McGraw-Hill Education 500 Review Questions for the MCAT_ Organic Chemistry and Biochemistry 2nd

🧭 Overview

🧠 One-sentence thesis

Valence bond theory explains molecular geometry through hybridization—the combination of atomic orbitals into new hybrid orbitals that determine bonding, structure, and properties of organic molecules.

📌 Key points (3–5)

• Hybridization creates new orbitals: atomic s, p, and d orbitals combine to form hybrid orbitals (sp, sp², sp³) that explain molecular geometry and bonding.
• Number of orbitals is conserved: hybridization never changes the total number of orbitals—four atomic orbitals always yield four hybrid orbitals.
• Hybrid orbitals blend parent character: each hybrid orbital contains features of its parent orbitals (e.g., sp hybrids have half s character and half p character).
• Common confusion—excited vs ground state: to achieve the electron arrangement needed for bonding, atoms may promote electrons from ground state to excited state before hybridization occurs.
• Hybridization determines structure and reactivity: the type of hybridization (sp, sp², sp³) affects bond angles, bond lengths, acidity, stability, and which orbitals interact during reactions.

🔬 Valence bond theory and hybridization mechanism

🔬 What valence bond theory does

Valence bond theory: uses hybridization to explain molecular geometry by redistributing electrons from s, p, and d atomic orbitals into new hybrid orbitals that can overlap and share electrons.

• The theory starts with atomic orbitals in their normal configuration.
• Electrons are rearranged to give a lower-energy (more favorable) arrangement.
• The new hybrid orbitals then overlap with orbitals from other atoms to form covalent bonds.

🧬 How hybridization works

• Definition:

  Hybridization: the combination of atomic orbitals into new hybrid orbitals.

• What changes: both the shape and the label of the orbitals change (e.g., s + p → sp).

• What does NOT change: the total number of orbitals remains constant.

• Example: if four atomic orbitals hybridize, exactly four hybrid orbitals form.

⚡ Electron promotion and excited states

• Sometimes the ground-state electron configuration does not match the bonding pattern shown in the Lewis structure.
• To achieve the correct arrangement, an electron may be promoted from a lower orbital (e.g., 2s) to a higher one (e.g., 2p).
• The resulting excited-state atom is less stable than the ground state, but the subsequent hybridization and bond formation release energy, making the overall process favorable.
• Example (BeI₂): beryllium's ground state is [He]2s². To form two separate bonds, one 2s electron is promoted to a 2p orbital, creating two unpaired electrons that can hybridize into two sp orbitals.

🧩 Types of hybrid orbitals and their properties

🧩 sp hybrids

• Formed from one s orbital and one p orbital.
• Each sp hybrid has half s character and half p character.
• Example: in BeI₂, the two sp hybrids allow beryllium to form two covalent bonds with iodine atoms.

🧩 sp² and sp³ hybrids

• sp²: formed from one s and two p orbitals → three hybrid orbitals (common in double bonds and aromatic systems).
• sp³: formed from one s and three p orbitals → four hybrid orbitals (common in single bonds, tetrahedral geometry).
• The excerpt emphasizes that hybrid orbitals "contain features of each parent orbital."

🔍 Unhybridized orbitals remain

• Not all atomic orbitals participate in hybridization.
• Orbitals that do not hybridize remain as normal p (or d) orbitals.
• These unhybridized orbitals can participate in π bonding or remain as lone pairs.

🔗 Bonding and molecular structure

🔗 σ and π bonds

• σ bonds: formed by direct overlap of hybrid orbitals (or hybrid with unhybridized orbitals).
• π bonds: formed by sideways overlap of unhybridized p orbitals.
• Example (from questions): a carbonyl group (C=O) has one σ bond and one π bond; the carbon is sp² hybridized.

📏 Bond length and character

• Bonds between sp² hybridized atoms are shorter than normal single bonds because sp² orbitals have more s character (s orbitals are closer to the nucleus).
• Example: in 1,3,5-hexatriene, the C(4)–C(5) bond is shorter than a typical alkane C–C bond because it is formed between two sp² carbons.
• More s character → shorter bond; more p character → longer bond.

🔄 Conjugation and resonance

• Conjugated systems (alternating single and double bonds) allow partial overlap of unhybridized p orbitals across nominally single bonds.
• This overlap gives the single bonds partial double-bond character, shortening them.
• Resonance stabilization (e.g., in amides, carboxylate ions) involves delocalization of lone pairs or π electrons, which affects stability and reactivity.

🧪 Functional groups and acidity

🧪 Identifying functional groups

The excerpt and questions reference many functional groups:

• Amide: contains C=O and C–N; stabilized by resonance involving the nitrogen lone pair.
• Nitrile: contains C≡N.
• Amine: contains N with lone pair.
• Ester: R–CO–O–R′.
• Acid anhydride: (RCO)₂O.
• Acetal: carbon with two –OR groups.
• Alcohol, ether, carboxylic acid, etc.

🔥 Acidity and stability of conjugate bases

• Stronger acids have more stable conjugate bases.
• Factors that stabilize the conjugate base (anion):
  • Resonance: delocalization of negative charge (e.g., carboxylate ion, phenoxide with electron-withdrawing groups).
  • Electron-withdrawing groups (e.g., –NO₂, –F): pull electron density away, stabilizing the anion.
  • Electronegativity: more electronegative atoms stabilize negative charge better.
• Factors that destabilize the conjugate base:
  • Electron-donating groups (e.g., –NH₂): push electron density toward the anion, destabilizing it.
• Example: ortho-nitrophenol is a stronger acid than phenol because the nitro group stabilizes the phenoxide anion by resonance.
• Example: para-aminophenol is a weaker acid than phenol because the amino group is electron-donating and destabilizes the conjugate base.

🚪 Leaving groups

• A good leaving group is a weak base (stable anion).
• Example: the triflate group (–OSO₂CF₃) is a better leaving group than hydroxyl (–OH) because the triflate anion is stabilized by resonance (the negative charge is delocalized over three oxygen atoms).

🏗️ Stereochemistry and isomerism

🏗️ Types of isomers

Isomers: compounds with the same molecular formula but different structural formulas.

• Structural (constitutional) isomers: different connectivity of atoms.
• Stereoisomers: same connectivity, different spatial arrangement.
  • Enantiomers: non-superimposable mirror images (require a chiral center).
  • Diastereomers: stereoisomers that are not mirror images.
• Conformers: different spatial arrangements due to rotation around single bonds (not true isomers, but interconvertible forms).
• Regioisomers: differ in the position of a substituent or functional group.

🪞 Chirality and optical activity

Chiral carbon (asymmetric carbon): a carbon atom with four different groups attached.

• Presence of one or more chiral carbons is sufficient to produce optical isomers.
• Optical isomers (enantiomers) are important in biology: one form may be active, the other inactive or even harmful.
• Example: many drugs exist in optically active forms with different biological effects.

🔄 Cis/trans and E/Z notation

• Cis/trans: used for simple disubstituted alkenes (cis = same side, trans = opposite sides).
• E/Z: used when Cahn-Ingold-Prelog priority rules are needed (E = higher-priority groups on opposite sides, Z = same side).
• Example: 3-hexene exists as cis and trans isomers; the cis isomer is less stable (higher heat of hydrogenation) due to steric crowding.

📐 Conformational analysis

• Conformers: different spatial arrangements from rotation around single bonds.
• Stability order (for substituted ethanes): anti > gauche > eclipsed.
• Example: for 1-chloropropane, the anti conformer is most stable, eclipsed is least stable.
• For cyclohexanes: equatorial substituents are more stable than axial (less steric crowding).

🧮 Degree of unsaturation and molecular formulas

🧮 What degree of unsaturation measures

• Degree of unsaturation (also called index of hydrogen deficiency): the number of rings and/or π bonds in a molecule.
• Formula (for CₙHₘ): degree of unsaturation = (2n + 2 − m) / 2.
• Each ring or π bond contributes 1 degree of unsaturation.
• For heteroatoms: nitrogen adds 1 to the hydrogen count; halogens replace hydrogen (subtract 1); oxygen does not change the count.

🧮 Using hydrogenation data

• Complete hydrogenation adds H₂ across each π bond.
• Example: if C₁₀H₁₄ absorbs 3 moles of H₂, it has 3 π bonds. The degree of unsaturation can be calculated; any additional unsaturation beyond the π bonds indicates rings.
• Example: C₁₀H₁₄ has degree of unsaturation = (20 + 2 − 14)/2 = 4. If 3 are π bonds, then 1 ring is present.

🧮 Functional group constraints

• Certain functional groups imply specific degrees of unsaturation.
• Example: an amide (R–CO–NH–R′) has one C=O π bond.
• Example: a nitrile (R–C≡N) has two π bonds.
• If a molecular formula does not allow the required unsaturation for a functional group, that group cannot be present.

📛 Nomenclature (IUPAC naming)

📛 General rules

• Number the longest carbon chain to give substituents the lowest numbers.
• Name substituents alphabetically (ignoring prefixes like di-, tri-).
• Indicate stereochemistry with cis/trans or E/Z.
• For alcohols: suffix -ol; for carboxylic acids: suffix -oic acid; for ketones: suffix -one; for alkynes: suffix -yne.

📛 Examples from the excerpt

• 5-hexyn-3-ol: a six-carbon chain with a triple bond starting at C-5 (or C-1, depending on numbering) and a hydroxyl at C-3.
• N-(4-hydroxyphenyl)acetamide: an amide where the nitrogen is attached to a para-hydroxyphenyl group and an acetyl group (Tylenol).
• 4-fluoropentyl propanoate: an ester formed from 4-fluoropentanol and propanoic acid.

📛 Common vs IUPAC names

• The excerpt mentions both (e.g., Tylenol, Demerol).
• IUPAC names are systematic; common names are historical or trivial.

⚗️ Oxidation states and redox reactions

⚗️ Oxidation state changes in sulfur

• Example: two cysteine molecules (each with –SH) undergo oxidation to form a disulfide linkage (–S–S–).
• Reduction reverses this: –S–S– + [H] → 2 –SH.
• Oxidation state change during reduction: sulfur goes from −1 (in disulfide) to −2 (in thiol, –SH).
• (Note: the excerpt's question lists options; the correct answer based on typical oxidation states is that each sulfur in a disulfide is −1, and in a thiol is −2, so the change is −1 to −2, but the question may have a typo or different convention.)

⚗️ Catalysis

• A catalyst speeds up a reaction without being consumed; it is regenerated.
• Example: base-catalyzed saponification of an ester uses a base that is regenerated, so it is a true catalyst (not consumed in a 1:1 ratio).
• The base alters the reaction pathway (stabilizes intermediates or transition states) but does not change the overall thermodynamics (heat of reaction).

🔢 Counting atoms and bonds

🔢 σ and π bond counting

• σ bonds: every single bond is one σ bond; every double bond has one σ bond; every triple bond has one σ bond.
• π bonds: each double bond has one π bond; each triple bond has two π bonds.
• Example: naphthalene (C₁₀H₈) has 19 σ bonds (10 C–C + 8 C–H + 1 for the shared bond in the fused ring system, counted carefully) and 5 π bonds (from the aromatic system).
• (Exact counts depend on careful structure analysis; the excerpt's questions test this skill.)

🔢 Hybridization counting

• Count how many atoms have each hybridization type.
• Example: in Demerol, count sp³ carbons (tetrahedral, four single bonds) vs sp² (trigonal planar, involved in double bonds or aromatic rings).

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Note: The excerpt is an introduction and a set of practice questions for MCAT organic chemistry. It covers foundational concepts (hybridization, bonding, isomerism, nomenclature, acidity, functional groups) without providing answer keys or detailed worked solutions. The questions test application of these principles to specific molecules.

🧭 Overview

🧠 One-sentence thesis

Isomers—compounds with identical molecular formulas but different structural arrangements—exhibit distinct physical properties and biological activities, especially when chiral centers create optical isomers that differ in their three-dimensional orientation.

📌 Key points (3–5)

• What isomers are: compounds sharing the same molecular formula but differing in structural formulas; many organic and biochemical compounds exist in multiple isomeric forms.
• Chirality and optical activity: a chiral (asymmetric) carbon atom with four different groups attached produces optical isomers; most biological molecules contain one or more chiral centers.
• Biological significance: different isomers often have different properties, especially in biological systems where one form may be active and the other inactive or even harmful.
• Common confusion: enantiomers vs. diastereomers vs. meso compounds—enantiomers are mirror images, diastereomers are non-mirror-image stereoisomers, and meso compounds have internal symmetry despite having chiral centers.
• Physical property relationships: enantiomers share identical boiling points and melting points, but diastereomers differ; hydrogen bonding, dipole moments, and molecular symmetry influence solubility and boiling points.

🔬 Chirality and optical isomers

🔬 What makes a molecule chiral

A chiral carbon atom has four different groups attached to it.

• Chirality means the molecule is not superimposable on its mirror image.
• The presence of even one asymmetric (chiral) carbon is sufficient to produce isomeric molecules.
• Example: if a carbon is bonded to H, OH, CH₃, and COOH, it is chiral because all four groups differ.

🔄 Optical activity and rotation

• Optical isomers rotate plane-polarized light.
• Levorotatory compounds rotate light counterclockwise (designated with a negative sign or (–)).
• Dextrorotatory compounds rotate light clockwise (designated with a positive sign or (+)).
• The (+) and (–) symbols describe observed rotation and are not directly equivalent to d and l symbols; each structure must be checked individually.
• Don't confuse: the sign of rotation ((+) or (–)) is independent of the absolute configuration (R or S); changing a substituent can reverse the rotation sign even if the configuration remains the same.

🧬 Biological importance

• The majority of biological molecules have one or more chiral carbon atoms, making them chiral.
• In living cells, one optical isomer may be biologically active while the other is inactive or produces entirely different biological effects.
• Example: thalidomide—one enantiomer acts as an antidepressant (treats morning sickness), while the other causes mutations in fetuses (birth defects).
• Many drugs exist in optically active forms; one form may have desired therapeutic results, the other may be harmful.

🧩 Types of isomers and their relationships

🧩 Enantiomers

• Enantiomers are mirror-image isomers that are not superimposable.
• They have identical physical properties (boiling point, melting point, solubility) except for the direction of optical rotation.
• If one enantiomer has a rotation of +15.6°, the other has –15.6°.
• Example: if compound A has an R configuration and positive rotation, its enantiomer has an S configuration and negative rotation.

🔀 Diastereomers

• Diastereomers are stereoisomers that are not mirror images.
• They have different physical properties: different melting points, boiling points, optical rotations, and chromatography retention times.
• When an S stereoisomer reacts with a racemic mixture (equal amounts of R and S), the products are diastereomers, not enantiomers, and can be easily separated.
• Don't confuse: diastereomers vs. enantiomers—diastereomers differ in multiple properties, while enantiomers differ only in optical rotation direction.

🪞 Meso compounds

• Meso compounds contain chiral centers but have an internal plane of symmetry, making them achiral overall.
• Despite having two or more chiral centers, meso compounds do not exhibit optical activity (rotation = 0°).
• They have a sharp melting point (unlike racemic mixtures).
• Example: a molecule with two chiral carbons in a symmetric arrangement (e.g., certain diols or dihalides) may be meso.

⚖️ Racemic mixtures

• A racemic mixture contains equal amounts (50:50) of two enantiomers.
• The optical rotation of a racemic mixture is 0° because the rotations cancel out.
• Laboratory synthesis of most chiral amino acids (except glycine, which is achiral) produces racemic mixtures.
• Don't confuse: a racemic mixture (two enantiomers) vs. a meso compound (one achiral molecule with internal symmetry)—both have zero rotation, but a racemic mixture can be separated into optically active components.

🔄 Other isomer types mentioned

• Conformers: different spatial arrangements due to rotation around single bonds (not permanent).
• Regioisomers: differ in the position of a functional group or substituent.
• Structural isomers: differ in the connectivity of atoms.

🧪 Absolute configuration and stereochemistry

🧪 R and S notation

• Absolute configuration describes the three-dimensional arrangement of groups around a chiral center.
• R and S are assigned using priority rules (Cahn-Ingold-Prelog).
• Example: a molecule may have (2R, 3S) configuration, meaning carbon 2 is R and carbon 3 is S.
• Don't confuse: R/S (absolute configuration) vs. (+)/(–) (observed rotation)—they are independent; an R compound can be (+) or (–) depending on the specific substituents.

🔄 Inversion and reactions

• An SN2 reaction at a chiral center inverts the configuration (Walden inversion).
• The optical rotation of the product cannot be predicted solely from the starting rotation; it depends on the new substituents.
• Example: if a chiral compound with +15.6° rotation undergoes SN2, the product's rotation is not predictable without knowing the new groups attached.

📐 Number of stereoisomers

• For a molecule with n chiral centers, the maximum number of stereoisomers is 2ⁿ.
• If meso forms are present, the actual number is less than 2ⁿ.
• Example: a molecule with 3 chiral centers can have up to 8 stereoisomers; if one is meso, the number is reduced.
• When two racemic mixtures (each with one chiral center) react, up to 4 stereoisomers can form (2 × 2).

🌡️ Physical properties and intermolecular forces

🌡️ Boiling points and hydrogen bonding

• Hydrogen bonding significantly raises boiling points.
• Compounds capable of hydrogen bonding (e.g., alcohols, amines, carboxylic acids) have higher boiling points than similar compounds without hydrogen bonding.
• Example: ethyl methyl amine (which has an N–H bond) has a higher melting point than trimethyl amine (which has no N–H bond) because ethyl methyl amine can form intermolecular hydrogen bonds.
• Carboxylic acids have higher boiling points than esters or ketones of similar molecular weight due to strong hydrogen bonding.

💧 Solubility in water

• Water solubility depends on the ability to form hydrogen bonds with water.
• Compounds with multiple –OH or –NH₂ groups are more soluble in water.
• Example: diols (two –OH groups) and diamines (two –NH₂ groups) are highly soluble in water; ionic compounds like Na₂SO₄ are also very soluble.
• Nonpolar compounds or those with large hydrocarbon portions are less soluble.

⚡ Dipole moments and polarity

• Dipole moment measures the separation of charge in a molecule.
• Symmetrical molecules (even with polar bonds) may have zero net dipole moment if the bond dipoles cancel.
• Example: a molecule with symmetrically arranged fluorine atoms may be nonpolar despite having polar C–F bonds.
• Asymmetrical molecules with polar bonds have nonzero dipole moments.
• Don't confuse: having polar bonds vs. being a polar molecule—symmetry can cancel out individual bond dipoles.

🔗 Resonance and stability

• Resonance stabilization affects physical properties and reactivity.
• Amides benefit greatly from resonance (the lone pair on nitrogen delocalizes into the carbonyl), making them very stable and less reactive.
• Conjugated systems (e.g., 1,3,5-hexatriene) have shorter C–C single bonds due to partial overlap of unhybridized p orbitals, giving partial double-bond character.
• Example: 1,3,5-cyclohexatriene (benzene) is highly stabilized by resonance, making it unreactive toward hydrogenation compared to 1,3,5-hexatriene.

🔄 Cis-trans isomerism and stability

🔄 Cis vs. trans isomers

• Cis and trans isomers differ in the spatial arrangement of groups around a double bond or ring.
• Trans isomers generally have less steric hindrance and are more stable.
• Cis isomers have groups on the same side, leading to greater steric crowding and higher energy.
• Example: trans-3-hexene is more stable than cis-3-hexene because the larger groups are farther apart.

🔥 Heat of hydrogenation

• Heat of hydrogenation measures the energy released when a double bond is hydrogenated.
• The less stable isomer releases more energy upon hydrogenation.
• Example: cis-3-hexene has a greater heat of hydrogenation than trans-3-hexene because the cis isomer is higher in energy (less stable).

🔒 Rigidity of double bonds

• Converting cis to trans (or vice versa) requires breaking the π bond, which is strong.
• Rotation around a double bond is not easy; it requires significant energy input.
• Don't confuse: π bonds (rigid, require breaking for isomerization) vs. σ bonds (allow free rotation, leading to conformers).

🧬 Special cases and applications

🧬 Glycine and achiral amino acids

• Glycine is achiral because it has two hydrogen atoms on the central carbon (no four different groups).
• Laboratory synthesis of glycine produces pure glycine.
• Laboratory synthesis of other natural amino acids (which are chiral) produces racemic mixtures (50% pure in terms of the desired enantiomer).

🧬 Nitrogen inversion

• Nitrogen compounds with three different groups and a lone pair can undergo inversion (the lone pair flips to the other side).
• This rapid inversion makes such nitrogen compounds achiral even though they appear to have four different groups.
• Example: a tertiary amine with three different substituents does not exhibit optical activity because of inversion.

🧬 Hybridization and geometry

• sp³ hybridized atoms (e.g., saturated carbons, amines) have tetrahedral geometry.
• sp² hybridized atoms (e.g., carbonyls, alkenes, aromatic nitrogens) have trigonal planar geometry.
• Example: in a nucleotide, one nitrogen may be sp³ (pyramidal, part of a saturated ring) and another sp² (planar, part of an aromatic ring or adjacent to a carbonyl).

🧬 Degree of unsaturation

• Degree of unsaturation (also called index of hydrogen deficiency) counts rings and π bonds.
• Formula: for CₙHₘNₓOᵧ, degree of unsaturation = (2n + 2 + x – m) / 2.
• Each ring or double bond contributes 1 degree of unsaturation; a triple bond contributes 2.
• Example: C₆H₁₀N₂O has degree of unsaturation = (2×6 + 2 + 2 – 10) / 2 = 3, indicating three rings/double bonds total.

📊 Summary table: Isomer relationships

Isomer type	Definition	Physical properties	Optical activity
Enantiomers	Mirror-image stereoisomers	Identical (except rotation direction)	Equal magnitude, opposite sign
Diastereomers	Non-mirror-image stereoisomers	Different (melting point, boiling point, etc.)	Different rotations
Meso compounds	Chiral centers + internal symmetry	Unique (not identical to enantiomers)	Zero (achiral overall)
Racemic mixture	50:50 mix of enantiomers	Mixture properties	Zero (rotations cancel)
Conformers	Rotation around single bonds	Same compound, different conformations	Same (if chiral)
Structural isomers	Different connectivity	Different	May or may not be chiral

🧭 Overview

🧠 One-sentence thesis

Substitution and elimination reactions—especially S_N2, S_N1, E1, and E2 mechanisms—are governed by factors such as nucleophile strength, substrate structure, solvent effects, and steric hindrance, which together determine reaction pathways, rates, and stereochemical outcomes.

📌 Key points (3–5)

• S_N2 vs S_N1 mechanisms differ fundamentally: S_N2 is one-step with second-order kinetics and inversion of configuration; S_N1 is two-step with first-order kinetics and racemization.
• Substrate structure controls mechanism preference: primary alkyl halides favor S_N2; tertiary favor S_N1; steric hindrance and carbocation stability are key.
• Solvent and nucleophile strength matter: polar protic solvents stabilize carbocations (favoring S_N1), while strong nucleophiles and polar aprotic solvents favor S_N2.
• Common confusion—S_N2 vs E2: both are second-order and one-step, but S_N2 involves nucleophilic attack at carbon with inversion, while E2 involves base-induced elimination forming a π bond.
• Free radical halogenation: a chain reaction (initiation, propagation, termination) that is nonregioselective and influenced by radical stability (3° > 2° > 1°).

🔬 S_N2 mechanism fundamentals

🔬 What defines S_N2

S_N2: a nucleophilic substitution reaction in which a nucleophile replaces a leaving group on an electrophile in a single concerted step.

• One-step mechanism: no intermediates; the nucleophile attacks the substrate carbon from the backside as the leaving group departs.
• Second-order kinetics: rate depends on both [nucleophile] and [substrate].
• Inversion of configuration: the stereochemistry at the reaction center flips (Walden inversion).
• Example: hydroxide ion attacking ethyl iodide displaces iodide and inverts the carbon center.

⚙️ Factors favoring S_N2

• Strong nucleophile: higher nucleophilicity increases attack rate.
• Primary > secondary > tertiary substrates: less steric hindrance at the reaction site speeds up backside attack.
• Polar aprotic solvents (e.g., DMSO, acetone): do not solvate the nucleophile strongly, leaving it more reactive.
• No carbocation intermediate: the reaction does not involve a carbocation, so carbocation stability is irrelevant.

🚫 What S_N2 is NOT

• Not a carbocation intermediate: there is a five-coordinate transition state, not a stable carbocation.
• Not retention of configuration: the product has inverted stereochemistry.
• Not first-order kinetics: doubling nucleophile concentration doubles the rate.
• Don't confuse: S_N2 has a transition state at the highest energy point on the reaction coordinate, not an intermediate.

🔄 S_N1 mechanism fundamentals

🔄 What defines S_N1

S_N1: a nucleophilic substitution reaction proceeding through a two-step mechanism with a carbocation intermediate.

• Two-step mechanism: (1) leaving group departs, forming a carbocation; (2) nucleophile attacks the carbocation.
• First-order kinetics: rate depends only on [substrate], not [nucleophile].
• Racemization: the planar carbocation can be attacked from either face, producing a racemic mixture if the carbon is chiral.
• Example: t-butyl iodide loses iodide to form a tertiary carbocation, then reacts with formate ion.

⚙️ Factors favoring S_N1

• Tertiary > secondary > primary substrates: more substituted carbocations are more stable (hyperconjugation and inductive effects).
• Polar protic solvents (e.g., water, alcohols): stabilize the carbocation intermediate and the leaving group.
• Weak nucleophile: since the rate-determining step is carbocation formation, nucleophile strength is less critical.
• Good leaving group: better leaving groups (e.g., I⁻, Br⁻) depart more easily.

🔍 Carbocation stability

• Order of stability: 3° > 2° > 1° > methyl.
• Why: more alkyl groups donate electron density through hyperconjugation and inductive effects, stabilizing the positive charge.
• Rearrangements possible: carbocations can rearrange (hydride or alkyl shifts) to more stable forms.
• Example: isopropyl bromide forms a secondary carbocation, which is more stable than a primary carbocation from propyl bromide.

🚫 What S_N1 is NOT

• Not dependent on nucleophile concentration: doubling [nucleophile] does not change the rate.
• Not inversion only: produces a racemic mixture (both enantiomers) at a chiral center.
• Not one-step: the carbocation intermediate is a distinct species.

🧪 E1 and E2 elimination mechanisms

🧪 E1 mechanism

E1: a two-step elimination reaction forming a π bond via a carbocation intermediate.

• Two-step: (1) leaving group departs, forming a carbocation; (2) base abstracts a proton, forming a double bond.
• First-order kinetics: rate depends only on [substrate].
• Competes with S_N1: both involve carbocation intermediates; product distribution depends on conditions.
• Example: dehydration of alcohols with phosphoric acid often proceeds via E1.

🧪 E2 mechanism

E2: a one-step elimination reaction in which a base abstracts a proton while the leaving group departs, forming a π bond.

• One-step, concerted: proton abstraction and leaving group departure occur simultaneously.
• Second-order kinetics: rate depends on [base] and [substrate].
• Anti-periplanar geometry preferred: the hydrogen and leaving group are typically on opposite sides of the molecule.
• No carbocation: unlike E1, there is no intermediate.
• Example: reaction of 3-chloro-3-ethylpentane with ethoxide can proceed via E2 if the base is strong.

🔀 E1 vs E2 vs S_N1 vs S_N2

Mechanism	Steps	Kinetics	Intermediate	Stereochemistry	Substrate preference
S<sub>N</sub>2	1	Second-order	Transition state	Inversion	1° > 2° > 3°
S<sub>N</sub>1	2	First-order	Carbocation	Racemization	3° > 2° > 1°
E2	1	Second-order	Transition state	Anti-periplanar	3° > 2° > 1°
E1	2	First-order	Carbocation	—	3° > 2° > 1°

• Common confusion: E2 and S_N2 are both one-step and second-order, but E2 forms a π bond (elimination) while S_N2 substitutes a group (substitution).

🔥 Free radical halogenation

🔥 Mechanism overview

Free radical halogenation: a chain reaction in which an alkane reacts with a halogen (Cl₂, Br₂) under light or heat to replace a hydrogen with a halogen.

• Three stages:
  1. Initiation: halogen molecule (e.g., Cl₂) absorbs light (hv) and homolytically cleaves into two radicals.
  2. Propagation: a halogen radical abstracts a hydrogen from the alkane, forming an alkyl radical; the alkyl radical then reacts with another halogen molecule, forming the product and regenerating a halogen radical.
  3. Termination: two radicals combine to form a stable molecule, ending the chain.
• Example: chlorination of methylpropane produces 1-chloro-2-methylpropane and 2-chloro-2-methylpropane.

⚛️ Radical stability

• Order of stability: 3° > 2° > 1° > methyl.
• Why: more substituted radicals are stabilized by hyperconjugation and inductive effects.
• Regioselectivity: the reaction preferentially forms the more stable radical intermediate, leading to the major product.
• Example: monobromination of isopentane favors formation of the tertiary radical, giving the tertiary bromide as the major product.

🔄 Propagation steps

• Definition: a free radical reacts to produce another free radical, continuing the chain.
• Example: Cl· + CH₃CH₂CH₃ → HCl + ·CH₂CH₂CH₃; ·CH₂CH₂CH₃ + Cl₂ → CH₂ClCH₂CH₃ + Cl·.
• Not initiation or termination: initiation produces radicals from a molecule; termination consumes radicals without regenerating them.

🎯 Selectivity and statistics

• Nonregioselective: the reaction can occur at multiple sites, producing a mixture of isomers.
• Influenced by:
  • Inductive effects: electron-donating groups stabilize radicals.
  • Statistical factors: more hydrogens at a given position increase the probability of abstraction there.
• Example: monochlorination of 2,4-dimethylpentane produces multiple products because there are several types of hydrogens.

🚫 What free radical halogenation is NOT

• Not stereoselective: does not preferentially produce one enantiomer.
• Not an S_N1 or S_N2 reaction: involves radicals, not carbocations or nucleophiles.
• Don't confuse: the initiation step is not a propagation step; initiation requires energy input (light) and does not regenerate a radical.

🧬 Stereochemistry and configuration

🧬 Inversion vs retention vs racemization

• S_N2 → inversion: backside attack flips the stereochemistry at the chiral center.
• S_N1 → racemization: planar carbocation is attacked from both faces, producing a 50:50 mixture of enantiomers (racemic mixture).
• E2 → anti-periplanar: the hydrogen and leaving group are on opposite sides, influencing the geometry of the product alkene.
• Example: reaction of (R)-3-bromo-2,3-dimethylpentane with NaI in acetone (S_N2) gives (S)-3-iodo-2,3-dimethylpentane.

🔬 Achiral substrate + chiral reagent

• If an achiral nucleophile attacks a chiral center in an S_N1 reaction, the product is a racemic mixture.
• If an achiral nucleophile attacks a chiral center in an S_N2 reaction, the product has inverted configuration.
• Don't confuse: the product of S_N1 is not achiral; it is a mixture of two enantiomers.

🧲 Nucleophilicity and leaving groups

🧲 What makes a good nucleophile

• Strong nucleophiles: species with high electron density and low solvation.
• Basicity vs nucleophilicity: in the same solvent, stronger bases are often better nucleophiles (e.g., NH₂⁻ > OH⁻).
• Polarizability: larger atoms (e.g., AsH₂⁻ > PH₂⁻) are more polarizable and better nucleophiles in polar aprotic solvents.
• Example: hydroxide ion is a strong nucleophile in S_N2 reactions.

🚪 What makes a good leaving group

• Stability after departure: the best leaving groups are weak bases that stabilize the negative charge well.
• Order: I⁻ > Br⁻ > Cl⁻ > F⁻; also, NH₃ > NH₂⁻ (neutral is better than anion).
• Diazonium groups: C–N₂⁺ is an excellent leaving group because N₂ gas is very stable.
• Example: in the reaction of an alkyl bromide, Br⁻ is a better leaving group than OH⁻.

🔄 Solvent effects

• Polar protic solvents (water, alcohols): solvate nucleophiles strongly, reducing their reactivity; stabilize carbocations, favoring S_N1.
• Polar aprotic solvents (DMSO, acetone, DMF): do not solvate nucleophiles strongly, leaving them more reactive; favor S_N2.
• Example: the reaction of t-butyl iodide with formate ion is faster in water (which stabilizes the carbocation) than in DMF.

🧮 Kinetics and rate laws

🧮 Second-order kinetics (S_N2, E2)

• Rate law: rate = k[substrate][nucleophile or base].
• Doubling [nucleophile]: doubles the rate.
• Doubling [substrate]: doubles the rate.
• Example: doubling the concentration of hydroxide ion in the reaction with ethyl iodide doubles the rate.

🧮 First-order kinetics (S_N1, E1)

• Rate law: rate = k[substrate].
• Doubling [nucleophile]: no change in rate.
• Doubling [substrate]: doubles the rate.
• Example: in the reaction of 3-chloro-3-ethylpentane via S_N1, doubling the ethoxide concentration does not change the rate.

🚫 Common confusion

• S_N2 is NOT first-order: the rate depends on both substrate and nucleophile.
• S_N1 is NOT second-order: the rate depends only on substrate concentration.
• Don't confuse: the rate-determining step in S_N1 is carbocation formation, not nucleophile attack.

🏗️ Steric and electronic effects

🏗️ Steric hindrance

• Definition: bulky groups around the reaction site slow down S_N2 reactions by blocking backside attack.
• Reactivity order for S_N2: methyl > 1° > 2° > 3°.
• Branching: increased branching on the alkyl halide decreases S_N2 rate.
• Example: ethyl chloride reacts faster than butyl chloride via S_N2 because the ethyl group causes less steric interference.

⚡ Inductive and resonance effects

• Inductive effects: electron-withdrawing groups (e.g., –CCl₃, –CN) stabilize negative charges and increase acidity; electron-donating groups destabilize negative charges.
• Resonance effects: delocalization of charge stabilizes intermediates and products.
• Example: trichloroacetate ion is a weaker base than acetate ion because the three chlorines withdraw electron density inductively, stabilizing the negative charge.

🔄 Carbocation rearrangements

• Why they occur: carbocations can rearrange via hydride or alkyl shifts to form more stable carbocations.
• Example: a secondary carbocation may rearrange to a tertiary carbocation if a neighboring carbon has a hydrogen or alkyl group that can shift.
• Don't confuse: rearrangements occur in S_N1 and E1 (which have carbocation intermediates), not in S_N2 or E2.

🧪 Specific reaction types

🧪 Alcohol reactions

• With HBr: tertiary and secondary alcohols react via S_N1; primary alcohols react via S_N2.
• Dehydration (E1): heating an alcohol with H₂SO₄ or H₃PO₄ removes water, forming an alkene; tertiary alcohols dehydrate fastest.
• With SOCl₂: converts alcohols to alkyl chlorides.
• Oxidation with K₂Cr₂O₇/H₂SO₄: primary alcohols → aldehydes → carboxylic acids; secondary alcohols → ketones; tertiary alcohols do not oxidize.
• Example: 2-methyl-1-pentanol dehydrates to form the most substituted alkene (Zaitsev's rule).

🧪 Ether synthesis (Williamson)

• Mechanism: an alkoxide ion (nucleophile) attacks a primary alkyl halide via S_N2.
• Role of alkoxide: it is the nucleophile, not the electrophile or catalyst.
• Example: sodium ethoxide reacts with methyl bromide to form ethyl methyl ether.

🧪 Halogenation selectivity

• Monochlorination: less selective; produces a mixture of isomers.
• Monobromination: more selective; favors the most stable radical (tertiary).
• Selectivity ratio: the ratio of products reflects both radical stability and the number of equivalent hydrogens.
• Example: monochlorination of methylpropane gives 64% 1-chloro-2-methylpropane and 36% 2-chloro-2-methylpropane; the selectivity ratio of 3° to 1° is approximately 5:1.

🧪 Conformational effects

• Cyclohexane conformations: bulky groups prefer equatorial positions to minimize steric repulsion.
• 1,2-Ethanediol: the conformation with both OH groups gauche is stabilized by intramolecular hydrogen bonding.
• Example: in 1-ethyl-3-t-butylcyclohexane, the t-butyl group is equatorial in the more stable conformation because it is larger and experiences more repulsion when axial.

🔬 Advanced topics

🔬 Hybridization in intermediates

• Carbocation: the positively charged carbon is sp² hybridized (planar, 120° bond angles).
• Alkyl radical: the radical carbon is approximately sp² hybridized (planar or nearly planar).
• Transition state in S_N2: the central carbon is approximately sp³ → sp² → sp³ during the reaction, with a five-coordinate transition state.

🔬 Optical activity and racemization

• Racemic mixture: an equimolar mixture of two enantiomers; optical rotation = 0°.
• Meso compound: a molecule with chiral centers but an internal plane of symmetry; optically inactive.
• S_N1 on a chiral center: produces a racemic mixture (optical rotation = 0°).
• Example: an unknown substance with optical rotation = 0° and sharp melting point could be a meso compound or a racemic mixture.

🔬 Activation energy and rate constants

• Activation energy (Ea): the energy barrier between reactants and the transition state.
• Rate constant (k): depends on Ea and temperature; lower Ea → higher k → faster reaction.
• Solvent effects on Ea: solvents that stabilize the transition state lower Ea and increase the rate.
• Example: increasing the size of R groups in RR'R"C-X increases steric hindrance, raising Ea and decreasing the S_N1 rate.

🔬 Regioselectivity in elimination

• Zaitsev's rule: elimination typically forms the more substituted (more stable) alkene.
• Example: dehydration of 2-methyl-1-pentanol gives primarily the more substituted alkene, not the terminal alkene.

🧭 Overview

🧠 One-sentence thesis

Electrophilic addition reactions allow unsaturated compounds (containing double or triple bonds) to react with electrophiles across those bonds to form saturated products, with regioselectivity often predicted by Markovnikov's rule.

📌 Key points (3–5)

• What addition reactions do: unsaturated compounds (double/triple bonds) react with groups that add across the bond to give saturated compounds.
• Symmetry matters: symmetrical compounds like ethylene yield a single product, but unsymmetrical compounds can form multiple products.
• Markovnikov's rule: predicts which product will predominate in unsymmetrical additions, summarized as "Them that has, gets."
• Common confusion: Markovnikov vs anti-Markovnikov—peroxides reverse the regioselectivity, leading to anti-Markovnikov products.
• Mechanism involves intermediates: most additions proceed through carbocation (carbonium ion) intermediates, with stability determining product distribution.

🔬 Core mechanism and intermediates

⚗️ What happens in addition reactions

Addition reaction: an unsaturated compound (containing double or triple bonds) reacts with another group that adds across that double or triple bond to give a saturated compound.

• The double or triple bond is broken, and new single bonds form to the added groups.
• Example: ethylene (symmetrical) + HBr → single product because both carbons are equivalent.
• Unsymmetrical alkenes can form more than one product depending on which carbon receives which part of the reagent.

🧪 Carbocation intermediates

• The passage describes intermediates as carbonium ions (carbocations) or carbanions.
• Questions distinguish between tertiary, secondary, and primary carbocations.
• Stability order: tertiary > secondary > primary.
• The more stable carbocation forms preferentially, determining the major product.
• Example: Markovnikov addition of water to an unsymmetrical alkene forms a tertiary carbonium ion intermediate (Question 180).

🔄 Peroxide effect (anti-Markovnikov)

• In the presence of peroxides, HBr addition follows a free radical mechanism instead of carbocation.
• This leads to the anti-Markovnikov product (Question 181).
• Don't confuse: normal HBr addition → Markovnikov; HBr + peroxide → anti-Markovnikov.

🎯 Markovnikov's rule and regioselectivity

📏 "Them that has, gets"

• Markovnikov's rule predicts which regioisomer will predominate when adding HX (hydrogen halides) or water to unsymmetrical alkenes.
• The hydrogen attaches to the carbon with more hydrogens already, and the other group (Br, OH, etc.) attaches to the more substituted carbon.
• This is because the more substituted carbocation intermediate is more stable.

🧩 Syn vs anti addition

• HCl addition to alkenes: both syn and anti addition occur, giving Markovnikov products (Question 182).
• Bromine in CCl₄ addition to cyclohexene: produces racemic 1,2-dibromocyclohexane (anti addition, Question 192).
• Hydroboration-oxidation (B₂H₆/THF then H₂O₂/NaOH): gives anti-Markovnikov alcohol with syn addition (Questions 195, 220).

Reagent	Addition type	Regioselectivity	Stereochemistry
HBr (no peroxide)	Electrophilic	Markovnikov	Syn + anti
HBr + peroxide	Free radical	Anti-Markovnikov	—
Br₂/CCl₄	Electrophilic	—	Anti (racemic)
B₂H₆ then H₂O₂/OH⁻	Hydroboration	Anti-Markovnikov	Syn

🧬 Specific addition reactions

💧 Hydration (adding water)

• Acid-catalyzed addition of water to alkenes yields the Markovnikov alcohol (Question 213).
• Mechanism: protonation of the double bond forms a carbocation, then water attacks.
• The acid is a catalyst because it is regenerated (Question 213).
• Example: hydration of 1-methylcyclohexene forms the tertiary alcohol.

🧊 Halogenation

• Br₂ or Cl₂ add across double bonds.
• Bromine in CCl₄ with cyclohexene → racemic 1,2-dibromocyclohexane (Question 192).
• The mechanism involves a bromonium ion intermediate, leading to anti addition.

🌀 Epoxidation and dihydroxylation

• mCPBA (meta-chloroperoxybenzoic acid) forms an epoxide (Questions 193, 207).
• OsO₄ followed by NaHSO₃ gives syn diols (Question 194).
• Basic permanganate with cyclohexene also gives syn diols (Question 208).
• Don't confuse: epoxide formation vs direct diol formation—epoxides can be opened with base to give trans diols.

⚡ Ozonolysis

• Ozone (O₃) followed by Zn/HOAc or Zn/H₂O cleaves double bonds to form carbonyl compounds (aldehydes or ketones) (Questions 196, 223, 232).
• Example: ozonolysis of a diene can yield multiple different carbonyl products.

🧲 Hydrogenation

• H₂ with Pt or Ni catalyst: complete reduction of alkenes to alkanes (Question 206).
• Lindlar catalyst: selective reduction of alkynes to cis alkenes (Question 197).
• Na/NH₃ (liquid): reduction of alkynes to trans alkenes (Question 236).

Reagent	Product	Stereochemistry
H₂/Pt or Ni	Alkane	—
H₂/Lindlar	cis-Alkene	Syn addition
Na/NH₃(l)	trans-Alkene	Anti addition

🔥 Electrophilic aromatic substitution

🎭 Aromatic reactivity

• Aromatic rings undergo electrophilic substitution rather than addition, preserving aromaticity.
• The aromatic ring acts as a nucleophile (Question 228).
• Aromatic systems are less reactive toward electrophilic substitution because of the stability of the aromatic system (Question 212).

🧭 Directing effects

• Substituents on the ring direct incoming groups to ortho, meta, or para positions.
• Electron-donating groups (e.g., -OH, -OCH₃, -CH₃) are ring-activating and direct ortho/para (Questions 189, 224).
• Electron-withdrawing groups (e.g., -NO₂, -COOH) are ring-deactivating and direct meta (Questions 203, 204).
• Example: nitration of benzoic acid gives meta-nitrobenzoic acid (Question 203).
• Example: nitrobenzene + Br₂/FeBr₃ → meta-brominated product (Question 204).

🔥 Common aromatic reactions

• Nitration: HNO₃/H₂SO₄ adds -NO₂ groups (Questions 198, 203, 216, 241).
• Sulfonation: SO₃/H₂SO₄ adds -SO₃H groups (Question 184).
• Halogenation: Cl₂/FeCl₃ or Br₂/FeBr₃ adds halogens (Questions 189, 190, 204, 224).
• Friedel-Crafts acylation: acyl chloride + AlCl₃ adds acyl groups (Questions 199, 237).
• Further substitution requires higher temperature if the first group is ring-deactivating (Question 216).

🧪 Stereochemistry and product mixtures

🪞 Enantiomers and racemic mixtures

• Addition of HBr to 1-butene forms 2-bromobutane as a racemic mixture (equal amounts of two enantiomers) (Questions 209, 210, 231).
• The carbocation intermediate is planar and can be attacked from either face.
• Specific rotation of a racemic mixture is 0° (Question 210).

🔀 Regioisomers

• Hydration of unsymmetrical alkenes can lead to two different compounds (regioisomers) depending on which carbon gets the -OH (Question 240).
• Example: 2-nitrotoluene and 4-nitrotoluene are constitutional isomers (Questions 184, 229).

➕ Diels-Alder reactions

• A diene + dienophile → cyclic product.
• Net change: –2 π bonds and +2 σ bonds (Question 185, corrected based on typical Diels-Alder).
• The most reactive dienophiles have electron-withdrawing groups (Question 218).
• Example: cyclopentadiene + a dienophile forms a bicyclic product (Question 238).

🧠 Mechanistic details and rate laws

⏱️ Rate-determining steps

• For HX addition to alkenes, the rate law is rate = k[HX][alkene] (Question 227).
• This indicates:
  • First-order in alkene.
  • First-order in HX.
  • Second-order overall.
  • The rate-determining step involves one molecule of each.
• However, the reaction is not a single step; it is a two-step process (Questions 211, 227).
• Don't confuse: the rate law tells us what is involved in the slow step, not that the entire reaction is one step.

🧲 Nucleophilic aromatic substitution

• Electron-withdrawing groups deactivate electrophilic substitution but promote nucleophilic aromatic substitution by stabilizing the negative intermediate (Question 225).
• Example: NaNH₂/NH₃ with chloro-anisole (Question 219).

🧪 Grignard reagents

• Grignard reagents (e.g., CH₃CH₂CH₂MgBr) react with carbonyl groups (Question 226).
• Reaction with formaldehyde (H₂C=O) → secondary alcohol.
• Reaction with other aldehydes → secondary or tertiary alcohols depending on the aldehyde.

🔍 Aromaticity

💍 Aromatic criteria

• Compounds must be planar and follow Hückel's rule (4n+2 π electrons) to be aromatic (Questions 200, 201).
• Examples of aromatic species: benzene, pyridine, cyclopentadienyl anion.
• Non-aromatic: cyclopentadienyl cation (does not meet electron count).

🧩 Aromatic vs non-aromatic

• Question 200: one of the planar molecules is not aromatic (likely violates electron count).
• Question 201: one species is not aromatic (e.g., a cation or anion with wrong electron count).

🧭 Overview

🧠 One-sentence thesis

Nucleophilic and cyclo addition reactions involve nucleophiles attacking electrophilic centers (especially carbonyl groups), leading to products such as aldols, acetals, esters, and polymers through mechanisms that include addition, elimination, and condensation steps.

📌 Key points (3–5)

• Aldol reactions: Enolate ions (nucleophiles) form from carbonyl compounds with α-hydrogens in base, then attack other carbonyl groups to produce β-hydroxy aldehydes/ketones; heating causes dehydration (aldol condensation) to form α,β-unsaturated products.
• Acyl derivatives reactivity: Acyl halides undergo addition-elimination reactions (halide is a good leaving group), while aldehydes/ketones undergo nucleophilic addition (no good leaving group).
• Acetal and hemiacetal formation: Aldehydes/ketones react with alcohols under acid catalysis; hemiacetals have one OR and one OH on the same carbon, acetals have two OR groups.
• Common confusion: Addition vs. addition-elimination—aldehydes/ketones stop at addition (no leaving group), but acyl derivatives (acid chlorides, esters, anhydrides) proceed through elimination because they have good leaving groups (Cl⁻, OR⁻, OCOR⁻).
• Grignard and reducing agents: Grignard reagents (RMgBr) act as nucleophiles adding to carbonyls; LiAlH₄ and NaBH₄ reduce carbonyl groups to alcohols; Zn(Hg) acts as a reducing agent.

🧪 Aldol reactions and condensations

🧪 Aldol reaction mechanism

Aldol reaction: The reaction of a carbonyl (aldehyde or ketone) with a base produces an enolate ion (a nucleophile), which attacks the carbonyl group of another aldehyde or ketone molecule to form a β-hydroxyaldehyde or ketone (an aldol).

• Step 1: Base removes an α-hydrogen (the hydrogen on the carbon adjacent to the carbonyl) to form an enolate ion.
• Step 2: The enolate ion (nucleophile) attacks the carbonyl carbon (electrophile) of another molecule.
• Result: A β-hydroxy carbonyl compound (aldol) forms—it contains both an aldehyde (or ketone) and an alcohol group in the β-position.
• Example: The passage states "the aldol formed in this reaction is 2-methyl-3-hydroxypentanal."
• Role of base: NaOH or other dilute base performs deprotonation (removal of the α-hydrogen).

🔥 Aldol condensation

• What happens: The aldol product, especially if heated, undergoes dehydration (loss of H₂O).
• Product: An α,β-unsaturated aldehyde (enal) or ketone.
• Driving force: Extended conjugation favors formation of the unsaturated product.
• Reversibility: Both aldol reaction and condensation are reversible.
• Don't confuse: Aldol reaction gives the β-hydroxy product; aldol condensation includes the subsequent dehydration step.

🌡️ Reaction conditions

• Temperature: Aldol addition typically occurs at low temperature (0–10°C) with dilute base (e.g., 10% NaOH/H₂O).
• Heating: Promotes dehydration and condensation.
• Acid vs. base: The passage focuses on basic conditions; acid-catalyzed aldol reactions also exist but are not detailed here.

🎯 Nucleophilic addition and addition-elimination

🎯 Nucleophilic addition to carbonyls

• Aldehydes and ketones: Undergo nucleophilic addition because they lack a good leaving group.
• Mechanism: Nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate; protonation gives the final product.
• Example: Reaction of an aldehyde with ammonia or an amine forms an imine; reaction with an alcohol forms a hemiacetal.

⚙️ Addition-elimination reactions

Why acyl halides react differently: Acyl halides react by addition-elimination reactions while aldehydes and ketones undergo nucleophilic addition reactions because the halide ions are good leaving groups.

• Acyl derivatives (acid chlorides, anhydrides, esters, amides) have a leaving group attached to the carbonyl.
• Mechanism: Nucleophile adds to the carbonyl carbon (addition step), then the leaving group departs (elimination step).
• Example: Benzoyl chloride reacts with water to produce benzoic acid—water is the nucleophile, benzoyl chloride is the electrophile, and the chloride ion is the leaving group.
• Reactivity order: Acid chlorides > acid anhydrides > esters > amides (better leaving group = more reactive).

🔄 Hemiacetal and acetal formation

Type	Structure	Formation conditions
Hemiacetal	One OR and one OH on the same carbon	Aldehyde/ketone + 1 equivalent alcohol + acid catalyst
Acetal	Two OR groups on the same carbon	Aldehyde/ketone + excess alcohol + acid catalyst

• Hemiacetal characteristics: Formation follows second-order kinetics; the transition state is not 5-coordinate (this is incorrect); it does not have a carbocation intermediate under typical conditions; it is not stereospecific.
• Example: Mixing a ketone with an alcohol in the presence of acid produces an acetal.

🧲 Grignard reagents and organometallic reactions

🧲 Grignard reagent behavior

• What they are: Compounds of the form RMgX (e.g., ethylmagnesium bromide, EtMgBr).
• Role: Strong nucleophiles that attack carbonyl carbons.
• Typical reaction: RMgBr + carbonyl → addition product; after acid workup (H⁺/H₂O), an alcohol forms.
• Example: Excess ethylmagnesium bromide reacting with ethyl butanoate followed by acid workup produces a tertiary alcohol.

🔧 Reducing agents

• LiAlH₄ (lithium aluminum hydride): Strong reducing agent; reduces carbonyl groups (aldehydes, ketones, esters, carboxylic acids) to alcohols.
• NaBH₄ (sodium borohydride): Milder reducing agent; reduces aldehydes and ketones to alcohols but does not reduce esters or carboxylic acids.
• Zn(Hg) (zinc amalgam): Acts as a reducing agent (e.g., in the Clemmensen reduction, converting nitro groups to amines in the presence of HCl).

🧪 Barbier reaction

Barbier reaction: The reaction of an alkyl halide with an aldehyde or ketone in the presence of Mg, Li, or Zn to form an alcohol.

• Type: Nucleophilic substitution (or addition, depending on perspective).
• Generic form: R–X + O=CR₂ → R–C(OH)R₂.

🧬 Ester and amide chemistry

🧬 Ester formation and hydrolysis

• Esterification: Carboxylic acid + alcohol (with acid catalyst) → ester + water.
• Example: Butyl alcohol + acetic anhydride → butyl acetate (more efficient than using acetic acid).
• Hydrolysis: Ester + water (acid or base) → carboxylic acid (or carboxylate) + alcohol.
• Example: Acid hydrolysis of an ester produces a carboxylic acid; base hydrolysis (saponification) produces a carboxylate salt.

🧼 Saponification

Saponification: The base-catalyzed hydrolysis of an ester (or fat) to produce a carboxylate ion and an alcohol.

• Traditional soap production: Heating fats with aqueous base produces the carboxylate ion of a fatty acid (soap).
• Products: R–COO⁻ (carboxylate) + R′OH (alcohol).

🧪 Amide reactions

• Acid-catalyzed hydrolysis of an amide: Produces an ammonium ion and a carboxylic acid.
• Example: Amide + H⁺/H₂O → R–COOH + NH₄⁺.
• Formation: Carboxylic acid derivatives (acid chlorides, anhydrides) react with ammonia or amines to form amides.
• Example: Benzoyl chloride + ethylamine → N-ethylbenzamide.

🔬 Acidity, tautomerism, and other key concepts

🔬 Acidity of carbonyl compounds

• α-Hydrogens are acidic: The hydrogen on the carbon adjacent to a carbonyl is more acidic than typical C–H bonds because the resulting enolate ion is stabilized by resonance.
• Effect of electron-withdrawing groups: Substituents like nitro (NO₂) or halogens increase acidity by stabilizing the conjugate base.
• Example: Nitrophenols are more acidic than phenol; ortho and para nitro groups have the strongest effect.

🔄 Tautomerism

Tautomerism: The process in which an aldehyde or ketone is in equilibrium with its enol form.

• Keto-enol tautomerism: The carbonyl form (keto) and the enol form (with C=C and OH) interconvert.
• Don't confuse with: Mutarotation (change in optical rotation due to ring-opening/closing in sugars), inversion of configuration, or hydrolysis.

🧪 Decarboxylation

• What it is: Loss of CO₂ from a carboxylic acid.
• Easiest to decarboxylate: β-keto acids (carboxylic acids with a carbonyl group in the β-position) because the transition state is stabilized by the adjacent carbonyl.

🧬 Peptide bond stabilization

• Resonance: Peptide bonds (amide linkages in proteins) are stabilized by resonance between the carbonyl oxygen and the nitrogen lone pair.
• Most effective resonance structure: The structure with a double bond between C and N and a negative charge on oxygen (C=N⁺–O⁻) is important, but the neutral structure and the structure with C–N and C=O also contribute.

🔗 Diels-Alder and cycloaddition reactions

🔗 Diels-Alder reaction

• What it is: A [4+2] cycloaddition reaction between a conjugated diene and a dienophile (typically an alkene with electron-withdrawing groups) to form a six-membered ring.
• Bond changes: During the reaction, two new σ bonds form and two π bonds are consumed.
• Overall change: +2 σ bonds and –2 π bonds.
• Example: The passage shows a generic Diels-Alder reaction forming a cyclohexene derivative.

🧪 Polymer formation

• Condensation polymerization: Compounds with two reactive groups (e.g., diacid chlorides and diamines) react to form polymers.
• Example: A compound with two acyl chloride groups reacts most easily with a diamine (H₂N(CH₂)₄NH₂) to form a polyamide.

🧪 Oxidation and reduction reactions

🧪 Oxidation of alcohols

• Primary alcohols: Oxidize to aldehydes (mild conditions) or carboxylic acids (strong conditions, e.g., acidic potassium dichromate, CrO₃).
• Secondary alcohols: Oxidize to ketones.
• Tertiary alcohols: Do not oxidize under typical conditions (no H on the carbon bearing the OH).
• Example: Oxidation of a primary alcohol with acidic potassium dichromate produces a carboxylic acid.

🔧 Reduction reactions

• LiAlH₄: Reduces carbonyl groups to alcohols; also reduces esters to two alcohols.
• NaBH₄: Reduces aldehydes and ketones to alcohols.
• Example: Reduction of 3-methylbutanoic acid with LiAlH₄ produces 3-methylbutanol.

🧪 Ozonolysis

• What it is: Cleavage of a C=C double bond by ozone (O₃) followed by workup to produce carbonyl compounds (aldehydes or ketones).
• Example: The passage shows a product of ozonolysis with two carbonyl groups; the starting material must have had a C=C bond at that position.

🧪 Miscellaneous reactions and concepts

🧪 Ether cleavage

• Reaction: Ethers react with hot concentrated hydrobromic acid (HBr) to cleave the C–O bond.
• Products: An alkyl bromide and an alcohol (or phenol if the ether is aromatic).
• Example: Isopropyl phenyl ether + HBr → isopropyl bromide + phenol.

🧪 Reaction with strong bases

• Compounds that react: Carbonyl compounds with α-hydrogens (aldehydes, ketones, esters) react with very strong bases (e.g., potassium t-butoxide) to form enolates.
• Compounds that do NOT react: Carbonyl compounds without α-hydrogens (e.g., benzaldehyde, formaldehyde) or compounds without acidic protons.

🧪 Infrared spectroscopy clues

• Carbonyl absorption: Strong IR absorption near 1700 cm⁻¹ indicates a C=O group (aldehyde, ketone, ester, acid).
• Alcohol absorption: Strong IR absorption near 3350 cm⁻¹ indicates an O–H group.
• Example: A compound with IR absorption at 1700 cm⁻¹ that reacts with a Grignard reagent to form an alcohol will show a new strong absorption at 3350 cm⁻¹.

🧭 Overview

🧠 One-sentence thesis

Spectroscopic methods (IR, UV-Vis, NMR) and separation techniques (extraction, chromatography, distillation) together enable chemists to identify molecular structures and isolate compounds from mixtures by exploiting differences in energy absorption and physical properties.

📌 Key points (3–5)

• Three main spectroscopic methods: IR detects bond vibrations (functional groups), UV-Vis detects electronic transitions (conjugation/multiple bonds), and NMR detects nuclear spin flips (chemical environment of H and C atoms).
• How NMR works: External magnetic field forces nuclei to align; energy absorption causes spin flipping; electron shielding affects the energy required (chemical shift measured in ppm).
• Separation by polarity: Extraction uses acid/base to move compounds between aqueous and organic layers; TLC separates by polarity (polar compounds stick to silica, nonpolar travel farther).
• Common confusion: IR absorption regions—1700 cm⁻¹ indicates C=O (carbonyl), 3400 cm⁻¹ indicates O–H (alcohol), 2100 cm⁻¹ indicates C≡C (alkyne); don't mix these up.
• Why it matters: These techniques are essential for monitoring reactions, confirming product identity, and purifying mixtures in organic chemistry labs.

🔬 Spectroscopic methods overview

🌈 Infrared (IR) spectroscopy

IR spectroscopy: absorption of infrared energy causes covalent bonds to vibrate (stretch and bend), revealing the types of bonds present.

• Energy requirement: Infrared light has enough energy to make bonds vibrate.
• When absorption occurs: Only if the vibration changes the molecule's dipole moment.
• Spectrum range: Most organic compounds absorb in the 4000–600 cm⁻¹ region.
• Fingerprint region: Below 1500 cm⁻¹; unique pattern for each molecule.
• Example: A compound with a strong absorption near 1700 cm⁻¹ likely contains a carbonyl group (C=O); one near 3400 cm⁻¹ likely has an O–H group (alcohol).

🌟 Ultraviolet-visible (UV-Vis) spectroscopy

UV-Vis spectroscopy: absorption of UV or visible light excites an electron from the ground state to an excited state, providing information about molecular orbital arrangement.

• Electronic transition: Usually from HOMO (highest occupied molecular orbital) to LUMO (lowest unoccupied molecular orbital).
• Best for: Compounds with multiple bonds (conjugated systems).
• What it reveals: Information about the electronic structure and conjugation in the molecule.
• Don't confuse: UV-Vis affects electronic structure, whereas IR affects vibrational structure and NMR affects nuclear spin states.

🧲 Nuclear magnetic resonance (NMR) spectroscopy

NMR spectroscopy: nuclei in an external magnetic field absorb radio-frequency energy to flip their spin alignment, revealing the chemical environment of specific nuclei (¹H and ¹³C).

• How it works:
  • Sample is placed in a large external magnetic field.
  • Nuclei align either with or against the field.
  • Absorbing energy causes nuclei to flip alignment (spin flipping = resonance).
• Energy region: Radio-frequency portion of the electromagnetic spectrum.
• Example: Ethanol has three types of hydrogen atoms—CH₃ (3 equivalent H), CH₂ (2 equivalent H), and OH (1 H)—so its ¹H NMR shows three signals.

🧪 NMR principles and interpretation

📍 Chemical shift and shielding

Chemical shift: the position of an NMR absorption signal, measured in parts per million (ppm) relative to a standard (TMS for ¹H NMR).

• Electron shielding: Electrons in the molecule create their own magnetic field that opposes the external field.
• Effect of shielding: Greater electron density around a nucleus → more shielding → lower energy required for resonance → signal appears upfield (lower ppm).
• Less shielding (deshielding): Electron-withdrawing groups reduce electron density → higher energy required → signal appears downfield (higher ppm).
• Typical ranges:
  • ¹H NMR: 0–15 ppm (relative to TMS, tetramethylsilane Si(CH₃)₄).
  • ¹³C NMR: 0–200 ppm.
• Example: A proton NMR signal at 9.8 ppm is more deshielded (less electron density) than one at 1.5 ppm.

🔢 Chemical equivalence and signal counting

Chemically equivalent nuclei: nuclei in identical environments that absorb at the same energy, producing a single NMR signal.

• How to count signals: Count the number of distinct chemical environments.
• Example: In ethanol (CH₃CH₂OH), the three CH₃ hydrogens are equivalent (one signal), the two CH₂ hydrogens are equivalent (one signal), and the OH hydrogen is unique (one signal) → three total ¹H NMR signals.
• Example: A molecule with high symmetry (e.g., all hydrogens equivalent) shows fewer signals; less symmetry shows more signals.

📊 Spin-spin coupling (splitting)

• What causes splitting: Nearby non-equivalent nuclei influence each other's magnetic environment.
• N+1 rule: A signal is split into N+1 peaks, where N is the number of adjacent equivalent hydrogens.
• Example: A triplet integrating for nine hydrogens indicates that those nine hydrogens are adjacent to two other hydrogens (N=2, so 2+1=3 peaks).
• Don't confuse: Singlets appear when there are no adjacent hydrogens or when all adjacent hydrogens are equivalent.

🗂️ Key IR absorption regions

🎯 Functional group identification by IR

The excerpt emphasizes specific wavenumber regions for common functional groups:

Functional group	IR absorption (cm⁻¹)	What it indicates
C=O (carbonyl)	~1700	Aldehydes, ketones, esters, carboxylic acids
O–H (alcohol)	~3400	Alcohols, phenols
C≡C (alkyne)	~2100	Terminal or internal alkynes
C–H stretch	~3000	Alkanes, alkenes, aromatic rings
Fingerprint region	<1500	Unique pattern for each molecule

🔍 Using IR to monitor reactions

• Appearance of a peak: Indicates formation of a new functional group.
  • Example: Appearance of a band near 1700 cm⁻¹ after oxidation of an alcohol suggests formation of a carbonyl (aldehyde or ketone).
• Disappearance of a peak: Indicates consumption of a functional group.
  • Example: Disappearance of the O–H stretch (~3400 cm⁻¹) when an alcohol is converted to an ether or alkene.
• Example: To confirm that diisopropyl ether formed (not propene), look for the absence of an O–H stretch and the absence of a C=C stretch.

🧴 Separation techniques

💧 Extraction with acid and base

Extraction: a separation method that moves compounds between an aqueous layer and an organic layer based on their acid-base properties.

• Principle: Acids dissolve in base (forming water-soluble salts); bases dissolve in acid (forming water-soluble salts); neutral compounds stay in the organic layer.
• Common reagents:
  • Sodium bicarbonate (NaHCO₃): Extracts carboxylic acids (pKₐ ~5) into the aqueous layer.
  • Sodium hydroxide (NaOH): Extracts carboxylic acids and phenols (pKₐ ~10) into the aqueous layer.
  • Hydrochloric acid (HCl): Extracts amines (basic) into the aqueous layer as ammonium salts.
• Example: A mixture of a carboxylic acid, an amine, and a neutral compound can be separated by first extracting with NaHCO₃ (removes acid), then with HCl (removes amine), leaving the neutral compound in the organic layer.
• Don't confuse: Sodium bicarbonate is weaker than sodium hydroxide; it will extract only carboxylic acids, not phenols.

🎨 Thin-layer chromatography (TLC)

TLC: a separation technique where compounds travel up a silica gel plate; polar compounds stick to the polar silica and travel less, while nonpolar compounds travel farther.

• Rₓ value: The ratio of the distance traveled by the compound to the distance traveled by the solvent front.
  • Higher Rₓ → compound is less polar (travels farther).
  • Lower Rₓ → compound is more polar (sticks to silica).
• Example: Toluene (nonpolar) travels farther than nitrotoluene (polar) on a normal silica gel TLC plate because toluene does not interact strongly with the polar silica.
• Monitoring reactions: TLC can show whether starting material remains or if products have formed by comparing Rₓ values.
• Don't confuse: On a normal silica gel plate, nonpolar compounds travel farther; on a reverse-phase plate, the opposite is true.

🌡️ Distillation and other methods

• Distillation: Separates compounds by boiling point; useful for large-scale separations of liquids with different boiling points.
  • Example: Fractional distillation can separate a 100-mL mixture of isomers with different boiling points.
• Chromatography: A general alternative to extraction when solvent extraction is inefficient.
• Electrophoresis: Separates charged molecules (e.g., amino acids) in an electric field based on their charge at a given pH.
  • Example: At pH = pI (isoelectric point), an amino acid has no net charge and does not migrate; at lower pH, it is positively charged and migrates toward the negative plate.
• Resolution: The specific technique for separating enantiomers (a racemic mixture).

🧬 Special tests and reactions

🧪 Iodoform test

• What it detects: Methyl ketones (compounds with a CH₃CO– group) or secondary alcohols that can be oxidized to methyl ketones.
• Reagent: Mixture of I₂ and NaOH.
• Result: Formation of a yellow precipitate (iodoform, CHI₃).
• Example: A compound that gives a positive iodoform test and has a strong IR absorption near 1700 cm⁻¹ but no signal in the 9–10 ppm region (aldehyde H) is likely a methyl ketone.

🧫 Tollens' test

• What it detects: Aldehydes (reducing sugars).
• Result: Positive test → aldehyde present; negative test → ketone or non-reducing compound.
• Example: A compound with IR absorption at 1700 cm⁻¹ and a negative Tollens' test is likely a ketone, not an aldehyde.

🔬 Ozonolysis

• Reaction: Alkenes react with ozone (O₃), then the ozonide is treated with zinc and water.
• Product: Carbonyl compounds (aldehydes and/or ketones).
• Characteristic IR peak: Every product will show an absorption near 1700 cm⁻¹ (C=O stretch).

🧩 Practical problem-solving strategies

🔎 Distinguishing isomers

• Best method: NMR spectroscopy, because it reveals the number and environment of hydrogen or carbon atoms.
  • Example: 2-nitrotoluene and 4-nitrotoluene (both C₇H₇NO₂) can be distinguished by ¹H NMR because the aromatic protons have different splitting patterns and chemical shifts.
• Why not other methods?:
  • Mass spectroscopy: Both isomers have the same molecular formula and mass.
  • IR spectroscopy: Both have the same functional groups (nitro, aromatic).
  • UV-Vis: Both have similar conjugation.

🧮 Counting NMR signals

• ¹H NMR: Count the number of distinct hydrogen environments; use symmetry to reduce the count.
  • Example: A compound with a plane of symmetry may have fewer signals than the total number of carbon atoms.
• ¹³C NMR: Count the number of distinct carbon environments.
  • Example: 1,3,5-trinitrobenzene has high symmetry, so all three carbons bearing nitro groups are equivalent, and all three carbons bearing hydrogens are equivalent → only 2 signals.
• Singlets: Appear when there are no adjacent non-equivalent hydrogens.
  • Example: A compound with three overlapping singlets, each integrating for three hydrogens, likely has three equivalent CH₃ groups with no adjacent hydrogens.

🧪 Monitoring reaction progress

• IR: Look for appearance or disappearance of characteristic peaks.
  • Example: Oxidation of an alcohol to a ketone → disappearance of O–H stretch (~3400 cm⁻¹) and appearance of C=O stretch (~1700 cm⁻¹).
• TLC: Compare Rₓ values of starting material and product.
  • Example: If the starting material is more polar, it will have a lower Rₓ value than the product.

🧴 Choosing the best separation method

• For acid/base mixtures: Use extraction with appropriate reagents (NaHCO₃, NaOH, HCl).
• For large volumes with different boiling points: Use distillation.
• For small amounts or similar boiling points: Use chromatography (TLC, column chromatography).
• For enantiomers: Use resolution (not extraction or distillation).
• Not useful for separation: IR, NMR, UV-Vis, and mass spectroscopy are analytical (identification) techniques, not separation techniques.

🧭 Overview

🧠 One-sentence thesis

Bioorganic chemistry integrates the structure, properties, and reactivity of amino acids, carbohydrates, proteins, and nucleic acids to explain how biological molecules behave in physiological systems and laboratory settings.

📌 Key points (3–5)

• Amino acid behavior at different pH: Amino acids exist in different ionization states (protonated, zwitterionic, deprotonated) depending on solution pH relative to their pKa and pI values.
• Electrophoresis and isoelectric point: At pH = pI, an amino acid carries no net charge and will not migrate in an electric field; migration direction depends on whether pH is above or below pI.
• Carbohydrate classification and reactivity: Monosaccharides are classified by functional group (aldose vs ketose), ring size, and anomeric configuration (α vs β); reducing sugars contain free or equilibrating aldehyde/hemiacetal groups.
• Protein structure and bonding: Peptide (amide) bonds link amino acids in primary structure; secondary structure arises from hydrogen bonding between peptide bonds; tertiary structure involves side-chain interactions.
• Common confusion—hemiacetal vs acetal in carbohydrates: A hemiacetal has one –OR and one –OH on the anomeric carbon (can revert to open-chain aldehyde); an acetal has two –OR groups (stable, non-reducing).

🧬 Amino Acids: Structure and Ionization

🧬 General structure and functional groups

Generic amino acid formula: a central carbon bearing an amino group (–NH₂), a carboxylic acid group (–COOH), a hydrogen, and a variable side chain (R).

• Every amino acid has at least two pKa values: one for the carboxylic acid group and one for the amino group.
• Some amino acids have a third pKa for an ionizable side chain (e.g., lysine has an extra amino group; serine's hydroxyl is not ionizable under normal conditions).
• Example: Asparagine has pKa = 2.0 (carboxylic acid) and pKa = 8.8 (amino group).

⚡ Zwitterion and isoelectric point (pI)

The isoelectric point (pI) is the pH at which an amino acid has no net charge; it is calculated as the average of the relevant pKa values.

• At pH = pI, the carboxylic acid group is deprotonated (–COO⁻) and the amino group is protonated (–NH₃⁺), giving a zwitterion with overall charge = 0.
• Example: For asparagine, pI = (2.0 + 8.8)/2 = 5.4. At pH 5.4, asparagine is zwitterionic and will not migrate in electrophoresis.
• Don't confuse: The zwitterion is not the only form at pI—it is the predominant form, but equilibria with other forms exist.

🔋 Electrophoresis and charge-dependent migration

• At pH < pI: The amino acid is protonated (net positive charge) and migrates toward the negative electrode.
• At pH = pI: No net charge → no migration.
• At pH > pI: The amino acid is deprotonated (net negative charge) and migrates toward the positive electrode.

pH condition	Charge state	Migration direction
pH < pI	Positive	Toward negative plate
pH = pI	Zero (zwitterion)	No migration
pH > pI	Negative	Toward positive plate

Example: At pH 1.0 (very acidic, < pI), asparagine is fully protonated (–COOH and –NH₃⁺) with net positive charge → migrates toward the negative plate. At pH 11.0 (very basic, > pI), asparagine is fully deprotonated (–COO⁻ and –NH₂) with net negative charge → migrates toward the positive plate.

🧪 Amphoteric nature of amino acids

Amphoteric: capable of acting as either an acid or a base.

• Amino acids can donate a proton (from –COOH or –NH₃⁺) or accept a proton (at –COO⁻ or –NH₂).
• This property allows amino acids to buffer solutions and participate in proton-transfer reactions.
• Don't confuse amphoteric with chirality: amphoteric refers to acid-base behavior, not stereochemistry.

🔬 Laboratory analysis: 2,4-dinitrofluorobenzene (DNFB)

• DNFB attaches to the N-terminal amino acid of a peptide (the amino acid with the free amino group).
• After hydrolysis, only the N-terminal residue carries the dinitrophenyl label.
• Mechanism: nucleophilic aromatic substitution (the amino group acts as a nucleophile attacking the aromatic ring, displacing fluoride).
• Example: Treatment of a hexapeptide with DNFB followed by hydrolysis labeled only glycine → glycine is the N-terminal amino acid.

🍬 Carbohydrates: Classification and Reactivity

🍬 Monosaccharide classification

• By functional group: Aldoses (contain an aldehyde group, –CHO) vs ketoses (contain a ketone group).
• By carbon count: Pentoses (5 carbons) vs hexoses (6 carbons).
• By stereochemistry: D vs L configuration (determined by the configuration of the chiral center farthest from the carbonyl).
• Example: Aldopentoses are five-carbon sugars with an aldehyde group (e.g., ribose).

🔄 Anomers and the anomeric carbon

The anomeric carbon is the carbonyl carbon in the open-chain form; upon ring closure, it becomes a new chiral center.

• Hemiacetal: The anomeric carbon in a cyclic sugar with one –OR and one –OH (can open to aldehyde form).
• Acetal: The anomeric carbon with two –OR groups (formed when a hemiacetal reacts with another alcohol; stable, does not open).
• α vs β anomers: In Fischer projections converted to Haworth, α has the –OH on the anomeric carbon on the opposite side of the ring from the CH₂OH; β has it on the same side.
• Example: In a disaccharide, the reducing end has a hemiacetal (can give a positive Benedict's test); the non-reducing end has an acetal.

🧪 Reducing sugars and Benedict's test

A reducing sugar contains a free or equilibrating aldehyde (or ketone) group that can reduce Cu²⁺ to Cu₂O (red precipitate).

• Only the open-chain (straight-chain) form of a sugar can reduce Benedict's reagent.
• Cyclic hemiacetal forms are in equilibrium with the open-chain aldehyde form.
• Example: Glucose in solution exists as ~43% open-chain and ~57% cyclic forms. Because the open-chain form can regenerate from the cyclic forms, 100% of the glucose will eventually react in Benedict's test (the equilibrium shifts as the open-chain form is consumed).
• Don't confuse: A sugar with only acetal linkages (e.g., sucrose) is non-reducing because it cannot open to an aldehyde.

🔗 Glycosidic linkages

A glycosidic linkage is an acetal bond connecting two monosaccharides (or a sugar to another molecule).

• Notation: α or β (anomeric configuration) followed by the carbon numbers involved (e.g., α-1,4 means the anomeric carbon of one sugar is linked to carbon 4 of the next in α configuration).
• Example: Maltose has an α-1,4 linkage between two glucose units. Lactose has a β-1,4 linkage between galactose and glucose.

🔄 Mutarotation

Mutarotation: the interconversion of α and β anomers in solution via the open-chain form.

• When a pure α or β anomer is dissolved, it equilibrates to a mixture of both anomers.
• This process involves opening the ring (hemiacetal → aldehyde) and re-closing in the opposite configuration.

🧬 Peptides and Proteins: Structure and Bonding

🧬 Peptide bond formation

A peptide bond (amide bond) links the carboxyl group of one amino acid to the amino group of the next, releasing water.

• The bond is planar and rigid due to resonance (partial double-bond character).
• Primary structure: the linear sequence of amino acids connected by peptide bonds.
• Example: A tripeptide gly-ala-ser has glycine at the N-terminus and serine at the C-terminus.

🔗 Determining peptide sequence

• Partial hydrolysis yields overlapping fragments (dipeptides, tripeptides).
• N-terminal analysis (e.g., DNFB) identifies the first amino acid.
• Composition analysis gives the total count of each amino acid.
• Example: From dipeptides gly-ala, ala-ser, ser-gly, ser-pro, gly-ser and tripeptides gly-ala-ser, gly-ser-pro, ala-ser-gly, plus DNFB labeling glycine, the sequence is gly-ala-ser-gly-ser-pro.

🌀 Secondary structure

Secondary structure: local folding patterns (α-helix, β-sheet) stabilized by hydrogen bonds between the carbonyl oxygen and amide hydrogen of the peptide backbone.

• These hydrogen bonds do not involve side chains.
• Example: An α-helix is a right-handed coil held together by hydrogen bonds between the C=O of residue n and the N–H of residue n+4.

🧩 Tertiary structure and side-chain positioning

Tertiary structure: the overall three-dimensional shape of a protein, determined by interactions between side chains (R groups).

• Polar side chains (e.g., serine, lysine) are typically on the exterior, interacting with the aqueous environment.
• Nonpolar side chains (e.g., valine, leucine) are typically in the interior, away from water.
• Example: Valine (with an isopropyl side chain) is nonpolar and would be found in a nonpolar region of a protein.
• Don't confuse tertiary structure (side-chain interactions) with secondary structure (backbone hydrogen bonding).

🔗 Quaternary structure

• The assembly of multiple polypeptide chains (subunits) into a functional protein complex.
• Stabilized by the same interactions as tertiary structure (hydrogen bonds, ionic interactions, disulfide bridges, hydrophobic interactions).

🧪 Special Topics in Bioorganic Chemistry

🧪 Degrees of unsaturation

• A measure of rings and/or double bonds in a molecule.
• Formula: (2C + 2 + N – H – X)/2, where C = carbons, N = nitrogens, H = hydrogens, X = halogens.
• Example: A carbohydrate sequence with multiple rings and double bonds can be analyzed to count degrees of unsaturation.

🧪 Chirality and optical activity

• Chiral amino acids: Most amino acids (except glycine) have a chiral center at the α-carbon.
• Achiral amino acids: Glycine (R = H) has no chiral center and is optically inactive.
• Meso compounds: Molecules with chiral centers but an internal plane of symmetry (e.g., alditol from D-allose reduction).
• Example: A racemic mixture (equal amounts of enantiomers) is optically inactive; separating enantiomers is important because they can have different biological activities (e.g., thalidomide).

🧪 Absolute configuration (R/S)

• Determined by Cahn-Ingold-Prelog priority rules.
• Example: Nicotine's chiral center can be assigned R or S configuration based on substituent priorities.
• For amino acids, the L/D system is used (based on glyceraldehyde reference), but R/S can also be assigned.

🧪 Biological pH and protonation states

• Physiological pH = 7.4 (slightly basic relative to pure water, pH 7.0).
• At pH 7.4, carboxylic acids (pKa ~2–5) are deprotonated (–COO⁻); amino groups (pKa ~9–10) are protonated (–NH₃⁺).
• At extreme pH (e.g., stomach pH ~1), even weakly basic groups are protonated; at pH 14, even weakly acidic groups are deprotonated.
• Example: At pH 1, the amino group of alitame is protonated; the carboxylic acid may still be protonated or partially deprotonated depending on its pKa.

🧪 Nucleotides and nucleic acids

A nucleotide is a nucleoside (sugar + nitrogenous base) plus one or more phosphate groups.

• DNA and RNA are polymers of nucleotides linked by phosphodiester bonds.
• The phosphate ester functional group forms when a phosphate group attaches to a sugar hydroxyl.
• Example: A nucleoside is adenosine (adenine + ribose); adding a phosphate makes it adenosine monophosphate (AMP), a nucleotide.

🧪 Lipids and saponification

Saponification: base-catalyzed hydrolysis of an ester (e.g., a fat) to produce glycerol and carboxylate salts (soap).

• Micelle structure: In aqueous solution, carboxylate salts (with long hydrocarbon chains) aggregate with hydrophilic carboxylate groups on the exterior (interacting with water) and hydrophobic hydrocarbon tails in the interior (away from water).
• Example: Soap molecules form micelles that can solubilize nonpolar substances (like grease) in water.

🧪 Partial hydrogenation of fats

• Natural unsaturated fats have cis double bonds; partial hydrogenation reduces some double bonds and can isomerize others to trans.
• Trans fats have straighter chains → stronger London dispersion forces → higher melting points.
• Conditions: H₂ with a metal catalyst (e.g., Ni) under controlled conditions to avoid complete saturation.

🧪 Enzyme specificity and stereochemistry

• Enzymes (e.g., α-glucosidases like sucrase and maltase) are stereospecific: they recognize and act on specific anomeric configurations.
• Example: Sucrase hydrolyzes sucrose (α-1,2 linkage between glucose and fructose) to glucose and fructose. Maltase hydrolyzes maltose (α-1,4 linkage between two glucose units) to two glucose molecules.
• Enantiomers can have vastly different biological activities because they interact differently with chiral enzyme active sites (e.g., thalidomide enantiomers).

🧪 Strecker synthesis of amino acids

• A laboratory method to synthesize amino acids from aldehydes using ammonia and HCN, followed by hydrolysis.
• Ammonia acts as a nucleophile, attacking the aldehyde carbonyl.
• The product is racemic (equal R and S) unless the starting material is chiral.
• Example: Synthesis from a symmetric aldehyde (e.g., formaldehyde) yields glycine, which is achiral.

🧪 Acid/base hydrolysis of proteins

• Complete hydrolysis: Hot concentrated acid (e.g., 6 M HCl) breaks all peptide bonds, yielding individual amino acids.
• Partial hydrolysis: Milder conditions yield peptide fragments (dipeptides, tripeptides) useful for sequencing.
• Enzymes (proteases) can also selectively cleave peptide bonds under physiological conditions.

🧪 Resonance and hybridization in biomolecules

• Amide resonance: The nitrogen in a peptide bond has partial double-bond character (sp² hybridization), restricting rotation.
• Aromatic bases: Purines (e.g., adenine) and pyrimidines have resonance structures; lone pairs on nitrogen can participate in resonance or act as bases.
• Example: In thalidomide, N-1 (in an amide) is sp² hybridized; N-2 (in an imide) is also sp² due to resonance with adjacent carbonyls.

🧪 Epimers and diastereomers

Epimers: diastereomers that differ in configuration at only one chiral center.

• Example: D-glucose and D-galactose are epimers (differ at C-4).
• Anomers (α and β forms of the same sugar) are also diastereomers (and a special case of epimers differing at the anomeric carbon).
• Don't confuse: All anomers are epimers, but not all epimers are anomers.

🧪 Fischer projections and D/L designation

In Fischer projections, the D/L designation is based on the configuration of the chiral center farthest from the carbonyl group.

• D-sugars have the –OH on the right; L-sugars have it on the left.
• Example: An L-monosaccharide in Fischer projection has the bottom chiral center's –OH on the left.

🧪 Isomerization reactions

Isomerization: rearrangement of atoms within a molecule without adding or removing atoms.

• Example: Epimerases catalyze isomerization of D-ribulose 5-phosphate to D-xylulose 5-phosphate (changing configuration at one carbon).
• Not isomerization: Hydrolysis of an acid chloride to a carboxylic acid (adds water, removes HCl).

🧪 Oxidation-reduction in biomolecules

• Disulfide bond formation: Two cysteine residues (each with a thiol, –SH) undergo oxidation to form cystine (with a disulfide, –S–S–).
• This is an oxidation-reduction reaction (thiols lose electrons).
• Example: Disulfide bonds stabilize protein tertiary structure by cross-linking different parts of the chain.

🧪 Buffer systems and pKa

• At pH = pKa, the protonated and deprotonated forms of an acid are present in equal concentrations (Henderson-Hasselbalch equation).
• Example: Valine at pH 2.3 (= pKa of carboxyl) has equal amounts of the fully protonated form and the zwitterion.
• At pH 12 (>> pKa of amino group), valine is fully deprotonated (–COO⁻ and –NH₂).

🧪 NMR splitting patterns

• The number of peaks in an NMR signal depends on the number of neighboring hydrogens (n+1 rule).
• Example: A methyl group (–CH₃) adjacent to a chiral carbon bearing one hydrogen will appear as a doublet (split by the one neighboring H).

🧪 Stability of functional groups in biological systems

• Acid chlorides and acid anhydrides are too reactive (rapidly hydrolyze in water) to exist in biological systems.
• Biological ester and amide synthesis uses activated intermediates (e.g., acyl phosphates, thioesters) and enzymes, not acid chlorides.
• Example: Fatty acid activation involves conversion to acyl-CoA (a thioester), not an acid chloride.

🧭 Overview

🧠 One-sentence thesis

This chapter presents a comprehensive set of practice problems covering fundamental organic chemistry concepts including stereochemistry, reaction mechanisms, functional groups, spectroscopy, and biomolecules to prepare students for standardized testing.

📌 Key points (3–5)

• Problem scope: Questions span hybridization, isomerism, reaction mechanisms, acid-base chemistry, polymers, and biological molecules.
• Stereochemistry emphasis: Multiple problems address chirality, enantiomers, diastereomers, conformers, and geometric isomers.
• Reaction mechanisms: Free radical reactions, substitution/elimination reactions, electrophilic aromatic substitution, and polymer formation are tested.
• Common confusion: Distinguishing between types of isomers (structural vs geometric vs enantiomers vs diastereomers) and recognizing hybridization states in different bonding environments.
• Biological applications: Amino acids, proteins, carbohydrates, and enzyme inhibition appear throughout, connecting organic chemistry to biochemistry.

🔬 Molecular Structure & Bonding

⚛️ Hybridization determination

• Hybridization depends on the number of groups (bonds + lone pairs) around an atom:
  • sp: 2 groups (linear, 180°)
  • sp²: 3 groups (trigonal planar, 120°)
  • sp³: 4 groups (tetrahedral, 109.5°)
• Example: In thalidomide (problem 429), nitrogen atoms must be analyzed based on their bonding patterns—triple bonds indicate sp, double bonds sp², and single bonds with lone pairs sp³.
• Don't confuse: A positively charged carbon (carbocation) with three bonds is sp², not sp³, because it has only three groups.

🔗 Sigma and pi bonds

• Every single bond is one σ bond; double bonds contain one σ and one π; triple bonds contain one σ and two π bonds.
• Problem 481 requires counting all bonds systematically in a complex molecule.
• The degree of unsaturation (problem 482) relates to the number of rings plus π bonds in a structure.

📐 Bond angles and geometry

• Problem 447 tests understanding that a positively charged carbon with three groups has sp² hybridization and 120° bond angles.
• Free radicals (problem 468) with three groups also adopt trigonal planar geometry with the unpaired electron in a p orbital.

🪞 Stereochemistry & Isomerism

🔄 Types of isomers

Isomer type	Relationship	Example from problems
Geometric (cis/trans)	Different spatial arrangement around double bond or ring	Problem 430: cis- vs trans-1,2-dibromopentene
Enantiomers	Non-superimposable mirror images	Problem 497 asks about fluorinated compounds
Diastereomers	Stereoisomers that are not mirror images	Geometric isomers are also diastereomers
Conformers	Different rotations around single bonds	Problem 485: butane conformations

🎯 Chiral centers

• A chiral center has four different groups attached to a tetrahedral carbon.
• Problems 480 and 483 require systematic identification of all chiral centers in complex molecules.
• Don't confuse: A carbon in a ring may or may not be chiral depending on whether the groups attached are different.

🔁 Mutarotation

• Problem 473 addresses carbohydrates that can change stereochemistry at the anomeric carbon (the carbon bonded to two oxygens in a ring).
• Only hemiacetal/hemiketal carbons can undergo mutarotation through ring opening and reclosing.

⚗️ Reaction Mechanisms & Kinetics

🔥 Free radical reactions

• Initiation: Formation of radicals (e.g., homolytic cleavage of Cl₂ or peroxides).
• Propagation: Radical reacts with stable molecule to form product and new radical.
• Termination: Two radicals combine.
• Problem 456: Attack of an alkene by a benzoyl radical is an initiation step for polymerization.
• Problem 452: Free radical bromination favors the most stable radical intermediate (tertiary > secondary > primary).

🎯 Substitution reactions

• Problem 446 shows the first step of an SN2 reaction: nucleophile approaches from the backside of the carbon bearing the leaving group.
• Problem 436: SN2 reactions invert stereochemistry, so (R)-2-bromopentane yields (S)-2-pentanol.
• Problem 492: Doubling both reactants in an SN2 reaction increases rate by a factor of 4 (rate = k[substrate][nucleophile]).

⚡ Energy diagrams

• Problem 444 requires identifying a two-step exothermic reaction where the first step is rate-determining (higher activation energy for step 1).
• Problem 445: The compound that forms more rapidly has a lower activation energy; the more stable product has lower overall energy.

🔬 Isotope effects

• Problem 434: Deuterium (D) forms stronger C–D bonds than C–H, so breaking a C–D bond is slower (kinetic isotope effect).

🧪 Functional Groups & Reactions

🏷️ Functional group identification

• Acid anhydride (problem 431): R–C(=O)–O–C(=O)–R
• Carboxylic acids extract into sodium bicarbonate (problem 435) because they are acidic enough to react with weak base.
• Problem 453: Acid chlorides and anhydrides are too reactive for biological systems; cells use alternative activation methods.

🔄 Oxidation and reduction

• Problem 434: Secondary alcohols oxidize to ketones using K₂Cr₂O₇.
• Problem 457: The Kolbe synthesis is an oxidation that removes CO₂ and couples alkyl groups.
• Problem 458: IR monitoring shows disappearance of C=C stretch (1650 cm⁻¹) and appearance of C=O stretch (1715 cm⁻¹).

💧 Dehydration reactions

• Problem 437: Alcohols that form more stable carbocations dehydrate faster (tertiary > secondary > primary).

🧬 Biological Molecules

🧬 Amino acids

• Amphoteric nature (problem 470): Amino acids can act as both acids and bases because they contain both –COOH and –NH₂ groups.
• Zwitterion form (problem 499): At intermediate pH, amino acids exist as ⁺H₃N–CHR–COO⁻ and can donate or accept protons.
• Problem 469: At pH 7, aspartic acid exists with both carboxylic acids deprotonated (pKₐ 2.1 and 3.9) and the amino group protonated (pKₐ 9.8).

🔗 Peptides and proteins

• Problem 476: Asparagine contains an amide side chain that can hydrolyze during protein hydrolysis.
• Problem 459: Amino acids with sulfur (cysteine) or nitrogen (histidine, lysine) in side chains can form coordinate covalent bonds to metal ions in metalloenzymes.

🍬 Carbohydrates

• Problem 473: Carbohydrate units attached to proteins can undergo mutarotation if they have a free anomeric carbon.

🧵 Polymers

• Condensation polymers (problem 474): Require two functional groups per monomer; form by eliminating small molecules (H₂O, HCl).
• Problem 477: Nylon-66 from diamine + diacid is a polyamide.
• Problem 494: Monomers need two reactive groups; CH₃CH₂CH₂OH has only one –OH and cannot form a polymer alone.

🎨 Aromatic Chemistry

⚡ Electrophilic aromatic substitution

• Problem 462: In nitration, the actual electrophile is NO₂⁺ (nitronium ion), not HNO₃.
• Problem 463: Methyl is an ortho/para director, so the nitro group adds at those positions.
• Problem 488: Carboxylic acid groups are deactivating (electron-withdrawing), requiring higher temperatures for substitution than activating alkyl groups.

🔄 Birch reduction

• Problem 455: Aromatic rings reduce to cyclohexadienes; the position of the remaining double bonds depends on whether substituents are electron-donating or electron-withdrawing (resonance stabilization).

🧲 Physical Properties

🌡️ Melting points

• Problem 443: Impurities lower and broaden melting points.
• Problem 464: Para-nitrobenzoic acid has a much higher melting point because it forms intermolecular hydrogen bonds; ortho isomers form intramolecular H-bonds and pack less efficiently.
• Problem 465: A mixture of isomers melts below the melting point of either pure compound.

💧 Solubility and intermolecular forces

• Problem 448: Compounds with more hydrogen bonding (carboxylic acids > alcohols > amines) have higher boiling points.
• Problem 449: Nonpolar compounds (hydrocarbons) are most soluble in nonpolar solvents like cyclohexane.
• Problem 496: Hydrogen bonding is important for molecules with N–H or O–H bonds (e.g., dimethylamine).

🔋 Acidity and basicity

• Problem 484: Acidity order: sp-hybridized C–H < alcohols < phenols < carboxylic acids < sulfonic acids.
• Problem 498: Trifluoroacetic acid is stronger than trichloroacetic acid because fluorine is more electronegative, creating a greater inductive effect that stabilizes the conjugate base.
• Problem 495: CH₃⁻ is the strongest base (weakest conjugate acid).