong nucleophiles tend to favor SN2 instead of SN1.
• Weaker nucleophiles are better for SN1: neutral substances like H₂O, ROH, and RCOOH are typically used.

🧪 Solvolysis reactions

Solvolysis reaction: a nucleophilic substitution in which the nucleophile is a molecule of the solvent as well.

• The term comes from "solvent + lysis" (cleavage by the solvent).
• In practice, neutral substances serve both as nucleophiles and solvents for SN1 reactions.
• An SN1 reaction is usually a solvolysis reaction.
• Example: H₂O acts as both nucleophile and solvent when tert-butyl bromide reacts with water.

📊 SN1 vs SN2 comparison

📊 Key differences table

Feature	SN1	SN2
Rate law	Rate = k[electrophile]	Rate = k[nucleophile]×[electrophile]
Mechanism	Multiple steps with carbocation intermediate	One step, concerted
Stereochemistry	Racemization at reaction center	Inversion at reaction center
Substrate reactivity	Tertiary > Secondary >> Primary/Methyl	Primary/Methyl > Secondary >> Tertiary
Nucleophile	Weak nucleophile, solvolysis	Strong nucleophile

📊 Why the differences matter

• The two mechanisms have opposite substrate preferences: what works best for one does not work for the other.
• Understanding these differences helps choose the proper reaction conditions for different substrates.
• Stereochemistry outcomes are completely different: racemization (SN1) vs inversion (SN2).

🧭 Overview

🧠 One-sentence thesis

The choice between SN1 and SN2 pathways depends primarily on substrate structure (primary, secondary, or tertiary) and is controlled by reaction conditions such as nucleophile strength and solvent polarity.

📌 Key points (3–5)

• Mechanistic differences: SN1 proceeds through a carbocation intermediate in multiple steps, while SN2 is a one-step concerted mechanism.
• Substrate preferences are opposite: tertiary substrates favor SN1, while primary and methyl substrates favor SN2; secondary substrates depend on conditions.
• Stereochemical outcomes differ: SN1 gives racemization at the reaction center, whereas SN2 gives inversion.
• Common confusion: polar protic solvents (water, alcohols) favor SN1 but hinder SN2 by solvating nucleophiles; polar aprotic solvents (DMSO, DMF) favor SN2.
• Practical control: choosing the appropriate solvent and nucleophile strength is an effective way to control which pathway the reaction follows.

⚖️ Side-by-side comparison of mechanisms

⚖️ Core mechanistic differences

Feature	SN1	SN2
Rate law	Rate = k [electrophile]	Rate = k [nucleophile]×[electrophile]
Mechanism	Multiple steps with carbocation intermediate	One step, concerted
Stereochemistry	Racemization on reaction center	Inversion on reaction center
Electrophilic substrate	tertiary 3° > secondary 2° > primary 1° and methyl	primary 1° and methyl > secondary 2° > tertiary 3°
Nucleophile	Weak nucleophile, solvolysis	Strong nucleophile

🔄 Why stereochemistry differs

• SN1 racemization: the carbocation intermediate is planar, so the nucleophile can attack from either face, producing a mixture of stereoisomers.
• SN2 inversion: the nucleophile attacks from the backside in a single concerted step, flipping the stereochemistry like an umbrella turning inside-out.
• Don't confuse: racemization does not mean "no stereochemistry"—it means equal amounts of both configurations are formed.

🧪 Substrate structure determines the pathway

• Primary and methyl substrates: predominantly undergo SN2 reactions because they cannot form stable carbocations.
• Tertiary substrates: go with the SN1 process because tertiary carbocations are stable and steric hindrance blocks backside attack.
• Secondary substrates: the reaction pathway mainly relies on the conditions applied (nucleophiles, solvents, etc.).
• Example: a tertiary substrate with a strong nucleophile in DMSO will still favor SN1 because the substrate structure dominates.

🧴 Solvent effects on reaction pathways

🧴 Three categories of solvents

Non-polar solvents: non-polar compounds (hexane, benzene, toluene, etc.).

Polar protic solvents: compounds containing OH or NH groups that are able to form hydrogen bonds; highly polar because of the OH or NH group.

Polar aprotic solvents: solvents with a medium range of polarity; polar because of polar bonds like C=O or S=O, but the polarity is not as high as the OH or NH group (acetone, DMSO, DMF, THF, CH₂Cl₂).

💧 Polar protic solvents favor SN1 reactions

• General guideline: SN1 reactions are favored by polar protic solvents (H₂O, ROH, etc.), and usually are solvolysis reactions.
• Why they work for SN1: the polar solvent can form hydrogen bonding with the leaving group in the transition state of the first step, thereby lowering the energy of the transition state that leads to the carbocation and speeding up the rate-determining step.
• Dual role: it is very common that polar protic solvents also serve as nucleophiles for SN1 reactions, so SN1 reactions are usually solvolysis reactions.
• Example: water or methanol stabilizes the leaving group as it departs, making carbocation formation easier.

🚫 Why polar protic solvents hinder SN2

• The solvation problem: the nucleophile anions will be surrounded by a layer of solvent molecules with hydrogen bonds—this is called the solvation effect.
• Result: the solvation effect stabilizes (or encumbers) the nucleophiles and hinders their reactivities in an SN2 reaction.
• Don't confuse: "polar" alone is not enough—protic solvents are polar but bad for SN2 because they trap the nucleophile in a hydrogen-bonding cage.

⚡ Polar aprotic solvents favor SN2 reactions

• Why they are needed: strong nucleophiles are usually negatively charged species (OH⁻, CH₃O⁻, CN⁻) that must stay with cations in a salt format like NaOH or CH₃ONa; salts are insoluble in non-polar solvents, so polar solvents are required.
• Why aprotic is better: polar aprotic solvents (acetone, DMSO, DMF) are polar enough to dissolve the salt format nucleophiles and they also do not interact as strongly with anions to hinder their reactivities.
• Result: the nucleophile anions still move around freely in polar aprotic solvents to act as nucleophiles.
• Dramatic rate increase: the excerpt provides data showing that the same SN2 reaction (CH₃I + Cl⁻ → CH₃Cl + I⁻) is 12.5 times faster in water than in methanol, and 1,200,000 times faster in DMF (polar aprotic) than in methanol.

Solvent	Relative rate
CH₃OH	1
H₂O	12.5
DMF	1,200,000

🎯 Choosing the reaction pathway

🎯 Decision framework

• The reaction pathway predominantly depends on the nature of the substrates (primary, secondary, or tertiary).
• The choice of proper reaction condition serves as a way to facilitate the process.

🔧 Practical tips for working with nucleophiles

• Anion vs salt format: strong nucleophiles can be shown as anions (OH⁻, CH₃O⁻, C₂H₅O⁻) or in a salt format (NaOH, KOH, CH₃ONa, C₂H₅ONa) in the reaction conditions—it is the same thing.
• Why both formats are used: the anion format is easy to identify and highlights the nature of these species; the salt format shows the actual chemical formula of the compound used in the reaction since anions must stay together with counter cations as salt.

🍷 Understanding RO⁻/ROH combinations

• Apparent contradiction: sometimes you may see a combination like CH₃ONa/CH₃OH, which combines CH₃O⁻ (strong nucleophile for SN2) together with its conjugate acid CH₃OH (a protic solvent that favors SN1).
• Why it works: CH₃ONa here still acts as a strong nucleophile and can be used for an SN2 reaction, and CH₃OH is the solvent for CH₃ONa.
• How it is prepared: CH₃ONa can be prepared by treating an alcohol with Na metal (or potassium metal K), or by reacting alcohol with NaH.
  • Example: 2 CH₃OH + 2 Na → 2 CH₃ONa + H₂
• Why alcohol is present: since alcohol is in excess in the above reactions, it is also a good solvent for the resulting alkoxide.
• Dual use: the RO⁻ in this combination can be used as a strong nucleophile for an SN2 reaction or a base in an elimination reaction.
• Don't confuse: the presence of protic solvent does not automatically mean SN1—if the nucleophile is strong and the substrate is primary, SN2 can still proceed.

🧭 Overview

🧠 One-sentence thesis

Nucleophilic substitution reactions can involve additional complexities such as carbocation rearrangements, intramolecular reactions, and strategies to convert poor leaving groups into good ones, but the fundamental SN1 and SN2 mechanisms still apply.

📌 Key points (3–5)

• Carbocation rearrangement: In SN1 reactions, carbocations may rearrange to more stable forms through 1,2-shifts of methyl groups or hydrogen atoms before the nucleophile attacks.
• Intramolecular reactions: A single bifunctional molecule can contain both a leaving group and a nucleophile, leading to cyclic products when the reaction occurs within the same molecule.
• Converting poor leaving groups: OH groups (poor leaving groups) can be converted to good leaving groups either by protonation with acid catalyst or by reaction with sulfonyl chlorides to form sulfonate esters.
• Common confusion: Seeing CH₃ONa/CH₃OH together may seem contradictory (strong nucleophile with protic solvent), but the alkoxide still acts as a strong nucleophile for SN2 while the alcohol serves as the solvent.
• Notation formats: Nucleophiles like hydroxide can be written as anions (OH⁻) or salts (NaOH)—both represent the same species, with the anion format highlighting reactivity and the salt format showing the actual compound used.

🔄 Carbocation rearrangement in SN1 reactions

🔄 What is carbocation rearrangement

Carbocation rearrangement: A phenomenon where a carbocation intermediate shifts to a more stable carbocation structure during an SN1 reaction.

• This rearrangement explains why products sometimes have the nucleophile attached to a different carbon than the one that originally held the leaving group.
• The rearrangement occurs between the leaving group departure and nucleophile attack steps.

🔀 Types of shifts

1,2-Methanide shift

• A methyl group (CH₃) moves with its bonding electron pair from one carbon to an adjacent positively charged carbon.
• "1,2-" refers to movement between two adjacent carbons, not necessarily C1 and C2.
• Example: A secondary carbocation can rearrange to a tertiary carbocation when a neighboring CH₃ group shifts over.

1,2-Hydride shift

• A hydrogen atom moves with its electron pair to an adjacent carbocation.
• Occurs when such a shift produces a more stable carbocation.

⚠️ Important rules about rearrangement

• Any reaction with a carbocation intermediate might have rearrangement (not limited to SN1).
• Not all carbocations rearrange—rearrangement only occurs if it leads to a more stable carbocation.
• Shifts are usually 1,2-shifts between adjacent carbons.
• Don't confuse: The product structure may look unexpected, but it follows from the rearranged carbocation, not the original one.

🔗 Intramolecular nucleophilic substitution

🔗 What is an intramolecular reaction

Intramolecular reaction: A reaction that occurs within the same molecule, where one part acts as the leaving group and another part acts as the nucleophile (Latin intra = "within").

• Produces cyclic products instead of two separate molecules.
• Requires a bifunctional molecule: a compound containing both a leaving group and a nucleophile.

🔬 Mechanism and stereochemistry

• The reaction still follows SN1 or SN2 mechanisms depending on the substrate and conditions.
• Example from the excerpt: A molecule with Br on a tertiary carbon (leaving group) and OH (nucleophile) undergoes intramolecular SN1.
  • Br leaves, forming a carbocation intermediate.
  • The OH group from the same molecule attacks the carbocation.
  • Result: A six-membered cyclic ether.
• Stereochemistry: If the mechanism is SN1, the carbocation is trigonal planar, so the nucleophile can attack from either side, producing a racemic mixture (optically inactive).

🔵 Ring size preference

• Intramolecular reactions that produce five- or six-membered rings are highly favored.
• Reason: These ring sizes have special stability compared to smaller or larger rings.

🔧 Converting poor leaving groups to good leaving groups

🔧 Why conversion is needed

• Substitution reactions require good leaving groups to proceed.
• OH groups are poor leaving groups (hydroxide is a strong base).
• Two main strategies exist to convert OH into a good leaving group.

🧪 Strategy 1: Acid catalyst (H⁺)

How it works

1. The OH group is protonated by acid (H⁺) in an acid-base reaction.
2. OH becomes H₂O⁺ (protonated alcohol).
3. H₂O is a much weaker base than OH⁻, making it a good leaving group.
4. H₂O departs, forming a carbocation (SN1 mechanism continues).
5. Nucleophile attacks the carbocation.
6. H⁺ is regenerated and can be reused—only a catalytic amount is needed.

Result: The product is a racemic mixture (characteristic of SN1).

Don't confuse: The first protonation step is essential—without it, the reaction cannot proceed because OH⁻ is too poor a leaving group.

🧪 Strategy 2: Sulfonyl chlorides

Formation of sulfonate esters

• Alcohols react with sulfonyl chlorides in the presence of a weak base.
• The OH group is converted to a sulfonate ester, which is an excellent leaving group.

Sulfonyl chloride	Abbreviation	Product ester	Abbreviation
p-toluenesulfonyl chloride	TsCl	p-toluenesulfonate	OTs (tosylate)
Methanesulfonyl chloride	MsCl	Methanesulfonate	OMs (mesylate)
Trifluoromethanesulfonyl chloride	TfCl	Trifluoromethanesulfonate	OTf (triflate)

Why sulfonate esters are good leaving groups

• They are conjugate bases of strong acids (e.g., TsOH is a strong acid).
• Very weak bases = excellent leaving groups.
• A weak base (like pyridine) is added to neutralize the side product HCl and drive the reaction to completion.

🔬 Multi-step synthesis example

The excerpt shows conversion of 1-butanol to butyl methyl ether:

Step 1: Convert OH to OTs

• 1-butanol + TsCl/pyridine → tosylate intermediate

Step 2: SN2 substitution

• Tosylate + CH₃O⁻ → butyl methyl ether (OTs is replaced by OCH₃)

Notation conventions

• Multi-step syntheses show only starting material and final product structures.
• Reaction conditions for each step are written above/below the arrow.
• Intermediate structures are not shown in the overall scheme.
• Steps must be labeled 1), 2), etc., in proper order and cannot be mixed.

🧪 Nucleophile and solvent notation

🧪 Anion vs salt format

• Nucleophiles can be written as anions (OH⁻, CH₃O⁻, C₂H₅O⁻) or salts (NaOH, KOH, CH₃ONa, C₂H₅ONa).
• Anion format: Easy to identify; highlights the reactive species.
• Salt format: Shows the actual chemical formula used in the reaction (anions must stay with counter cations).
• Both represent the same thing.

🧪 Solvent combinations

Polar aprotic solvents favor SN2

• Examples: OH⁻/DMSO, CH₃ONa/DMF
• Strong nucleophile + polar aprotic solvent = SN2 conditions

Alkoxide with its conjugate acid

• Example: CH₃ONa/CH₃OH
• May seem contradictory: strong nucleophile (for SN2) combined with protic solvent (typically for SN1).
• Explanation: CH₃ONa still acts as a strong nucleophile for SN2; CH₃OH serves as the solvent.
• Why this combination exists: Alkoxides (RONa) are prepared by treating alcohols with Na metal or NaH, leaving excess alcohol that serves as a convenient solvent.

Preparation reactions

• Alcohol + Na metal → RONa + hydrogen gas
• Alcohol + NaH → RONa + hydrogen gas
• Since alcohol is in excess, it becomes the solvent for the resulting alkoxide.

Don't confuse: The RO⁻/ROH combination can be used for SN2 reactions (RO⁻ as nucleophile) or elimination reactions (RO⁻ as base).

🧭 Overview

🧠 One-sentence thesis

E2 is a single-step, second-order elimination mechanism that requires anti-coplanar geometry and produces alkenes whose substitution pattern depends on base size—small bases favor more substituted alkenes (Zaitsev's rule) while bulky bases favor less substituted alkenes (Hofmann's rule).

📌 Key points (3–5)

• E2 mechanism: a concerted, single-step bimolecular elimination where the base attacks β-hydrogen, the C-H bond becomes a π bond, and the leaving group departs simultaneously; rate depends on both substrate and base concentration (second-order).
• Regioselectivity rules: small bases (OH⁻, CH₃O⁻, EtO⁻) favor more substituted alkenes (Zaitsev's rule) due to product stability; bulky bases (t-BuO⁻, LDA) favor less substituted alkenes (Hofmann's rule) due to steric hindrance.
• Stereochemistry requirement: the leaving group and β-hydrogen must be anti-coplanar (same plane, opposite sides) to allow proper orbital overlap and maintain a stable staggered transition state.
• Common confusion: Zaitsev vs Hofmann—the difference is not about mechanism but about which product forms preferentially based on base size; small bases → thermodynamic control (more stable product), bulky bases → kinetic control (less hindered pathway).
• Stereoselectivity: the anti-coplanar requirement leads to preferential formation of one stereoisomer (E or Z) over the other.

🔬 E2 Mechanism and Kinetics

⚙️ How E2 works

E2 mechanism: the bimolecular elimination mechanism where the reaction rate depends on the concentration of both the substrate and base.

• Single-step concerted process: multiple electron-pair transfers happen simultaneously, similar to SN2.
• Three simultaneous events:
  1. The base (e.g., OH⁻) uses its electron pair to attack a β-hydrogen on the β-carbon and starts forming a bond.
  2. The β C-H σ bond begins to move in to become the π bond of a double bond.
  3. The leaving group (e.g., Br) begins to depart by taking the bonding electrons with it.
• A transition state forms with partially breaking and partially forming bonds.
• At completion: C=C double bond and H₂O molecule are fully formed, with Br⁻ leaving completely.

📊 Second-order kinetics

• Why second-order: both the substrate (halide) and the base are involved in the single-step mechanism.
• Rate law depends on concentrations of both reactants.
• Example: elimination of 2-bromo-2-methylpropane with OH⁻ → rate = k[substrate][base].

🧪 Why HX doesn't appear in the equation

• Dehydrohalogenation produces HX as a side product.
• With excess base (OH⁻) present, HBr reacts with OH⁻ to give H₂O and Br⁻.
• The mechanism shows the base directly attacking the β-hydrogen, so HX is neutralized immediately.

🗺️ Regioselectivity: Which alkene forms?

🌿 When multiple products are possible

• If the substrate has identical β-carbons (e.g., 2-bromo-2-methylpropane), only one elimination product forms (2-methylpropene).
• If the substrate has different β-carbons (e.g., 2-bromo-2-methylbutane), more than one product can form.
• Example: 2-bromo-2-methylbutane can produce:
  • 2-methyl-2-butene (trisubstituted alkene)
  • 2-methyl-1-butene (monosubstituted alkene)

📏 Alkene stability hierarchy

General relative stability of alkenes: tetrasubstituted > trisubstituted > disubstituted > monosubstituted > ethene

• More alkyl groups bonded to the double bond carbons → more stable alkene.
• This stability difference drives regioselectivity with small bases.

✅ Zaitsev's rule (small bases)

Zaitsev's rule: when a small base is applied, the more substituted alkene is obtained preferably.

• Small bases: OH⁻, CH₃O⁻, EtO⁻
• Why: the relative stability of the product is the key factor in determining the major product.
• The more substituted alkene is more stable, so it forms preferentially.
• Example: 2-bromo-2-methylbutane + small base → 2-methyl-2-butene (trisubstituted) is the major product.

🧱 Hofmann's rule (bulky bases)

Hofmann's rule: when the elimination yields the less substituted alkene, it is said to follow Hofmann's rule.

• Bulky bases: t-BuO⁻, LDA (lithium diisopropylamide)
• Why: steric hindrance makes it difficult for the big, bulky base to approach hydrogens from the β-carbon bonded with more substituents.
• The hydrogen of the methyl group (less substituted position) is much easier to access.
• Example: 2-bromo-2-methylbutane + t-BuO⁻ → 2-methyl-1-butene (monosubstituted) is the major product.

🔄 Don't confuse: Zaitsev vs Hofmann

Rule	Base size	Major product	Reason
Zaitsev's	Small (OH⁻, CH₃O⁻, EtO⁻)	More substituted alkene	Product stability (thermodynamic control)
Hofmann's	Bulky (t-BuO⁻, LDA)	Less substituted alkene	Steric hindrance (kinetic control)

• Both follow the E2 mechanism; the difference is which β-hydrogen is attacked based on accessibility.

🎯 Stereochemistry Requirements

🔗 Anti-coplanar conformation requirement

Anti-coplanar conformation: the leaving group and H must be in the same plane and in anti-position to each other.

• Two requirements:
  1. The bond connected with the leaving group and the bond connected with the H must be in the same plane.
  2. The leaving group and H must be in anti-position to each other (opposite sides).

🧬 Why anti-coplanar is necessary

• Orbital overlap: allows proper orbital overlapping of the two carbons in the formation of the π bond of the alkene product.
• Transition state stability: the anti-position allows the transition state to be in the more stable staggered conformation.
• Staggered conformation helps to lower the energy level of the transition state and speed up the reaction.

🎭 Stereoselectivity

Stereoselectivity: one stereoisomer will be produced preferably over the other because of the anti-coplanar conformation requirement.

• Example: elimination of (2S,3S)-2-bromo-3-phenylbutane produces the E isomer specifically, not the Z isomer at all.
• Why: when H is in an anti-position to the leaving group Br, the whole compound is in a staggered conformation, and the other groups retain their relative position in elimination, which leads to the E isomer.
• Don't confuse: the stereochemistry of the product is determined by the geometry of the starting material and the anti-coplanar requirement, not by product stability.

🧪 Bases Used in E2 Reactions

🔬 Common bases

• Most commonly applied: hydroxide OH⁻ and alkoxide RO⁻.
• Typical combinations: base with the corresponding alcohol is used broadly.
  • CH₃ONa/CH₃OH
  • C₂H₅ONa/C₂H₅OH

📦 Small vs bulky bases

Category	Examples	Structure notes
Small bases	OH⁻, CH₃O⁻, C₂H₅O⁻, NH₂⁻	Compact, can access hindered positions
Bulky bases	t-BuO⁻, LDA (lithium diisopropylamide)	Large, sterically hindered

• Small bases → Zaitsev's rule (more substituted alkene)
• Bulky bases → Hofmann's rule (less substituted alkene)

🧭 Overview

🧠 One-sentence thesis

E1 reactions are first-order elimination reactions that proceed through a carbocation intermediate in multiple steps, with the rate depending only on substrate concentration and favored by weak bases.

📌 Key points (3–5)

• Mechanism: E1 is a multiple-step process with a carbocation intermediate, unlike the one-step E2.
• Rate law: first-order kinetics—rate depends only on substrate concentration, not base concentration.
• Reaction conditions: weak bases (like H₂O or ROH) and polar protic solvents favor E1; strong bases favor E2 instead.
• Product selectivity: more substituted (more stable) alkenes are the major products, similar to E2 with small bases.
• Common confusion: E1 vs E2—E1 requires weak base and goes through a carbocation; E2 requires strong base and is concerted (one step).

⚙️ E1 mechanism and kinetics

⚙️ Two-step process

The E1 mechanism involves two distinct steps:

Step 1 (rate-determining):

• The substrate (e.g., t-butyl bromide) dissociates, and the leaving group (bromide) departs.
• A carbocation intermediate forms (e.g., tertiary 3° carbocation).
• This is a slow bond-breaking step and determines the overall reaction rate.
• Because only the substrate is involved in this step, the rate law is first-order: Rate = k × [substrate].
• The base concentration does not affect the rate.

Step 2 (fast):

• The hydrogen on the β-carbon (the carbon next to the positively charged carbon) becomes acidic due to the adjacent positive charge.
• The base (e.g., ethanol) removes this β-H.
• The electrons from the C-H bond move to form a C=C π bond, producing the alkene.

📉 First-order kinetics

E1 rate law: Rate = k × [substrate]

• The reaction rate depends only on the concentration of the substrate, not the base.
• This is because the rate-determining step involves only the substrate breaking apart to form the carbocation.
• Example: Doubling the concentration of ethanol does not change the reaction rate; doubling the substrate concentration doubles the rate.

🧪 Reaction conditions and solvolysis

🧪 Weak base requirement

• E1 reactions use weak bases such as neutral molecules (H₂O, ROH like ethanol).
• Strong bases favor the E2 mechanism instead, so a weak base discourages E2 and allows E1 to proceed.
• In the example given, ethanol is both a weak base and the solvent.

💧 Solvolysis

Solvolysis: a reaction where the solvent also acts as the base.

• In E1 reactions, the solvent (e.g., ethanol) often serves as the weak base.
• This is typical for E1 because polar protic solvents stabilize the carbocation intermediate.

🎯 Product formation and rearrangements

🎯 Major product: more substituted alkenes

• When multiple alkene products are possible, the major product is the more substituted (more stable) alkene.
• This is similar to E2 reactions with small bases (Zaitsev's rule).
• Reason: More substituted alkenes are thermodynamically more stable.

🔄 Carbocation rearrangement

• Because E1 involves a carbocation intermediate, rearrangement can occur if it leads to a more stable carbocation.
• Example from the excerpt: A secondary carbocation undergoes a 1,2-methanide shift to form a more stable tertiary benzylic carbocation, which then leads to the final elimination product.
• Don't confuse: E2 does not have carbocation intermediates, so rearrangements do not occur in E2.

📊 E1 vs E2 comparison

📊 Key differences

Feature	E1	E2
Rate law	Rate = k × [substrate] (first-order)	Rate = k × [substrate] × [base] (second-order)
Mechanism	Multiple steps with carbocation intermediate	One step, concerted
Base	Weak base (H₂O, ROH)	Strong base (OH⁻, RO⁻, etc.)
Product	More substituted alkenes (more stable)	Small base: more substituted (Zaitsev); bulky base: less substituted (Hoffmann)
Substrate preference	3° > 2° > 1° (no E1 for primary)	3° > 2° > 1°
Carbocation rearrangement	Possible (carbocation intermediate exists)	Not possible (no carbocation)

🧩 Substrate reactivity

• Tertiary (3°) substrates: most reactive for E1; can form stable carbocations easily.
• Secondary (2°) substrates: can undergo E1, but less favorable than tertiary.
• Primary (1°) substrates: do not undergo E1 because primary carbocations are too unstable to form.

🔍 When to expect E1 vs E2

• E1 is favored by: weak base, polar protic solvent (which can also be the base, i.e., solvolysis).
• E2 is favored by: high concentration of strong base, polar aprotic solvent.
• In practice, E1 and E2 require different reaction conditions, so the choice is usually clear from the conditions given.
• The excerpt notes that competition between E1 and E2 is less of an issue than competition between elimination (E1/E2) and substitution (SN1/SN2).

🧭 Overview

🧠 One-sentence thesis

E1 and E2 are two distinct elimination mechanisms that differ in kinetics, mechanism steps, and reaction conditions, with the choice between them determined primarily by substrate structure and base strength.

📌 Key points (3–5)

• Rate law difference: E1 is first-order (depends only on substrate concentration); E2 is second-order (depends on both substrate and base concentration).
• Mechanism difference: E1 is a multi-step process with a carbocation intermediate; E2 is a one-step concerted process.
• Base strength determines pathway: weak bases (H₂O, ROH) favor E1; strong bases (OH⁻, RO⁻) favor E2.
• Substrate structure matters: primary substrates can only do E2 (no E1 possible); secondary and tertiary substrates can do either depending on conditions.
• Common confusion: E1 vs E2 is not usually a practical problem because they require very different reaction conditions—the real competition is between elimination and substitution.

🔬 E1 mechanism details

🔬 What E1 is

E1 mechanism: a multiple-step elimination mechanism that includes a carbocation intermediate and shows first-order kinetics.

• Similar to substitution reactions, some eliminations follow first-order kinetics.
• The "1" in E1 refers to the first-order rate law.

⚙️ How E1 works (two steps)

Step 1: Carbocation formation (rate-determining)

• The leaving group (e.g., bromide) dissociates, forming a carbocation.
• This is a slow bond-breaking step and determines the overall reaction rate.
• Only the substrate is involved in this step, so the rate law is first-order: Rate = k × [substrate].
• The rate does not depend on base concentration.

Step 2: Proton removal and π bond formation

• The hydrogen on the β-carbon (the carbon next to the positively charged carbon) becomes acidic due to the adjacent positive charge.
• The base (e.g., ethanol) removes this β-H.
• The C–H bond electron pair moves in to form the C–C π bond (the double bond).

🧪 E1 reaction conditions

• Weak base: Neutral molecules like ethanol (EtOH) are very weak bases; weak bases favor E1 by discouraging E2.
• Solvolysis: The base often acts as the solvent as well (e.g., ethanol is both base and solvent).
• Don't confuse: A strong base would favor E2, so E1 requires deliberately weak bases.

🔄 Carbocation rearrangement in E1

• Because E1 involves a carbocation intermediate, rearrangement can occur if it leads to a more stable carbocation.
• Example: A secondary carbocation can undergo a 1,2-methanide shift to form a more stable tertiary benzylic carbocation, which then leads to the final elimination product.

🏆 E1 product selectivity

• If more than one alkene can form, the major product is the more substituted alkene (like E2).
• Reason: More substituted alkenes are more stable.

📊 E1 vs E2 comparison

📊 Side-by-side comparison table

Feature	E1	E2
Rate law	Rate = k × [substrate] (first-order)	Rate = k × [substrate] × [base] (second-order)
Mechanism	Multiple steps with carbocation intermediate	One step, concerted
Product	More substituted, more stable alkenes	Small base: more substituted alkenes (Zaitsev's rule); Bulky base: less substituted alkenes (Hoffmann rule)
Substrate	tertiary 3° > secondary 2° > primary 1° (no E1)	tertiary 3° > secondary 2° > primary 1°
Base	Weak base (H₂O, ROH)	Strong base (OH⁻, RO⁻, etc.)

🎯 Which substrates can do which mechanism

• Primary (1°) substrates: E2 only—no E1 possible because primary carbocations are too unstable to form.
• Secondary (2°) and tertiary (3°) substrates: Can do either E1 or E2, depending on reaction conditions.

🔑 How to choose between E1 and E2

For secondary and tertiary substrates:

• E2 is favored by: high concentration of a strong base (OH⁻, RO⁻, or NH₂⁻) and a polar aprotic solvent.
• E1 is favored by: a weak base and a polar protic compound (H₂O, ROH) that can be both base and solvent (solvolysis).

Don't confuse: In study contexts, comparing E1 and E2 helps understand the mechanisms; in practice, they require very different conditions, so the competition is rarely an issue.

🧩 Practical perspective

🧩 The real competition

• The comparison between E1 and E2 is useful for understanding the mechanisms in depth.
• However, in practice, E1 and E2 require rather different reaction conditions, so competition between them is not usually an issue.
• More important: The real competition is between elimination and substitution—that is, between SN1, SN2, E1, and E2.

🔜 What comes next

• The excerpt indicates that detailed discussions on the comparison and competition between all four types of reactions (SN1, SN2, E1, and E2) will follow.
• Understanding E1 vs E2 is a foundation for understanding the broader competition among all four mechanisms.

🧭 Overview

🧠 One-sentence thesis

The structural nature of a substrate (methyl, primary, secondary, or tertiary) is the most critical factor in determining which reaction pathway it follows, with reaction conditions playing a key role when multiple pathways are possible.

📌 Key points (3–5)

• Substrate structure is decisive: methyl and primary substrates cannot form carbocations, so they exclude SN1/E1 pathways entirely.
• Secondary substrates are most complex: all four pathways (SN1, SN2, E1, E2) are possible, making reaction conditions critical for determining the outcome.
• Strong base vs good nucleophile: distinguishing whether a reagent acts as a base (favoring E2) or nucleophile (favoring SN2) depends on basicity, size, and polarizability.
• Common confusion: all nucleophiles are potential bases and vice versa because both use lone pair electrons; the distinction lies in relative strength and molecular properties.
• Tertiary substrates avoid SN2: steric hindrance blocks SN2, leaving only E2 (with strong base) or SN1/E1 (with neutral conditions).

🔬 Substrate-based pathway determination

🔬 Methyl substrates

• Only pathway: SN2 reaction
• Why: elimination is impossible (no β-hydrogen), and CH3+ is too unstable for SN1
• This is the simplest case with no competition between mechanisms

🔬 Primary (1°) substrates

Primary substrates cannot undergo unimolecular reactions (no SN1/E1) because primary carbocations are too unstable to form.

• Predominant pathway: SN2 when a good nucleophile is used
• Exception: E2 becomes major when a big bulky base/nucleophile is used
• Why SN2 dominates: primary substrates are very good candidates for backside attack with minimal steric hindrance
• Example scenario: a primary substrate with a small nucleophile → SN2; with a bulky base → E2

🔬 Secondary (2°) substrates

• Most challenging case: all four pathways are possible
• Reaction conditions become the key determining factor
• Three distinct pathway groups:
  • E2: favored by strong base
  • SN2: favored by good nucleophile (relatively weaker base)
  • SN1/E1: both occur together under neutral conditions; increasing temperature favors E1 over SN1

🔬 Tertiary (3°) substrates

Tertiary substrates do not undergo SN2 reactions because of steric hindrance.

• Two possible pathways:
  • E2 with strong base
  • SN1/E1 with neutral conditions (poor nucleophile/weak base)
• Practical note: E2 is preferred for synthesizing alkenes because E1 always combines with SN1, making it nearly impossible to avoid substitution products
• E2 and E1 theoretically give the same elimination product

🧪 Reaction conditions and reagent properties

🧪 Separating SN2/E2 from SN1/E1

• Relatively easy distinction:
  • SN2 and E2 require strong nucleophile or strong base (usually negatively charged species)
  • SN1/E1 require neutral conditions (weak base/poor nucleophile like H2O or ROH)
• Solvolysis: when H2O or ROH acts as both base and solvent

🧪 Distinguishing strong bases from good nucleophiles

All nucleophiles are potential bases, and all bases are potential nucleophiles because the reactive part of both is lone pair electrons.

What makes the difference:

• Basicity, size, and polarizability determine whether an anion acts as base or nucleophile
• General rule: relatively stronger bases tend to act as bases; relatively weaker bases with small size and good polarizability tend to act as nucleophiles

Classification:

• Strong bases: OH⁻, RO⁻ (R: small alkyl group), NH2⁻
• Good nucleophiles (relatively weaker bases): Cl⁻, Br⁻, I⁻, RS⁻, N3⁻, CN⁻, RCO2⁻, RNH2

🧪 Bulky bases special case

• Bulky bases such as t-BuO⁻ and LDA always favor E2
• Why: they are too big to perform backside attack required for SN2
• They generate elimination products following Hofmann's rule

📊 Summary table and strategy

📊 Pathway preferences by substrate type

Substrate	Preferred Reaction Pathways
Methyl	SN2 reaction only
Primary	Predominantly SN2; Exception: E2 with bulky base
Secondary	SN2 with good nucleophile (e.g., RS⁻, RCO2⁻); E2 with strong base (e.g., OH⁻, OR⁻); SN1/E1 with neutral conditions (e.g., H2O, ROH)
Tertiary	E2 with strong base (e.g., OH⁻, OR⁻); SN1/E1 with neutral conditions (e.g., H2O, ROH)

📊 Analysis strategy

• Step 1: Identify substrate type (methyl, primary, secondary, tertiary)
• Step 2: Eliminate impossible pathways based on substrate structure
• Step 3: If multiple pathways remain, examine reaction conditions (base/nucleophile strength, solvent, temperature)
• Recommendation: Follow the logical analysis rather than just memorizing the table; practice is essential for mastery

📊 Temperature effects

• For secondary substrates under neutral conditions (SN1/E1 competition): increasing temperature favors E1 over SN1
• Don't confuse: SN1 and E1 are hard to separate completely because both go through carbocation intermediates

🧭 Overview

🧠 One-sentence thesis

Bonds can break in two fundamentally different ways—heterolytic cleavage transfers both electrons to one fragment, while homolytic cleavage splits the electron pair evenly to produce radicals—and this distinction determines whether reactions proceed through charged intermediates or radical intermediates.

📌 Key points (3–5)

• Heterolytic cleavage: one fragment takes both electrons from the bond, producing ions (e.g., carbocation and anion).
• Homolytic cleavage: each fragment takes one electron, producing radicals (species with unpaired electrons).
• Arrow notation difference: double-barbed arrows show electron-pair transfer (heterolytic); single-barbed (fishhook) arrows show single-electron transfer (homolytic).
• Common confusion: heterolytic cleavage occurs in polar bonds and produces charged species; homolytic cleavage occurs mainly in non-polar bonds and requires heat or light to produce neutral radicals.
• Why it matters: radicals are highly reactive intermediates that enable reactions (like halogenation) that alkanes otherwise cannot undergo.

🔀 Two pathways for bond breaking

⚡ Heterolytic bond cleavage

Heterolytic bond cleavage: the process in which one part of the bond takes both electrons (the electron pair) away when the bond breaks.

• The σ bond breaks unevenly.
• One fragment gets both electrons → becomes negatively charged (anion).
• The other fragment gets zero electrons → becomes positively charged (cation).
• Example: In an S_N1 reaction, the leaving group Br takes the electron pair to form Br⁻ and leaves behind a carbocation intermediate.

Arrow notation:

• A double-barbed arrow is used to show the transfer of the electron pair.
• This is the familiar curved-arrow notation from earlier reactions.

🪝 Homolytic bond cleavage (homolysis)

Homolytic cleavage (homolysis): the bond-breaking process in which each part of the σ bond takes one electron away.

• The electron pair separates evenly.
• Each fragment receives one electron → both products contain a single unpaired electron.
• These species are called radicals (or free radicals).

Arrow notation:

• A single-barbed arrow (shaped like a fishhook) is used to show single-electron transfer.
• Each arrow represents the movement of one electron, not a pair.

🔥 Conditions for homolysis

• Homolysis occurs mainly for non-polar bonds (e.g., C–C, C–H, Cl–Cl).
• Heat or light is needed to provide enough energy to initiate the process.
  • Δ is the symbol for heat.
  • hν is used to represent light.
• Example: Chlorine molecules (Cl₂) dissociate homolytically under heat or light to produce two chlorine radicals (Cl•).

🧪 Radicals as reactive intermediates

🧪 What is a radical?

Radical (or free radical): a species that contains one or more single (unpaired) electrons.

• Radicals are produced from homolytic cleavage.
• They are highly reactive reaction intermediates because they lack a complete octet.
• The unpaired electron makes them seek to form new bonds.

⚙️ Why radicals matter for alkanes

• Alkanes are normally inert in most organic reactions because:
  • They contain only C–C and C–H σ bonds, which are strong and stable (formed by head-to-head orbital overlap).
  • Both C–C and C–H bonds are non-polar, so no nucleophile or electrophile sites exist.
• Exception: Alkanes can undergo halogenation substitution via a radical mechanism.
• Radicals enable reactions that would otherwise be impossible for unreactive alkanes.

🆚 Comparing the two cleavage types

Feature	Heterolytic cleavage	Homolytic cleavage
Electron distribution	Both electrons go to one fragment	One electron to each fragment
Products	Ions (cation + anion)	Radicals (neutral, unpaired electrons)
Arrow notation	Double-barbed arrow (electron pair)	Single-barbed arrow (one electron)
Bond type	Often polar bonds	Mainly non-polar bonds
Conditions	Typical reaction conditions	Requires heat (Δ) or light (hν)
Example	S_N1: Br leaves as Br⁻, carbocation forms	Cl₂ → 2 Cl• under light

🚫 Don't confuse

• Heterolytic = uneven split → charged species (ions).
• Homolytic = even split → neutral species with unpaired electrons (radicals).
• The arrow style directly reflects the electron movement: double-barb = pair; single-barb = single electron.

🧭 Overview

🧠 One-sentence thesis

Halogenation of alkanes proceeds through a chain-reaction mechanism involving radical intermediates, with chlorine and bromine being the practical halogens due to their moderate reactivity.

📌 Key points (3–5)

• What halogenation is: a substitution reaction where hydrogen atoms in alkanes are replaced by halogen atoms (Cl or Br) under heat or light, producing alkyl halides.
• How the mechanism works: a three-stage chain reaction—initiation (radical formation), propagation (product formation and radical regeneration), and termination (radical consumption).
• Why propagation is self-sustaining: the second propagation step regenerates the halogen radical, allowing the cycle to repeat hundreds or thousands of times with only a small initial amount of radical.
• Reactivity order and practical choice: fluorine is too reactive (dangerous), iodine is unreactive (positive enthalpy), so Cl₂ and Br₂ are used; Cl₂ is more reactive than Br₂.
• Common confusion: don't confuse the overall reaction (simple substitution) with the mechanism (multi-step radical chain); the mechanism explains why only catalytic amounts of radicals are needed.

🔬 The halogenation reaction

🔬 General reaction equation

Halogenation: a reaction where alkanes react with halogen (Cl₂ or Br₂) under heat or light, replacing hydrogen with halogen to produce alkyl halides.

• General form: alkane + halogen → alkyl halide + hydrogen halide.
• Example: methane + Cl₂ → chloromethane (CH₃Cl) + HCl.
• The reaction is also called substitution (hydrogen is substituted) or halogenation (halogen is introduced); both terms are used interchangeably.
• Heat (Δ) or light (hν) is required to provide energy to start the process.

⚛️ Radicals and homolytic cleavage

Radical (or free radical): a species containing one or more unpaired single electrons, produced from homolytic cleavage.

• Homolytic cleavage: a bond breaks so that each atom takes one of the bonding electrons (shown with single-barb "fishhook" arrows).
• Occurs mainly for non-polar bonds; requires heat or light for energy.
• Radicals are highly reactive intermediates because they lack an octet.
• Example: Cl₂ under heat or light → two Cl• radicals, each with one unpaired electron.

🔄 The chain-reaction mechanism

🔄 Three stages overview

The halogenation mechanism is a chain reaction: a step-wise sequence where each step generates intermediates that cause the next step to occur.

Stage	What happens	Role
Initiation	Radicals are produced	Starts the chain
Propagation	Product forms and radicals regenerate	Self-sustaining core
Termination	Radicals combine and are consumed	Stops the chain

🚀 Initiation: production of radicals

• Chlorine molecules (Cl₂) dissociate homolytically under heat or light.
• Each chlorine atom takes one bonding electron → two Cl• radicals.
• This step provides the initial radicals needed to start propagation.

♻️ Propagation: formation of product and regeneration of radicals

Propagation involves two sub-steps that repeat cyclically:

Step 1: Hydrogen abstraction

• Cl• takes a hydrogen atom from methane (CH₄).
• The C–H bond breaks homolytically; each atom keeps one electron.
• Products: HCl (side product) and CH₃• (methyl radical, the critical intermediate).

Step 2: Chlorine abstraction

• CH₃• takes a chlorine atom from Cl₂.
• Products: CH₃Cl (final product) and Cl• (regenerated radical).
• The regenerated Cl• attacks another methane molecule, repeating step 1, then step 2, and so on.

Why propagation is self-sustaining:

• The regeneration of Cl• in step 2 is particularly significant.
• The cycle repeats hundreds or thousands of times.
• Only a small amount of Cl• is required at the beginning to initiate the process.

🛑 Termination: consumption of radicals

• When two radicals meet, they combine to form a stable molecule.
• This decreases the number of radicals available to propagate the reaction.
• The reaction slows and eventually stops.
• Examples of termination: Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → CH₃CH₃.
• Other combinations are possible as well.

🌡️ Energy considerations

🌡️ Energy level diagram

• Step 1 (hydrogen abstraction): endothermic (absorbs energy).
• Step 2 (chlorine abstraction): exothermic (releases energy).
• Overall propagation: exothermic; the energy absorbed in step 1 is offset by step 2.
• Products are at a lower energy level than reactants.

🧮 Calculating reaction enthalpy

Reaction heat (enthalpy) for each propagation step can be calculated using homolytic bond dissociation energies:

• Bond breaking: energy absorbed, use "+" sign.
• Bond forming: energy released, use "−" sign.

Example: Mono-chlorination of methane propagation

Step 1: H–CH₃ + •Cl → CH₃• + H–Cl

• H–CH₃ bond broken: +440 kJ
• H–Cl bond formed: −432 kJ
• ΔH₁ = +440 + (−432) = +8 kJ (endothermic)

Step 2: Cl–Cl + CH₃• → CH₃–Cl + •Cl

• Cl–Cl bond broken: +243 kJ
• CH₃–Cl bond formed: −352 kJ
• ΔH₂ = +243 + (−352) = −109 kJ (exothermic)

Overall propagation: ΔH = +8 + (−109) = −101 kJ (exothermic)

⚖️ Reactivity comparison of halogens

⚖️ Enthalpy data for different halogens

The enthalpy changes for halogenation with different halogens (mono-halogenation of methane):

Reaction step	F₂	Cl₂	Br₂	I₂
Step 1: H–CH₃ + •X → CH₃• + H–X	−130	+8	+74	−142
Step 2: X–X + CH₃• → CH₃–X + •X	−322	−109	−100	−89
Overall propagation	−452	−101	−26	+53

(All values in kJ/mol)

🎯 Reactivity order and practical choice

Reactivity order: fluorine > chlorine > bromine > iodine

Why each halogen is or isn't used:

• Fluorine (F₂): most reactive, extremely high exothermicity (−452 kJ/mol); vigorous and even dangerous with a lot of heat released; hard to control, not practical for applications.
• Iodine (I₂): least reactive, overall positive enthalpy (+53 kJ/mol); very unreactive, does not react with alkane at all.
• Chlorine (Cl₂) and bromine (Br₂): reactivity in the medium range; used for halogen substitutions of alkanes.
• Cl₂ vs Br₂: Cl₂ is more reactive than Br₂; this leads to different selectivity and applications (discussed further in section 9.4).

Don't confuse: high reactivity is not always beneficial—fluorine's extreme reactivity makes it dangerous and impractical, while iodine's low reactivity makes it useless for this reaction.

🧪 Stability of alkyl radicals

🧪 Why radical stability matters

• The alkyl radical is the key intermediate for halogenation.
• The relative stability of radicals determines the relative reactivity.
• Based on the energy diagram: the alkane that generates the more stable carbon radical exhibits higher reactivity.

📊 Stability trend

Alkyl radicals with different structures show different stabilities:

Stability order: tertiary > secondary > primary > methyl

This follows the same trend as the stability of carbocations.

🔍 Two reasons for the stability trend

1. Hyperconjugation effect of alkyl (R) groups:

• Alkyl groups are electron-donating through hyperconjugation.
• Hyperconjugation: electron density of C–C or C–H σ bonds overlaps with the half-filled p orbital of the carbon radical.
• Similar to carbocations, carbon radicals are electron-deficient species.
• The electron-donating effect of alkyl groups helps stabilize the radical.
• More alkyl groups → more stable radical.

2. Homolytic bond dissociation energy comparison:

• (The excerpt ends here; this point is introduced but not elaborated.)

🧭 Overview

🧠 One-sentence thesis

The stability of alkyl radicals determines the reactivity of alkanes in halogenation reactions, with tertiary radicals being the most stable due to hyperconjugation and lower bond dissociation energy, while benzylic and allylic radicals gain even greater stability through resonance effects.

📌 Key points (3–5)

• Why radical stability matters: the more stable the carbon radical formed, the higher the reactivity of the parent alkane in halogenation reactions.
• Stability order for simple alkyl radicals: tertiary > secondary > primary ≈ methyl, following the same trend as carbocations.
• Two main stabilizing factors: hyperconjugation from electron-donating alkyl groups and lower homolytic bond dissociation energy for more substituted C–H bonds.
• Resonance trumps substitution: benzylic and allylic radicals are more stable than even tertiary alkyl radicals because resonance disperses electron density across multiple contributors.
• Common confusion: don't assume all primary radicals are equally unstable—benzylic and allylic primary radicals are exceptionally stable due to resonance, not substitution.

🔗 Why radical stability controls reactivity

🔗 The connection to halogenation

• Alkyl radicals are the key intermediates in halogenation reactions of alkanes.
• The relative stability of these radicals determines the relative reactivity of the starting alkane.
• Energy diagram principle: an alkane that generates a more stable carbon radical will exhibit higher reactivity.
• Example: if two alkanes can form radicals, the one producing the more stable radical will react faster under the same conditions.

🏗️ Stability order and hyperconjugation

🏗️ General stability trend

The excerpt establishes a clear hierarchy:

Radical type	Stability rank	Reason
Tertiary (3°)	Most stable	Most alkyl groups attached
Secondary (2°)	Intermediate	Moderate alkyl substitution
Primary (1°)	Least stable	Fewest alkyl groups
Methyl	Least stable	No alkyl groups

• This trend mirrors the stability of carbocations.

⚡ Hyperconjugation effect

Hyperconjugation effect of the alkyl (R) group: alkyl groups are electron-donating groups through the hyperconjugation effect, which is the electron density of C–C or C–H σ bond overlap with the half-filled p orbital of carbon radicals.

• Carbon radicals are electron-deficient species (like carbocations).
• Alkyl groups donate electron density through overlap of their σ bonds with the radical's half-filled p orbital.
• More alkyl groups → more hyperconjugation → more stable radical.
• Example: a tertiary radical has three alkyl groups donating electron density, while a primary radical has only one.

🔨 Bond dissociation energy evidence

🔨 What homolytic cleavage reveals

• Homolytic cleavage of a C–H bond produces a carbon radical.
• Different C–H bonds have different bond dissociation energies.
• The excerpt compares primary vs secondary C–H bonds in propane:
  • The 1° C–H bond has 10 kJ/mol higher dissociation energy than the 2° C–H bond.
  • Higher bond energy means more energy is required to break the bond.
  • Higher energy input → higher energy radical produced → less stable radical.
• Therefore, the secondary radical is more stable than the primary radical.

🧮 Interpreting the energy difference

• Since both radicals come from the same compound (propane), the comparison is direct.
• The lower the bond dissociation energy, the easier it is to form the radical, and the more stable that radical is.
• Don't confuse: lower bond energy does not mean the radical itself is "weaker"—it means the radical is more stable and easier to form.

🌀 Resonance effect: benzylic and allylic radicals

🌀 Benzylic radicals

Benzylic position: the position right next to the benzene ring.
Benzylic radical: a radical at the benzylic position.

• A benzylic radical is technically a primary radical (one alkyl group attached to the radical carbon).
• However, it is more stable than regular tertiary radicals.
• Why? The presence of the benzene ring allows resonance.
• The benzylic radical has a total of five resonance contributors.
• Resonance principle: more resonance contributors → better dispersion of electron density → more stable species.

🌊 Allylic radicals

Allylic position: the carbon right next to the C=C double bond.

• An allylic radical also benefits from resonance.
• The excerpt shows resonance structures for an allylic radical example.
• Like benzylic radicals, allylic radicals are more stable than tertiary alkyl radicals because of resonance effects.

🔍 Resonance vs substitution

• Key distinction: resonance stabilization is more powerful than hyperconjugation from alkyl substitution.
• A primary benzylic or allylic radical (with resonance) is more stable than a tertiary alkyl radical (without resonance).
• Don't confuse: the number of alkyl groups attached to the radical carbon is not the only factor—resonance can override substitution effects.
• Example: a benzylic primary radical (one alkyl group, but five resonance structures) beats a tertiary radical (three alkyl groups, no resonance).

📋 Summary of stabilizing effects

Effect	Mechanism	Impact on stability
Hyperconjugation	Alkyl groups donate electron density via σ bond overlap	More alkyl groups → more stable
Bond dissociation energy	Lower energy to break C–H bond → more stable radical	3° < 2° < 1° bond energy
Resonance	Electron density dispersed over multiple atoms	Benzylic and allylic radicals > tertiary radicals

• All three effects work together, but resonance is the strongest stabilizing factor when present.

🧭 Overview

🧠 One-sentence thesis

Bromination is more selective and synthetically useful than chlorination because bromine's lower reactivity allows it to discriminate among different hydrogen types and produce one major product, whereas chlorine's high reactivity leads to mixtures of isomers and multi-substitution problems.

📌 Key points (3–5)

• Chlorine's low selectivity: Cl₂ does not discriminate greatly among primary, secondary, or tertiary hydrogens, producing mixtures of isomeric products.
• Predicting chlorination products: the amount of each product depends on both the number of hydrogens available (probability) and the relative reactivity of each position.
• Multi-substitution problem: chlorination reactions continue beyond one substitution, creating mixtures of di-, tri-, and tetra-chlorinated products.
• Bromination's high selectivity: Br₂ reacts selectively at the most reactive position (forming the most stable radical), giving one major product.
• Common confusion: reactivity vs selectivity—higher reactivity (chlorine) means lower selectivity; lower reactivity (bromine) means higher selectivity and better synthetic utility.

⚗️ Chlorination characteristics

⚗️ Low selectivity of chlorine

• Chlorine is a "rather reactive reagent" that shows "relatively low selectivity."
• This means Cl₂ does not discriminate greatly among different types of hydrogen atoms (1°, 2°, or 3°) in an alkane.
• Result: a mixture of isomeric monochlorinated products is obtained.
• Example: propane chlorination gives both 1-chloropropane (45%) and 2-chloropropane (55%).

🧮 Predicting product ratios

To predict the relative amount of different chlorination products, consider two factors simultaneously: reactivity and probability.

Amount of a certain type of product = number of that type of hydrogens × relative reactivity

Relative reaction rates for chlorination:

• Primary (1°) hydrogen: 1.0
• Secondary (2°) hydrogen: 3.8
• Tertiary (3°) hydrogen: 5.0

Calculation example for propane:

• 1-chloropropane: 6 (number of 1° H) × 1.0 = 6.0
• 2-chloropropane: 2 (number of 2° H) × 3.8 = 7.6
• Yield %: 6.0/13.6 = 44% vs 7.6/13.6 = 56%

🔄 Multi-substitution problem

• Although we may assume chlorination occurs once, "this is not the actual case."
• Multiple substitution always happens: monochloride → dichloride → trichloride → tetrachloride.
• Example: methane + Cl₂ produces chloromethane, dichloromethane, trichloromethane, and tetrachloromethane.
• The mechanism for multi-chlorination is similar to monochlorination, going through similar propagation steps.

🛠️ Minimizing multi-chlorination

Methods to reduce (but not completely avoid) multi-substitution:

• Use high concentration of alkane relative to Cl₂ to decrease the possibility of multi-chlorination.
• Control reaction time: stop after a "short" time to favor the monochlorination product.

🎯 Bromination advantages

🎯 High selectivity of bromine

• Bromine's "relative lower reactivity" makes it exhibit "much greater selectivity."
• Less reactive means it reacts more slowly, giving it the chance to differentiate between different types of hydrogens.
• Bromine selectively reacts with the most reactive hydrogen (the one forming the most stable radical).

Relative reaction rates for bromination:

• Primary (1°): 1
• Secondary (2°): 82
• Tertiary (3°): 1640

The huge differences in reactivity mean the reactivity factor becomes predominant for determining the product.

✅ Selective product formation

• Bromination usually occurs selectively on the most reactive position.
• This gives one major product exclusively.
• Example: isobutane bromination produces only the tertiary bromide (at the most substituted carbon).
• Don't confuse: chlorination gives mixtures; bromination gives one major product.

🔬 Synthetic utility

Bromination has the greatest utility for synthesis of alkyl halide.

Halogenation	Selectivity	Product mixture	Synthetic usefulness
Chlorination	Low	Multiple isomers + multi-substitution	Not useful for specific products
Bromination	High	One major product	Greatest utility for synthesis

🔺 Stereochemistry of halogenation

🔺 Radical geometry

• Carbon radical has three bonds and one single electron.
• Experimental evidence shows most alkyl radicals have a trigonal planar shape.
• Carbon is in sp² hybridization with one unpaired electron in the unhybridized 2p orbital.

🪞 Racemic mixture formation

When a stereocenter is generated during halogenation, a racemic mixture is obtained.

Why racemic?

• The carbon radical intermediate is trigonal planar.
• The halogen can attack from either side of the plane.
• Both sides are identical, so the probability is the same for each side.
• Equal amounts of R and S enantiomers are produced.

Example: (±)-3-methylhexane + Br₂ → racemic mixture of R and S 3-bromo-3-methylhexane.

🔄 Similarity to SN1

• The stereochemistry of radical substitution is similar to that of the SN1 reaction.
• Both carbon radical and carbocation are in a trigonal planar shape.
• Both produce racemic mixtures when forming new stereocenters.

🧭 Overview

🧠 One-sentence thesis

When halogenation of alkanes creates a stereocenter, the reaction produces a racemic mixture because the planar radical intermediate can react equally from either side.

📌 Key points

• Radical geometry: Carbon radicals adopt a trigonal planar shape (sp² hybridization) with an unpaired electron in a 2p orbital, not tetrahedral as VSEPR might suggest.
• Racemic product formation: When a stereocenter is generated, equal amounts of R and S enantiomers form because both sides of the planar radical are identical.
• Mechanism explanation: The propagation steps show that bromine can attack either face of the planar radical with equal probability.
• Common confusion: Don't confuse radical geometry with VSEPR prediction—experimental evidence shows most alkyl radicals are planar, not tetrahedral.
• Analogy to other reactions: Radical substitution stereochemistry resembles SN1 reactions because both involve planar intermediates (radical vs carbocation).

🔬 Structure of carbon radicals

🧪 Predicted vs actual geometry

• VSEPR prediction: A carbon radical has three bonds plus one single electron = four electron groups → should be tetrahedral.
• Experimental reality: Most alkyl radicals are actually trigonal planar.

Carbon radical structure: Carbon has three bonds and one single electron; the carbon adopts sp² hybridization in a trigonal planar shape with the unpaired electron in an unhybridized 2p orbital.

⚠️ Why this matters

• The planar geometry is key to understanding why racemic mixtures form.
• The 2p orbital with the unpaired electron is perpendicular to the plane of the three bonds.
• Example: In a radical intermediate, the carbon center lies flat with three substituents in a plane, not pyramidal.

🎯 Stereochemical outcome: racemic mixtures

🧬 The bromination of (±)-3-methylhexane example

The excerpt uses bromination of (±)-3-methylhexane to illustrate stereochemistry:

• Product: Racemic mixture of R and S 3-bromo-3-methylhexane.
• Key observation: Equal amounts of both enantiomers are obtained.

🔄 Why equal amounts form

Step 1 (propagation): A carbon radical forms → trigonal planar shape.

Step 2 (propagation): The radical reacts with bromine (Br₂).

• Bromine can attack from either side of the planar radical.
• Both sides of the plane are identical (the radical is achiral/symmetric).
• Therefore, the probability of attack from either side is 50:50.
• Result: Equal amounts of R and S enantiomers → racemic mixture.

📐 Visual understanding

• Imagine the planar radical as a flat disk with three groups attached.
• The bromine molecule can approach from above or below the disk.
• Since both approaches are equally likely, both stereoisomers form in equal amounts.
• Don't confuse: This is not selective—it's statistical equality due to symmetry.

🔗 Comparison with other reactions

🔀 Similarity to SN1 mechanism

The excerpt explicitly compares radical substitution to SN1:

Reaction type	Intermediate	Geometry	Stereochemical result
Radical substitution	Carbon radical	Trigonal planar	Racemic mixture
SN1	Carbocation	Trigonal planar	Racemic mixture

Why they're similar: Both intermediates are planar, allowing equal attack from either face when a stereocenter is formed.

Key insight: The planar geometry of the intermediate determines the stereochemical outcome in both cases.

🎓 Practical implication

• When designing a halogenation reaction that creates a stereocenter, expect a racemic product.
• You cannot use radical halogenation to selectively make one enantiomer.
• Example: If you need a single enantiomer, radical halogenation is not the right synthetic method.

🧭 Overview

🧠 One-sentence thesis

Retrosynthetic analysis is a backward-working strategy that helps chemists design multi-step synthesis routes by starting from the target molecule and identifying precursors step-by-step until reaching the starting material.

📌 Key points (3–5)

• Core strategy: Work backward from the product rather than forward from the starting material to visualize complex multi-step syntheses.
• The process: Identify precursor 1 that makes the target in one step, then find the precursor for precursor 1, repeating until you reach the starting material.
• Presentation convention: Open arrows indicate backward analysis; reagents/conditions are typically not specified until writing the forward synthesis route.
• Common confusion: The retrosynthetic analysis shows the "thinking process" backward, but the final answer must be written as a forward synthesis route with all reagents and conditions.
• Multiple pathways: Different precursors may lead to multiple valid routes; evaluate efficiency by comparing benefits and disadvantages of each path.

🎯 Why synthesis matters

🎯 The goal of organic synthesis

Building larger, complex organic molecules from smaller, simple molecules is the goal of organic synthesis.

• Organic synthesis has importance across multiple domains:
  • Testing newly developed reaction mechanisms or methods
  • Replicating molecules found in living nature
  • Producing new molecules with potential applications in energy, materials, or medicine

🧗 The challenge of multi-step synthesis

• Most syntheses require multiple steps—anywhere from a few to 20 or more steps.
• Visualizing all necessary steps from the beginning is challenging.
• This difficulty motivates the need for a systematic strategy like retrosynthetic analysis.

🔄 The retrosynthetic method

🔄 Working backward instead of forward

• Traditional approach problem: Looking at the starting material and deciding the first step can be overwhelming.
• Retrosynthetic approach: Look at the product and decide how to do the last step first.
• This reversal makes complex synthesis planning more manageable.

➡️ Visual representation

The analysis uses open arrows (not regular reaction arrows) to indicate backward thinking:

• Target compound ⟹ Precursor 1 ⟹ Precursor 2 ⟹ ... ⟹ Starting material
• Each arrow represents one synthetic step in reverse.

📝 Key conventions

• During analysis: Reagents and conditions are typically not specified.
• Final answer: Once the retrosynthetic analysis is complete, write the synthesis route in the forward direction with all reagents and conditions included.
• Don't confuse: The analysis is backward (for thinking), but the solution is forward (for execution).

🔬 Applying retrosynthetic analysis

🔬 The worked example approach

The excerpt provides an example synthesizing methoxymethylbenzene from toluene:

Step 1 (backward thinking):

• Target is an ether
• S_N_2 reaction between CH₃O⁻ (nucleophile) and a halide gives the desired ether
• Therefore, precursor 1 = a halide

Step 2 (backward thinking):

• The halide can be made from toluene through halogenation
• Therefore, precursor 2 = toluene (the starting material)

Final solution (forward direction):

• Write the actual synthesis route going forward
• Include all reagents and conditions for each step

🧠 Prerequisites for success

To design synthesis routes effectively, you must:

• Be very familiar with all reaction types
• Understand how functional groups transform in each reaction
• Know what reagents and conditions are involved
• Sometimes consider reaction features like stereochemistry

🛤️ Multiple pathways

• You may identify multiple valid routes with different precursors.
• Evaluate each path by comparing:
  • Possible benefits of each route
  • Disadvantages of each route
• Select the most efficient synthesis route based on this evaluation.

🧪 Context: Available reactions

🧪 Reactions learned so far

The excerpt mentions three major reaction types available at this point:

1. Nucleophilic substitution
2. Elimination
3. Halogenation of alkanes (radical substitution)

These reactions form the toolkit for designing synthesis routes in the examples provided.

🔗 Connection to the example

• The example uses halogenation (to make the halide precursor) and S_N_2 substitution (to introduce the methoxy group).
• This demonstrates how multiple reaction types combine in a synthesis route.

🧭 Overview

🧠 One-sentence thesis

Alkenes can be synthesized primarily through elimination reactions—either E2 dehydrohalogenation of alkyl halides or acid-catalyzed dehydration of alcohols—with the mechanism and substrate structure determining the product distribution.

📌 Key points (3–5)

• Two main synthesis routes: E2 elimination of alkyl halides and dehydration of alcohols both produce alkenes by removing small molecules.
• Mechanism matters: alkyl halides undergo E2 elimination, while alcohols undergo E1 (or E2 for primary) dehydration through carbocation intermediates.
• Substrate and base choice control products: secondary/tertiary substrates prefer E2; small bases give more substituted alkenes (Zaitsev), bulky bases give less substituted (Hofmann).
• Common confusion: dehydration of alcohols can involve carbocation rearrangement to more stable intermediates, changing the carbon skeleton of the product.
• Reactivity trend for alcohols: tertiary > secondary > primary, because stability of the carbocation intermediate determines the rate.

🧪 Dehydrohalogenation of alkyl halides

🔧 How E2 elimination works

E2 elimination reaction of alkyl halide: one of the most useful methods for synthesizing alkene.

• The alkyl halide loses a halogen (X) and a hydrogen from an adjacent carbon in a single concerted step.
• The double bond forms where the H and X were removed.
• Example: an alkyl halide treated with strong base produces an alkene.

🎯 Practical hints for E2 synthesis

Substrate choice:

• Prefer secondary or tertiary substrates—they naturally favor E2 over substitution.
• If you must use a primary substrate, choose a bulky base like t-BuO⁻ to avoid competing SN2 substitution (since OH⁻ is also a good nucleophile and would cause substitution).

Reaction conditions:

• High concentration of strong base + elevated temperature favors E2.

Product control:

• Small base → Zaitsev's product (more substituted, more stable alkene).
• Bulky base → Hofmann's product (less substituted alkene).
• Don't confuse: the base size determines which hydrogen is removed, not just the strength.

🌡️ Dehydration of alcohols

💧 What dehydration means

Dehydration: elimination reaction where an alcohol loses an OH group and an H atom from an adjacent carbon, eliminating a water molecule overall.

• Requires strong acid (concentrated H₂SO₄ or H₃PO₄) with heat.
• The OH group itself is a poor leaving group, but protonation by acid converts it to H₂O, a good leaving group.
• Example: tert-butyl alcohol treated with acid and heat produces an alkene.

⚙️ Mechanism for tertiary and secondary alcohols (E1-type)

Step-by-step process:

1. Protonation: Strong acid protonates the OH group, converting it from a poor leaving group (OH) to a good leaving group (H₂O).
2. Carbocation formation: H₂O leaves, forming a carbocation intermediate (rate-determining step).
3. Deprotonation: A base removes a β-hydrogen, forming the double bond.

Key insight: This is essentially an E1 mechanism, but it starts with a protonation step. The overall reaction can be viewed as E1 elimination of a protonated alcohol.

📊 Reactivity trend

Alcohol type	Reactivity	Reason
Tertiary (3°)	Highest	Forms most stable carbocation
Secondary (2°)	Moderate	Forms moderately stable carbocation
Primary (1°)	Does not undergo E1	Primary carbocation too unstable

• The rate-determining step is carbocation formation, so relative carbocation stability defines reactivity.

🔄 Carbocation rearrangement

Why rearrangement happens:

• The mechanism involves carbocation intermediates.
• If a more stable carbocation can form through rearrangement (hydride or methyl shift), it will.

Example from the excerpt:

• Dehydration of 3,3-dimethyl-2-butanol produces two alkenes, both with different carbon skeletons than the starting material.
• The initially formed secondary carbocation undergoes 1,2-methanide shift to produce a more stable tertiary carbocation.
• Two β-hydrogens are available for removal; the more substituted (more stable) alkene is the major product.

Don't confuse: Not all dehydrations rearrange—only when a more stable carbocation can form.

🔬 Primary alcohol dehydration (E2-type)

Different mechanism:

• Primary alcohols cannot form primary carbocations (too unstable).
• Instead, they undergo E2 elimination after protonation.

Two-step process:

1. Protonation of OH to convert it to a good leaving group (H₂O).
2. Concerted E2 elimination of the protonated alcohol.

Example: Ethanol treated with concentrated acid and heat produces ethene through this E2 pathway.

🧩 Comparing the two methods

Method	Substrate	Mechanism	Key feature	Product control
Dehydrohalogenation	Alkyl halide	E2	Base size matters	Small base → Zaitsev; bulky base → Hofmann
Dehydration	Alcohol	E1 (3°, 2°) or E2 (1°)	Requires acid + heat; may rearrange	More substituted alkene usually favored

When to use which:

• Use dehydrohalogenation when you have an alkyl halide and want control over regioselectivity through base choice.
• Use dehydration when starting from an alcohol, but watch for possible rearrangements with 2° and 3° alcohols.

🧭 Overview

🧠 One-sentence thesis

The addition of hydrogen halides to alkenes produces alkyl halides through an electrophilic addition mechanism that favors the formation of the more stable carbocation intermediate, explaining the regioselective outcome known as Markovnikov's rule.

📌 Key points (3–5)

• Why alkenes are reactive: the π bond is weak (formed by side-by-side overlapping) and the π electrons are loosely held, making the double bond electron-rich and attracted to electrophiles.
• Two-step electrophilic addition mechanism: (1) π electrons attack the electrophilic H, forming a carbocation and halide ion; (2) the halide ion attacks the carbocation to form the product.
• Regioselectivity and Markovnikov's rule: the hydrogen adds to the double bond carbon with more hydrogen atoms because this produces the more stable carbocation intermediate.
• Common confusion—two mechanisms for HBr: electrophilic addition (no peroxide) gives Markovnikov product; radical addition (with peroxide) gives anti-Markovnikov product.
• Why mechanism matters: memorizing products is overwhelming; understanding the carbocation stability principle unifies many reactions and makes learning easier.

🔬 Why alkenes undergo addition reactions

🔬 The reactivity center of alkenes

The double bond is the reactivity center of alkene; this is mainly because of the relatively loosely held π electrons of the double bond.

• The π bond is formed by side-by-side overlapping, a relatively weak overlapping mode.
• This makes the π bond weak and highly reactive.
• The π electrons make the double bond carbons electron-rich and have the tendency to be attracted to an electrophile.

⚙️ What is an addition reaction

In addition reaction, a small molecule is added to multiple bonds, and one π bond is converted to two σ bonds (unsaturation degree decreases) as a result of the addition.

• An addition reaction is the opposite process of elimination.
• The general equation: a small molecule adds across the C=C double bond.
• Example: adding HBr to an alkene produces an alkyl halide.

🧪 Electrophilic addition of HX to alkenes

🧪 The two-step mechanism

The mechanism involves two steps:

Step 1: Formation of carbocation

• The π electrons of the alkene act as nucleophiles and are attracted to the partially positively charged hydrogen (electrophile) of HX.
• As the π electrons move toward the hydrogen, the H-X bond breaks.
• The halogen (X) moves away with the bonding electrons.
• A new σ bond forms between one double bond carbon and hydrogen.
• A carbocation and a halide ion (X⁻) are formed.

Step 2: Nucleophilic attack

• The halide ion (X⁻) reacts with the positively charged carbocation to give the final product.
• This step is similar to the second step of the S_N1 reaction, in which a nucleophile reacts with an electrophile (carbocation).

🎯 Why it's called electrophilic addition

Because the first step of the above reaction is the addition of an electrophile (H⁺) to the alkene, the reaction is called an electrophilic addition reaction.

• An electrophilic addition reaction is a characteristic type of reaction of alkenes.
• Several other addition reactions also belong to this category.

📐 Regioselectivity and Markovnikov's rule

📐 What is regioselectivity

For the reaction in which two or more constitutional isomers could be obtained as products, but one predominates, the reaction is said to be a regioselective reaction.

• Regio comes from the Latin word regionem, which means direction.
• Example: adding HBr to 2-methylpropene could theoretically produce 2-bromo-2-methylpropane or 1-bromo-2-methylpropane, but only the first is the major product.

📏 Markovnikov's rule

In the addition of HX to an alkene, the hydrogen atom adds to the double bond carbon that has the greater number of hydrogen atoms.

• This rule was summarized by Russian chemist Vladimir Markovnikov.
• The rule describes the regioselectivity trend of the electrophilic addition of HX to alkenes.

🧠 The underlying reasoning: carbocation stability

• When the new bond forms between the double bond carbon and hydrogen in the first step, the hydrogen could bond with either carbon.
• This produces carbocations with different structures.
• The more stable carbocation (e.g., tertiary > secondary > primary) is formed preferably.
• Example: in the addition of HBr to 2-methylpropene, path (a) forms a tertiary carbocation and path (b) forms a primary carbocation; the tertiary carbocation is much more stable, so 2-bromo-2-methylpropane is the major product.

Don't confuse: It may seem easy to just memorize the rule or the product structure, but memorization becomes overwhelming with many reactions. The proper way is to learn and understand the mechanism and unify the principles based on mechanisms.

🔄 Radical addition of HBr to alkenes

🔄 Anti-Markovnikov product via radical mechanism

• Electrophilic addition of HX to alkene gives addition products that follow Markovnikov's rule.
• However, HBr can also be added to alkene in a way that gives an anti-Markovnikov product.
• The anti-Markovnikov product is obtained through a different mechanism: the radical mechanism.

Mechanism	Conditions	Product
Electrophilic addition	No peroxide	Markovnikov product
Radical addition	With peroxide	Anti-Markovnikov product

🧨 Role of peroxide as radical initiator

To initiate the radical mechanism, peroxide must be involved to generate the radical in the initiation step of the mechanism.

• The O-O bond of peroxide is weak (bond energy about 150 kJ/mol).
• It undergoes homolytic cleavage readily with heat to produce alkoxyl radicals.
• The peroxide acts as a radical initiator by generating radicals.
• The addition is called radical addition.

⚙️ Radical addition mechanism

The mechanism has three stages:

Initiation (two steps)

1. Peroxide undergoes homolytic cleavage to generate alkoxyl radicals.
2. The alkoxyl radical reacts with H-Br to generate the bromine radical (Br·), which reacts with an alkene to initiate the chain reaction.

Propagation

• The order of addition is reversed compared to electrophilic addition.
• The bromine radical (Br·) is added to the double bond first, followed by the abstraction of hydrogen atoms (H).
• This produces the anti-Markovnikov product.

Don't confuse: Only HBr proceeds with radical addition in the presence of peroxide, not HCl or HI.

🔍 Why the order reverses

• In electrophilic addition: H⁺ adds first → carbocation forms → X⁻ attacks → Markovnikov product.
• In radical addition: Br· adds first → carbon radical forms → H abstraction → anti-Markovnikov product.
• The stability of the intermediate (tertiary radical is more stable than primary radical) still governs which carbon the Br· attacks, but because Br· adds first, the regioselectivity is reversed.

🧭 Overview

🧠 One-sentence thesis

Alkenes react with water or alcohols in the presence of acid catalysts to form alcohols or ethers through a mechanism that follows Markovnikov's rule and is the reverse of dehydration reactions.

📌 Key points (3–5)

• Why acid is needed: Pure water cannot react with alkenes because it is not acidic enough; a small amount of acid (catalyst) enables the reaction.
• What products form: Water addition produces alcohols (hydration); alcohol addition produces ethers.
• Regioselectivity follows Markovnikov's rule: The hydrogen adds to the carbon with more hydrogens; the OH (or OR) adds to the carbon with fewer hydrogens.
• Common confusion: Hydration vs. dehydration—these are reverse reactions; excess water favors alcohol formation, while high acid concentration and heat favor alkene formation.
• Carbocation rearrangement is possible: Since the mechanism involves carbocation intermediates, rearrangement may occur if a more stable carbocation can form.

💧 Hydration of alkenes (addition of water)

💧 Why alkenes need acid to react with water

• Pure water alone does not react with alkenes because water is not acidic enough.
• The hydrogen in water cannot act as an electrophile without activation.
• A small amount of acid (catalyst) makes the reaction possible.

🔬 Common acid catalysts

Acid	Description
Dilute H₂SO₄	Sulfuric acid dissociates completely in water to generate hydronium ion (H₃O⁺)
TsOH (tosyl acid)	A strong organic acid sometimes used as an alternative

• The hydronium ion (H₃O⁺) participates in the reaction as the electrophile.

🍷 Product of hydration

Hydration: the acid-catalyzed addition of water to an alkene, producing an alcohol.

• A water molecule (H—OH) is added across the double bond.
• This reaction has great utility in large-scale industrial production of low-molecular-weight alcohols.
• Example: An alkene reacts with water in the presence of acid to form an alcohol.

⚙️ Mechanism of acid-catalyzed hydration

⚙️ Three-step mechanism

The mechanism is essentially the same as hydrogen halide (HX) addition to alkenes:

Step 1: Electrophilic attack

• The hydronium ion (H₃O⁺) attacks the double bond.
• A carbocation intermediate forms at the more substituted carbon (more stable carbocation).

Step 2: Nucleophilic attack

• Water molecule attacks the carbocation.

Step 3: Deprotonation

• A proton is removed to generate the neutral alcohol product.
• The hydronium ion (H₃O⁺) is regenerated, confirming that acid acts as a catalyst (only a small amount is needed).

🎯 Regioselectivity: Markovnikov's rule

• The hydrogen atom connects to the double bond carbon that has more hydrogen atoms.
• The OH group adds to the carbon that has fewer hydrogen atoms.
• Why: The first step forms the more stable carbocation (at the more substituted carbon).
• Example: In 1-methylcyclohexene hydration, the OH adds to the more substituted carbon.

🔄 Hydration vs. dehydration: reverse reactions

🔄 Understanding the equilibrium

• Hydration (alkene → alcohol): addition reaction.
• Dehydration (alcohol → alkene): elimination reaction.
• These are reverse reactions of each other.

🔄 How to favor each direction

Goal	Conditions	Why
Produce alcohol from alkene	Excess water	Pushes equilibrium toward alcohol product
Produce alkene from alcohol	High acid concentration + elevated temperature	Favors elimination; distill product as it forms to push equilibrium toward alkene

• Don't confuse: The same reactants can give different products depending on reaction conditions.

🧪 Addition of alcohol to alkenes

🧪 Formation of ethers

• With acid present, an alcohol can be added to an alkene in the same way water can.
• The product is an ether instead of an alcohol.
• Example: Methanol adds to 2-methyl-1-butene in the presence of acid to form an ether.

🧪 Mechanism for alcohol addition

The mechanism parallels water addition:

Step 1: Electrophilic attack of H₃O⁺ to the alkene → carbocation intermediate formed.

Step 2: Alcohol molecule (acting as nucleophile) reacts with the carbocation.

Step 3: Deprotonation to get the neutral ether product.

⚠️ Important note: carbocation rearrangement

For any reaction involving a carbocation intermediate, rearrangement is always an option.

• If a more stable carbocation can form through hydride or alkyl shift, rearrangement may occur.
• The addition of water or alcohol to alkenes may involve carbocation rearrangement if possible.
• Don't assume the carbocation stays in its initial position—check whether rearrangement would produce a more stable intermediate.