Organic Chemistry

Organic Chemistry for the MCAT

Part 1 of 7 — Functional Groups & Stereochemistry

Must-Know Functional Groups

Stereochemistry

• Chirality: 4 different groups on a carbon → chiral center
• Enantiomers: Non-superimposable mirror images (same physical properties except optical rotation)
• Diastereomers: Stereoisomers that are NOT mirror images (different physical properties)
• Meso compounds: Have chiral centers but an internal plane of symmetry → optically inactive

R/S Assignment (Cahn-Ingold-Prelog)

1. Assign priority by atomic number (highest = 1)
2. Orient lowest priority group away from you
3. 1→2→3 clockwise = R; counterclockwise = S

Stereochemical Relationships You Must Distinguish

• Constitutional isomers: same formula, different connectivity
• Stereoisomers: same connectivity, different 3D arrangement
• Conformational isomers: interconvert by bond rotation (usually not isolated)

On the MCAT, many questions hide stereochemistry inside a passage about receptor binding where only one stereoisomer is biologically active.

Functional Groups & Stereochem 🎯

Key Takeaways — Part 1

• Know ALL functional groups instantly — they appear in every MCAT passage
• $2^n$ rule for maximum stereoisomers
• Enantiomers: mirror images, same properties (except rotation). Diastereomers: different properties.
• Amide = peptide bond — this connects to biochemistry
• Always classify relationship first: constitutional vs stereoisomer vs conformer.

Worked Examples — Functional Groups & Stereochemistry

Question: Assign R or S to this chiral carbon:

Where: 1=Cl, 2=$CH_{3}$, 3=OH, 4=H

Solution:

1. Assign atomic numbers: Cl (17) > OH (8) > $CH_{3}$ (6) > H (1)

   • Priority 1 = Cl
   • Priority 2 = OH
   • Priority 3 = $CH_{3}$
   • Priority 4 = H

2. Orient H (priority 4) away from viewer ✓ (already shown)

3. Trace 1→2→3:

   • 1 (Cl) → 2 (OH) → 3 $(CH_{3})$
   • This traces counterclockwise = S-configuration

MCAT Strategy: Draw/mentally rotate to get H in back. Then trace 1→2→3 path. Clockwise=R, counterclockwise=S. Practice this 10× before test day.

Question: How many stereoisomers exist for 2,3,4-trihydroxybutanal?

Solution:

1. Identify chiral centers: C2 and C3 have 4 different groups each → 2 chiral centers
2. Maximum stereoisomers = $2^n = 2^2 = 4$
3. These are: 2R,3R / 2R,3S / 2S,3R / 2S,3S

MCAT Strategy: $2^n$ rule doesn't account for meso compounds (which reduce the number). For this molecule, check if any stereoisomer has an internal plane of symmetry. (It doesn't, so answer = 4.)

Question: Molecules A and B both have formula $C_{4}H_{8}Cl_{2}$. A has both Cl atoms on C1 (geminal), and B has Cl atoms on C1 and C2. What is their relationship?

Solution:

1. Connectivity differs: A = 1,1-dichlorobutane; B = 1,2-dichlorobutane
2. Different connectivity means constitutional isomers (not stereoisomers)
3. They are NOT related by stereochemistry alone

MCAT Strategy: If the atoms are in different positions (different connectivity), don't even look at stereochemistry—they're constitutional isomers.

If they had same connectivity, different 3D arrangement? Then determine stereoisomer relationship: enantiomers (mirror images) or diastereomers (not mirror images).

Question: Is this molecule chiral?

(Assuming this is drawn with internal plane symmetry)

Solution:

1. It has 2 chiral centers (both carbons have 4 different groups)
2. BUT if you draw it in 3D, the molecule has an internal plane of symmetry
3. The left half is the mirror image of the right half
4. Result: Chiral centers exist, but molecule is achiral overall (meso compound)
5. The molecule does NOT rotate plane-polarized light

MCAT Strategy: Meso compounds are rare on the MCAT, but they're a "gotcha." Always ask: "Does this have a plane of symmetry?" If yes, it's achiral despite chiral centers.

Organic Chemistry for the MCAT

Part 2 of 7 — SN1, SN2, E1, E2 Reactions

Substitution vs. Elimination Decision Tree

Key Points

• SN2: Rate = $k[\text{substrate}][\text{nuc}]$. Backside attack → inversion. Sterically hindered substrates slow it.
• SN1: Rate = $k[\text{substrate}]$. Carbocation intermediate → racemization. Favored by 3° substrates, polar protic solvents.
• E2: Strong BULKY base (t-BuOK) favors elimination over substitution. Anti-periplanar geometry required.
• E1: Shares carbocation intermediate with SN1. Heat favors elimination.

Fast Test-Day Decision Rules

• If you see methyl/1 degree + strong nucleophile + aprotic solvent, think SN2 first.
• If you see 3 degree substrate + protic solvent, think SN1/E1 competition.
• If you see strong bulky base + heat, think E2 (often Hofmann product favored).

Reaction Mechanisms 🎯

Key Takeaways — Part 2

• SN2: strong nuc + methyl/1° + polar aprotic → inversion
• SN1: weak nuc + 3° + polar protic → racemization
• E2: strong bulky base + 2°/3° → Zaitsev product
• The SN1/E1/SN2/E2 decision chart is GUARANTEED on the MCAT
• Tie mechanism choice to substrate class first, then reagent/solvent/temperature.

Worked Examples — Substitution & Elimination

Question: Predict the main product(s) when 2-bromopropane is treated with potassium ethoxide (KOEt) in ethanol at 75°C:

Solution:

1. Substrate class: 2° (secondary) – can undergo SN2, E2, SN1, or E1
2. Nucleophile/Base: KOEt is a strong base (not bulky)
3. Solvent: Ethanol – polar protic (the conjugate-acid solvent of the base)
4. Temperature: 75°C – high temperature favors elimination

Decision:

• Strong base + heat on a 2° substrate → E2 mechanism (elimination dominates)
• SN2 is only a minor competitor, because heat and a strong base favor elimination at a 2° carbon
• Product: Mainly propene (the only alkene this substrate can form)

MCAT Strategy: A strong base plus heat on a 2° substrate is shorthand for E2, even when the base is not bulky; a bulky base (t-BuOK) pushes further toward elimination and the less-substituted (Hofmann) alkene. Prioritize substrate class first, then base strength, bulk, and temperature.

Question: 1-bromo-1-methylcyclohexane is dissolved in water at room temperature. What is the major product?

Solution:

1. Substrate class: 3° – strongly favors SN1/E1
2. Nucleophile: Water – weak nucleophile, weak base
3. Solvent: Water – polar protic (stabilizes carbocation)
4. Temperature: Room temperature – doesn't favor elimination overly

Decision:

• Carbocation forms readily → SN1/E1 competition
• Water attacks the carbocation → SN1 product (3° alcohol) dominates at room temp
• E1 product (alkene) forms as minor by-product

Products: ~70% (1-methylcyclohexan-1-ol); ~30% (1-methylcyclohexene)

MCAT Strategy: At room temperature in water, nucleophilic attack dominates. Alcohol is major product. If the question said "heat," elimination would increase.

Question: (1S,2S)-1-bromo-2-methylcyclohexane is treated with KCN (strong nucleophile) in DMSO. What happens?

Solution:

1. Substrate: 2° but IN A RING, with a methyl group on the adjacent carbon (C2)
2. Nucleophile: $CN^{-}$ – strong, small, polar aprotic solvent
3. Expected mechanism: SN2 (good nucleophile + aprotic solvent)

BUT: The ring + adjacent methyl makes backside attack slower. SN2 is retarded but still wins, because $CN^{-}$ is a weak base and DMSO does not support a 2° carbocation.

• Major: SN2 (nitrile product with config inversion at C1)
• Minor: E2 (competes more than on an unhindered substrate)

MCAT Strategy: Steric hindrance strongly disfavors SN2 at 3° carbons and slows it at hindered 2° centers. With a strong, weakly basic nucleophile in a polar aprotic solvent, a hindered 2° substrate still reacts mainly by SN2.

Question: (R)-2-iodooctane in aqueous $AgNO_{3}$ (SN1 conditions) produces a 50:50 mixture of (R) and (S) alcohols. Why not 100% inversion or 100% retention?

Solution:

1. $Ag^{+}$ abstracts $I^{-}$ → carbocation forms
2. Carbocation is planar $(sp^{2} hybridized)$
3. Water can attack from either face (above or below the plane)
4. ~50% attack from above → (S) enantiomer
5. ~50% attack from below → (R) enantiomer
6. Result: Racemic mixture (although sometimes slightly favor one direction)

Why not 100% one product?

• The carbocation has no stereochemistry (planar, achiral)
• Once it forms, the stereochemical information is lost
• Product ratio depends on attack from both faces

MCAT Strategy: SN1 → racemization. Inversion suggests SN2. Retention is chemically rare (rearrangement-driven).

Organic Chemistry for the MCAT

Part 3 of 7 — Carbonyl Chemistry

Carbonyl Reactivity

The $C=O$ is polar: carbon is electrophilic (attacked by nucleophiles).

Aldol Condensation

$\text{Enolate} + \text{Aldehyde} \to \beta\text{-hydroxy carbonyl} \xrightarrow{\text{heat}} \alpha,\beta\text{-unsaturated carbonyl}$

Key Carbonyl Reactions

MCAT High Yield: Reducing Agents

• NaBH$_4$: mild, reduces aldehydes and ketones only
• LiAlH$_4$: strong, reduces ALL carbonyls (including esters, carboxylic acids, amides)

Nucleophilic Addition vs Acyl Substitution

• Aldehydes/ketones: nucleophilic addition (no leaving group on carbonyl carbon)
• Carboxylic acid derivatives: nucleophilic acyl substitution (tetrahedral intermediate + leaving group)

Recognizing whether a leaving group is present is often enough to choose the mechanism class.

Carbonyl Chemistry 🎯

Key Takeaways — Part 3

• NaBH$_4$: mild (aldehydes/ketones only). LiAlH$_4$: strong (everything).
• PCC: mild oxidation (1° ROH → aldehyde). Jones/CrO$_3$: full oxidation.
• Fischer esterification: carboxylic acid + alcohol + acid catalyst → ester + water
• Amide bonds (peptides) are resistant to hydrolysis — that's why enzymes are needed!
• Ask first: addition (aldehyde/ketone) or acyl substitution (derivative with leaving group)?

Worked Examples — Carbonyl Chemistry

Question: You want to reduce ONLY the ketone in this compound without touching the ester:

Which reagent should you use?

Options: $NaBH_{4}$ / $LiAlH_{4}$ / $H_{2}SO_{4}$ / Zn(Hg)/HCl

Solution:

1. Identify functional groups: ketone (left C=O) and ester (right C=O with $OCH_{3}$)
2. Goal: Reduce only ketone to 2° alcohol
3. $NaBH_{4}$: Reduces aldehydes & ketones → leaves esters alone ✓
4. $LiAlH_{4}$: Reduces BOTH ketones AND esters → not selective
5. $H_{2}SO_{4}$: Catalyst for esterification, not reduction
6. Zn(Hg)/HCl: Clemmensen reduction (reduces the ketone C=O all the way to $CH_{2}$, not to an alcohol)

Answer: $NaBH_{4}$ (selective for ketone)

MCAT Strategy: Always memorize: $NaBH_{4}$ = selective for aldehydes/ketones. $LiAlH_{4}$ = no selectivity (overkill). This question pattern appears every 3-4 years on the MCAT.

Question: Start with 1-pentanol. You want to make pentanal (not pentanoic acid). Which oxidizing agent is correct?

Solution:

1. Target: 1° alcohol → aldehyde (oxidation level +1)
2. Wrong: $Jones/CrO_{3}$ (over-oxidizes to carboxylic acid)
3. Wrong: $KMnO_{4}$ (too strong, over-oxidizes)
4. Correct: PCC (pyridinium chlorochromate) — stops at aldehyde
5. Alternative: DMP (Dess-Martin Periodinane), IBX, Swern oxidation — all work

Mechanism:

• Mild oxidation (PCC) → Aldehyde
• Strong oxidation (Jones) → Carboxylic acid

Product: Pentanal $(CH_{3}CH_{2}CH_{2}CH_{2}CHO)$ with no over-oxidation

MCAT Strategy: "1° alcohol needs to be an aldehyde" → PCC is your friend. Practice this memorization: 1° → aldehyde (PCC), 2° → ketone (any oxidizing agent), carboxylic acid (strong oxidation or Jones).

Question: What is the major product when acetaldehyde $(CH_{3}CHO)$ undergoes an aldol condensation followed by dehydration?

Solution:

1. Enolate formation: α-H on acetaldehyde deprotonated by base → enolate ($CH_{2}$=$CHO^{-}$)
2. Nucleophilic attack: Enolate attacks the carbonyl carbon of a second acetaldehyde molecule
3. Aldol adduct: <pre> $CH_{3}-CHOH-CH_{2}-CHO$ (3-hydroxybutanal / aldol adduct) </pre>
4. Dehydration (heat + acid): Water leaves from aldol adduct
5. Final product: 2-butenal (crotonaldehyde), $CH_{3}-CH$=CH-CHO (α,β-unsaturated aldehyde)

Key: The double bond forms between the α-carbon and the carbonyl-bearing carbon

MCAT Strategy: Aldol condensations create a new C-C bond and introduce an α,β-unsaturated carbonyl (which stabilizes via conjugation). The product is usually smaller molecules (acetaldehyde) condensing to form crotonaldehyde.

Question: Predict whether acetone $(CH_{3}COCH_{3})$ reacts via nucleophilic addition (NA) or nucleophilic acyl substitution (NAS) when treated with methylamine $(CH_{3}NH_{2})$:

Solution:

1. Structure of acetone: $CH_3-C(=O)-CH_3$
2. Key question: Is there a leaving group on the carbonyl carbon?
   • CHO carbon is bonded to: C, C, O (no leaving group like Cl, $OCH_{3}$, OAc, etc.)
3. Answer: Nucleophilic addition (NA)
4. Mechanism:
   • Methylamine acts as nucleophile, attacks C=O
   • Intermediate: C-OH tetrahedral intermediate
   • Lone pair on N attacks C, OH leaves
   • Product: Imine ($CH_{3}-N$=$C(CH_{3})_{2}$ or iminium salt initially)
5. Final product: N-methylpropan-2-imine or acetone methyl imine

Compare to acyl substitution:

MCAT Strategy: Ketones/aldehydes → NA (no leaving group). Esters/acid halides/anhydrides → NAS (leaving group present). This is a fundamental distinction tested every year.

Organic Chemistry for the MCAT

Part 4 of 7 — Carboxylic Acid Derivatives

Reactivity Order (most reactive → least)

$\text{Acid halide} > \text{Anhydride} > \text{Ester} > \text{Amide} > \text{Carboxylate}$

Why? The better the leaving group, the more reactive.

• Acid halide: Cl$^-$ is excellent leaving group
• Amide: NH$_2^-$ is terrible leaving group → most stable

Key Interconversions

• Acid halide + ROH → Ester
• Acid halide + RNH$_2$ → Amide (how peptide bonds form in lab!)
• Ester + H$_2$O (acid/base) → Carboxylic acid + ROH (hydrolysis)
• Ester + NaOH → Carboxylate + ROH (saponification = soap making!)

Biochemistry Connection

Thioester (CoA derivatives) are key metabolic intermediates — more reactive than regular esters due to weak C-S bond.

Core Mechanistic Pattern

Most derivative reactions proceed through:

1. Nucleophilic attack on carbonyl carbon
2. Tetrahedral intermediate formation
3. Collapse and leaving-group departure

The best leaving group generally determines the direction and feasibility of interconversion.

Carboxylic Acid Derivatives 🎯

Key Takeaways — Part 4

• Reactivity of acid derivatives: halide > anhydride > ester > amide
• Saponification = base hydrolysis of an ester → soap (carboxylate salt)
• Thioesters (e.g., acetyl-CoA) are biologically activated intermediates
• Peptide bond = amide bond — resistant to hydrolysis (needs enzymes)
• Reaction prediction improves if you compare leaving-group quality first.

Worked Examples — Carboxylic Acid Derivatives

Question: Compare the reactivity of these compounds toward nucleophilic attack:

Solution:

1. Leaving group quality comparison:

   • (A) $Cl^{-}$ = excellent leaving group → most reactive
   • (B) $CH_{3}COO^{-}$ = good leaving group (stabilized by acetyl C=O) → 2nd most reactive
   • (C) $CH_{3}O^{-}$ = moderate leaving group → less reactive than B
   • (D) $NH_{2}^{-}$ = terrible leaving group (basic) → least reactive

2. Resonance effects (secondary factor):

   • (D) $NH_{2}$ donates electron density to C=O, reducing electrophilicity
   • (A) Cl is electron-withdrawing by induction, increasing C=O electrophilicity

3. Reactivity order: (A) > (B) > (C) > (D)

MCAT Strategy: If you forget the exact order, remember: the better the leaving group, the more reactive. Cl > O-C(=O) > O-alkyl > $NH_{2}$.

Question: Methyl acetate $(CH_{3}COOCH_{3})$ reacts with excess aniline $(C_{6}H_{5}NH_{2})$ to form an amide. Show the product and explain why the ester is displaced:

Solution:

1. Nucleophilic acyl substitution mechanism:

   • Aniline (nucleophile) attacks the ester carbonyl
   • Tetrahedral intermediate forms
   • Methoxide $(CH_{3}O^{-})$ is a weaker leaving group than aniline is as a nucleophile
   • But we must compare: Which is the better leaving group?

2. Leaving group comparison:

   • $CH_{3}O^{-}$ $pK_{a}(conjugate acid)$ ≈ 15 (methanol) → decent leaving group
   • $NH_{2}^{-}$ $pK_{a}(conjugate acid)$ ≈ 35 (ammonia) → terrible leaving group

3. Why does the ester react?

   • Aniline is a strong nucleophile (aromatic amine with lone pair on N)
   • Although $NH_{2}^{-}$ is a poor leaving group, the ester is reactive enough
   • Once the amide forms, it is resistant to further reaction (poor leaving group protects it)

4. Product: $CH_{3}-C$(=O)$-NH-C_{6}H_{5}$ (N-phenylacetamide or N-acetylaniline)

Biochemistry connection: Peptides are made this way in lab (activating carboxylic acids), and the resulting amide bonds resist hydrolysis because $NH_{2}^{-}$ is a terrible leaving group. This stability is why enzymes are required to break peptide bonds in the body.

MCAT Strategy: Ester → Amide is thermodynamically favorable because the resulting amide is so unreactive (stability of product).

Question: A triglyceride (fat with 3 ester groups) is treated with excess NaOH in ethanol (soap-making process). What happens?

Solution:

1. Reagent: NaOH = strong base + nucleophile ($OH^{-}$ is attacking)

2. Mechanism: Nucleophilic acyl substitution (ester hydrolysis)

   • $OH^{-}$ attacks the ester C=O
   • Tetrahedral intermediate forms
   • Alkoxide $(R-CH-O^{-})$ leaves via C-O bond cleavage
   • Carboxylate ion $(RCOO^{-})$ is formed

3. Stoichiometry: 1 triglyceride + 3 NaOH → 1 glycerol $(HOCH_{2}CHOH CH_{2}OH)$ + 3 sodium carboxylates (RCOONa, "soap")

4. Products:

   • Glycerol: 3-carbon backbone (reusable for biodiesel)
   • Sodium carboxylates: Long-chain salts (e.g., sodium stearate $C_{17}H_{35}COONa$) = SOAP!
   • pH increases (basic product)

Why saponification is irreversible:

• Carboxylate $(RCOO^{-})$ is stabilized by delocalization
• Reverse reaction would require $O^{2}$ form attack at C=O (bad nucleophile)
• Very thermodynamically favorable

MCAT Strategy: Saponification = ester hydrolysis in basic conditions. Always produces carboxylate salt + alcohol. Often appears with triglycerides or phospholipids in biochemistry passages.

Question: Explain why peptide bonds (amides) require enzymatic hydrolysis in the body, while esters can be hydrolyzed chemically. Use resonance to justify your answer.

Solution:

1. Amide resonance:

   • Nitrogen lone pair ON THE SAME ATOM as the C=O
   • N can donate electron density into π system of C=O
   • Two resonance forms:
     • Form 1: C=O with negative charge on O
     • Form 2: C-O with positive charge on N (C=$N^{+}$)
   • Result: Partial double-bond character in the C-N bond (~40%)

2. Consequence of amide resonance:

   • C=O electrophilicity decreases (less polarized)
   • C-N bond becomes stronger (shorter, more rigid)
   • Nucleophile cannot easily attack C=O
   • Even if attack occurs, $NH_{2}^{-}$ is an extremely poor leaving group

3. Ester comparison:

   • Oxygen lone pair is FARTHER from C=O (on neighboring carbon)
   • Resonance is WEAK
   • C=O remains strongly electrophilic
   • Chemical hydrolysis (mild acid/base) works fine

4. Why peptides need enzymes:

   • Proteases stabilize the tetrahedral intermediate
   • Enzymes position water & catalytic residues
   • Lower activation energy enough to make hydrolysis significant at body temp

MCAT Strategy: "Peptide bonds are resistant to hydrolysis" = amide resonance makes them resistant. This is a key concept linking organic chemistry to biochemistry. If a question asks "Why don't enzymes need help to break ester/phosphoester bonds?" → Esters are already labile, don't need enzyme stabilization.

Organic Chemistry for the MCAT

Part 5 of 7 — Aromatic Chemistry & Lab Techniques

Aromaticity Rules (Huckel)

Must have: planar ring, conjugated $\pi$ system, $4n + 2$ $\pi$ electrons ($n = 0, 1, 2...$)

• Benzene: 6 $\pi$ electrons ($n = 1$) ✓
• Cyclopentadienyl anion: 6 $\pi$ electrons ✓
• Cyclooctatetraene: 8 $\pi$ electrons → anti-aromatic (if planar)

Electrophilic Aromatic Substitution (EAS)

Lab Techniques on the MCAT

• Distillation: Separates by boiling point
• Extraction: Separates by solubility (aqueous vs. organic layer)
• Chromatography: Separates by polarity (TLC, column)
• Recrystallization: Purifies by differential solubility at different temps

EAS Logic Shortcuts

• Activating groups stabilize the sigma complex and speed substitution.
• Deactivating groups destabilize it and slow substitution.
• Halogens are the classic exception: deactivating by induction but ortho/para directing by resonance.

Aromatics & Lab 🎯

Key Takeaways — Part 5

• Aromaticity: planar + conjugated + $4n+2$ $\pi$ electrons
• Activators → ortho/para; Deactivators → meta (except halogens: deactivating but ortho/para)
• Know lab separation techniques — the MCAT loves "which technique would you use to..." questions
• Treat substitution patterns as resonance/inductive effects, not memorization alone.

Worked Examples — Aromatic Chemistry & Lab Techniques

Question: Chlorobenzene is nitrated with HNO$_3$/H$_2$SO$_4$. Which major products form?

Solution:

1. Chlorine is deactivating by induction but ortho/para directing by resonance.
2. Nitration therefore occurs mainly at ortho and para positions.
3. Sterics favor para over ortho.

Major products: o-nitrochlorobenzene and p-nitrochlorobenzene (para usually higher).

MCAT tip: Halogens are the classic exception: deactivating yet ortho/para directing.

Question: A mixture contains benzoic acid, anisole, and toluene. Best method to isolate benzoic acid?

Solution:

1. Benzoic acid is acidic and can be converted to water-soluble benzoate.
2. Add NaHCO$_3$ or NaOH to extract benzoic acid into the aqueous layer.
3. Separate layers, then acidify aqueous phase to precipitate benzoic acid.

Best method: acid-base extraction.

MCAT tip: If one component is acidic or basic, extraction is often superior to distillation/TLC for bulk separation.

Question: Is cycloheptatriene aromatic?

Solution:

1. Aromaticity needs planarity, full conjugation, and $4n+2$ pi electrons.
2. Neutral cycloheptatriene contains an sp$^3$ carbon, so conjugation is interrupted.
3. Therefore it is nonaromatic.

Related high-yield contrast: the tropylium cation (C$_7$H$_7^+$) is aromatic with 6 pi electrons.

MCAT tip: Huckel count alone is not enough; check continuous p-orbital overlap.

Organic Chemistry for the MCAT

Part 6 of 7 — Spectroscopy (NMR, IR, Mass Spec)

IR Spectroscopy — Key Absorptions

$^1$H NMR — Quick Guide

• Chemical shift ($\delta$): TMS = 0 ppm (reference)
• Alkyl: 0.5-2.0 ppm
• Next to C=O: 2.0-2.5 ppm
• Next to O or N: 3.0-4.0 ppm
• Aromatic: 6.5-8.0 ppm
• Aldehyde H: 9.0-10.0 ppm
• Carboxylic acid H: 10-12 ppm

Splitting (n+1 rule)

A proton with $n$ equivalent neighboring protons splits into $n + 1$ peaks.

• Triplet: 2 neighbors
• Quartet: 3 neighbors

Integration and Signal Counting

• Integration gives relative proton counts for each signal.
• Number of unique proton environments gives number of distinct $^1$H NMR signals.
• Symmetry can reduce the number of observed signals.

Spectroscopy 🎯

Key Takeaways — Part 6

• IR: Broad O-H (3200-3600 for alcohol, 2500-3300 for acid) and sharp C=O (~1715)
• NMR: Chemical shift tells you environment, splitting tells you neighbors
• n+1 rule: number of peaks = neighbors + 1
• Mass spec: molecular ion peak (M$^+$) gives molecular weight
• Use all clues together: IR functional groups + NMR environment + mass constraints.

Worked Examples — Spectroscopy (NMR, IR, Mass Spec)

Question: A spectrum shows a strong peak near 1715 cm$^{-1}$ and a very broad band from 2500 to 3300 cm$^{-1}$. Which functional group is most likely present?

Solution:

1. A strong 1715 cm$^{-1}$ signal suggests C=O.
2. A very broad 2500 to 3300 cm$^{-1}$ signal is characteristic of acidic O-H.
3. Together, these strongly indicate a carboxylic acid.

MCAT tip: Carbonyl plus very broad low O-H region is the classic COOH fingerprint.

Question: A molecule gives two signals: quartet integrating to 2H at 3.6 ppm, and triplet integrating to 3H at 1.2 ppm. What fragment is present?

Solution:

1. Quartet (2H) means that proton set sees 3 neighboring equivalent H.
2. Triplet (3H) means that proton set sees 2 neighboring equivalent H.
3. Combined pattern is the classic ethyl group CH$_3$-CH$_2$.
4. The 3.6 ppm shift for CH$_2$ suggests it is next to oxygen.

Likely fragment: CH$_3$CH$_2$O-.

MCAT tip: Triplet/quartet with 3H/2H integration is a fast ethyl identifier.

Question: Formula C$_4$H$_{10}$O. NMR shows a 9H singlet near 1.2 ppm and a broad 1H signal that disappears with D$_2$O. Identify the compound.

Solution:

1. 9H singlet indicates three equivalent methyl groups attached to one carbon.
2. D$_2$O-exchangeable 1H indicates an O-H proton.
3. Structure that matches is (CH$_3$)$_3$C-OH.

Compound: tert-butanol (2-methyl-2-propanol).

MCAT tip: D$_2$O disappearance confirms exchangeable protons like O-H or N-H.

Question: A spectrum has M$^+$ at m/z 92 and a strong fragment at m/z 91. What structural motif is likely present?

Solution:

1. m/z 91 is the tropylium/benzyl cation signal.
2. This peak strongly suggests a benzyl-containing structure.
3. A common case is toluene-like or alkylbenzene compounds that form the stable C$_7$H$_7^+$ ion.

MCAT tip: m/z 91 is one of the highest-yield aromatic fragmentation clues.

Organic Chemistry for the MCAT

Part 7 of 7 — Review & MCAT Strategy

Highest-Yield MCAT Organic Topics

1. SN1/SN2/E1/E2 — almost guaranteed
2. Functional group recognition — in every passage
3. Amino acid chemistry — bridges to biochemistry
4. Carbonyl chemistry — reduction/oxidation
5. Stereochemistry — R/S, enantiomers vs diastereomers
6. Lab techniques — separation and purification

Amino Acid Side Chain Chemistry (bridges to Biochem)

Test-Day Organic Workflow

1. Identify the functional group(s) first.
2. Decide mechanism class (substitution, elimination, addition, acyl substitution).
3. Check stereochemical consequence (inversion, racemization, retention).
4. Use reagent strength and solvent to resolve competing pathways.

Worked Example — Predicting an Amino Acid's Charge at a Given pH

Scenario: A passage asks for the net charge on the side chain of lysine (side-chain p$K_a \approx 10.5$) in blood at pH 7.4, and how that compares with glutamate (side-chain p$K_a \approx 4.1$).

Step 1 — Recall the rule. Compare the solution pH to each ionizable group's p$K_a$:

• If pH < p$K_a$, the group is mostly protonated.
• If pH > p$K_a$, the group is mostly deprotonated.

Step 2 — Apply it to lysine. The lysine side chain is a basic amine. At pH 7.4 < p$K_a$ 10.5, it is protonated as $-\text{NH}_3^+$, so it carries a +1 charge.

Step 3 — Apply it to glutamate. The glutamate side chain is a carboxylic acid. At pH 7.4 > p$K_a$ 4.1, it is deprotonated as $-\text{COO}^-$, so it carries a –1 charge.

Step 4 — Connect to function. At physiological pH, basic residues (Lys, Arg) tend to be positive and acidic residues (Asp, Glu) negative; these opposite charges form salt bridges that stabilize protein structure and mediate substrate binding — a recurring biochemistry link.

MCAT takeaway: Charge follows the pH-vs-p$K_a$ comparison. Below p$K_a$ → protonated; above p$K_a$ → deprotonated. At pH 7.4, Lys/Arg are typically +1, Asp/Glu are –1, and His (p$K_a \approx 6$) sits near the transition, which is why it is a flexible catalytic residue.

Final Review 🎯

Organic Chemistry — Complete! ✅

Master the reaction decision chart, functional groups, and stereochemistry. These connect directly to amino acid and enzyme chemistry in Biochemistry.

The strongest MCAT performance comes from mechanism-first thinking, not memorizing isolated reactions.