MCAT Organic Chemistry Review_ New for MCA

🧭 Overview

🧠 One-sentence thesis

The MCAT is not a content-memorization test but rather an assessment of your ability to apply basic science knowledge to new situations and reason under pressure, requiring strategic preparation that differs fundamentally from typical college exam studying.

📌 Key points (3–5)

• What the MCAT really tests: ability to apply knowledge to new situations and reason logically under timed pressure, not just content recall.
• Test structure: 7.5-hour computer-based exam with four sections (Chem/Phys, CARS, Bio/Biochem, Psych/Soc), mostly passage-based questions.
• Scoring approach: scaled exam with no guessing penalty; you can void scores within 5 minutes after completing the test.
• Common confusion: treating the MCAT like a college science test that rewards memorization—medical schools already see your transcripts; they want to know how you think.
• Strategic preparation: review basics, then practice applying them to MCAT-style questions; use process of elimination and pacing strategies.

🎯 What the MCAT actually measures

🧪 Not a typical college science test

The MCAT is really a test of your ability to apply basic knowledge to different, possibly new, situations, and a test of your ability to reason out and evaluate arguments.

• Most test-takers mistakenly approach it as a fact-regurgitation exam.
• Medical school admission committees can see your content knowledge from transcripts.
• What they want to see: how you think, especially under pressure.
• Your MCAT score tells them about your reasoning ability, not just what you've memorized.

📚 Content knowledge still matters—but differently

• You absolutely need to know science content, but not at the level most test-takers assume.
• The depth of knowledge required is more about understanding basics that can be applied flexibly.
• Science knowledge won't help on the Critical Analysis and Reasoning Skills (CARS) section at all.
• Don't confuse: needing to know content vs. needing to memorize every detail—the MCAT emphasizes the former.

🏗️ Test structure and format

🖥️ Computer-based test characteristics

Feature	Details
Format	Computer-based, not adaptive (linear/fixed-form)
Length	7.5 hours including breaks
Frequency	28 times per year (Jan, Mar–Sep)
Registration	Online via AAMC, up to 6 months in advance
Security	Photo ID, electronic fingerprint, signature verification

• Different test-takers see different versions on the same day.
• Questions are in predetermined order and don't change based on your answers.
• No proctoring; test administrator checks you in and assigns your computer seat.

📋 Four sections breakdown

1. Chemical and Physical Foundations of Biological Systems (Chem/Phys): 59 questions, 95 minutes
2. Critical Analysis and Reasoning Skills (CARS): 53 questions, 90 minutes
3. Biological and Biochemical Foundations of Living Systems (Bio/Biochem): 59 questions, 95 minutes
4. Psychological, Social, and Biological Foundations of Behavior (Psych/Soc): 59 questions, 95 minutes

• Optional 10-minute breaks between sections.
• Longer lunch break (30–45 minutes) available.
• Total test center time includes check-in (up to 40 minutes), tutorial (10 minutes), and post-test survey (10 minutes).

📝 Question distribution

• Passage-based questions: approximately 3/4 of science sections (all 53 CARS questions).
• Freestanding questions (FSQs): about 1/4 of science sections, appearing in groups of 4–5 questions.
• Passages include paragraphs plus equations, tables, graphs, figures, and experiments in science sections.
• CARS passages come from social sciences, humanities, ethics, philosophy, cultural studies, and population health—no content knowledge tested.

🎲 Scoring and the void option

📊 How scoring works

• Scaled exam: raw score converted to scaled score accounting for question difficulty.
• No "magic number" of correct answers needed for a particular score.
• No guessing penalty: always answer every question.
• Some questions are "experimental" and don't count toward your score.
• Different test versions have different scales to account for varying difficulty.

⚠️ The void decision

After completing the final section, you have 5 minutes to choose:

• "I wish to have my MCAT exam scored" OR
• "I wish to VOID my MCAT exam"

Void consequences:

• No numerical score ever (cannot "unvoid" later).
• Medical schools won't know you took the test.
• No refunds granted.
• If you don't choose within 5 minutes, test automatically scores.

🎯 What's a good score?

• Depends on your overall application package: GPA, coursework, letters of recommendation, medical field experience, extracurriculars, personal statement.
• Low GPA requires higher MCAT scores to compensate; strong GPA allows for lower scores.
• Medical schools want a complete package, not just scores and GPA.

🛠️ Test-taking tools and strategies

💻 Available CBT tools

Tool	Function	Persistence
Highlighting	Click-drag in passage text	Does NOT persist after leaving passage
Strike-outs	Click answer choices to eliminate	DOES persist after leaving passage
Mark button	Flag questions for later review	Available throughout section
Review button	See all questions' status (answered/unanswered/marked)	Access anytime during section
Exhibit button	Open periodic table (resizable window)	Available throughout
Scratch paper	4 pages (8 faces) provided; can request more	First set collected before receiving new

📄 Organizing scratch paper effectively

• Indicate passage number in a box at top of work.
• Circle question numbers to the left of notes.
• Draw a line under work when changing passages.
• Don't erase or scribble—draw one line through incorrect work and start fresh.
• Keep organized; you may need to review questions later.

⏱️ Pacing strategies

Science sections:

• About 1 minute 35 seconds per question if completing all.
• At passage start: note allotted time and starting timer value.
• Write on scratch paper what timer should say at passage end.
• If running out of time: guess on remaining questions, make notes, mark questions, move on.

CARS section:

• About 1 minute 40 seconds per question.
• Most test-takers maximize score by NOT completing every passage.
• Good strategy: complete all but one passage, randomly guess on the last.
• Spend about 10 minutes per passage (adjust for difficulty and question count).
• Work passages in order of comfort—skip difficult ones first time through.

✂️ Process of Elimination (POE)

Since there's no guessing penalty, POE increases probability of correct answers:

1. Strike out choices you're sure are incorrect.
2. Jot notes on scratch paper to clarify thoughts.
3. Use "Mark" button to flag for review (but in CARS, generally don't return to rethink).
4. Don't leave blank—if you've spent >60 seconds, pick from remaining choices.
5. If three choices eliminated, the fourth must be correct—click it and move on.
6. MCAT only cares that you selected the right answer, not that you understand why.

📚 Question categories

🔬 Science section question types

Type	Description	Percentage
Memory	Answered from prior knowledge	~25%
Explicit	Answer explicitly stated in passage (definitions, graphs, simple connections)	~35%
Implicit	Apply knowledge to new situations; answer implied by passage	~40%

• Implicit questions often start with "If… then…" constructions.
• Example: "If we modify the experiment like this, then what result would we expect?"

📖 CARS section question types

1. Specific questions: Retrieval (facts from passage) or Inference (deduce what's likely true).
2. General questions: Summarize themes (main idea, primary purpose) or evaluate author's opinion (tone/attitude).
3. Reasoning questions: Describe purpose/support for statements (Structure) or judge argument quality (Evaluate).
4. Application questions: Apply new information from question stem or answer choices to passage (New Information, Strengthen, Weaken, Analogy).

🎒 Test day preparation

📅 Before test day

• Visit test center 1–2 days before to find location, gauge traffic, check parking.
• Don't study heavily the day before.
• Get good sleep in nights leading up to test.
• Experiment with foods during practice tests to find what gives steady energy.
• Good snacks: sports drinks, peanut-butter crackers, trail mix.

🏢 Test day logistics

• Arrive at least 30 minutes before start time.
• Check-in order: arrival order.
• Assigned locker/secure area for personal items (no textbooks/notes allowed).
• ID checked, fingerprint taken digitally, sign in.
• Receive scratch paper and pencils (or request paper if given whiteboard).
• Assigned computer (cannot choose).
• Only allowed at station: photo ID, locker key, factory-sealed ear plugs.

☕ During breaks

• Fingerprint checked again when leaving/returning.
• Sign in and out.
• Can access locker items except notes and cell phones.
• Take the breaks—walk around, clear your head, get blood flowing.
• Request new scratch paper if needed.
• Scratch paper collected and shredded at test end.

🧭 Overview

🧠 One-sentence thesis

The MCAT contains three science sections—Chemical and Physical Foundations, Biological and Biochemical Foundations, and Psychological, Social, and Biological Foundations—each with distinct content distributions, mostly passage-based questions, and 95 minutes of testing time.

📌 Key points (3–5)

• Three science sections: Chem/Phys (third section), Bio/Biochem (first section), and Psych/Soc (fourth section), each testing different subject combinations.
• Content distribution varies: Chem/Phys emphasizes General Chemistry and Physics with math; Bio/Biochem focuses on biology and biochemistry with minimal calculation; Psych/Soc draws mainly from psychology and sociology with basic statistics.
• Question format: approximately 75% passage-based (9–10 passages with 5–7 questions each) and 25% freestanding questions (FSQs in groups of 4–5).
• Common confusion: Math requirements differ—Chem/Phys requires solid math fundamentals (no calculus), while Bio/Biochem and Psych/Soc need only basic statistics understanding.
• Timing: 95 minutes per section, roughly 1 minute 25 seconds per question.

📚 The three science sections

🧪 Chemical and Physical Foundations of Biological Systems (Chem/Phys)

• Position: third section on the test.
• Content breakdown:
  • General Chemistry: ~35%
  • Physics: ~25%
  • Organic Chemistry: ~15%
  • Biochemistry: ~25%
• Context: questions often test chemical and physical concepts within biological settings.
  • Example: pressure and fluid flow in blood vessels.
• Math requirements: solid grasp of fundamentals is required—arithmetic, algebra, graphs, trigonometry, vectors, proportions, and logarithms.
  • No calculus-based questions.

🧬 Biological and Biochemical Foundations of Living Systems (Bio/Biochem)

• Position: first section on the test.
• Content breakdown:
  • Biology: ~65%
  • Biochemistry: ~25%
  • Organic and General Chemistry: ~10%
• Math requirements: calculations are generally not required.
  • A basic understanding of statistics as used in biological research is helpful.

🧠 Psychological, Social, and Biological Foundations of Behavior (Psych/Soc)

• Position: fourth and final section on the test.
• Content breakdown:
  • Psychology: ~60%
  • Sociology: ~30%
  • Biology: ~10%
• Math requirements: calculations are generally not required.
  • A basic understanding of statistics as used in research is helpful.

📝 Question format and structure

📖 Passage-based questions

• Proportion: about 75% of questions in each science section.
• Number of passages: likely about 9 or 10 per section.
• Passage content: a few paragraphs of information including equations, reactions, graphs, figures, tables, experiments, and data.
• Questions per passage: 5 to 7 questions associated with each passage.

🔢 Freestanding questions (FSQs)

• Proportion: remaining 25% of questions in each science section.
• Format: appear in groups interspersed between the passages.
• Group size: each group contains 4 to 5 questions.

⏱️ Timing

• Total time per section: 95 minutes.
• Time per question: approximately 1 minute and 25 seconds.
• Don't confuse: this is an average—passages may take longer, FSQs may be faster.

🔬 Science passage types

📋 Three main categories

The excerpt introduces that passages fall into one of three main categories:

Category	Description (from excerpt)
Information and/or Situation Presentation	Present straightforward scientific information
Experiment/Research Presentation	(Not detailed in this excerpt)
Persuasive Reasoning	(Not detailed in this excerpt)

Note: The excerpt ends before fully describing all three passage types.

🧭 Overview

🧠 One-sentence thesis

The science sections of the MCAT use three main passage types—Information/Situation Presentation, Experiment/Research Presentation, and Persuasive Reasoning—each testing different reasoning skills through passage-based and freestanding questions.

📌 Key points (3–5)

• Three passage categories: Information/Situation Presentation (straightforward facts or scenarios), Experiment/Research Presentation (data and procedures), and Persuasive Reasoning (hypotheses and arguments).
• What each type tests: Information passages test basic facts and predictions; Experiment passages test data interpretation and inference; Persuasive passages test hypothesis evaluation.
• Passage distribution: about 75% of questions are passage-based (9–10 passages per section, 5–7 questions each); 25% are freestanding questions in groups of 4–5.
• Common confusion: Persuasive Reasoning passages are less common than the other two types in the science sections.
• Time constraint: 95 minutes per science section, approximately 1 minute 25 seconds per question.

📚 Information and/or Situation Presentation passages

📚 What they present

Information and/or Situation Presentation passages: passages that either present straightforward scientific information or describe a particular event or occurrence.

• These passages lay out facts or describe a specific scenario without emphasizing experimental design or competing arguments.
• The content is typically descriptive rather than analytical.

🎯 What questions test

• Basic science facts: recall or apply fundamental concepts directly stated in the passage.
• Predictions with new variables: given new information or changed conditions, determine what outcome would occur.
• Example: A passage describes a smoke detector mechanism with capacitor plates and radioactive decay; questions might ask what happens if smoke enters or if the battery voltage drops.

🔍 How to recognize them

• Look for passages that explain a mechanism, device, or phenomenon without presenting multiple experiments or competing viewpoints.
• The smoke detector example in the excerpt describes how americium-241 ionizes air to allow current flow, and how smoke interrupts this flow to trigger an alarm—this is a straightforward situation description.

🔬 Experiment/Research Presentation passages

🔬 What they present

Experiment/Research Presentation passages: passages that present the details of experiments and research procedures, often including data tables and graphs.

• These passages describe one or more experiments, including methods, materials, and results.
• They frequently include tables (e.g., Table 1 and Table 2 in the estrogen receptor example) and may describe multiple related experiments (Experiments 1–4 in the excerpt).

🎯 What questions test

• Interpret data: read tables and graphs to understand what the results show.
• Draw conclusions: determine what the data imply about the underlying biology, chemistry, or physics.
• Make inferences: extend findings to new situations or predict outcomes of modified experiments.
• Example: The estrogen receptor passage shows fluorescence results in different cell types and conditions; questions might ask what the presence or absence of fluorescence reveals about receptor binding or cellular mechanisms.

🔍 How to recognize them

• Look for numbered experiments, data tables, and procedural details (e.g., "cells were grown for 48 hours," "exposed to a 10 mg/mL solution").
• The excerpt's estrogen example includes four experiments with systematic variations (different cell types, pesticide exposure, cytoplasmic extracts) and two results tables.

💬 Persuasive Reasoning passages

💬 What they present

Persuasive Reasoning passages: passages that typically present a scientific phenomenon along with a hypothesis that explains the phenomenon, and may include counter-arguments as well.

• These passages describe a phenomenon and then offer one or more explanations or arguments.
• They may present competing viewpoints (e.g., Chemist #1 vs. Chemist #2 in the acidity example).
• The excerpt notes these are less common than the other two types in the science sections.

🎯 What questions test

• Evaluate hypotheses: assess whether an explanation is consistent with the evidence.
• Evaluate arguments: determine the strengths, weaknesses, or logical structure of competing claims.
• Example: The acidity passage presents two chemists' explanations—one emphasizing bond polarization, the other emphasizing resonance stabilization of the conjugate base; questions might ask which explanation better accounts for observed pKa trends.

🔍 How to recognize them

• Look for multiple viewpoints or hypotheses presented side-by-side.
• The excerpt's acidity example explicitly labels "Chemist #1" and "Chemist #2," each offering a different rationale for acid strength.
• Don't confuse: a single hypothesis tested by experiments is usually an Experiment passage, not a Persuasive Reasoning passage; Persuasive passages emphasize argument evaluation rather than data interpretation.

📊 Passage and question structure

📊 Distribution and format

Feature	Detail
Passage-based questions	About 75% of questions in each science section
Number of passages	About 9–10 per section
Questions per passage	5–7
Freestanding questions (FSQs)	About 25% of questions, in groups of 4–5 between passages
Time per section	95 minutes
Time per question	Approximately 1 minute 25 seconds

• Passages include paragraphs of text plus equations, reactions, graphs, figures, tables, experiments, and data.
• FSQs appear in groups interspersed between passages and are not tied to any passage.

🧪 Science section composition

The excerpt briefly mentions the content breakdown for context (though this is not the main focus of section 2.2):

• Chem/Phys section (third section): General Chemistry (~35%), Physics (~25%), Organic Chemistry (~15%), Biochemistry (~25%); requires math fundamentals but no calculus.
• Bio/Biochem section (first section): Biology (~65%), Biochemistry (~25%), Organic and General Chemistry (~10%); math calculations generally not required, but basic statistics helpful.
• Psych/Soc section (fourth section): Psychology (~60%), Sociology (~30%), Biology (~10%); calculations generally not required, but basic statistics helpful.

🧭 Overview

🧠 One-sentence thesis

MCAT science questions fall into three main types—Memory, Explicit, and Implicit—that differ in how much they rely on passage information versus prior knowledge and the complexity of reasoning required.

📌 Key points (3–5)

• Three question types: Memory (prior knowledge only), Explicit (passage + simple connections), and Implicit (passage + prior knowledge + new application).
• Frequency breakdown: Memory ~25%, Explicit ~35–40%, Implicit ~35–40% of science questions.
• Memory questions: answerable directly from prior knowledge with no need to reference the passage.
• Common confusion: Explicit vs Implicit—both use the passage, but Explicit requires simpler retrieval/connections while Implicit demands applying information to new situations and drawing logical conclusions.
• Where they appear: Memory questions usually appear as freestanding questions (FSQs) but can also be embedded in passages.

🧠 Memory Questions

🧠 What they are

Memory questions: can be answered directly from prior knowledge, with no need to reference the passage or question text.

• You do not need to look at any passage information.
• They test pure recall of concepts, mechanisms, or facts you already know.
• Example: "Which of the following acetylating conditions will convert diethylamine into an amide at the fastest rate?" with answer choices listing different reagents—this requires knowing reaction rates from prior study, not passage data.

📍 Where to find them

• Usually appear as freestanding questions (FSQs).
• Can also be tucked into a passage, but still answerable without reading the passage.

📊 Frequency

• Represent approximately 25 percent of the science questions on the MCAT.

📖 Explicit Questions

📖 What they are

Explicit questions: can be answered primarily with information from the passage, along with prior knowledge.

• Require you to retrieve data, analyze graphs, or make a simple connection between passage content and what you already know.
• The passage provides the key information; you just need to find it and apply straightforward reasoning.
• Example: "The sensor device D shown in Figure 1 performs its function by acting as: A) an ohmmeter. B) a voltmeter. C) a potentiometer. D) an ammeter."—you look at Figure 1, identify the device's function, and match it to your knowledge of instruments.

🔍 How they differ from Memory questions

• Memory: no passage needed at all.
• Explicit: passage is the primary source, but the connection or retrieval is straightforward.

📊 Frequency

• Make up approximately 35–40 percent of the science questions on the MCAT.

🧩 Implicit Questions

🧩 What they are

Implicit questions: require you to take information from the passage, combine it with your prior knowledge, apply it to a new situation, and come to some logical conclusion.

• Demand more complex connections than Explicit questions.
• Often involve hypothetical scenarios or variations on the passage experiment.
• May still require data retrieval or graph analysis, but the reasoning step is deeper.
• Example: "If Experiment 2 were repeated, but this time exposing the cells first to Pesticide A and then to Pesticide B before exposing them to the green fluorescent-labeled estrogen and the red fluorescent probe, which of the following statements will most likely be true?"—you must understand the passage experiment, predict the outcome of a modified protocol, and draw a logical conclusion.

🔍 How they differ from Explicit questions

Feature	Explicit	Implicit
Passage use	Primary source; simple retrieval or connection	Primary source; complex reasoning and application
Reasoning	Straightforward	Apply to new situation; draw logical conclusion
Understanding needed	Basic passage comprehension	Solid understanding of passage information

• Don't confuse: both use the passage, but Implicit questions ask "what if we changed X?" or "what does this imply about Y?" rather than "what does the passage say about X?"

📊 Frequency

• Make up approximately 35–40 percent of the science questions on the MCAT.

📋 Summary Table

Question Type	Definition	Passage needed?	Complexity	Frequency
Memory	Answerable from prior knowledge alone	No	Recall only	~25%
Explicit	Passage info + simple connection	Yes (primary)	Straightforward retrieval/analysis	~35–40%
Implicit	Passage + prior knowledge + new application	Yes (essential)	Complex reasoning, logical conclusion	~35–40%

🧭 Overview

🧠 One-sentence thesis

Organic chemistry makes up a small fraction of the MCAT (roughly 15% of one section and 5% of another), focuses heavily on carbonyl chemistry and lab techniques, and requires a passage-reading strategy that prioritizes chemical structures and data over text.

📌 Key points (3–5)

• Distribution: O-Chem represents ~15% of the Chemical and Physical Foundations section and ~5% of the Biological and Biochemical Foundations section.
• Format: Expect 2–4 freestanding questions plus either one longer passage (6–7 questions) or two short passages (4 questions each) in the Chemical section; in the Biological section, O-Chem questions appear as FSQs or mixed into biology/biochemistry passages.
• Content focus: Covers roughly two semesters of college material but emphasizes carbonyl chemistry and laboratory techniques.
• Common confusion: Unlike other subjects, O-Chem passage text is rarely useful—the critical information lives in chemical structures, schemes, tables, graphs, and figures.
• Reading strategy: Skim text quickly, read titles of figures/schemes for the big picture, then move to questions; avoid getting bogged down in detailed syntheses or mechanisms on the first pass.

📊 Question distribution and format

📊 Chemical and Physical Foundations section

• Overall volume: Around 9–10 passages and ~17 freestanding questions per science section.
• O-Chem share: Roughly 15% of this section.
• Typical format:
  • 2–4 freestanding questions (FSQs)
  • Either one longer passage with 6–7 questions, or
  • Two very short passages with ~4 questions each

🧬 Biological and Biochemical Foundations section

• O-Chem share: Only about 5% of questions.
• Typical format:
  • 3–4 O-Chem questions total
  • Likely to appear as FSQs or embedded within biology/biochemistry passages

🧪 Content scope and emphasis

🧪 Breadth vs depth

• Coverage: Spans roughly two college semesters of organic chemistry material.
• Focus areas: The MCAT concentrates most heavily on:
  • Carbonyl chemistry
  • Laboratory techniques
• Don't confuse: while the scope is broad, not all topics are equally weighted—carbonyl reactions and lab methods are disproportionately important.

🧬 Biological context

• O-Chem passages typically present a biologically important compound or reaction as context.
• The passage text may include biological concepts or facts, but the presence of many chemical structures signals an O-Chem passage rather than a Biology passage.

🗺️ Reading strategy for O-Chem passages

🗺️ What to prioritize

The most important information in O-Chem passages appears in chemical structures from synthetic or mechanistic schemes, or experimental data from tables, graphs, or figures.

• Low priority: The text of the passage.
  • There is "hardly ever information within the text" that you will need to answer passage-based questions.
• High priority:
  • Chemical structures
  • Synthetic schemes
  • Mechanistic schemes
  • Tables, graphs, and figures

⚠️ What to avoid

• Don't get bogged down: Complicated syntheses and mechanisms can be intimidating and slow you down if you pay too much attention during your first pass.
• Don't read every detail: The excerpt warns that detailed schemes "can slow you down considerably if you pay too much attention to this information during your first run through the passage."

✅ Recommended approach

1. Read titles of figures or schemes to grasp the big picture.
2. Skim the text quickly (it's usually not critical).
3. Jump into answering questions rather than trying to master every mechanistic detail upfront.

• Example: If a passage shows a multi-step synthesis with many intermediates, note the overall transformation and the types of reactions, then refer back to specific steps only when a question requires it.

🔍 Distinguishing O-Chem from other passages

🔍 Visual cues

• Sure sign of an O-Chem passage: Chemical structures, "usually lots of them."
• Even if the passage discusses biological topics, abundant structures indicate you should use the O-Chem reading strategy.

🔍 Strategy contrast

Subject	Text importance	Where to focus
Biology, Biochemistry, Physics, General Chemistry	Text often contains key information	Read text carefully, then examine figures/data
Organic Chemistry	Text rarely needed for questions	Skim text; prioritize structures, schemes, and experimental data

• Don't confuse: An O-Chem passage may look biological because it discusses a biologically important molecule, but your reading approach should still emphasize structures and data over narrative text.

🧭 Overview

🧠 One-sentence thesis

Organic chemistry passages on the MCAT require a different reading strategy—skim the text quickly and focus on structures, schemes, and data, because most passage-based questions can be answered by referencing figures rather than reading dense paragraphs.

📌 Key points (3–5)

• Where the real information lives: chemical structures, synthetic schemes, mechanisms, and experimental data (tables/graphs) contain the useful details, not the text.
• How to read O-Chem passages: skim headings and titles to get the big picture, but don't get bogged down in details until a question directs you to a specific structure or step.
• Passage-based questions often act like freestanding questions: they typically require only a reference to a structure or data point in the passage, not deep comprehension of the text.
• Common confusion: resisting the urge to highlight or analyze every detail during the first read—save detailed analysis for when questions point you to specific figures.
• Mapping strategy: jot down minimal notes (paragraph topics, key definitions) on scratch paper as you answer questions, not before; label passages and record important conclusions to avoid re-doing work.

📊 What to expect from O-Chem on the MCAT

📊 Test distribution and content focus

• Organic chemistry makes up roughly 15% of the Chemical and Physical Foundations section and only about 5% of the Biological and Biochemical Foundations section.
• In Chemical/Physical Foundations: 2–4 freestanding questions plus either one longer passage (6–7 questions) or two short passages (4 questions each).
• In Biological/Biochemical Foundations: 3–4 O-Chem questions appear as freestanding or mixed into biology/biochemistry passages.
• Content spans about two college semesters but focuses most on carbonyl chemistry and laboratory techniques.

🧪 Passage context

• Passages typically present a biologically important compound or reaction as context.
• The sure sign of an O-Chem passage (vs. a Biology passage): chemical structures, usually lots of them.
• Text may include biological concepts, but the structures and data are what matter for answering questions.

🗂️ Types of O-Chem passages

🗂️ Information and/or Situation Presentation (most common)

This type presents one of the following:

What the passage shows	What questions ask
Multistep synthetic scheme, novel reaction, or atypical reaction outcomes	Analyze or classify steps; use lab techniques to analyze intermediates; justify exceptions to rules
A class of biologically important molecules	Analyze with lab techniques; examine structure or relationships; predict reactivity with a given reagent
A biochemical process or mechanism	Explain intermediate stability; predict products from new reactants using mechanistic steps

🔬 Experiment/Research Presentation

• Presents details of an experiment or mechanistic study.
• Often includes spectroscopy data (IR or NMR) in lists or tables.
• Questions ask you to interpret data, identify reaction pathways, identify compounds, or choose appropriate purification/identification techniques.

🧐 Persuasive Reasoning (least common)

• Compares two mechanisms attempting to explain a reaction outcome.
• Questions ask you to evaluate arguments, often relating to stability of intermediates.

📖 How to read an O-Chem passage

📖 Skim, don't read

You should never really read much of the text of an O-Chem passage, but rather, just skim through the text.

• Most important information is in structures and data, not paragraphs.
• Many passage-based questions are essentially freestanding questions that require only a reference to a structure.
• Problem: you can't highlight or mark structures with on-screen tools, so you won't know in advance which details matter.

🎯 What to focus on while skimming

• Read titles and headings of figures and tables to get a general sense of importance.
• Don't get bogged down in figure details—wait until a question directs you to a specific step or structure.
• Look for and highlight:
  • New italicized terms (definitions).
  • Unexpected outcomes of experiments.
  • Exceptions to rules.
• The MCAT asks you to apply fundamentals to novel situations, so flag anything out of the ordinary.

⚠️ Don't confuse

• Don't treat O-Chem passages like Biology passages—Biology passages require careful reading of text; O-Chem passages require quick skimming and focus on visuals.
• Resist the urge to make lots of highlights in the text; the text often just reproduces information already shown in flowcharts or schemes.

🗺️ Mapping an O-Chem passage

🗺️ Minimal mapping before questions

• Your passage map will develop as you answer questions, not before.
• Before jumping into questions, you'll have very little to jot down.
• Focus on structures (flowcharts, reaction schemes, mechanisms) rather than text.

📝 What to write on scratch paper

• Label each passage with a number and an identifying title summarizing the main point.
• Jot down brief paragraph summaries (e.g., "P1 – what dyes are," "P2 – definitions").
• Record important conclusions you reach while answering questions—other questions may need this info, and a note saves time vs. re-confirming your conclusion.
• Keep scratch paper organized so it's useful when checking answers at the end.

📝 Example passage map

For a passage about alkenals from milkweed bugs:

• P1 – alkenals
• P2 – separation and identification of alkenals
• P3 – NMR data

For a passage about dyes:

• P1 – what dyes are and how they work
• P2 – Definitions: mordant dye vs. direct dyes, fiber types dyed
• P3 – Structure requirements for diazocoupling

⚠️ Don't confuse

• Don't try to analyze spectroscopic data (NMR, IR) in detail before a question asks—just note where it is.
• Don't reproduce information from figures into your notes; just note the topic and location.

🎯 Tackling the questions

🎯 Order of attack

1. Answer all freestanding questions first in any science section before reading passages.
   • Memory questions jump-start your brain, earn quick points, and ensure you've answered easy questions before time runs out.
2. Consider doing O-Chem passages before other subjects—O-Chem passage-based questions are some of the most straightforward and quickest to answer, banking extra time for wordier Biology passages.
3. Alternative strategy: start with whichever subject you feel most comfortable with, saving your most difficult subject for last.
4. Within a subject: do all passages in that subject before switching.
5. Within passages: tackle easier questions first, leaving the most time-consuming ones for last.

🧠 O-Chem Memory Questions

These questions can be answered directly from prior knowledge.

• Often recognizable by one- or two-word answer choices—a sign that the answer is already in your head.
• Freestanding questions are typically Memory questions.

⚠️ Don't confuse

• Don't assume all passage-based questions require deep passage analysis—many are essentially freestanding and need only a quick reference to a structure.

🧭 Overview

🧠 One-sentence thesis

Organic Chemistry questions on the MCAT are among the most straightforward and quickest to answer, and success depends on strategic question ordering, recognizing question types (Memory, Explicit, Implicit), and applying fundamental concepts rather than memorizing trivia.

📌 Key points (3–5)

• Strategic ordering: Answer freestanding questions first, do O-Chem passages before wordier Biology passages to bank time, and tackle easier questions within each passage before harder ones.
• Three question types: Memory (direct recall, short answers), Explicit (stated in passage, simple lookup), and Implicit (apply knowledge to new situations, longer answers with explanations).
• Four content categories: Structure, Stability, Laboratory practices, and Predict the product—focus on fundamentals of structure and stability rather than exceptions or trivia.
• Common confusion: Don't pick answers that are merely true statements; the answer must actually answer the question being asked.
• Process of Elimination is paramount: Use strikeout tools, work strategically with Roman numeral and ranking questions, and never leave questions blank.

📋 Question-answering strategy

🎯 Overall approach

• Start with freestanding questions in any science section before reading passages—they are often Memory questions that earn quick points and warm up your brain.
• Do O-Chem passages first (or your strongest subject first) because they are straightforward and fast, banking extra time for wordier Biology passages.
• Within a subject, do passages in order of comfort: leave your most difficult topic or the hardest-looking passage for last.
• Within each passage, tackle easier questions first, saving the most time-consuming ones for last.
• Example: O-Chem passages often have "hidden" freestanding questions associated with them, so get to the questions quickly rather than getting stuck in passage details.

⏱️ Time management

• Answer straightforward questions first; leave questions requiring analysis of experiments and graphs for later.
• Math questions (rare in O-Chem) should be left for last; always round numbers and estimate.
• Never leave a question blank—there is no penalty for guessing.

🧩 Three question types

🧠 Memory questions

Memory questions can be answered directly from prior knowledge.

• How to recognize: One- or two-word answer choices are a good indication.
• Where they appear: Commonly as freestanding questions; also "hidden" within O-Chem passages.
• Approach: You already have the answer in your head; no passage lookup needed.

Example scenario:

• Question: "Which acetylating condition converts diethylamine into an amide at the fastest rate?" with choices like acetic acid/HCl, acetic anhydride, acetyl chloride, ethyl acetate.
• Strategy: Identify the reaction type (nucleophilic addition-elimination), then recall relative reactivities of carboxylic acid derivatives (amide < ester < anhydride < acid halide). Consider that amines are both nucleophilic and basic, so they will be protonated by HCl and acetic acid, making them non-nucleophilic. The acid chloride derivative reacts fastest.

📖 Explicit questions

Explicit questions have answers that are explicitly stated in the passage.

• What they require: Finding a definition, reading a graph, or making a simple connection.
• Common in O-Chem: Often ask you to identify features in chemical structures (e.g., number of chiral centers, presence of functional groups).
• Approach: Go back to the passage to retrieve the highlighted information.

Example scenario:

• Question: "Which biologically important molecule is most likely labeled by a mordant dye?" with choices glycogen, chromatin, cholesterol, starch.
• Strategy: Recognize "mordant" as a highlighted term; passage states mordants bind to protein-based fibers. Eliminate carbohydrates (glycogen, starch) and lipids (cholesterol); chromatin contains proteins, so it's the answer.

🔍 Implicit questions

Implicit questions require you to apply knowledge to a new situation or make a complex connection; the answer is typically implied by the passage.

• How to recognize: Longer answer choices, often in two parts where the second half explains the first.
• Most common type in O-Chem questions.
• Approach: Use Process of Elimination on the first part of each answer; if any part is false, eliminate the entire statement. Then apply fundamental concepts to distinguish remaining choices.

Example scenario:

• Question: "Why is the diazonium coupling reaction faster than most electrophilic substitutions of benzene?"
• Strategy: Eliminate choices with obviously false first parts (e.g., "diazonium ion is a good nucleophile" is false because it's positively charged and electron deficient). Then apply fundamentals: benzene has six π electrons and is electron rich, so it behaves as a nucleophile, not an electrophile. The correct answer must explain benzene's nucleophilic behavior, not its electrophilic behavior.

Don't confuse: A true statement vs. an answer that actually answers the question—always make sure your choice addresses what is being asked.

🗂️ Four content categories

🏗️ Structure

• Questions about functional groups, stereochemistry, isomers, electron density (nucleophiles vs. electrophiles), and nomenclature.
• Focus on recognizing and understanding structural features.

⚖️ Stability

• Questions about stability of products or reaction intermediates.
• Topics include inductive effects, resonance, steric strain, torsional strain, ring strain.
• Key principle: Generalize as much as possible; focus on fundamentals of structure and stability rather than exceptions.

🔬 Laboratory practices

• Identify appropriate separation techniques (extraction, chromatography, distillation) for a given mixture.
• Interpret or predict results of separation procedures.
• Choose appropriate spectroscopic techniques (IR, NMR, mass spec, UV-vis) to identify a compound, or interpret spectroscopic data.

🧪 Predict the product

• Given a starting material and reaction conditions, choose the major product.
• Important: Only one-step synthesis; no multi-step processes will be presented.
• Generally associated with a passage that explains a reaction type in detail, rather than a freestanding question.

🛠️ Ten strategic techniques

🎯 Process of Elimination (POE)

• Most important strategy: Use the strikeout tool to eliminate answer choices; improves guessing odds if you can't narrow to one choice.
• If any part of an answer choice is false, eliminate the entire statement.

🔢 Special question formats

Format	Strategy
I-II-III questions	Work between the Roman numeral statements and answer choices. Once a statement is determined true or false, strike out choices that don't fit. Use scratch paper since you can't strike out Roman numerals.
Ranking questions	Look for an extreme (least or greatest), then eliminate choices. Often you can immediately eliminate two choices. Then examine the other end of the ranking.
2×2 style questions	Answer choices like "A because X, B because X, A because Y, B because Y." Tackle one piece of information at a time to quickly eliminate two choices.
LEAST/EXCEPT/NOT questions	Write T or F next to each choice on scratch paper. The one that stands out as different is correct. Don't get tricked by questions asking for what doesn't fit.

📄 Using the passage

• If you don't know how to answer a question, look to the passage for help.
• The passage likely contains pertinent information, either in text or experimental data.
• Chemical structures are the most common source of referenced information in O-Chem passages.

🧮 Calculations

• Leave math questions for last (rare in O-Chem).
• Always round numbers and estimate on scratch paper.

📚 Study approach

❌ Why flashcards are NOT the best method

• The MCAT is different: Most exams require regurgitation; the MCAT requires applying basic scientific knowledge to unfamiliar situations.
• Flashcards help memorize facts but won't help you apply knowledge to new contexts.
• The most challenging aspect of the MCAT is not memorizing fine details but applying basic knowledge to unfamiliar situations.

✅ When flashcards ARE useful

• Only for deficient basic content knowledge: If you don't know definitions (e.g., types of isomers), functional groups, or basic classifications (e.g., polar vs. nonpolar amino acids).
• Making your own is key: Customized for personal weak areas, and the act of writing helps stick information in your brain.
• Better alternative: Do and analyze practice passages rather than carrying flashcards for most of your study time.

🧭 Overview

🧠 One-sentence thesis

Organic chemistry nomenclature is primarily a memorization task, but mastering the terminology—especially IUPAC names and abbreviated line structures—is essential for reading and applying organic chemistry concepts on the MCAT.

📌 Key points (3–5)

• What this section covers: fundamentals of organic chemistry nomenclature, including IUPAC and some common names, plus abbreviated drawing conventions.
• Study approach: memorization is the primary technique, but the goal is comfort with reading, hearing, and using the terminology.
• Abbreviated line structures: a shorthand drawing system that omits C—H bonds and represents carbons as vertices to simplify large molecules.
• Common confusion: hydrogens bonded to carbon are assumed and not drawn; hydrogens bonded to any other atom (N, O, etc.) must be shown explicitly.
• Why it matters: fluency with nomenclature and structures is necessary for interpreting MCAT organic chemistry problems efficiently.

📚 Study strategy context

📚 When to use flashcards

• Flashcards are useful only for basic fact-memorization in personal weak areas.
• Examples of appropriate flashcard topics:
  • Definitions of isomer types
  • Functional group recognition
  • Amino acid classifications (polar vs. nonpolar, acidic vs. basic)
• Making your own flashcards helps because:
  • They are customized to your weaknesses.
  • The act of writing information helps retention.

🚫 What flashcards won't do

• The MCAT requires applying basic knowledge to unfamiliar situations, not just memorizing details.
• Beyond straight fact-memorization, practice passages and analysis are more effective than flashcards.

🏷️ Basic nomenclature elements

🏷️ Carbon chain prefixes and alkane names

• The excerpt references a table (not fully shown) that lists names for continuous carbon chains.
• These names form the backbone of IUPAC nomenclature.

🔄 Cyclo- prefix

For an all-carbon ring, the prefix cyclo- is added.

• Example: a six-membered ring of −CH₂− units is called cyclohexane.

🧩 Substituent names

The excerpt provides a table of common substituents:

Substituent	Name
−CH₃	methyl
−CH₂CH₃	ethyl
−CH₂CH₂CH₃	propyl
(branched three-carbon)	isopropyl
−CH₂CH₂CH₂CH₃	butyl (or n-butyl)
(branched four-carbon)	sec-butyl
(branched four-carbon)	tert-butyl (or t-butyl)

• These names describe alkyl groups attached to a main chain or functional group.
• The excerpt mentions "Common Functional Groups" but does not provide the table content.

✏️ Abbreviated line structures

✏️ Why use abbreviated structures

• Organic molecules often contain many C—H bonds.
• Drawing every C—H bond for large molecules (steroids, polymers) would be impractical.
• Abbreviated line structures simplify drawing and reading.

📐 Three core rules

1. Carbons are vertices: each bend or endpoint in a line represents a carbon atom.
2. C—H bonds are not drawn: hydrogens attached to carbon are assumed, not shown.
3. Hydrogens on other atoms must be shown: if H is bonded to N, O, S, etc., it must be drawn explicitly.

🔍 How to interpret vertices and assumed hydrogens

• Each carbon must have four bonds (carbon's valency).
• Count the bonds shown to other atoms; the remaining bonds are assumed to be C—H.
• Example from the excerpt (pentane):
  • Three middle carbons are each bonded to two other carbons → assume two hydrogens on each.
  • Two end carbons are each bonded to one other carbon → assume three hydrogens on each.
• Drawing out all assumed hydrogens recovers the full Lewis structure.

🧪 Hydrogens on heteroatoms

• Example: dimethyl amine (two methyl groups attached to nitrogen).
  • Hydrogens on the methyl carbons are assumed (not drawn).
  • The hydrogen on nitrogen must be shown explicitly.
• Lone pairs of electrons are often omitted in abbreviated structures.

🔄 Don't confuse

• C—H hydrogens: assumed, not drawn.
• Heteroatom hydrogens (N—H, O—H, etc.): must be drawn.
• Forgetting this distinction leads to incorrect structure interpretation.

🧮 Nomenclature of alkanes

🧮 General approach

• Alkanes are named by a set of simple rules.
• The excerpt begins to illustrate the process with a specific example but is cut off.

🔢 Step 1: Identify the longest continuous carbon chain

• The longest chain determines the base name (from the "Carbon Chain Prefixes and Alkane Names" table).
• This is the first step in systematic IUPAC naming.
• The excerpt does not provide further steps, but this foundation is critical for building more complex names.

🧭 Overview

🧠 One-sentence thesis

Abbreviated line structures simplify organic molecule drawings by representing carbons as vertices and omitting C—H bonds, while hydrogens bonded to non-carbon atoms must always be shown.

📌 Key points (3–5)

• Why abbreviated structures exist: drawing every C—H bond in large molecules (steroids, polymers) would be impractical, so chemists use a shorthand system.
• Three core rules: carbons are vertices; C—H bonds are not drawn; hydrogens on non-carbon atoms must be shown.
• Assumed hydrogens: the number of hydrogens needed to complete carbon's valency (four bonds total) is automatically assumed even though not drawn.
• Common confusion: hydrogens bonded to carbon can be omitted, but hydrogens bonded to nitrogen, oxygen, or other heteroatoms must be explicitly drawn.
• Lone pairs: often omitted in abbreviated structures.

📐 The three drawing rules

📐 Rule 1: Carbons as vertices

• In abbreviated line structures, each carbon atom is represented simply as a corner or endpoint (a vertex) in the drawing.
• You do not write "C" at each position; the vertex itself stands for carbon.
• Example: a chain of five carbons (pentane) appears as a zigzag line with four bends (five vertices total).

📐 Rule 2: C—H bonds are not drawn

• Bonds between carbon and hydrogen are omitted entirely from the drawing.
• The hydrogen atoms themselves are also not written out.
• This rule applies only to hydrogens attached to carbon.
• Example: pentane's full Lewis structure shows all C—H bonds explicitly, but the abbreviated structure shows only the carbon backbone.

📐 Rule 3: Hydrogens on non-carbon atoms must be shown

• If a hydrogen is bonded to any atom other than carbon (e.g., nitrogen, oxygen, sulfur), you must draw it explicitly.
• Example: dimethyl amine has two methyl groups (C—H bonds omitted) but the N—H bond must be shown.
• Don't confuse: hydrogens on carbon = assumed and hidden; hydrogens on heteroatoms = must be visible.

🧮 How to infer assumed hydrogens

🧮 Carbon's valency requirement

• Carbon always forms four bonds total.
• When you see a carbon vertex bonded to fewer than four other atoms in the drawing, the "missing" bonds are assumed to be C—H bonds.
• Example in pentane:
  • The three middle carbons are each bonded to two other carbons → two bonds used → two hydrogens assumed on each.
  • The two end carbons are each bonded to one other carbon → one bond used → three hydrogens assumed on each.

🧮 Verifying the structure

• If you add back all the assumed hydrogens, you should recover the full Lewis structure.
• The excerpt demonstrates this by showing that the abbreviated pentane structure, when expanded, matches the full Lewis structure with all C—H bonds drawn.

🔬 Special cases and conventions

🔬 Lone pairs of electrons

• Lone pairs (non-bonding electron pairs on atoms like nitrogen or oxygen) are often omitted in abbreviated structures.
• This is a convention for simplicity; the lone pairs are still understood to be present.

🔬 Functional groups and heteroatoms

• When a molecule contains atoms other than carbon (e.g., nitrogen in dimethyl amine), those atoms and their attached hydrogens must be shown.
• Example: dimethyl amine's abbreviated structure shows "NH" explicitly, while the two methyl groups attached to nitrogen are drawn as simple lines (C—H bonds assumed).

🔄 Translating between full and abbreviated structures

🔄 From Lewis to abbreviated

• Identify all carbon atoms and replace them with vertices (corners or endpoints).
• Remove all C—H bonds and hydrogen labels on carbon.
• Keep all heteroatoms (N, O, S, etc.) and their attached hydrogens visible.
• Example: the excerpt provides four Lewis structures (a–d) and their corresponding abbreviated forms.

🔄 From abbreviated to Lewis

• Place a carbon atom at each vertex.
• For each carbon, count how many bonds are already drawn to other atoms.
• Add enough hydrogens to bring the total to four bonds per carbon.
• Restore any lone pairs if needed (though often omitted in abbreviated form).
• Example: the excerpt provides four abbreviated structures (a–d) and their corresponding full Lewis structures.

Direction	Key step	Watch out for
Lewis → Abbreviated	Remove all C—H bonds and H labels on carbon	Keep H on heteroatoms visible
Abbreviated → Lewis	Add hydrogens to complete carbon's four bonds	Don't add extra H to heteroatoms unless shown

🧭 Overview

🧠 One-sentence thesis

Alkanes are named systematically by identifying the longest carbon chain, numbering it to give substituents the lowest numbers, and alphabetizing substituents with appropriate prefixes.

📌 Key points (3–5)

• Core process: find the longest continuous carbon chain, identify substituents, number the chain to minimize substituent positions, then alphabetize.
• Numbering rule: assign carbon numbers so that substituents appear on the lowest-numbered carbons possible.
• Grouping identical substituents: use prefixes di-, tri-, tetra-, penta- and separate their positions with commas.
• Common confusion: when alphabetizing, ignore prefixes like di-, tri-, n-, sec-, tert-, but do not ignore "iso" because it is part of the substituent name itself.
• Extension to haloalkanes: halogen substituents (F, Cl, Br, I) follow the same rules, using prefixes fluoro-, chloro-, bromo-.

📐 The five-step naming process

📐 Step 1: Identify the longest continuous carbon chain

The longest continuous carbon chain determines the base name of the alkane, using the prefixes given in the carbon chain table (e.g., 7 carbons = heptane, 10 carbons = decane).

• "Continuous" means the chain does not have to be drawn in a straight line; it can zigzag.
• Example: a molecule with a longest chain of 7 carbons is based on heptane.
• This chain forms the backbone; everything else is a substituent.

🔢 Step 2: Identify substituents

• Substituents are groups attached to the main chain (e.g., methyl, ethyl, isopropyl).
• The excerpt lists common hydrocarbon substituents in a table (not fully reproduced here, but examples include methyl, isopropyl, ethyl, sec-butyl).
• Halogens (F, Cl, Br, I) are also substituents, named with prefixes fluoro-, chloro-, bromo-.

🔢 Step 3: Number the main chain

• Number the carbons of the main chain so that substituents get the lowest possible numbers.
• Example: if numbering from one end gives substituents at positions 2, 3, 3, 4 and numbering from the other end gives higher numbers, choose the first direction.
• Each substituent is then associated with its carbon number (e.g., 2-methyl, 4-isopropyl).

🔢 Step 4: Group identical substituents

• When the same substituent appears multiple times, use a prefix:
  • di- for two
  • tri- for three
  • tetra- for four
  • penta- for five
• List all their positions separated by commas.
• Example: three methyl groups at positions 2, 3, and 3 become 2,3,3-trimethyl.

🔢 Step 5: Alphabetize and assemble the name

• Alphabetize substituents by their base name, ignoring the prefixes di-, tri-, tetra-, penta-, n-, sec-, tert-.
• Do not ignore "iso" because it is part of the substituent name.
• Separate numbers from words with a hyphen; separate numbers from numbers with a comma.
• Example: 4-isopropyl-2,3,3-trimethylheptane (isopropyl comes before trimethyl alphabetically, even though "t" comes after "i"; the "tri-" is ignored).

🧪 Worked examples from the excerpt

🧪 Example: 4-isopropyl-2,3,3-trimethylheptane

1. Longest chain: 7 carbons → heptane.
2. Substituents: three methyl groups and one isopropyl group.
3. Numbering gives positions 2, 3, 3, 4.
4. Group methyls: 2,3,3-trimethyl.
5. Alphabetize: isopropyl before trimethyl → 4-isopropyl-2,3,3-trimethylheptane.

🧪 Example: 3,5-diethyl-6-methyldecane

1. Longest chain: 10 carbons → decane.
2. Substituents: two ethyl groups and one methyl group.
3. Numbering gives ethyls at 3 and 5, methyl at 6.
4. Group ethyls: 3,5-diethyl.
5. Alphabetize: diethyl before methyl (ignore "di-") → 3,5-diethyl-6-methyldecane.

🧪 Additional examples from the excerpt

The excerpt provides five practice problems with solutions:

• (a) 2,3-dimethylbutane
• (b) 2,3-dimethylpentane
• (c) 4-isopropyl-4-methylheptane
• (d) 5-sec-butyl-2,7,7-trimethylnonane (note: sec- is ignored in alphabetizing)
• (e) 3-ethyl-5,5-dimethyloctane

🧂 Extension: Haloalkanes

🧂 Naming halogen substituents

• Haloalkanes are alkanes with halogen substituents (F, Cl, Br, I).
• They follow the same five-step rules as simple alkanes.
• Halogen prefixes:

Halogen	Prefix
Fluorine	fluoro-
Chlorine	chloro-
Bromine	bromo-

• Example: a chlorine at position 2 on a pentane chain would contribute 2-chloro to the name.
• Alphabetize halogens with other substituents (e.g., bromo- comes before methyl).

🧂 Don't confuse: "iso" vs other prefixes

• Ignore when alphabetizing: di-, tri-, tetra-, penta-, n-, sec-, tert-.
• Do not ignore: "iso" (it is part of the substituent name, e.g., isopropyl).
• Example: 4-isopropyl-2,3,3-trimethylheptane lists isopropyl first because "i" comes before "m" (the "tri-" in trimethyl is ignored).

🧭 Overview

🧠 One-sentence thesis

Haloalkanes follow the same naming rules as simple alkanes, with halogens treated as substituents using specific prefixes (fluoro-, chloro-, bromo-, iodo-) that are alphabetized alongside other substituents.

📌 Key points (3–5)

• Core principle: Haloalkanes use identical nomenclature rules to simple alkanes; halogens are simply another type of substituent.
• Halogen prefixes: Each halogen has a specific prefix—fluorine → fluoro-, chlorine → chloro-, bromine → bromo-, iodine → iodo-.
• Alphabetization rule: Halogen prefixes are alphabetized with other substituents; numerical prefixes (di-, tri-, etc.) are ignored when alphabetizing.
• Common confusion: Unlike "iso" (which is part of the substituent name and counts in alphabetization), prefixes like di-, tri-, sec-, tert-, and n- are ignored.
• Numbering priority: The main chain is numbered to give substituents (including halogens) the lowest possible numbers, following the same logic as for alkanes.

🧪 Core naming framework

🔗 The five-step process

The excerpt demonstrates that haloalkane naming follows the same systematic approach as alkane naming:

1. Identify the longest continuous carbon chain (the parent alkane).
2. Identify all substituents on that chain (including halogens).
3. Number the carbons so substituents get the lowest numbers.
4. Group identical substituents using di-, tri-, tetra-, penta- prefixes with comma-separated position numbers.
5. Alphabetize substituents (ignoring numerical prefixes) and assemble the name with hyphens between numbers and words, commas between numbers.

🧩 What makes haloalkanes different

Haloalkanes: alkanes with halogen (F, Cl, Br, I) substituents.

• The only new element is the halogen substituent itself.
• Halogens are not treated as a special functional group requiring a suffix; they are simply named as substituents with their own prefixes.
• Example: A chlorine atom attached to carbon 2 of butane → "2-chlorobutane."

🔤 Halogen naming conventions

🏷️ The four halogen prefixes

Halogen	Prefix
Fluorine	fluoro-
Chlorine	chloro-
Bromine	bromo-
Iodine	iodo-

• These prefixes are used exactly like hydrocarbon substituent names (methyl, ethyl, isopropyl, etc.).
• Multiple identical halogens are grouped: two bromines → dibromo-, three chlorines → trichloro-.

📝 Verified examples from the excerpt

The excerpt provides three structures with names to verify:

• 2-chlorobutane: A 4-carbon chain with one chlorine on carbon 2.
• 2-chloro-1-fluoro-4-methylpentane: A 5-carbon chain with chlorine at 2, fluorine at 1, and a methyl group at 4.
• 2,2-dibromo-3-iodo-4-methylhexane: A 6-carbon chain with two bromines at carbon 2, one iodine at 3, and one methyl at 4.

Example: If you have a 4-carbon chain with a chlorine on the second carbon, the name is "2-chlorobutane" (not "chloro-2-butane").

🔢 Alphabetization and numbering rules

🔤 What to ignore when alphabetizing

• Ignore: di-, tri-, tetra-, penta- (numerical multipliers).
• Ignore: n-, sec-, tert- (structural descriptors).
• Do NOT ignore: "iso" (it is part of the substituent name, not a prefix).

Example: In "4-isopropyl-2,3,3-trimethylheptane," "isopropyl" comes before "methyl" alphabetically because "iso" is part of the name, while "tri-" is ignored.

🔢 Numbering to minimize position numbers

• Number the main chain so that the first substituent gets the lowest possible number.
• If there is a tie, the next substituent should have the lowest number, and so on.
• Halogens and hydrocarbon substituents are treated equally in this numbering priority.

Don't confuse: The chain direction is chosen to minimize substituent numbers overall, not to favor any particular type of substituent (halogen vs. hydrocarbon).

📚 Worked examples

🧪 Example structures and names

The excerpt provides several practice problems with solutions:

Example 3-4 solutions:

• (a) 1,1,1-tribromo-2,2-dimethylpropane: Three bromines on carbon 1, two methyls on carbon 2, 3-carbon chain.
• (b) 2-fluoro-2,3-dimethylpentane: One fluorine and one methyl on carbon 2, one methyl on carbon 3, 5-carbon chain.
• (c) 2,3,4,4-tetrachloro-3-isopropylhexane: Four chlorines (at positions 2, 3, 4, 4) and one isopropyl at 3, 6-carbon chain.
• (d) 4-bromo-3-ethyl-4-fluoro-3-iodo-2,2-dimethylhexane: Multiple halogens (bromo, fluoro, iodo) and hydrocarbon substituents (ethyl, dimethyl) on a 6-carbon chain.

✍️ Drawing from names (Example 3-5)

The excerpt also shows the reverse process: given a name, draw the structure.

• (a) 3-chloro-2,2-dimethylbutane: Start with a 4-carbon chain, place a chlorine on carbon 3, and two methyls on carbon 2.
• (c) 2,3-dibromo-1,1-diiodopropane: A 3-carbon chain with two bromines (at 2 and 3) and two iodines (both at 1).

Example: "2,3-dibromo-1,1-diiodopropane" means a propane (3 carbons) with bromines at positions 2 and 3, and two iodines both at position 1.

🔗 Connection to broader nomenclature

🧬 Consistency across functional groups

• The excerpt notes that alcohols (covered in section 3.5) "also follow many of the same nomenclature rules as alkanes."
• Haloalkanes are an intermediate step: they add substituents (halogens) but do not change the parent alkane suffix.
• This consistency means mastering the alkane naming framework (longest chain, numbering, alphabetization) applies across many organic compound types.

🧠 Why this matters

• Systematic naming allows chemists to communicate structures unambiguously.
• The same five-step process works whether substituents are hydrocarbon groups, halogens, or (as hinted) other functional groups.
• Don't confuse: Halogens are substituents (prefixes), while some other functional groups (like alcohols) use suffixes that replace part of the parent alkane name.

🧭 Overview

🧠 One-sentence thesis

Alcohols follow alkane naming rules but use the suffix −ol to denote the hydroxyl group, with numbering prioritized to give the hydroxyl group the lowest possible position number.

📌 Key points (3–5)

• Suffix rule: The −ol suffix replaces the final −e in the corresponding alkane name to indicate a hydroxyl group (−OH).
• Position numbering: When the hydroxyl position must be specified, the number is placed after the main chain name and before −ol, separated by hyphens.
• Numbering priority: The carbon chain is numbered to give the hydroxyl group the lowest possible number.
• Common confusion: If no number is given before −ol, the position "−1−" is assumed (e.g., "pentanol" means "pentan-1-ol").
• Multiple hydroxyl groups: Use suffixes like −diol for two hydroxyl groups, with position numbers for each.

🏗️ Basic naming pattern

🏗️ How the suffix works

The suffix −ol replaces the last −e in the name of the alkane to denote a hydroxyl group (−OH).

• Start with the alkane name (methane, ethane, propane, butane, etc.).
• Drop the final −e and add −ol.
• Example: methane → methanol; ethane → ethanol; propane → propanol; butane → butanol.

Comparison table:

Alkane	Structure	Alcohol	Structure
methane	CH₄	methanol	CH₃OH
ethane	CH₃CH₃	ethanol	CH₃CH₂OH
propane	CH₃CH₂CH₃	propanol	CH₃CH₂CH₂OH
butane	CH₃CH₂CH₂CH₃	butanol	CH₃CH₂CH₂CH₂OH

🔢 When position matters

• For straight-chain alcohols with the hydroxyl at the end (position 1), no number is needed in simple cases.
• When the hydroxyl group is not at position 1, or when ambiguity exists, specify the position number.
• Format: place the number after the main chain name and before −ol, separated by hyphens.
• Example: "hexan-2-ol" means the hydroxyl is on carbon 2 of a six-carbon chain.

🎯 Numbering priority rules

🎯 Lowest number for hydroxyl

• The main carbon chain is numbered to give the hydroxyl group the lowest possible number.
• This is the priority rule: hydroxyl position takes precedence over other substituents when deciding which end of the chain to start numbering from.
• Example: If a hydroxyl could be at position 2 or position 5 depending on which end you count from, number so it is at position 2.

🔄 Assumed position

• Don't confuse: When no number is given (e.g., "pentanol"), the position "−1−" is assumed.
• This means the hydroxyl is on carbon 1.
• Example: "4,4-dichloro-2-methylpentanol" is understood as "4,4-dichloro-2-methylpentan-1-ol."

🔗 Multiple hydroxyl groups and complex cases

🔗 Multiple −OH groups

• When two hydroxyl groups are present, use the suffix −diol.
• Specify the position of each hydroxyl group with numbers.
• Example: "propan-1,2-diol" (or "1,2-propandiol") means hydroxyl groups on carbons 1 and 2 of propane.
• The excerpt shows both formats are acceptable: "propan-1,2-diol" and "1,2-propandiol."

🧩 Combining with other substituents

• Other functional groups (halogens, alkyl groups) are named as prefixes, following the same rules as for alkanes.
• The hydroxyl group still determines the main chain numbering priority.
• Example: "2-chloro-2-fluoro-3-methylbutan-1,1-diol" has:
  • A four-carbon main chain (butan-)
  • Two hydroxyl groups both on carbon 1 (−1,1-diol)
  • Chloro and fluoro on carbon 2
  • Methyl on carbon 3

📋 Example interpretations from the excerpt

• "6-chloro-4-ethylhexan-2-ol": six-carbon chain, hydroxyl on carbon 2, chloro on carbon 6, ethyl on carbon 4.
• "4,4-dichloro-2-methylpentanol": five-carbon chain, hydroxyl on carbon 1 (assumed), two chloro groups on carbon 4, methyl on carbon 2.

🔍 Connection to broader nomenclature

🔍 Consistency with alkane rules

• Alcohols follow "many of the same nomenclature rules as alkanes."
• The main difference is the suffix change (−e → −ol) and the priority given to the hydroxyl group in numbering.
• Other substituents (halogens, alkyl groups) are still named and numbered as prefixes.

🔍 Scope note

• The excerpt notes that "other organic functional groups have small nuances to their nomenclature."
• This introduction to alcohol nomenclature is intended to allow interpretation of chemical names on the MCAT.
• The principles here (suffix for functional group, numbering priority, position specification) extend to other functional groups with minor variations.

🧭 Overview

🧠 One-sentence thesis

Understanding the stability of organic molecules and their intermediates—through saturation, hybridization, inductive effects, resonance, and ring strain—allows chemists to predict reactivity, acidity, and how reactions will proceed.

📌 Key points (3–5)

• Saturation vs. unsaturation: Saturated molecules have no π bonds or rings; unsaturated molecules have at least one π bond or ring, and the degree of unsaturation formula reveals how many.
• Stability determines reactivity: More stable molecules are less reactive; higher-energy species are more reactive—this inverse relationship is a central theme in organic chemistry.
• Two major stabilization mechanisms: Inductive effects stabilize charge through σ bonds; resonance effects stabilize charge by delocalizing electrons through π bonds.
• Common confusion—hybridization and resonance: An atom that appears sp³ (four σ bonds/lone pairs) cannot participate in resonance; only atoms with p orbitals (sp² or sp) can delocalize electrons into adjacent π systems.
• Acidity follows stability: The strength of an acid depends on how well the conjugate base stabilizes its negative charge—via electronegativity, resonance, or inductive effects.

🔢 Degree of Unsaturation

🔢 What saturation means

Saturated: an organic molecule with no π bonds and no rings.
Unsaturated: a molecule with at least one π bond or a ring.

• A saturated compound with n carbons has exactly 2n + 2 hydrogens.
• An unsaturated compound with n carbons has fewer than 2n + 2 hydrogens.

🧮 The degree of unsaturation formula

The formula is:
d = [(2 × number of carbons + 2) − x] / 2

• x = number of hydrogens plus any monovalent atoms (F, Cl, Br, I).
• Oxygen is ignored (divalent atoms don't affect the count).
• For nitrogen: replace each N with 1 C and 1 H before calculating.

What the number means:

• d = 1 → one π bond or one ring.
• d = 2 → two π bonds (two separate double bonds or one triple bond), or one π bond + one ring, or two rings.
• d = 0 → the molecule is saturated.

🧪 How heteroatoms affect the calculation

Atom type	How to treat it	Example
Halogen (F, Cl, Br, I)	Count as one hydrogen	C₄H₇Cl: d = [(2×4+2)−(7+1)]/2 = 1
Oxygen (or divalent)	Ignore it	C₃H₆O: d = [(2×3+2)−6]/2 = 1
Nitrogen (or trivalent)	Replace with 1 C + 1 H	C₃H₇N → treat as C₄H₈: d = 1

Example: C₆H₈ has d = 3; C₃H₈O has d = 0 (saturated).

🔬 Hybridization

🔬 How to determine hybridization

Hybridization: the mixing of atomic orbitals (s, p, d, f) to house electron pairs around an atom.

Counting rule:
Count all electron pairs (bonding + lone pairs) around the atom, then assign orbitals in order: s, p, p, p, d, d, …

• 4 pairs → s + p + p + p = sp³ (e.g., methane CH₄).
• 3 pairs → s + p + p = sp².
• 2 pairs → s + p = sp.

📐 Geometry and bond angles

Hybridization	Number of orbitals	Ideal bond angle	Geometry
sp³	4	109°	Tetrahedral
sp²	3	120°	Trigonal planar
sp	2	180°	Linear

Don't confuse: If a lone pair is delocalized into a π system (e.g., nitrogen in aniline), the atom may be sp² rather than sp³, even though it "looks" like it has four pairs.

⚡ Reaction Intermediates

⚡ Carbocations (carbonium ions)

Carbocation: a positively charged species with a full positive charge on carbon.

• On the MCAT, carbocations are always sp² hybridized with an empty p orbital.
• Stability determines reactivity: more stable carbocations are less reactive.

⚡ Carbanions

Carbanion: a negatively charged species with a full negative charge on carbon.

• Stability also determines reactivity.
• Generally less stable than carbocations because carbon is not very electronegative.

🔑 The stability–reactivity inverse relationship

Core principle: Reactivity is inversely related to stability.

• More stable → less reactive.
• Higher energy → more reactive.

This theme recurs throughout organic chemistry and helps predict reaction pathways.

🔌 Inductive Effects

🔌 What inductive effects are

Inductive effect: stabilization of charge by sharing electrons through σ bonds.

• Electron-withdrawing groups (EWGs): pull electrons toward themselves through σ bonds; more electronegative than carbon.
• Electron-donating groups (EDGs): push electrons away through σ bonds; less electronegative than carbon.
• On the MCAT, alkyl substituents are always electron-donating.

🔌 How they stabilize intermediates

• EDGs stabilize electron-deficient intermediates (carbocations) by donating electron density.
• EWGs stabilize electron-rich intermediates (carbanions) by withdrawing electron density.

🔌 Distance and strength matter

• Distance: Inductive effects decrease with distance from the charge.
• Strength: More electronegative substituents have stronger inductive effects.

Example: Trichloroacetic acid (pKₐ = 0.6) is much more acidic than acetic acid (pKₐ = 4.8) because the three Cl atoms withdraw electrons, weakening the O–H bond and stabilizing the conjugate base (trichloroacetate anion).

🌀 Resonance Stabilization

🌀 What resonance is

Conjugated system: three or more atoms each bearing a p orbital, aligned parallel, allowing electron delocalization.
Localized electrons: confined to one orbital.
Delocalized electrons: spread over multiple atoms via overlapping p orbitals.

• Resonance occurs when electrons can interact with orbitals on adjacent atoms.
• Multiple Lewis structures (resonance structures) represent the same molecule; the true structure is a resonance hybrid (average of all contributors).

🌀 Key examples

Allyl cation:
The π bond electrons interact with the empty p orbital on the positively charged carbon, delocalizing both electrons and positive charge over three carbons.

Benzene (C₆H₆):
All six carbons are sp² hybridized with parallel p orbitals. Two equivalent resonance structures show π electrons delocalized over the entire ring, resulting in identical C–C bond lengths.

Thiophene:
One lone pair on sulfur (in a p orbital) delocalizes into the cyclic π system, imparting aromatic stability. The sulfur is sp² hybridized.

Aniline:
The nitrogen lone pair can delocalize into the benzene ring, though this disrupts aromaticity. The nitrogen hybridization is intermediate between sp² and sp³.

🌀 Three principles of resonance

1. No resonance through sp³ atoms: An atom with four σ bonds/lone pairs (sp³) cannot participate in resonance.
2. Electrons must be adjacent to a π bond or p orbital: Resonance requires electrons one atom away from a π system or empty p orbital.
3. Lowest-energy structures are most important:
   • a. Octet rule satisfied for all atoms (highest priority).
   • b. Minimize charge separation (formal charge).
   • c. Negative charge on more electronegative atoms; positive charge on less electronegative atoms.

Don't confuse: Resonance structures are not different molecules; they are multiple representations of one real structure (the hybrid).

🍋 Acidity

🍋 What makes an acid strong

Brønsted-Lowry acid: a molecule that donates a proton (H⁺), forming a conjugate base (usually negatively charged).

• Acid strength = degree of dissociation in solution.
• Strong acids dissociate completely; weak acids (all MCAT organic acids) do not.
• Key principle: Acid strength is determined by how well the conjugate base stabilizes its negative charge.

🍋 Electronegativity effects

More electronegative atoms stabilize negative charge better.

Example: Alcohols (R–OH) are more acidic than alkanes (R–H) because the alkoxide ion (R–O⁻) is more stable than the carbanion (R–C⁻)—oxygen is more electronegative than carbon.

🍋 Resonance effects on acidity

Carboxylic acids vs. alcohols:
Carboxylic acids (R–COOH) are much more acidic than alcohols because the carboxylate ion (R–COO⁻) has two equivalent resonance structures with the negative charge delocalized over two oxygens. The alkoxide ion (R–O⁻) has no resonance, so the charge is localized and less stable.

Phenols:
Phenols (benzene ring with –OH) are more acidic than alkyl alcohols because the phenoxide ion can delocalize the negative charge into the benzene ring (four resonance structures). However, phenols are less acidic than carboxylic acids because three of the four resonance structures place the negative charge on carbon (less electronegative).

🍋 General acidity ranking (MCAT)

From most acidic to least acidic:

Functional group	Approximate pKₐ range
Carboxylic acids	~3–5
Phenols	~10
Alcohols	~15–18
Alkynes (terminal)	~25
Alkanes	~50

🍋 Inductive effects on acidity

Electron-withdrawing groups increase acidity by stabilizing the conjugate base.

• Distance matters: The closer the EWG to the acidic proton, the greater the effect.
• Strength matters: More electronegative substituents have stronger effects.

Example order (increasing acidity):
Acetic acid (pKₐ 4.8) < Chloroacetic acid < Dichloroacetic acid < Trichloroacetic acid (pKₐ 0.6).

Example order for fluorobutanoic acid isomers:
4-fluorobutanoic acid < 3-fluorobutanoic acid < 2-fluorobutanoic acid (F closest to COOH is most acidic).

🍋 Substituent effects on phenols

• Electron-withdrawing groups (e.g., –NO₂) increase phenol acidity by stabilizing the phenoxide ion via resonance.
• Electron-donating groups (e.g., –OCH₃, –NH₂) decrease phenol acidity by destabilizing the phenoxide ion (donate electrons, placing negative charge on adjacent carbon).

Example: para-Nitrophenol is more acidic than phenol; para-methoxyphenol is less acidic than phenol.

🎯 Nucleophiles and Electrophiles

🎯 Nucleophiles

Nucleophile: a species with unshared electron pairs or π bonds, often with a negative or partial negative charge; "nucleus-seeking" or "nucleus-loving."

• Nucleophiles are Lewis bases (electron pair donors).
• Examples: OH⁻, NH₃, Cl⁻, H₂O, alkenes.

🎯 Nucleophilicity trends

Nucleophilicity: a measure of how strong a nucleophile is.

1. Nucleophilicity increases as negative charge increases: NH₂⁻ > NH₃.
2. Nucleophilicity increases going down a group (related to polarizability—larger atoms are more polarizable): F⁻ < Cl⁻ < Br⁻ < I⁻.
3. Nucleophilicity increases going left across a period (related to electronegativity—less electronegative atoms hold charge less well): NH₂⁻ > OH⁻.

Don't confuse: Trend #2 applies within a column; Trend #3 applies across a row.

🎯 Electrophiles

Electrophile: an electron-deficient species with a full or partial positive charge; "electron-loving."

• Electrophiles are Lewis acids (electron pair acceptors).
• Often have an incomplete octet.
• Examples: H⁺, carbocations, carbonyl carbons (C=O), alkyl halides.

🎯 How they react

In most organic reactions (except free-radical and pericyclic), nucleophiles donate an electron pair to electrophiles, forming a new covalent bond:

Nucleophile + Electrophile → New bond

🚪 Leaving Groups

🚪 What makes a good leaving group

Leaving group: a group that dissociates from a substrate during a reaction.

Core principle: Good leaving groups are stable in solution after they leave.

• Resonance-stabilized groups are excellent leaving groups (e.g., tosylate, mesylate, acetate).
• Weak bases are good leaving groups because their negative charge is stabilized by large size (e.g., I⁻ > Br⁻ > Cl⁻).
• Strong bases (HO⁻, RO⁻, NH₂⁻) are poor leaving groups because they cannot stabilize negative charge well and are very reactive.

🚪 Improving bad leaving groups

Strong bases can be converted into better leaving groups by protonation.

Example: The –OH group of an alcohol is a poor leaving group (would leave as OH⁻). Treating with acid protonates the oxygen to –OH₂⁺, which can leave as neutral H₂O (no negative charge to stabilize). This is why many organic reactions are acid-catalyzed.

🔗 Ring Strain

🔗 What ring strain is

Ring strain: instability arising when bond angles between ring atoms deviate from the ideal angle predicted by hybridization.

• Unlike inductive and resonance effects, ring strain destabilizes molecules.
• Strained rings are higher energy and more reactive.

🔗 Examples of strained rings

Cyclopropane (C₃H₆):

• Bond angles approach 60°, far from the ideal 109° for sp³ carbons.
• Very high strain; weakens C–C bonds.
• Unusually reactive for an alkane—can undergo hydrogenation (C–C bond cleavage).

Cyclobutane (C₄H₈):

• Bond angles ~88° (one carbon bent out of plane to minimize eclipsing).
• Still strained; also more reactive than typical alkanes.
• Can undergo hydrogenation.

Cyclopentane and cyclohexane:

• Low or no ring strain (bond angles near 109°).
• React like straight-chain alkanes; do not undergo hydrogenation under normal conditions.

Don't confuse: While C=C double bonds readily undergo hydrogenation, typical alkane C–C single bonds do not—except in highly strained small rings like cyclopropane and cyclobutane.

🧭 Overview

🧠 One-sentence thesis

Isomerism describes how molecules with the same molecular formula can differ in connectivity (constitutional isomers), rotation around bonds (conformational isomers), or spatial arrangement of atoms (stereoisomers), with stereoisomers being especially important in biological chemistry because they include chiral molecules that rotate plane-polarized light and exist as non-superimposable mirror images.

📌 Key points (3–5)

• Constitutional isomers have the same molecular formula but different atom connectivity; conformational isomers have the same connectivity but differ by rotation around σ bonds; stereoisomers have the same connectivity but different spatial arrangements that cannot be interconverted by rotation.
• Chirality arises when a carbon has four different groups (a chiral center), making the molecule non-superimposable on its mirror image; such molecules rotate plane-polarized light.
• Enantiomers are non-superimposable mirror images with opposite absolute configurations at all chiral centers; they rotate light equally but in opposite directions, so a 50:50 mixture (racemic mixture) is optically inactive.
• Diastereomers are stereoisomers that are not mirror images (at least one, but not all, chiral centers differ); unlike enantiomers, diastereomers have different physical properties with no predictable relationship between their optical rotations.
• Common confusion: meso compounds have chiral centers and an internal mirror plane, so they are achiral and optically inactive despite having stereocenters; epimers are diastereomers differing at only one chiral center; geometric isomers (cis/trans or Z/E) are diastereomers around rings or double bonds.

🔗 Constitutional Isomerism

🔗 Definition and connectivity

Constitutional (structural) isomers: compounds that have the same molecular formula but have their atoms connected together differently.

• The key is connectivity: the order in which atoms are bonded.
• Example: pentane (C₅H₁₂) has three constitutional isomers—n-pentane, isopentane, and neopentane—all with the same formula but different carbon skeletons.
• Constitutional isomers are fundamentally different molecules with different chemical and physical properties.

🔗 How to identify

• Write the molecular formula.
• Check if atoms are bonded in the same sequence; if not, they are constitutional isomers.
• Example: hexane (C₆H₁₄) has five constitutional isomers, each with a different branching pattern.

🔄 Conformational Isomerism

🔄 Definition and rotation

Conformational isomers: compounds that have the same molecular formula and the same atomic connectivity, but differ from one another by rotation about a σ bond. In truth, they are the exact same molecule.

• Conformational isomers are not truly different molecules; they are different shapes of the same molecule.
• Rotation around single (σ) bonds is usually free, so conformations interconvert rapidly at room temperature.

🔄 Staggered vs eclipsed conformations

• Staggered conformation: a σ bond on one carbon bisects the angle formed by two σ bonds on the adjacent carbon; bonds are as far apart as possible.
• Eclipsed conformation: a σ bond on one carbon directly lines up with a σ bond on an adjacent carbon; bonds are aligned.
• Staggered is more stable than eclipsed for two reasons:
  • Electronic repulsion: covalent bonds (made of electrons) repel each other; staggered keeps them farther apart.
  • Steric hindrance: atoms attached to the bonds are roomier (60° apart) in staggered, crowded (0° apart) in eclipsed.

🔄 Visualizing conformations: Newman projections

• Flagged bond notation and Newman projection are two ways to represent conformations.
• In a Newman projection, you look down a C–C bond; the front carbon is a dot, the back carbon is a circle.
• Example: ethane (C₂H₆) can be drawn in staggered or eclipsed form using either notation.

🔄 Anti, gauche, and syn conformations (n-butane example)

• Anti conformation: the two largest groups on

🧭 Overview

🧠 One-sentence thesis

Separation techniques exploit differences in physical and chemical properties—such as solubility, polarity, charge, size, and volatility—to isolate individual compounds from complex mixtures.

📌 Key points (3–5)

• Extraction principle: compounds distribute between two immiscible solvents based on their relative solubilities (partition coefficient).
• Acid-base extraction strategy: adjusting pH converts organic compounds (amines, carboxylic acids, phenols) into charged salts that become water-soluble and separable.
• Chromatography core mechanism: all chromatography methods partition compounds between a mobile phase and a stationary phase; separation depends on differing affinities.
• Common confusion—polarity and elution order: in normal-phase chromatography (TLC, column), polar compounds move slower; in reverse-phase HPLC, polar compounds elute first.
• Distillation vs. GC: both separate by volatility/boiling point, but distillation handles bulk quantities while gas chromatography analyzes small samples.

🧪 Extraction methods

🧪 Liquid-liquid extraction basics

Liquid-liquid extraction: separating a compound by shaking a solution with a second immiscible solvent; the compound distributes between the two phases based on its solubility in each.

• The partition (distribution) coefficient is the ratio of solubilities in the two solvents.
• "Like dissolves like": polar solutes dissolve in polar solvents; nonpolar solutes dissolve in nonpolar solvents.
• Example: hydrocarbons (nonpolar) have very low solubility in water (polar).

💧 Simple water extraction

• Removes highly polar or charged substances: inorganic salts, strong acids/bases, and small polar compounds (< 5 carbons) like alcohols, amines, carboxylic acids.
• The aqueous phase extracts these compounds; the organic phase retains nonpolar compounds.

⚗️ Acid-base extraction strategy

The excerpt describes a systematic separation using pH-selective extractions:

Compound type	Extraction reagent	What happens	Result
Carboxylic acids (e.g., benzoic acid)	5% NaHCO₃ (weak base)	Deprotonates COOH → anionic salt	Extracted into aqueous layer
Phenols (e.g., para-cresol)	10% NaOH (strong base)	Deprotonates phenol OH → anionic salt	Extracted into aqueous layer; NaOH also extracts carboxylic acids
Amines (e.g., aniline)	10% HCl (acid)	Protonates amine → cationic salt	Extracted into aqueous layer
Hydrocarbons (e.g., naphthalene)	None	No reaction	Remains in organic layer

Don't confuse: NaHCO₃ is weak enough to deprotonate carboxylic acids but not phenols; NaOH is strong enough to deprotonate both.

🔬 Separatory funnel procedure

• The mixture and extraction solvent are added to a separatory funnel, mixed, then allowed to separate into two layers.
• Each layer is collected separately by opening the stopcock at the bottom.
• Example walkthrough: a mixture of para-cresol, benzoic acid, aniline, and naphthalene in diethyl ether is separated by sequential extractions (NaHCO₃ → NaOH → HCl), isolating each compound in turn.

📊 Chromatography principles

📊 What all chromatography has in common

• Purpose: separate mixtures; some methods are for identification, others for purification.
• Core mechanism: compounds partition between a mobile phase (moves) and a stationary phase (fixed).
• Separation occurs because different compounds have different affinities for each phase.

🎨 Thin-layer chromatography (TLC)

TLC: a solid-liquid partitioning technique where a mobile liquid phase ascends a thin layer of polar absorbent (usually silica, SiO₂) coated on a glass plate.

• How it works:
  • A tiny sample (~1 µL) is spotted near the base of the plate.
  • The plate is placed upright in a sealed container with shallow solvent.
  • Solvent ascends by capillary action, carrying compounds at different rates.
• Separation logic:
  • More polar compounds interact more with the polar stationary phase → travel slower.
  • Less polar compounds have greater affinity for the solvent → travel faster.
• Visualization: UV light, iodine vapor, or chemical staining (most compounds are not colored).
• Rₓ value ("ratio to front"): distance traveled by compound ÷ distance traveled by solvent front.
  • Always between 0 and 1.
  • Example: if a compound travels 4.9 cm and the solvent front travels 10 cm, Rₓ = 0.49.

🏛️ Column (flash) chromatography

• Uses the same principles as TLC but for bulk isolation of compounds.
• A column is packed with silica gel saturated with solvent; the mixture is added to the top.
• Solvent is periodically added; separated compounds are collected from the bottom.
• Elution order: least polar to most polar (polar compounds spend more time adsorbed on the polar stationary phase).

⚡ Ion exchange chromatography

• Stationary phase: polymeric resin functionalized with charged groups (e.g., anionic sulfonate groups).
• How it works:
  • Positively charged species displace sodium ions and bind to anionic groups on the resin → retained.
  • Negatively charged and neutral species pass through quickly → elute first.
  • Retained species are later eluted with a concentrated sodium solution.
• Application: separating proteins by charge state at a given pH.
  • If pH < pI of a protein → protein is positively charged.
  • Choice of anionic vs. cationic resin determines which proteins are retained.

🚀 High performance liquid chromatography (HPLC)

• Key difference: mobile phase is forced through at very high pressure → faster and more efficient than column chromatography.
• Reverse-phase HPLC (most common for organic compounds):
  • Stationary phase: nonpolar (e.g., silica bonded to octadecylsilane).
  • Mobile phase: more polar than stationary phase.
  • Elution order: polar compounds elute first (high affinity for mobile phase); nonpolar compounds elute last (interact with nonpolar stationary phase).
  • Don't confuse: this is the reverse of TLC/column chromatography elution order.
• For amino acids: often uses ion exchange column; elution order depends on side-chain intermolecular forces.

📏 Size exclusion chromatography

• Separation basis: molecular size.
• Stationary phase: chemically inert, porous polymer beads.
• How it works:
  • Large molecules are excluded from pores → take a direct path → elute first.
  • Small molecules permeate the beads → take a longer, more complicated path → elute last.
• Applications: separating large polymers from small oligomers, or full proteins from peptide chains.
• Limitation: not effective for separating compounds of similar size.

🧲 Affinity chromatography

• Purpose: purify proteins or nucleic acids from complex mixtures (e.g., cell lysates, blood).
• Principle: highly specific binding interactions between macromolecules.
• How it works:
  1. Target molecule binds to a functionalized solid resin (e.g., antibody-coated beads or Protein A/G/L-linked beads).
  2. Unwanted components are washed away.
  3. Target is eluted in a highly purified state.
• Isolation methods:
  • Centrifugation: heavy resin settles; supernatant is decanted.
  • Magnetic beads: magnet holds beads against tube wall; solution is decanted.
• Affinity tags (e.g., His tags): small molecular tags added to proteins; bind specific ions (e.g., nickel) for purification.

💨 Gas chromatography (GC)

Gas chromatography: partitioning between a mobile gas phase and a stationary liquid phase; separates compounds based on volatility.

• How it works:
  • Sample is injected, vaporized by a heater, and carried by inert gas (helium) through a column.
  • Column is packed with particles coated with a liquid absorbent.
  • Components equilibrate between gas and liquid phases many times.
• Separation logic:
  • More volatile (lower boiling point) components spend more time in the gas phase → elute first.
  • Less volatile (higher boiling point) components spend more time in the liquid phase → elute last.
• Detection: components are burned as they exit; resulting ions are detected electrically and recorded.
• Application: analyzing small amounts of material; similar principle to distillation but for analytical purposes.

🌡️ Physical properties and intermolecular forces

🌡️ Melting and boiling points

Melting point (mp) and boiling point (bp): indicators of how strongly identical molecules attract each other.

• Higher intermolecular forces → more energy needed to melt or boil → higher mp and bp.
• For nonpolar molecules (e.g., hydrocarbons): London dispersion forces (van der Waals forces) are the main attractive force.

🌿 Factors affecting mp and bp in hydrocarbons

Factor	Effect	Reason
Branching	Lowers mp and bp	Reduces surface area for intermolecular interaction
Molecular weight	Raises mp and bp	More surface area → more van der Waals interactions

• Trend by size:
  • 1–4 carbons: gases at room temperature.
  • 5–16 carbons: liquids at room temperature.

  • 16 carbons: waxy solids at room temperature.

• Example: n-octane (unbranched) has higher mp and bp than 2,4-dimethylhexane (branched isomer).

🔗 Hydrogen bonding

Hydrogen bond: electrostatic interaction between a hydrogen-bond donor (H covalently bonded to N, O, or F) and a hydrogen-bond acceptor (lone pair on N, O, or F in another molecule).

• Not a covalent bond; it is an intermolecular interaction.
• Dramatically increases mp and bp.
• Example: n-butanol (can form H-bonds) has bp = 117°C; diethyl ether (same formula, cannot form H-bonds) has bp = 34.6°C.
• Intramolecular vs. intermolecular H-bonding:
  • Intermolecular (e.g., 4-nitrophenol): H-bonds between different molecules → higher mp/bp.
  • Intramolecular (e.g., 2-nitrophenol): H-bond within the same molecule → reduces intermolecular H-bonding → lower mp/bp.
• Rule of thumb: more H-bond donors and acceptors → higher mp and bp.

🔥 Distillation techniques

🔥 What distillation does

• Process: heating a liquid until it vaporizes, then condensing the vapor back to liquid and collecting it.
• Separates compounds based on boiling point (volatility).

🧪 Simple distillation

• When to use: removing trace impurities from a relatively pure compound, or separating compounds with significantly different boiling points.
• Example: purifying drinking water from saltwater—water (volatile) boils away and is collected; salts (nonvolatile) remain behind.

🏗️ Fractional distillation

• When to use: separating compounds with similar boiling points.
• How it works:
  • Column is packed with material (e.g., glass beads, stainless steel sponge).
  • Mixture undergoes many vaporization-condensation cycles as it moves up the column.
  • Vapor becomes progressively enriched in the lower-boiling component.
  • Nearly pure vapor of the lower-boiling component reaches the condenser at the top and is collected.
• Don't confuse: if simple distillation gives incomplete separation (e.g., GC shows two peaks in the distillate), switch to fractional distillation.

🧭 Overview

🧠 One-sentence thesis

Spectroscopy techniques—including mass spectrometry, UV/Vis, IR, and ¹H NMR—allow chemists to deduce molecular structure by measuring how molecules absorb or interact with different forms of electromagnetic radiation.

📌 Key points (3–5)

• Core principle of absorption spectroscopy: molecules absorb photons matching the energy gap between ground and excited states, then release energy as they relax back.
• Different radiation types reveal different features: mass spectrometry measures mass; UV/Vis detects conjugation; IR identifies functional groups; ¹H NMR maps hydrogen environments.
• IR spectroscopy key: always look first for the carbonyl stretch (~1700 cm⁻¹) and the broad O—H stretch (3600–3200 cm⁻¹).
• ¹H NMR four essentials: number of signals = number of nonequivalent H sets; splitting pattern = neighboring H count (n+1 rule); integration = relative H numbers; chemical shift = electronic environment.
• Common confusion: in NMR, "equivalent" hydrogens have identical electronic environments (interchangeable by rotation or symmetry), not just the same number of neighbors.

🔬 Absorption spectroscopy fundamentals

🔬 Ground state and excited state

Ground state: the lowest energy form in which molecules normally exist.

Excited state: the higher-energy condition when a molecule absorbs a photon whose energy matches the gap between two fixed electronic energy levels.

• Molecules prefer the ground state; to return from an excited state, they must lose the absorbed energy.
• Energy loss occurs by emitting heat or (less commonly) light.
• Scientists measure this released energy to reveal structural features.

🌈 The electromagnetic spectrum and transitions

• Many forms of light exist (radio, infrared, visible, UV, X-ray, etc.).
• Different forms induce different transitions (vibrational, electronic, nuclear spin).
• Each transition type provides distinct structural information about the molecule.

📏 Mass spectrometry

📏 How it works

Mass spectrometry: a technique that determines the mass of compounds by ionizing molecules in high vacuum (usually with high-energy electrons), then using a magnetic field to alter their flight path based on mass-to-charge ratio (M/e).

• Most ions have charge e = +1, so peaks represent molecular mass (in amu).
• The detector translates the degree of path change into a mass readout.

📊 Reading the spectrum

Feature	What it shows	Why
Main peak (molecular ion)	Molecular weight	The intact ionized molecule
Peaks > molecular weight	Isotope variants	Molecules containing ¹³C (1.1% natural abundance) or deuterium (0.015%) instead of ¹²C or ¹H
Peaks < molecular weight	Molecular fragments	High-energy electrons break bonds, creating smaller pieces

Example: n-nonane (MW = 128 g/mol) shows a peak at 113 (lost terminal CH₃, mass 15) and at 99 (lost CH₂CH₃, mass 29).

🧪 Characteristic isotope patterns

• Bromine: two isotopes (⁷⁹Br and ⁸¹Br) of nearly equal abundance → two major peaks of nearly equal height, 2 amu apart.
• Chlorine: ³⁵Cl (75%) and ³⁷Cl (25%) → a second peak 2 amu heavier than the main peak, about one-third its height.
• Don't confuse: these patterns are due to natural isotope ratios, not fragmentation.

🌞 UV/Vis spectroscopy

🌞 What it measures

• UV and visible wavelengths are energetic enough to promote ground-state valence electrons into excited states (electronic excitation).
• Used for two main applications:
  1. Monitoring transition metal complexes (closely spaced d-orbitals allow easy electron promotion, giving bright colors).
  2. Studying highly conjugated organic systems.

🎨 Conjugation and wavelength

• Conjugated π-systems form many bonding, non-bonding, and anti-bonding orbitals close together in energy.
• The more extensive the conjugated system, the longer the wavelength of maximum absorption (λ_max).

Example: polycyclic aromatic hydrocarbons—adding each aromatic ring extends conjugation and increases λ_max.

🎨 Predicting color

• The color a compound absorbs (λ_max) is complementary to the color it appears.
• A compound absorbing only UV reflects all visible wavelengths → appears white or colorless.
• A compound absorbing blue light appears orange (complementary colors on opposite sides of the color wheel).

🔴 Infrared (IR) spectroscopy

🔴 Core principle

• IR radiation (wavelength 2.5–20 μm; wavenumber range typically 4000–1000 cm⁻¹ on MCAT) causes covalent bonds to vibrate at distinct energy levels.
• When a bond absorbs IR at a specific frequency, that frequency is not recorded by the detector → appears as a peak (low transmittance = absorbance).

Wavenumber (ῡ): the reciprocal of wavelength; directly proportional to frequency and energy. Higher wavenumber = higher frequency = greater energy.

🔍 Key stretching frequencies to memorize

🔍 Carbonyl (C=O) stretch

• Location: ~1700 cm⁻¹
• Characteristics: very strong and very intense (sharp "V" shape).
• Why it matters: always look for this first. If absent, eliminate aldehydes, ketones, carboxylic acids, acid chlorides, esters, amides, anhydrides. If present, one of these groups is in the molecule.

🔍 C=C double bond stretch

• Location: ~1650 cm⁻¹
• Slightly lower than carbonyl.

🔍 Triple bond stretches

• C≡C: ~2150 cm⁻¹
• C≡N: ~2250 cm⁻¹
• Few molecules have these, so easy to spot when present.

🔍 O—H stretch

• Location: 3600–3200 cm⁻¹
• Characteristics: strong and very broad (wide "U" shape, not a sharp spike).
• Why broad: hydrogen bonding.
• Always look for this stretch; amines also have stretches in this region but vary in intensity.

🔍 C—H stretches

• Location: 3300–2850 cm⁻¹
• Almost always present (most organic compounds have C—H bonds).
• Aliphatic C—H: slightly below 3000 cm⁻¹
• Aromatic C—H: slightly above 3000 cm⁻¹

📋 Summary table

Bond	Wavenumber (cm⁻¹)	Characteristics
C=O	~1700	Strong, intense, sharp "V"
C=C	~1650	—
C≡C	~2150	—
C≡N	~2250	—
O—H	3600–3200	Strong, broad "U"
C—H (aliphatic)	<3000	—
C—H (aromatic)	>3000	—

🧲 ¹H Nuclear Magnetic Resonance (NMR) spectroscopy

🧲 Overview

• Uses radio-frequency light to induce energy absorptions.
• Theory is beyond MCAT scope; focus on interpretation.
• Four essential features: (1) number of signals, (2) splitting pattern, (3) integration, (4) chemical shift.
• Most important for MCAT: features 1 and 2.

🔢 Feature 1: Chemically equivalent hydrogens

🔢 What "equivalent" means

Equivalent hydrogens: protons with identical electronic environments; they can be interchanged by free rotation or a symmetry operation (mirror plane or rotational axis).

• Equivalent H's appear at the same location in the ¹H NMR spectrum (same signal/resonance).
• Nonequivalent H's appear at different locations (different signals).
• Number of signals = number of nonequivalent H sets.

🔢 Examples

Example 1: A hydrocarbon C₅H₁₂ shows only one peak.

• Degrees of unsaturation d = [2(#C) + 2 − (#H)]/2 = 0 → no double bonds or rings.
• One peak → all protons equivalent → structure is neopentane (all 12 H's equivalent by symmetry).

Example 2: C₅H₁₀ shows only one peak.

• d = 1 → one double bond or ring.
• All C₅H₁₀ alkenes have more than one type of proton.
• Cyclopentane has a five-fold axis of symmetry → all H's equivalent.

✂️ Feature 2: Splitting (spin-spin splitting)

✂️ The n+1 rule

Splitting: nonequivalent neighboring hydrogens interact, causing a signal to split into n+1 lines, where n = number of nonequivalent neighboring (interacting) protons.

• The magnetic field felt by a proton is influenced by surrounding protons.
• This effect extends over two adjacent carbons.

n (# of neighboring H's)	Splitting pattern
0	Singlet (1 line)
1	Doublet (2 lines)
2	Triplet (3 lines)
3	Quartet (4 lines)
4	Quintet or multiplet (5 lines)
5	Sextet or multiplet (6 lines)

✂️ Example: CH₃CH₂I

• α-hydrogens (CH₂): have 3 neighboring H's on CH₃ → split into quartet.
• β-hydrogens (CH₃): have 2 neighboring H's on CH₂ → split into triplet.

Important for MCAT: neighboring H's do not have to be equivalent to each other to add them together for n.

Example: A proton H_b has 3 H's on one side (H_a) and 2 H's on the other side (H_c). Even if H_a and H_c are not equivalent to each other, n = 5 → H_b appears as a sextet (or multiplet).

✂️ Practice scenarios

• Molecule (a): Three signals. C1 protons split by 2 H's on C2 → triplet. C2 protons split by 5 H's total (C1 + C3) → sextet/multiplet. C3 protons split by 2 H's on C2 → triplet.
• Molecule (b): One signal; all equivalent → no splitting.
• Molecule (c): Four signals. C1 → triplet (2 neighbors on C2). C2 → quartet (3 neighbors total from C1 and C3). C3 → multiplet (8 neighbors). C4 and C5 → doublet (1 neighbor on C3).
• Molecule (d): Three signals. H on C3 (bearing Br) → quartet (3 H's on C4). H's on C4 → doublet (1 H on C3). tert-butyl H's → singlet (no neighbors, 9 equivalent H's).

📐 Feature 3: Integration

📐 What it measures

Integration: a mathematical calculation measuring the area under each absorption peak; proportional to the relative number of protons giving rise to that peak.

• Indicates the relative number of protons in each set.
• Example: if one signal integrates to 3 and another to 2, the ratio of protons is 3:2.

📍 Feature 4: Chemical shift

📍 What it indicates

Chemical shift: the location (in ppm) of a resonance in the ¹H NMR spectrum, reflecting the electronic environment of the protons.

• Shielded protons (electron-rich environment) appear upfield (to the right, lower ppm).
• Deshielded protons (electron-poor environment) appear downfield (to the left, higher ppm).

📍 Factors affecting chemical shift

📍 Electronegativity

• Electronegative atoms nearby decrease electron density → deshield the proton → downfield shift.
• Example: In methyl acetate, both methyl groups are near electronegative O or C=O → both shifted downfield.

📍 Hybridization

• Greater s-orbital character → less electron density on H → more deshielded → downfield shift.
• Order (by s-character alone): sp (50%) > sp² (33%) > sp³ (25%).
• Exception: alkyne protons appear more upfield than expected due to a complex phenomenon beyond MCAT scope.
• Key shifts to know:
  • Aromatic H: δ = 6.5–8 ppm
  • Alkene H: δ = 5–6 ppm

📍 Acidity and hydrogen bonding

• Protons on heteroatoms (O, N) are quite deshielded.
• Carboxylic acid H: extreme downfield shift (very acidic).
• Alcohol H: variable (δ = 2–5 ppm), depends on H-bonding, solvent, and temperature.

📍 Common chemical shifts to memorize

Proton type	Chemical shift (ppm)
Carboxylic acid (COOH)	~10–12
Aromatic	6.5–8
Alkene	5–6
Alcohol (variable)	2–5
Aliphatic (near electronegative groups)	Higher ppm
Aliphatic (far from electronegative groups)	Lower ppm

Don't confuse: chemical shift (location in spectrum) vs. splitting (number of lines in a signal) vs. integration (relative number of protons).

🧭 Overview

🧠 One-sentence thesis

Aldehydes and ketones are carbonyl compounds whose chemistry is governed by the electrophilic carbonyl carbon and the acidic α-protons, enabling nucleophilic addition reactions and enolate formation that underpin many organic transformations.

📌 Key points (3–5)

• Formation: Aldehydes and ketones are formed by oxidation of alcohols; anhydrous oxidants (PCC) stop at the aldehyde, while aqueous oxidants overoxidize primary alcohols to carboxylic acids.
• Carbonyl polarization: The C=O bond is highly polarized (oxygen pulls electrons), making the carbon electrophilic (δ+) and susceptible to nucleophilic attack.
• α-Proton acidity: Protons adjacent to the carbonyl are acidic because deprotonation yields a resonance-stabilized enolate ion, which is nucleophilic at the α-carbon.
• Common confusion: Enolate nucleophilicity resides predominantly at the α-carbon, not the oxygen, despite the negative charge being delocalized onto oxygen.
• Key reactions: Nucleophilic additions (hydride reduction, Grignard reagents, acetal/imine/enamine formation) and aldol condensation (combining enolate nucleophilicity with carbonyl electrophilicity).

🔬 Formation and Structure

🧪 Oxidation of alcohols to carbonyls

• Primary alcohols → aldehydes (with anhydrous oxidants) or carboxylic acids (with aqueous oxidants).
• Secondary alcohols → ketones (with any oxidant).
• Tertiary alcohols → no reaction (no hydrogen at the reactive carbon).

Oxidizing agents: Compounds that absorb electrons and are reduced; they remove hydrogen from the alcohol carbon.

Common oxidants:

Type	Examples	Behavior
Aqueous	H₂CrO₄, CrO₄²⁻, Cr₂O₇²⁻, MnO₄⁻, CrO₃	Overoxidize primary alcohols to carboxylic acids
Anhydrous	PCC (Pyridinium Chlorochromate)	Stops at aldehyde; does not overoxidize

⚡ Carbonyl polarization and reactivity

Carbonyl polarization: Oxygen is much more electronegative than carbon, so it pulls the π electrons of the C=O bond toward itself, leaving the carbon electron-deficient (δ+).

• Resonance structures show the carbon bearing a partial positive charge and the oxygen bearing a partial negative charge.
• This polarization drives two main reactivities:
  1. Acidic α-protons (next to the carbonyl).
  2. Electrophilic carbonyl carbon (attracts nucleophiles).

🔋 Acidity and Enolate Chemistry

🔑 α-Proton acidity

• Protons on the carbon adjacent (α) to the carbonyl are acidic enough to be removed by strong bases (OH⁻, OR⁻).
• Why acidic? The electrons left behind on the α-carbon can delocalize into the carbonyl π system, placing the negative charge on the electronegative oxygen.

Enolate ion: A resonance-stabilized carbanion formed by deprotonation of an α-proton; it is negatively charged and nucleophilic.

• Key point: The nucleophilic character of the enolate resides predominantly at the α-carbon, not the oxygen.
• Example: A proton α to two carbonyl groups is much more acidic than one α to only one carbonyl (greater resonance stabilization).

🔄 Keto-enol tautomerism

Tautomers: Readily interconvertible constitutional isomers in equilibrium, differing only in the position of a proton and a double bond.

• Keto form: The carbonyl structure (C=O with α-H).
• Enol form: The alcohol-alkene structure (C=C–OH).
• Conversion: Deprotonation of the α-carbon and protonation of the carbonyl oxygen (or vice versa).

⚠️ Racemization via enolization

• If a chiral center exists at the α-carbon, oxidation to the ketone followed by enolization destroys stereochemistry.
• Mechanism: The enol tautomer has an sp²-hybridized, planar α-carbon; protonation can occur from either face, yielding a racemic mixture (both R and S configurations).
• Don't confuse: A single chiral alcohol → ketone → racemic mixture at the α-carbon.

🎯 Nucleophilic Addition Reactions

🧩 General mechanism

Nucleophilic addition: A reaction in which a π bond in the carbonyl is broken and two σ bonds are formed in the product.

• The electrophilic carbonyl carbon attracts nucleophiles.
• Generic mechanism: Nu:⁻ attacks the carbonyl carbon → tetrahedral intermediate → protonation.

🔻 Hydride reduction to alcohols

• Reducing agents: NaBH₄ (sodium borohydride), LiAlH₄ (lithium aluminum hydride).
• These reagents donate hydride (H⁻, a hydrogen with a pair of electrons) to the carbonyl carbon.
• Result: Aldehydes and ketones are converted to alcohols.
• Why they work: Strong reducing agents have many hydrogens attached to elements with low electronegativity, making them electron-rich.

🛠️ Organometallic reagents (Grignard and lithium)

Organometallic reagent: Structure R−M⁺; acts as an electron-rich (anionic) carbon, functioning as a strong base or nucleophile.

• Grignard reagents: Made by reacting an alkyl or acyl halide with magnesium metal (e.g., CH₃MgBr).
• Reaction conditions: Aprotic solvent (e.g., diethyl ether) to avoid protonating the very basic Grignard reagent.
• Mechanism: The Grignard reagent (acting as Nu:⁻) attacks the carbonyl carbon → alkoxide intermediate → aqueous acidic workup protonates the alkoxide → alcohol product.
• Caution: The substrate must not contain other acidic hydrogens (e.g., −OH, −COOH) or they will react with the Grignard reagent first; protect such groups if necessary.

🛡️ Protecting groups: Mesylates and tosylates

• Purpose: Temporarily convert reactive −OH (or −NH₂) groups into inert groups to prevent unwanted reactions.
• Mesylate: Methanesulfonyl group (CH₃−SO₃−).
• Tosylate: Toluenesulfonyl group (CH₃C₆H₄−SO₃−).
• Formation: React the alcohol with mesyl chloride or tosyl chloride in the presence of a base (triethylamine or pyridine) to neutralize HCl.
• Removal: After the desired reaction, the protecting group is removed (generally under reductive conditions) to regenerate the −OH or −NH₂.
• Bonus: Mesylates and tosylates are also good leaving groups, making nucleophilic displacement easier (whereas −OH is a poor leaving group).

🧬 Acetals, Hemiacetals, and Related Compounds

🔍 Identifying acetals and hemiacetals

Hemiacetal: A carbon bonded to one −OR group, one −OH group, and one H or R group (from the original carbonyl).

Acetal: A carbon bonded to two −OR groups and one H or R group (from the original carbonyl).

• Key distinction: Hemiacetals have one −OR and one −OH; acetals have two −OR groups.
• The terms "ketal" and "hemiketal" (for ketone-derived compounds) are obsolete.

⚙️ Acetal formation mechanism

• Overall: Aldehyde or ketone + alcohol (in acidic conditions) → hemiacetal → acetal + water.
• Mechanism:
  1. Protonate the carbonyl oxygen (makes the carbon more electrophilic).
  2. Alcohol attacks the carbonyl carbon → positively charged oxygen on the alcohol.
  3. Deprotonate the alcohol oxygen → hemiacetal.
  4. Protonate the hemiacetal −OH (converts poor leaving group to good one).
  5. Second alcohol attacks → positively charged oxygen.
  6. Deprotonate → acetal product.
• Equilibrium: Aldehyde/ketone ⇌ hemiacetal ⇌ acetal; water is lost in forming the acetal.
• Prediction tip: The R groups originally on the carbonyl carbon remain attached; two −OR groups from the alcohol are added.

🧪 Cyanohydrin formation

• Nucleophile: Cyanide ion (⁻C≡N).
• Product: Cyanohydrin (a tetrahedral compound with −OH and −C≡N on the same carbon).
• Mechanism: Similar to hemiacetal formation (cyanide attacks carbonyl carbon → alkoxide → protonation).
• Equilibrium: Heavily favors the product (essentially a one-way reaction), unlike hemiacetal formation.

🧫 Reactions with Amines

🧬 Amine structure and reactivity

Amines: Organic compounds with the general structure R−NH₂; contain nitrogen with a lone pair of electrons.

• Classification:
  • Alkyl amines: Nitrogen bonded to sp³-hybridized carbon.
  • Aryl amines: Nitrogen bonded to sp²-hybridized aromatic carbon.
• Further classification: Primary (RNH₂), secondary (R₂NH), tertiary (R₃N), quaternary ammonium (R₄N⁺).
• Geometry: sp³-hybridized nitrogen with pyramidal geometry (~109° bond angles).
• Reactivity: The lone pair makes amines act as Brønsted-Lowry bases or nucleophiles.

🔗 Imine formation (primary amines)

• Reactants: Aldehyde or ketone + primary amine (RNH₂) under weakly acidic conditions (pH ~4–5).
• Product: Imine (C=N double bond) + water.
• Mechanism:
  1. Protonate carbonyl oxygen (increases electrophilicity).
  2. Primary amine attacks carbonyl carbon → tetrahedral intermediate.
  3. Deprotonate nitrogen, protonate oxygen (converts −OH to −OH₂⁺).
  4. Water leaves → carbocation stabilized by nitrogen lone pair (iminium ion).
  5. Deprotonate nitrogen → imine.
• Key: The R groups on the carbonyl carbon remain attached; water is liberated.

🌿 Enamine formation (secondary amines)

• Reactants: Aldehyde or ketone + secondary amine (R₂NH) under similar acidic conditions.
• Product: Enamine (C=C−NR₂ structure).
• Mechanism: Identical to imine formation until the final step; since the amine is secondary, the iminium ion cannot be deprotonated at nitrogen. Instead, an α-proton (now more acidic due to the positive charge on nitrogen) is removed → enamine.
• Structure: Resembles an enol, but with a secondary amine replacing the oxygen.
• Resonance: Enamine ⇌ iminium; partial double-bond character in C−N bond → sp²-hybridized nitrogen and hindered rotation.
• Reactivity: The α-carbon is nucleophilic (more so than neutral enols, less than enolates) and reacts with electrophiles (e.g., alkyl halides).
• Hydrolysis: Enamines can be hydrolyzed back to the carbonyl compound under aqueous acidic conditions (reversible reaction).

🔄 Aldol Condensation

🧩 Overview and mechanism

Aldol condensation: A reaction in which the enolate anion of one carbonyl compound attacks the carbonyl carbon of another, forming a β-hydroxy carbonyl compound.

• Combines both reactivities: α-proton acidity (enolate formation) and carbonyl electrophilicity (nucleophilic attack).
• Mechanism:
  1. Strong base (OH⁻ or RO⁻) removes an α-proton → resonance-stabilized enolate anion.
  2. The α-carbon of the enolate (nucleophilic) attacks the carbonyl carbon of another aldehyde/ketone molecule.
  3. Alkoxide ion forms → protonated by water → β-hydroxy carbonyl product.
• Key points:
  • Requires a strong base.
  • One carbonyl acts as the enolate source (nucleophile); the other is the electrophile.
  • The two carbonyls can be the same or different (crossed aldol condensation).

🔀 Crossed aldol condensation

• Definition: The two carbonyl compounds are different.
• Strategy to avoid mixtures: Choose one carbonyl without acidic α-protons (cannot form enolate) → it must be the electrophile.

⚖️ Kinetic vs. thermodynamic control

• Thermodynamic control (room temperature, unencumbered base):
  • Favors the more substituted enolate (more stable double bond with fewer vinyl hydrogens).
  • Deprotonation occurs at the more sterically crowded α-carbon.
• Kinetic control (low temperature or bulky base like LDA):
  • Favors the less substituted enolate (kinetically more accessible α-carbon).
  • The bulky base or low energy prevents access to the more hindered α-carbon.
• Don't confuse: Same starting ketone can yield different enolates (and hence different aldol products) depending on conditions.

🔙 Retro-aldol and dehydration

• Retro-aldol: Strong base deprotonates the β-hydroxyl group → reverse of the aldol condensation, breaking the molecule back into two carbonyl compounds.
• Dehydration (heating the β-hydroxy carbonyl):
  • Elimination reaction removes water.
  • Product: α,β-unsaturated carbonyl compound (new C=C bond conjugated with C=O, which stabilizes the molecule).
• Tip: Any β-hydroxy aldehyde or ketone can be traced back to its aldol precursors by working through the retro-aldol mechanism.

🧭 Overview

🧠 One-sentence thesis

Carboxylic acids are biologically fundamental molecules that form strong hydrogen bonds, can be reduced to primary alcohols with LiAlH₄, and undergo decarboxylation when a carbonyl group is positioned β to the carboxylate.

📌 Key points (3–5)

• Hydrogen bonding: carboxylic acids form stable dimers because the carboxylate group contains both a hydrogen bond donor (acidic proton) and acceptor (carbonyl oxygen lone pair), resulting in high melting and boiling points.
• Reduction selectivity: LiAlH₄ reduces carboxylic acids to primary alcohols, but NaBH₄ does not—aluminum's higher electropositivity makes the Al–H bond more polarized and reductive.
• β-Keto acid instability: carboxylic acids with a carbonyl group β to the carboxylate undergo decarboxylation (loss of CO₂) through a cyclic transition state.
• Common confusion: LiAlH₄ vs NaBH₄—both reduce ketones and aldehydes, but only LiAlH₄ is strong enough to reduce carboxylic acids.
• Biological importance: fatty acids (long-chain carboxylic acids) play key roles in cellular structure and metabolism.

🔗 Hydrogen bonding and physical properties

🔗 Why carboxylic acids form strong hydrogen bonds

Carboxylic acids form strong hydrogen bonds because the carboxylate group contains both a hydrogen bond donor and a hydrogen bond acceptor.

• The acidic proton acts as the hydrogen bond donor.
• A lone pair of electrons on the carbonyl oxygen acts as the hydrogen bond acceptor.
• Example: acetic acid demonstrates intermolecular hydrogen bonding using both features.

🔄 Formation of stable dimers

• Because carboxylic acids can both donate and accept hydrogen bonds, they form stable hydrogen-bonded dimers.
• This dimerization leads to high melting and boiling points compared to molecules of similar molecular weight.
• Don't confuse: the high boiling point is not due to molecular weight alone, but to the dual hydrogen-bonding capability.

⚗️ Reduction of carboxylic acids

⚗️ Selective reduction with metal hydrides

• Carboxylic acids can be reduced to primary alcohols.
• The excerpt states: "LiAlH₄ is effective, but NaBH₄ is not."

Reagent	Effectiveness	Reason
LiAlH₄	Reduces carboxylic acids to primary alcohols	Aluminum is more electropositive than boron
NaBH₄	Does NOT reduce carboxylic acids	Boron is less electropositive; B–H bond is less polarized and less reductive

🔬 Why aluminum hydride is stronger

• Aluminum is slightly more electropositive than boron.
• This makes the Al–H bond more highly polarized.
• The greater polarization makes LiAlH₄ more reductive and capable of performing "more challenging reductions."
• Example: reducing a ketone or aldehyde is easier; reducing a carboxylic acid requires the stronger LiAlH₄.

🔥 Decarboxylation of β-keto acids

🔥 What makes β-keto acids unstable

Carboxylic acids that have carbonyl groups β to the carboxylate are unstable because they are subject to decarboxylation.

• β-keto acid: a carboxylic acid with a carbonyl group on the carbon that is β (two carbons away) from the carboxylate.
• These molecules are unstable and lose carbon dioxide (CO₂).

🔄 Mechanism of decarboxylation

• The reaction proceeds through a cyclic transition state.
• The result is loss of carbon dioxide from the β-keto acid.
• Don't confuse: this is not a simple elimination; the cyclic transition state is key to the mechanism.
• Example: a β-keto acid heated or under certain conditions will spontaneously release CO₂, leaving behind a simpler carbonyl compound.

🧬 Biological significance

🧬 Fatty acids in biological systems

• The excerpt states: "Carboxylic acids are of fundamental importance in many biological systems."
• Fatty acids are long-chain carboxylic acids.
• They play important roles in:
  • Cellular structure (e.g., membrane components).
  • Metabolism (e.g., energy storage and release).
• The excerpt refers to further detail in Section 7.4, indicating that fatty acids are a major application of carboxylic acid chemistry in biology.

🧭 Overview

🧠 One-sentence thesis

Carboxylic acid derivatives undergo nucleophilic addition-elimination reactions with reactivity determined primarily by the basicity of their leaving groups, decreasing in the order: acid halides > anhydrides > esters > amides.

📌 Key points (3–5)

• Core reaction mechanism: carboxylic acid derivatives undergo addition-elimination (not simple addition like aldehydes/ketones) because they contain a leaving group.
• Reactivity order: acid halide > acid anhydride > ester > amide, determined by leaving group basicity—better (less basic) leaving groups make the derivative more reactive.
• Synthesis pathways: derivatives are mostly interconverted from one to another; acid halides come from carboxylic acids + SOCl₂ or PX₃; esters from acids + alcohols (esterification); amides from acid halides/anhydrides/esters + amines (but not directly from acids).
• Common confusion: amides cannot be made directly from carboxylic acids and amines because the acid-base reaction (forming a salt) is much faster than the desired addition-elimination.
• Hydrolysis stability: acid chlorides and anhydrides hydrolyze readily in water; esters require acidic/basic conditions and heat; amides need acidic conditions, high temperature, and long reaction times.

🧪 Core reaction mechanism

🎯 Addition-elimination vs simple addition

Carboxylic acids and their derivatives undergo addition-elimination reactions rather than simple additions because they contain a leaving group.

• Why the difference from aldehydes/ketones: the presence of an electronegative leaving group (like –OR, –NR₂, –Cl) attached to the carbonyl carbon allows elimination after nucleophilic attack.
• The two-step process:
  1. Nucleophile attacks the electrophilic carbonyl carbon → forms a tetrahedral intermediate with an oxy-anion.
  2. The tetrahedral intermediate eliminates the leaving group (as an alkoxide, halide, etc.) → regenerates the carbonyl (or forms a new functional group).
• Example: In soap-making, hydroxide ions attack the carbonyl carbons of triacylglycerides three times; each attack forms a tetrahedral intermediate that then eliminates the –OR portion as an alkoxide, ultimately yielding glycerol and three fatty acids.

🧼 Real-world application: soap-making

• Ancient soap-making treats animal fat with lye (NaOH or KOH) over fire.
• What's happening chemically: basic hydrolysis of a triacylglyceride → one glycerol molecule + three fatty acid molecules.
• Why soap works: fatty acids are amphipathic (hydrophilic carboxylate head + hydrophobic hydrocarbon tail) → form micelles in water with tails inside (excluding water) and charged heads outside → grease/fats are adsorbed by the hydrophobic interior and washed away.

🔧 Synthesis of carboxylic acid derivatives

🧪 Acid halides

• How to make them: carboxylic acid + SOCl₂ or PX₃ (where X = Cl, Br).
• These are the most reactive derivatives.

🔗 Acid anhydrides

• Method 1: condensation of two carboxylic acids with loss of water (hence "anhydride" = "without water").
• Method 2: carboxylic acid (or carboxylate ion) + corresponding acid halide.

🍷 Esters

• Primary method: carboxylic acid + alcohol → esterification (the reaction seen earlier in the text).
• Alternative methods: acid halide, anhydride, or another ester + alcohol.
• Esters are moderately reactive.

🧬 Amides

• How to make them: acid halide, anhydride, or ester + desired amine.
• Critical restriction: amides cannot be prepared directly from carboxylic acid + amine.
• Why not: amines are very basic and carboxylic acids are very acidic → acid-base reaction (forming a salt) occurs much faster than the desired addition-elimination reaction.
• Don't confuse: the passage notes that by using excess amine, high temperatures, and long reaction times, the salt can be converted to the amide, but this method is not regularly used because other derivatives (acid halides, anhydrides, esters) react more easily with amines.

♻️ Regenerating carboxylic acids

• Carboxylic acids can be prepared from any of the derivatives by heating the derivative in acidic aqueous solutions.

⚡ Relative reactivity and leaving groups

📊 Reactivity order

Derivative	Leaving group	Reactivity rank
Acid halide	Halide (Cl⁻, Br⁻)	Highest (1)
Acid anhydride	Carboxylate (RCOO⁻)	High (2)
Ester	Alkoxide (RO⁻)	Moderate (3)
Amide	Amide ion (R₂N⁻)	Lowest (4)

• The key principle: reactivity decreases with increasing basicity of the leaving group.
• Better (less basic, more stable) leaving groups → more reactive derivative.
• Halides are excellent leaving groups (low basicity, stable anions) → acid halides are most reactive.
• Amide ions are very basic (poor leaving groups) → amides are least reactive.

💧 Hydrolysis stability

• Acid chlorides and anhydrides: readily hydrolyzed in water (high reactivity).
• Esters: require either acidic or basic conditions and elevated temperatures to hydrolyze.
• Amides: generally only hydrolyzed under acidic conditions, high temperatures, and long reaction times (very stable, low reactivity).
• Example: This stability difference explains why amides are so prevalent in biological systems (peptide bonds in proteins)—they don't fall apart easily in aqueous environments.

🔍 Common confusion: why leaving group basicity matters

• Don't confuse "good leaving group" with "nucleophile quality."
• A good leaving group is stable after it leaves (low basicity, can stabilize negative charge).
• A good nucleophile is electron-rich and reactive (often basic).
• In carboxylic acid derivatives, the leaving group's basicity determines how easily the tetrahedral intermediate can collapse back to reform a carbonyl by expelling that group.
• More basic leaving groups (like amide ions) are less willing to leave → slower reaction → lower reactivity.

🧭 Overview

🧠 One-sentence thesis

Amino acids are the building blocks of proteins, each with a unique side chain that determines its chemical properties, and their acid-base behavior governs their charge state, reactivity, and role in protein structure.

📌 Key points (3–5)

• Structure: All 20 amino acids share the same nitrogen-carbon-carbon backbone; the unique side chain (R-group) gives each amino acid its distinct properties.
• Chirality: All animal amino acids have the L-configuration (amino group on the left in Fischer projection); glycine is the only achiral amino acid.
• Acid-base behavior: Amino acids are amphoteric (both acidic and basic); their charge state depends on pH relative to the pKₐ values of their functional groups.
• Isoelectric point (pI): The pH at which an amino acid has no net charge (zwitterion); for simple amino acids, pI equals the average of the two backbone pKₐ values.
• Common confusion: Don't confuse the three classification systems—(+)/(−) describes optical activity, R/S describes absolute configuration, and D/L describes the precursor molecule (glyceraldehyde).

🧱 Structure and classification

🧱 Core amino acid structure

All amino acids share a generic structure: a central α-carbon bonded to an amino group (−NH₂), a carboxyl group (−COOH), a hydrogen atom, and a variable side chain (R-group).

• The α-carbon is a stereocenter in all amino acids except glycine (which has two hydrogens).
• The side chain determines each amino acid's unique physical and chemical properties: shape, charge, hydrogen bonding ability, and acid-base behavior.
• These properties are critical for protein structure and function.

🔬 Chirality and stereochemistry

All animal amino acids have the L-configuration, meaning the amino group is drawn on the left in Fischer projection notation.

• Fischer projections: Horizontal lines project toward the viewer; vertical lines project away. The most oxidized carbon is at the top.
• Glycine is the only achiral amino acid because its α-carbon has two identical hydrogen atoms.
• D-amino acids (amino group on the right) occur in specialized structures like bacterial cell walls but not in animal proteins.

Example: If you crash-land on Mars where life evolved with L-carbohydrates (opposite of Earth's D-carbohydrates), you cannot metabolize them because your enzymes are specific to D-carbohydrates.

🔤 Three classification systems

The excerpt describes three systems that do not predict each other:

System	What it describes	Example
(+) and (−) (or d and l)	Optical activity: direction of plane-polarized light rotation	A (+) molecule rotates light clockwise
R and S	Absolute configuration: actual 3D arrangement of groups	Describes the spatial structure at a chiral center
D and L	Precursor molecule: derived from D- or L-glyceraldehyde	All animal amino acids are L; all animal carbohydrates are D

• Important rule: All L-amino acids are S (except cysteine, where the side chain has higher priority than COOH). All D-sugars are R at the penultimate carbon.
• Don't assume a molecule with R configuration is (+) or that a (+) molecule is D—these systems are independent.

🧪 Categories by side chain properties

⚡ Acidic amino acids

• Glutamic acid and aspartic acid have carboxyl groups in their side chains (pKₐ ≈ 4).
• These amino acids have three functional groups that can act as acids or bases: the two backbone groups plus the R-group.
• The anionic (unprotonated) forms are called glutamate and aspartate.

⚡ Basic amino acids

• Lysine (pKₐ = 10), arginine (pKₐ = 12), and histidine (pKₐ = 6.5) have basic side chains.
• Histidine is unique: its pKₐ (6.5) is close to physiological pH (7.4), so it can be either protonated or deprotonated at physiological pH.
• This makes histidine a readily available proton donor or acceptor, explaining why it is common at protein active sites.
• Mnemonic: "His goes both ways."
• Don't confuse histidine (an amino acid) with histamine (a small molecule involved in allergic responses).

💧 Hydrophobic (nonpolar) amino acids

• Have either aliphatic (alkyl) or aromatic side chains.
• Aliphatic: glycine, alanine, valine, leucine, isoleucine.
• Aromatic: phenylalanine, tyrosine, tryptophan.
• Hydrophobic residues associate with each other rather than water, so they are found in the interior of folded proteins, away from water.
• Larger hydrophobic groups have stronger hydrophobic forces.

💧 Polar amino acids

Polar amino acids have R-groups that are polar enough to hydrogen bond with water but not polar enough to act as acids or bases.

• They are hydrophilic and interact with water.
• Serine, threonine, and tyrosine have hydroxyl groups that can be modified by phosphate attachment (via kinase enzymes), changing protein structure and activity.
• This phosphorylation is an important regulatory mechanism.

🔗 Sulfur-containing amino acids

• Cysteine contains a thiol (sulfhydryl) group and is fairly polar.
• Methionine contains a thioether and is fairly nonpolar.

🔄 Proline

• Unique: its amino group is covalently bound to part of the side chain, creating a secondary α-amino group and a distinctive ring structure.
• This has important consequences for protein folding.

⚗️ Synthesis methods

⚗️ Strecker synthesis

• Transforms aldehydes into α-amino acids using ammonium and cyanide salts.
• Mechanism:
  1. Aldehyde reacts with ammonia to form an imine.
  2. Protonation by HCN makes the imine more electrophilic.
  3. Cyanide attacks the imine carbon, forming an α-amino nitrile.
  4. Acid-catalyzed hydrolysis yields the α-amino acid.
• Drawback: Produces a racemic mixture (both D and L forms), unlike naturally occurring amino acids which are stereochemically pure (L-enantiomers).
• Advantage: Can synthesize both natural and non-natural amino acids depending on the aldehyde used.

⚗️ Gabriel-malonic ester synthesis

• Converts the nitrogen in phthalimide into a primary amine.
• Steps:
  1. Phthalimide is deprotonated with KOH to form a resonance-stabilized anion.
  2. The anion displaces bromide from α-bromomalonic ester.
  3. Enolization with a strong base creates a nucleophilic carbon.
  4. The enolate reacts with the desired side-chain precursor.
  5. Acid hydrolysis and heat-induced decarboxylation yield the racemic amino acid.
• Also produces racemic mixtures.

🔋 Acid-base chemistry and reactivity

🔋 Amphoteric nature

Amino acids are amphoteric: they have both acidic (carboxyl group) and basic (amino group) activity.

• Carboxyl group: pKₐ ≈ 2 (stronger acid).
• Ammonium group: pKₐ ≈ 9–10 (weaker acid).
• Whether a group is protonated depends on the pH of the solution and the pKₐ of the group.

🔋 Henderson-Hasselbalch equation

The relationship between pH, pKₐ, and equilibrium is given by:

pH = pKₐ + log([A⁻]/[HA]) = pKₐ + log([base]/[acid])

Key rules:

• Low pH means high [H⁺] (more acidic).
• Lower pKₐ (higher Kₐ) describes a stronger acid.
• If pH > pKₐ, the site is mostly deprotonated.
• If pH < pKₐ, the site is mostly protonated.

Example: At pH 4.7, acetic acid (pKₐ = 4.7) has a 1:1 ratio of acetate to acetic acid.

🔋 Charge state at different pH values

• At pH 1.0 (very acidic):
  • Carboxyl group (pKₐ ≈ 2): protonated (−COOH).
  • Amino group (pKₐ ≈ 9): protonated (−NH₃⁺).
  • Net charge: positive.
• At pH 6.0 (between the two pKₐ values):
  • Carboxyl group: deprotonated (−COO⁻).
  • Amino group: protonated (−NH₃⁺).
  • Net charge: zero (zwitterion).
• At pH 12 (very basic):
  • Carboxyl group: deprotonated (−COO⁻).
  • Amino group: deprotonated (−NH₂).
  • Net charge: negative.

⚖️ Isoelectric point (pI)

The isoelectric point (pI) is the pH at which an amino acid has no net charge (exists as a zwitterion with balanced positive and negative charges).

• For a molecule with two functional groups (like glycine, with no acidic or basic side chain): pI = average of the two pKₐ values.
• Example: Glycine has pKₐ values of ~2 (carboxyl) and ~9 (amino). Its pI = (2 + 9)/2 = 5.5.
• At the pI, the amino acid is a dipolar ion or zwitterion (German "zwitter" = hybrid).

🧲 Gel electrophoresis separation

🧲 Principle

Gel electrophoresis separates amino acids based on their charge in an electric field.

• Amino acids are loaded onto a gel held at constant pH.
• When the gel pH ≠ pI, each amino acid bears an overall charge.
• Amino acids migrate through the gel based on their charge and the electric field.

🧲 Migration rules

Condition	Charge	Migration direction
pH < pI	Positive (more protonated)	Toward negative (−) electrode
pH = pI	Zero (zwitterion)	No migration
pH > pI	Negative (more deprotonated)	Toward positive (+) electrode

Example: Glycine (pI = 5.5) at pH 6.0 has a slight negative charge → migrates toward the positive electrode.

Example: Aspartic acid (pKₐ values: 2.1, 3.9, 9.8) at pH 7.4 (physiological):

• Both carboxyl groups are deprotonated (−COO⁻).
• Amino group is protonated (−NH₃⁺).
• Net charge: negative (two − charges, one + charge) → migrates toward the positive electrode.

Example: Aspartic acid at pH 1.0:

• All groups are protonated.
• Net charge: positive → migrates toward the negative electrode.

🧲 Protein location prediction

• Hydrophobic amino acids (alanine, phenylalanine, leucine, isoleucine) are found in the interior of proteins at pH 7.0.
• Polar amino acids (serine) are found on the exterior of proteins at pH 7.0, where they can interact with water.
• Charged amino acids (glutamic acid) are also typically on the exterior.

🔗 Peptide bond formation

🔗 Natural peptide bonds

A peptide bond is an amide bond formed between the carboxyl group of one amino acid and the α-amino group of another, with the loss of water.

• The mechanism is nucleophilic addition-elimination (same as for carboxylic acid derivatives).
• Important: Simple condensation is not thermodynamically favorable in nature.
• In cells, peptide bond formation during translation requires enzyme catalysis, RNA direction, and co-factors.

🔗 DCC coupling (laboratory synthesis)

• DCC (dicyclohexyl carbodiimide) is used to synthesize peptides artificially.
• Mechanism:
  1. DCC converts the −OH of a carboxyl group into a good leaving group ("activated" amino acid).
  2. The amino group of a second amino acid attacks the carbonyl carbon.
  3. DCC leaves with the oxygen.
• Protecting groups are used to ensure amino acids are added in the correct order and direction.

🧭 Overview

🧠 One-sentence thesis

Proteins fold into unique three-dimensional structures through four hierarchical levels—primary sequence, secondary hydrogen-bonded motifs, tertiary hydrophobic/hydrophilic interactions, and quaternary subunit assembly—and this folding is essential for function, while denaturation disrupts shape without breaking peptide bonds.

📌 Key points (3–5)

• Two main covalent bonds in proteins: peptide bonds link amino acids into chains; disulfide bridges form between cysteine residues.
• Peptide bond formation is thermodynamically unfavorable in isolation but occurs in cells via enzyme catalysis; lab synthesis uses DCC coupling.
• Four levels of structure: 1° = sequence (peptide bonds), 2° = backbone H-bonds (α-helix, β-sheet), 3° = R-group interactions (hydrophobic/hydrophilic), 4° = subunit assembly.
• Common confusion—denaturation vs hydrolysis: denaturation disrupts shape without breaking peptide bonds (caused by urea, pH, temperature, salt); hydrolysis cleaves peptide bonds (acid or protease).
• Disulfide bonds stabilize structure and form under oxidizing conditions (extracellular), not in the reducing environment inside cells.

🔗 Covalent bonds in proteins

🔗 The peptide bond (amide bond)

A peptide bond is formed between the carboxyl group of one amino acid and the α-amino group of another amino acid with the loss of water.

• It is the same mechanism as forming any carboxylic acid derivative (nucleophilic addition-elimination, Section 6.3).
• A peptide bond is simply an amide bond between two amino acids.
• Example: glycine + alanine → dipeptide + water.

Thermodynamics vs kinetics:

• The simple condensation diagram is misleading: peptide bond formation is not thermodynamically favorable (requires energy input).
• In cells, the reaction is enzyme-catalyzed, RNA-directed, and co-factor mediated during translation.
• At equilibrium, individual amino acids are favored over the dipeptide (lower free energy).
• Yet peptide bonds are maintained inside cells because enzymes and cellular machinery drive the reaction forward and prevent spontaneous hydrolysis.

🧪 DCC coupling (lab synthesis)

• DCC (dicyclohexyl carbodiimide) converts the carboxylate OH into a good leaving group.
• The amino group of a second amino acid attacks the activated carbonyl carbon.
• DCC leaves with the oxygen.
• Protecting groups ensure only one carboxyl and one amino group react, so amino acids add in the correct order.

🧬 Backbone and terminology

• The backbone is the repeating N−C−C−N−C−C pattern from the amino acids.
• A residue is an individual amino acid within a polypeptide chain.
• Amino terminus (N-terminus) is made first; carboxy terminus (C-terminus) is made last.
• By convention, the N-terminal residue is always written first.
• Example: in Phe-Glu-Gly-Ser-Ala, Phe has the free α-amino group; Ala has the free α-carboxyl group.

🔒 Planarity and rigidity

• The peptide bond is planar and rigid because resonance delocalizes the nitrogen's electrons to the carbonyl oxygen, giving substantial double-bond character to the C−N bond.
• No rotation around the peptide bond.
• Don't confuse: the entire polypeptide is not rigid—rotation can occur around other bonds in the backbone (the bonds flanking the α-carbon), allowing the chain to fold.

⚗️ Hydrolysis of the peptide bond

Hydrolysis: any reaction in which water is inserted into a bond to cleave it.

Thermodynamics vs kinetics:

• Hydrolysis of the peptide bond (amide → free amine + carboxylic acid) is thermodynamically favored (products have lower free energy).
• But it is kinetically slow (high activation energy).

Two ways to accelerate hydrolysis (destroy proteins):

Method	Mechanism	Specificity
Acid hydrolysis	Strong acid + heat cleaves all peptide bonds	Non-specific; breaks protein into constituent amino acids
Proteolytic cleavage	Protease (proteolytic enzyme) cuts specific bonds	Specific; many enzymes cleave only adjacent to certain amino acids

Examples of protease specificity (do not memorize):

• Trypsin cleaves on the carboxyl side of basic residues (arginine, lysine).
• Chymotrypsin cleaves adjacent to hydrophobic residues (phenylalanine).
• Example: Ala-Gly-Glu-Lys-Phe-Phe-Lys cleaved by trypsin → cuts after each Lys → new N-terminus is Phe (after first Lys), and three fragments result.

Key distinction:

• After acid hydrolysis of Gly-Phe-Ala, you cannot determine the original order of residues—only the amino acid composition.
• Proteolytic cleavage is specific and can be used to map sequences.

🔗 The disulfide bond

• Cysteine has a reactive thiol (−SH) in its side chain.
• Two cysteines react to form a covalent disulfide bond (S−S).
• The cysteines may be in the same or different polypeptide chains.
• Once bonded, the residue is called cystine instead of cysteine.

Oxidation state:

• Cystine (disulfide) is more oxidized than cysteine (thiol).

Where disulfide bonds form:

• The cell interior is a reducing environment (antioxidants prevent oxidation).
• Disulfide bridges are more likely in extracellular proteins under oxidizing conditions, not inside cells.

🏗️ Four levels of protein structure

🧱 Primary (1°) structure: sequence

The linear ordering of amino acid residues in the polypeptide chain.

• Primary structure = sequence.
• The bond that determines 1° structure is the peptide bond (links one amino acid to the next).

🌀 Secondary (2°) structure: backbone hydrogen bonds

Initial folding into shapes stabilized by hydrogen bonds between backbone NH and CO groups.

Two common motifs:

Motif	Description	Key features
α-helix	Right-handed coil	5 Å wide, 1.5 Å rise per residue, 3.6 residues per turn; α-carboxyl O of one residue H-bonds to α-amino H of residue three positions away
β-pleated sheet	Extended backbone with H-bonds between distant residues or separate chains	Side groups above/below the plane; can be parallel (same direction) or antiparallel (opposite direction)

α-helix details:

• All polar NH and CO groups in the backbone are hydrogen-bonded inside the helix.
• This makes α-helices favorable for hydrophobic transmembrane regions: polar backbone groups don't interact with the hydrophobic membrane interior; hydrophobic R-groups radiate outward to interact with the membrane.
• Proline forces a kink in the chain, so it never appears within an α-helix.

β-sheet details:

• Hydrogen bonding occurs between residues distant in the chain or on separate chains.
• If a single polypeptide folds once and forms a β-sheet with itself, it is antiparallel (the two strands run in opposite directions).

Effect of disrupting H-bonds:

• Molecules like urea disrupt hydrogen bonding → disrupt secondary (and higher) structure → denature the protein.

🌐 Tertiary (3°) structure: R-group interactions

Interactions between amino acid residues located distantly in the chain; driven by hydrophobic/hydrophilic interactions.

Key principle:

• Hydrophobic R-groups fold into the interior (away from water).
• Hydrophilic R-groups are exposed on the surface (in contact with water).
• This folding is spontaneous under the right conditions, driven by hydrophobic avoidance of water and hydrogen bonding, reaching the lowest energy conformation.

Anfinsen experiment (ribonuclease):

• Ribonuclease has 8 cysteines forming 4 disulfide bonds.
• Reducing agent (β-mercaptoethanol) breaks disulfide bonds → disrupts tertiary structure.
• If disulfides only lock in place a structure that forms first on its own, the reducing agent would not prevent correct folding (the protein would refold correctly once the reducing agent is removed).
• Correct refolding order: (1) reduce disulfides, (2) denature with urea, (3) remove urea (allow refolding), (4) remove reducing agent (allow disulfides to reform → correct structure).
• Incorrect order (1, 2, 4, 3): if you reform disulfides before removing urea, the protein is still denatured and disulfides form randomly → incorrect structure.

Examples of tertiary structure:

• van der Waals interactions between distant Phe R-groups (yes).
• Hydrogen bonds between backbone groups (no—that's secondary structure).
• Disulfide bonds between distant cysteines (yes, though disulfides are covalent, they form after 2° and before 4°, so are considered part of 3°).

Effect of hydrophobic solvent:

• Dissolving a globular protein in hexane (hydrophobic) would reverse the normal folding: hydrophobic R-groups would move to the surface (to interact with hexane), and hydrophilic R-groups would cluster in the interior → disrupts tertiary structure.

🧩 Quaternary (4°) structure: subunit assembly

The arrangement of multiple polypeptide subunits in a multisubunit complex.

• A subunit is a single polypeptide chain in a large complex.
• Example: mammalian RNA polymerase II has 12 subunits; hemoglobin has 4 subunits (cooperative oxygen binding).

Bonds involved:

• Generally the same as 2° and 3°: non-covalent (hydrogen bonds, van der Waals).
• Covalent bonds (e.g., disulfide bridges in antibodies) may also stabilize quaternary structure.
• Key distinction: the peptide bond is never involved in quaternary structure—it defines sequence (1° structure).

Disulfide in 4° vs 3°:

• Tertiary: disulfide between cysteines in the same polypeptide chain.
• Quaternary: disulfide between cysteines in different polypeptide chains (holding subunits together).

🔥 Denaturation vs hydrolysis

🔥 Denaturation

Disruption of a protein's shape without breaking peptide bonds.

Causes:

• Urea (disrupts hydrogen bonding).
• Extremes of pH.
• Extremes of temperature.
• Changes in salt concentration (tonicity).

Result:

• Loss of secondary, tertiary, and quaternary structure.
• Protein becomes non-functional.
• Primary structure (sequence) remains intact.

✂️ Hydrolysis

• Breaks peptide bonds (cleaves the backbone).
• Acid hydrolysis: non-specific, breaks all peptide bonds → amino acids.
• Proteolytic cleavage: specific, cuts at certain residues → fragments.

Don't confuse:

• Denaturation = shape disruption, peptide bonds intact.
• Hydrolysis = peptide bond cleavage, backbone broken.

🧭 Overview

🧠 One-sentence thesis

Carbohydrates serve as the principal energy source for cellular metabolism and as structural building blocks, with their function determined by stereochemistry and the specific types of glycosidic linkages between sugar units.

📌 Key points (3–5)

• What carbohydrates are: chains of hydrated carbon atoms (C_n H_2n O_n) that can be oxidized to release large amounts of energy.
• Structural hierarchy: monosaccharides (single sugars) join to form disaccharides, oligosaccharides, and polysaccharides through glycosidic linkages.
• Stereochemistry matters: all sugars in our bodies have the D configuration; enzyme specificity for α vs β linkages explains why we can digest starch but not cellulose.
• Common confusion: hemiacetals vs acetals—hemiacetals are in constant equilibrium with the open carbonyl form and can mutarotate; acetals are stable and cannot mutarotate until an enzyme breaks the bond.
• Why linkage type matters: the same monosaccharide (glucose) forms digestible starch (α-1,4 linkages) or indigestible cellulose (β-1,4 linkages), demonstrating the biological significance of stereochemistry.

🧱 Basic structure and nomenclature

🧱 What carbohydrates are

Carbohydrates: chains of hydrated carbon atoms with the molecular formula C_n H_2n O_n, usually beginning with an aldehyde or ketone and continuing as a polyalcohol in which each carbon has a hydroxyl substituent.

• Produced by photosynthesis in plants and biochemical synthesis in animals.
• Can be broken down to CO₂ in oxidation (burning/combustion), releasing large amounts of energy.
• Serve as the principle energy source for cellular metabolism.
• Glucose polymers form cellulose, the building block of wood and cotton.

🏷️ Naming monosaccharides

Monosaccharides receive a two-part name:

Part	Meaning	Examples
First part	"aldo" (aldehyde) or "keto" (ketone)	Aldose vs ketose
Second part	Number of carbons	Triose (3), tetrose (4), pentose (5), hexose (6), heptose (7)

• Example: glucose is an aldohexose (six-carbon chain beginning in an aldehyde).
• "Glucose" and "fructose" are common names; IUPAC nomenclature is not usually used because systematic names are too long.
• Carbons are numbered beginning with carbon #1 at the most oxidized end (the aldehyde or ketone end).

🔄 Hierarchy of carbohydrate polymers

• Monosaccharide: a single carbohydrate molecule (simple sugar)
• Disaccharide: two monosaccharides bonded together
• Oligosaccharide: several monosaccharides bonded together
• Polysaccharide: many monosaccharides bonded together

• Strong acid hydrolyzes these polymers to monosaccharides, which are not further hydrolyzed.

🔬 Stereochemistry and configuration

🪞 D vs L configuration

• Carbohydrates contain chiral carbons and are classified as D or L based on the configuration of the last chiral carbon in the chain (farthest from the aldehyde or ketone).
• This configuration is determined by comparison with glyceraldehyde.
• If a monosaccharide's last chiral carbon matches the chiral carbon of D-glyceraldehyde, it receives the "D" label.
• The sugars in our bodies have the D configuration.
• Remember: we have only L-amino acids and only D-sugars in our bodies.

Drawing convention (Fischer projection):

• Put the aldehyde or ketone on top and the CH₂OH group (last carbon) on the bottom.
• The last chiral carbon will have its OH on the Left for L monosaccharides.

🧬 Stereoisomer relationships

For a given class of monosaccharide, there are 2^n different stereoisomers, where n is the number of chiral carbons.

Venn diagram hierarchy (from broadest to most specific):

• Isomers: same atoms but different bonds (unless also stereoisomers)
• Stereoisomers: same atoms and bonds, but different bond geometries
  • Enantiomers: mirror-image stereoisomers
  • Diastereomers: non-mirror-image stereoisomers
    • Epimers: a subset of diastereomers (differ at one chiral center)

Don't confuse: The excerpt notes these terms were discussed in Chapter 4 and provides this concise categorization.

🔄 Cyclic structures and ring formation

🔄 Why sugars form rings

• In solution, hexoses and pentoses spontaneously form five- and six-membered rings.
• Cyclic structures are thermodynamically favored; only a small percentage exist in the open chain form.
• Pyranoses: six-membered ring structures (resembling pyran)
• Furanoses: five-membered ring structures (resembling furan)

🧪 Hemiacetal formation mechanism

The ring forms when the OH on C5 nucleophilically attacks the carbonyl carbon (C1), forming a hemiacetal.

• Hemiacetal: has one −OR group and one −OH group; in constant equilibrium with the carbonyl form
• Acetal: has two −OR groups; quite stable, requiring an enzyme to react

Key difference: The hemiacetal is in constant equilibrium with the carbonyl form; the acetal form is stable and does not equilibrate without enzymatic catalysis.

Example: In glucose, the C5-hydroxyl group attacks the C1 carbonyl carbon, creating a six-membered ring (glucopyranose).

🔀 Anomers and mutarotation

• Anomers: two different ring structures (α and β) formed depending on which face of the carbonyl the C5-hydroxyl group attacks
• Anomeric carbon: always the carbonyl carbon (C1 in aldoses, C2 in ketoses)
• Mutarotation: the interconversion between the two anomers

How to distinguish α from β:

• If the attack comes from one face, the carbonyl oxygen becomes an equatorial hydroxyl group.
• If from the other face, it becomes an axial hydroxyl group.
• Memory aid: "It's always better to βE up (happy)!" → in β-D-Glucose, the anomeric hydroxyl group is up (equatorial).

Chair conformation details:

• Groups on the left in Fischer notation are above the ring in chair notation.
• Axial substituents point straight up or down.
• Equatorial substituents point out from the ring (less steric hindrance, thermodynamically more favorable).

📐 Drawing conventions

Haworth notation conversion (from Fischer projection):

1. Draw the basic structure of the sugar.
2. If D-sugar: place −CH₂OH above the ring on the carbon to the left of the oxygen. If L-sugar: place it below.
3. For α-sugar: place −OH below the ring on the carbon to the right of the ring oxygen. For β-sugar: place it above.
4. −OH groups on the right go below the ring; those on the left go above, using the −CH₂OH group as the reference point.

🔗 Glycosidic linkages and polysaccharides

🔗 How sugars bond together

Glycosidic linkage: the covalent bond between two sugar molecules, formed in a dehydration reaction that requires enzymatic catalysis.

• Typically joins C1 of one pyranose or furanose to C4 (sometimes C2 or C6) of another through an oxygen atom.
• Once formed, the anomeric carbon is in an acetal configuration (two −OR constituents).
• Significance: The glycosidic linkage stays in the α or β configuration until an enzyme breaks the bond, because the acetal is stable.
• Once a monosaccharide has attacked another sugar to form a glycosidic linkage, it is no longer free to mutarotate.

Naming convention: Linkages are named by which carbon in each sugar comprises the linkage, plus the configuration (α or β).

• Example: lactose has a galactose-β-1,4-glucose linkage.

🌾 Important polysaccharides

Polysaccharide	Composition	Linkage type	Function	Found in
Glycogen	Thousands of glucose units	α-1,4 (with α-1,6 branches)	Energy storage	Animals
Starch	Glucose units	α-1,4 (branches slightly different)	Energy storage	Plants
Cellulose	Cellobiose polymer	β-1,4	Structural (long, straight, fibrous)	Wood, cotton

Common disaccharides (do NOT memorize these linkages for the MCAT):

• Sucrose (table sugar): Glc-α-1,2-Fru
• Lactose (milk sugar): Gal-β-1,4-Glc
• Maltose: Glc-α-1,4-Glc
• Cellobiose: Glc-β-1,4-Glc

🔪 Hydrolysis of glycosidic linkages

• Polysaccharides must be hydrolyzed into monosaccharides to enter metabolic pathways (e.g., glycolysis) and be used for energy.
• Hydrolysis: water is the nucleophile, and one sugar is the leaving group (the one that was the attacker during bond formation).
• The cleavage reaction is precisely the reverse of the formation reaction.
• Hydrolysis is thermodynamically favored (releases energy) but does not occur at a significant rate without enzymatic catalysis.
• Enzymes are named for the sugar they hydrolyze (e.g., maltase hydrolyzes maltose into two glucose monosaccharides).

Enzyme specificity is critical:

• Each enzyme is highly specific for its linkage.
• This specificity is a great example of the significance of stereochemistry.

🐄 Why we can't digest cellulose but cows can

• A cotton T-shirt is pure sugar (cellulose = glucose polymer).
• We can't digest it because mammalian enzymes can't hydrolyze β-glycosidic linkages that make cellobiose from glucose.
• Cellulose is the energy source in grass and hay.
• Cows depend on bacteria in an extra stomach (rumen) to digest cellulose for them.

Exception: Humans can digest lactose (which has a β linkage) because we have a specific enzyme, lactase.

• This is an exception to the rule that mammalian enzymes cannot hydrolyze β-glycosidic linkages.
• People without lactase are lactose malabsorbers; lactose ends up in the colon, potentially causing gas and diarrhea (lactose intolerance).
• Most adults naturally stop making lactase after childhood.

Don't confuse: The inability to digest cellulose is about β-linkages in general; lactase is a specific exception for lactose only.

🧪 Reducing sugars and Benedict's test

🧪 What Benedict's test detects

Benedict's test: a chemical assay that detects the carbonyl units of sugars; distinguishes hemiacetals from acetals (only hemiacetals are in equilibrium with the carbonyl/open-chain form).

How it works:

• Benedict's reagent (oxidized copper, Cu²⁺) oxidizes a sugar's aldehyde or ketone to the corresponding carboxylic acid.
• Yields a reddish precipitate.
• The Cu²⁺ is reduced to Cu⁺ while the sugar is oxidized.

🔴 What is a reducing sugar

Reducing sugar: any carbohydrate that can be oxidized by Benedict's reagent (it reduces the Cu²⁺ to Cu⁺ while itself being oxidized).

Which sugars are reducing:

• All monosaccharides are reducing sugars.
• More generally, all aldehydes, ketones, and hemiacetals give a positive result.
• Acetals give a negative result because they do not react with Cu²⁺ and are not in equilibrium with the open-chain (carbonyl) form.

🔬 Practical application

Benedict's test distinguishes free monosaccharides from polysaccharides:

• Free glucose (monosaccharide form): many hemiacetals → strongly positive Benedict's test.
• Glycogen polymer: all glucose units tied up in acetal linkages except the very first one in the chain → weakly positive Benedict's test.

Key principle: Once a monosaccharide has attacked another sugar to form a glycosidic linkage, its anomeric carbon is in an acetal configuration and is thus no longer free to mutarotate nor to reduce Benedict's reagent.

🧮 Equilibrium and oxidation

If 98% of a monosaccharide is present as the ring form at equilibrium in solution, all of the sugar can still be oxidized in Benedict's reaction.

• The excerpt implies that as the open-chain form reacts, the equilibrium shifts to replace it, allowing complete oxidation over time.

🥛 Reducing vs non-reducing disaccharides

• Lactose (Gal-β-1,4-Glc): The glucose unit retains a free anomeric carbon (hemiacetal) → reducing sugar.
• Sucrose (Glc-α-1,2-Fru): Both anomeric carbons are involved in the glycosidic linkage (both in acetal form) → non-reducing sugar.

Don't confuse: A disaccharide is reducing if at least one anomeric carbon remains as a hemiacetal (free to equilibrate with the carbonyl form); if both anomeric carbons are locked in the glycosidic bond, it is non-reducing.

🧭 Overview

🧠 One-sentence thesis

Lipids are hydrophobic molecules that serve three key physiological roles—forming membrane barriers, storing energy as fats, and building steroid hormones—all driven by their water-fearing nature.

📌 Key points (3–5)

• Cardinal characteristic: hydrophobicity (water-fearing) is the defining feature of all lipids; they contain mostly nonpolar C–C and C–H bonds.
• Three physiological roles: phospholipids form cellular membranes, triglycerides store energy in adipose cells, and cholesterol is the precursor for steroid hormones.
• Energy storage advantage: fats store more energy than carbohydrates because they pack more densely (no water of solvation) and are more reduced (release more energy when oxidized).
• Common confusion: hydrophobic = lipophilic (lipid-loving); hydrophilic = lipophobic (lipid-fearing)—these are synonyms, not opposites.
• Membrane fluidity determinants: degree of saturation, fatty acid tail length, and cholesterol content all modulate how fluid or solid a membrane is.

🧪 Hydrophobicity and the nature of lipids

💧 What hydrophobic means

Hydrophobic means water-fearing.

• Water is very polar, so polar substances dissolve well (hydrophilic = water-loving).
• Carbon–carbon and carbon–hydrogen bonds are nonpolar, so substances with only C and H do not dissolve in water.
• Example: table sugar dissolves in water, but cooking oil floats or forms droplets; cotton T-shirts (glucose polymer cellulose) get wet, but nylon jackets (nonpolar covalent atoms) do not.

🔄 Synonyms to remember

• Hydrophobic = lipophilic (lipid-loving).
• Hydrophilic = lipophobic (lipid-fearing).
• Don't confuse: these pairs are equivalent terms, not opposing concepts.

🧬 Fatty acids and their structures

🧬 Basic fatty acid structure

Fatty acids are composed of long unsubstituted alkanes that end in a carboxylic acid.

• Typically 14 to 18 carbons long.
• Synthesized two carbons at a time from acetate, so only even-numbered fatty acids are made in human cells.

🔗 Saturated vs unsaturated fatty acids

Type	Definition	Key feature
Saturated	No carbon–carbon double bonds	Every carbon bound to the maximum number of hydrogens
Unsaturated	One or more double bonds in the tail	Double bonds are almost always (Z) or cis

• Nomenclature: position of a double bond is denoted by Δ and the number of the first carbon in the bond (counting from the carboxylic acid carbon). Example: a (Z) double bond between carbons 3 and 4 is called (Z)-Δ³ or cis-Δ³.

🌀 Shape differences

• Unsaturated fatty acids have kinks due to cis double bonds, disrupting orderly packing.
• Saturated fatty acids are straight and pack more tightly.

🧼 Micelles and hydrophobic interactions

• When fatty acids are mixed into water, they associate into a micelle: a spherical structure with hydrophobic tails hidden inside and polar carboxylic acid heads facing the water.
• Hydrophobic interaction: water molecules form an orderly solvation shell around hydrophobic substances to maximize water–water interaction and minimize water–lipid interaction.
• Forming a solvation shell decreases entropy (ΔS < 0), which is unfavorable; micelles minimize this by hiding tails inside a sphere.

🧼 Soaps

Soaps are the sodium salts of fatty acids (RCOO⁻ Na⁺). They are amphipathic, which means both hydrophilic and hydrophobic.

• Soap helps remove grease by solubilizing oils (hydrophobic tails interact with grease) while remaining water-soluble (hydrophilic heads interact with water).

🔋 Triglycerides: energy storage

🔋 Triglyceride structure

The technical name for fat is triacylglycerol or triglyceride.

• Composed of three fatty acids esterified to a glycerol molecule.
• Glycerol is a three-carbon triol (HOCH₂–CHOH–CH₂OH) with three hydroxyl groups.
• Fatty acids are stored as triglycerides because free fatty acids are reactive chemicals.

⚡ Why fats are efficient energy storage

Fats are more efficient than carbohydrates for two reasons:

1. Packing: Hydrophobicity allows fats to pack much more closely than carbohydrates. Carbohydrates carry water-of-solvation (water molecules hydrogen-bonded to hydroxyl groups), so the amount of carbon per unit weight is much greater in a fat droplet than in dissolved sugar.
2. Energy content: Fats are much more reduced than carbohydrates. Since energy metabolism begins with oxidation of foodstuffs, and carbohydrates are more oxidized to start with, oxidizing them releases less energy.

• Animals store most energy as fat, with only a small amount as glycogen.
• Plants (e.g., potatoes) commonly store a large percentage as carbohydrates (starch).

🧪 Saponification

• Saponification: base-catalyzed hydrolysis of triglycerides into fatty acid salts (soaps).
• Triglycerides undergo reactions typical of esters, such as base-catalyzed hydrolysis.
• Lipases: enzymes that hydrolyze fats.

🧱 Phospholipids and membrane bilayers

🧱 Phospholipid structure

Membrane lipids are phospholipids derived from diacylglycerol phosphate (DG-P).

• Example: phosphatidyl choline is formed by esterification of a choline molecule to the phosphate group of DG-P.
• Phospholipids are detergents: substances that efficiently solubilize oils while remaining highly water-soluble (stronger than soaps).

🧱 Lipid bilayer formation

• Phospholipids spontaneously form a lipid bilayer to minimize interactions with water.
• Hydrophobic interactions drive formation; van der Waals forces between long tails stabilize the bilayer.
• The bilayer acts like a seal around the cell, impermeable to charged particles (e.g., Na⁺); protein gateways (ion channels) are required for ions to enter or exit.

🌡️ Membrane fluidity

Three structural determinants control membrane fluidity:

1. Degree of saturation: Unsaturation (double bonds) increases fluidity by disrupting orderly packing of hydrophobic tails, decreasing the melting point. Saturated fatty acids have more van der Waals interactions with neighboring chains, making membranes less fluid.
2. Tail length: Shorter fatty acid tails increase fluidity.
3. Cholesterol content: At low temperatures, cholesterol increases fluidity (acts as membrane antifreeze); at high temperatures, it reduces fluidity. Overall, cholesterol keeps fluidity at an optimum level.

• Don't confuse: the same membrane can be more or less fluid depending on temperature and composition.

📡 Membrane proteins

• Proteins integrated into membranes transmit signals from outside to inside the cell.
• Example: peptide hormones cannot pass through the membrane due to their charged nature; instead, protein receptors bind them and transmit a signal via a second messenger cascade.

🧬 Terpenes and terpenoids

🧬 Terpene definition

A terpene is a member of a broad class of compounds built from isoprene units (C₅H₈) with a general formula (C₅H₈)ₙ.

• Terpenes may be linear or cyclic.
• Classified by the number of isoprene units:
  • Monoterpenes: two isoprene units
  • Sesquiterpenes: three isoprene units
  • Diterpenes: four isoprene units
  • Triterpenes: six isoprene units (e.g., squalene, used in steroid biosynthesis)

🧬 Terpenoids

Terpenoids are functionalized terpenes built from an isoprene skeleton and functionalized with other elements (O, N, S, etc.).

• Example: Vitamin A (C₂₀H₃₀O) is a terpenoid.

🧪 Steroids and cholesterol

🧪 Steroid structure

• All steroids have a basic tetracyclic ring system based on cholesterol, a polycyclic amphipath (several rings; both hydrophilic and hydrophobic characteristics).
• Steroids are included with lipids because of their hydrophobicity.

🧪 Cholesterol roles

• Membrane component: an important modulator of lipid bilayer fluidity (see above).
• Obtained from diet and synthesized in the liver.
• Carried in the blood packaged with fats and proteins into lipoproteins; one type implicated in atherosclerotic vascular disease (cholesterol plaque build-up in blood vessels).

💊 Steroid hormones

• Steroid hormones are made from cholesterol.
• Examples: testosterone (androgen/male sex hormone) and estradiol (estrogen/female sex hormone).
• Key difference from peptide hormones:
  • Steroid hormones are highly hydrophobic, so they diffuse right through the lipid bilayer into the cytoplasm; receptors are located within cells.
  • Peptide hormones (e.g., insulin) are charged and cannot pass through the membrane; they bind to receptors on the cell surface and trigger second messenger cascades.
• Don't confuse: no receptors for steroid hormones exist on the cell surface; they act intracellularly.

🧭 Overview

🧠 One-sentence thesis

Nucleotides—the building blocks of DNA and RNA—are composed of a sugar, a nitrogenous base, and phosphate groups, with ATP serving as the universal short-term energy storage molecule through its high-energy phosphate bonds.

📌 Key points (3–5)

• Phosphate bonds store energy: Pyrophosphate (two linked phosphates) contains high-energy bonds that release ~7 kcal/mol (or ~12 kcal/mol in cells) when hydrolyzed.
• Why phosphate bonds are high-energy: Three reasons—negative charge repulsion, more resonance forms in free orthophosphate, and better water interaction after hydrolysis.
• Nucleotide structure: Each nucleotide has three parts—a ribose (or deoxyribose) sugar, a purine or pyrimidine base attached to carbon 1, and one to three phosphate units attached to carbon 5.
• ATP's dual role: ATP is both an RNA precursor and the universal energy currency that stores energy from food oxidation for immediate cellular use.
• Common confusion: Don't confuse the number of phosphates—ATP has three phosphates (triphosphate), and the bonds between them (phosphoanhydride bonds) are what store the energy, not the bond to the sugar.

🔋 Phosphorus compounds and energy storage

⚡ Phosphoric acid basics

Phosphoric acid: An inorganic acid (no carbon) that can donate three protons, with pKₐ values of 2.1, 7.2, and 12.4.

• At physiological pH (~7.4), phosphoric acid is significantly dissociated and exists largely as a negatively charged anion (phosphate).
• This anionic form is also called orthophosphate.

🔗 Pyrophosphate formation

Pyrophosphate: Two orthophosphates linked together via an anhydride linkage (P–O–P bond).

• The P–O–P bond is a high-energy phosphate bond.
• "High-energy" means hydrolysis (breaking the bond with water) is thermodynamically very favorable.
• The ΔG° for pyrophosphate hydrolysis is about −7 kcal/mol under standard conditions, and about −12 kcal/mol in the cell (even more favorable).

🌀 Why phosphate bonds store so much energy

Three reasons make phosphate anhydride bonds act like "compressed springs":

1. Charge repulsion: When phosphates are linked, their negative charges strongly repel each other, creating instability.
2. Resonance stabilization: Free orthophosphate (after hydrolysis) has more resonance forms, lowering its free energy compared to linked phosphates.
3. Solvation: Orthophosphate interacts more favorably with water than linked phosphates do.

Key mental image: Think of linked phosphates as compressed springs waiting to fly open and release energy for enzyme-catalyzed reactions.

Example: When a cell needs energy to drive a reaction, breaking one phosphate bond from ATP provides the "push" needed.

🧬 Nucleotide structure

🧱 Three components of nucleotides

Nucleotides: The building blocks of nucleic acids (RNA and DNA), each containing three parts.

Component	Details
Sugar	Ribose (in RNA) or deoxyribose (in DNA)
Base	Purine or pyrimidine, attached to carbon 1 of the sugar ring
Phosphate(s)	One, two, or three phosphate units attached to carbon 5 of the sugar ring

• The sugar provides the backbone structure.
• The base (purine or pyrimidine) carries genetic information.
• The phosphate groups provide energy storage (when multiple phosphates are present).

🔢 Naming by phosphate number

• One phosphate: nucleotide monophosphate
• Two phosphates: nucleotide diphosphate
• Three phosphates: nucleotide triphosphate (like ATP)

Don't confuse: The number refers to how many phosphate groups are attached to carbon 5 of the sugar, not the total number of phosphorus atoms in the molecule.

⚡ ATP as the universal energy currency

💰 ATP structure and function

ATP (Adenosine Triphosphate): The universal short-term energy storage molecule, consisting of adenine (base), ribose (sugar), and three phosphate groups.

• ATP is both a nucleotide (RNA precursor) and an energy carrier.
• Energy extracted from oxidizing food is immediately stored in ATP's phosphoanhydride bonds.
• This stored energy later powers cellular processes or is used to synthesize longer-term storage molecules (glucose, fats).

🌍 Universal across life

• All living organisms use ATP, from bacteria to humans.
• Even some viruses carry ATP with them outside the host cell (though viruses cannot make their own ATP).
• This universality makes ATP the "energy currency" of life—every cell "spends" ATP to do work.

Example: When a muscle cell contracts, it breaks down ATP to release energy; when a plant cell builds glucose during photosynthesis, it uses ATP energy to drive the synthesis reaction.

🔄 Energy flow summary

1. Food oxidation releases energy.
2. Energy is captured in ATP's phosphate bonds.
3. ATP is broken down (hydrolyzed) to release energy when needed.
4. Energy drives cellular work or builds storage molecules (glucose, fats).

Don't confuse: ATP is for short-term storage; glucose and fats are for long-term storage. ATP is like cash in your wallet; glucose/fats are like money in the bank.