Alcohols

Alcohols are everywhere the MCAT looks: carbohydrates, amino acids (serine, threonine, tyrosine), glycerol, nucleotides, and most metabolic pathways. The whole topic flows from one idea — a hydroxyl group on an $sp^3$ carbon — and how that group drives reactivity.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Alcohol Structure and Classification

Must know

An alcohol is a hydroxyl group ($\ce{-OH}$) bonded to a saturated ($sp^3$) carbon. The C–O bond is polar (O is far more electronegative), which underpins almost every reaction and property.

Alcohols are classified by the substitution of the carbinol carbon (the one bearing –OH):

Class	Carbinol C bonded to	Example
Primary (1°)	1 other carbon	Ethanol, $\ce{CH3CH2OH}$
Secondary (2°)	2 other carbons	Isopropanol, $\ce{(CH3)2CHOH}$
Tertiary (3°)	3 other carbons	tert-Butanol, $\ce{(CH3)3COH}$

A diol (glycol) has two –OH groups (ethylene glycol); a polyol has three or more (glycerol). Classification is your first move on any alcohol problem: it controls oxidation product, SN1 vs SN2, and ease of dehydration.

Quick check: Is the carbon bearing –OH in lactic acid ($\ce{CH3CH(OH)COOH}$) primary, secondary, or tertiary?
Answer: Bonded to $\ce{CH3}$ and $\ce{COOH}$ → secondary. (Lactate dehydrogenase oxidizes this 2° alcohol to the ketone pyruvate.)

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Nomenclature

Must know

IUPAC rules: longest chain that includes the carbinol carbon; number to give –OH the lowest locant; replace the alkane "-e" with -ol plus a locant if needed (propan-2-ol). Diols take -diol (ethane-1,2-diol).

Common names to know: methanol ($\ce{CH3OH}$), ethanol ($\ce{CH3CH2OH}$), isopropanol/propan-2-ol ($\ce{(CH3)2CHOH}$), glycerol/propane-1,2,3-triol, ethylene glycol/ethane-1,2-diol.

Phenols are NOT alcohols — the –OH sits directly on an aromatic ring, giving them very different acidity (below).

Quick check: Name $\ce{CH3CH(OH)CH2CH3}$ by IUPAC rules.
Answer: Longest chain with –OH is 4 carbons; –OH at C-2 → butan-2-ol.

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Physical Properties: Hydrogen Bonding and Acidity

Hydrogen Bonding and Boiling Points

Must know

The –OH group lets alcohols hydrogen bond (both donor and acceptor), giving much higher boiling points than alkanes of similar weight (ethanol 78 °C vs propane −42 °C). As the chain lengthens, boiling point rises but water solubility falls; short alcohols (through butanol) are miscible, longer ones are dominated by the hydrophobic tail.

Acidity

Must know

Alcohols are slightly weaker acids than water (water pKa ≈ 15.7; aliphatic ROH ≈ 16–18, getting weaker $1° \to 2° \to 3°$).

$\text{water} > 1° \text{ ROH} > 2° \text{ ROH} > 3° \text{ ROH}$

Why 3° is weakest: more electron-donating alkyl groups destabilize the negative charge on the alkoxide → less stable conjugate base → weaker acid.

Phenol is the big exception (pKa ≈ 10): the phenoxide charge delocalizes into the ring (resonance), making phenol ~$10^6$× more acidic than a normal alcohol — roughly weak-carboxylic-acid territory.

Fully deprotonating an alcohol needs a strong base (NaH, $n$-BuLi, or Na metal). The resulting alkoxide ($\ce{RO^-}$) is a strong base and good nucleophile.

Williamson ether synthesis is the key alkoxide reaction: $\ce{RO^-}$ does $S_N2$ on an alkyl halide.

$\ce{RO^- + R'X -> R-O-R' + X^-}$

Because it is $S_N2$, the halide partner must be methyl or primary (bulky/tertiary eliminate instead) — always pair the alkoxide with the less hindered partner.

Quick check: Would $\ce{NaOH}$ (conjugate acid water, pKa $\approx 15.7$) fully deprotonate ethanol (pKa $\approx 16$)?
Answer: No. Ethanol is slightly weaker, so $\ce{EtOH + NaOH <=> EtO^- + H2O}$ lies slightly left. You need a stronger base (NaH, Na metal).

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Oxidation of Alcohols

The Logic

Must know

Oxidation adds bonds to oxygen. Primary alcohols climb a ladder:

$\underbrace{\ce{R-CH2OH}}_{\text{1° alcohol}} \xrightarrow{[O]} \underbrace{\ce{R-CHO}}_{\text{aldehyde}} \xrightarrow{[O]} \underbrace{\ce{R-COOH}}_{\text{carboxylic acid}}$

Secondary alcohols → ketones (and stop, no C–H left to remove). Tertiary alcohols do not oxidize — no H on the carbinol carbon. This is a major MCAT discriminator.

Which Oxidant

Must know

PCC (mild, anhydrous) stops a 1° alcohol at the aldehyde; any aqueous strong oxidant (Jones = $\ce{CrO3}/\ce{H2SO4}$, $\ce{K2Cr2O7}/\ce{H2SO4}$, hot $\ce{KMnO4}$) takes it to the carboxylic acid. All of them take 2° → ketone.

The biological version: alcohol dehydrogenase oxidizes ethanol → acetaldehyde, then aldehyde dehydrogenase → acetic acid, using $\ce{NAD+}$ as the hydride acceptor. Knowing NAD$^+$ is the biological alcohol oxidant is the testable point.

Quick check: Can you oxidize 2-methylpropan-2-ol ($\ce{(CH3)3COH}$) with Jones reagent?
Answer: No — it is tertiary, no C–H on the carbinol carbon. No standard oxidant works.

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Substitution Reactions: SN1 or SN2

–OH Is a Bad Leaving Group

Know the logic

Neutral –OH won't leave ($\ce{OH^-}$ is a strong base). You must either protonate it (acid) to make $\ce{OH2^+}$ (leaves as water), or convert it to a better leaving group (halide, mesylate, tosylate).

With Hydrogen Halides (HX)

Know the logic

$\ce{ROH + HX -> RX + H2O}$: protonate –OH, then either lose water to a carbocation (SN1) or take backside attack by $\ce{X^-}$ (SN2).

• SN1 for 3° and 2° (carbocation stabilized) → expect racemization.
• SN2 for 1° / methyl (no stable cation) → expect inversion.
• Reactivity: $3° > 2° > 1°$; and $\ce{HI} > \ce{HBr} > \ce{HCl}$.

SOCl₂ and PBr₃ — Cleaner Routes

Must know

$\ce{SOCl2}$ (→ Cl) and $\ce{PBr3}$ (→ Br) convert 1°/2° alcohols to alkyl halides without a free carbocation, so they avoid rearrangement and generally give inversion.

$\ce{ROH + SOCl2 -> RCl + SO2 + HCl} \qquad \ce{3 ROH + PBr3 -> 3 RBr + H3PO3}$

Passage-level

The Lucas test ($\ce{ZnCl2}/\ce{HCl}$) distinguishes alcohol class by turbidity speed (3° immediate, 2° minutes, 1° none at RT) — recognize it, don't drill it.

Dehydration (Elimination)

Know the logic

Under strong acid + heat, alcohols undergo E1 to alkenes (protonate, lose water to carbocation, lose adjacent H). Product follows Zaitsev (more substituted alkene); ease $3° > 2° > 1°$. Higher temperature favors elimination over substitution.

Watch for carbocation rearrangements: because SN1 and E1 both go through a carbocation, a 1,2-hydride or methyl shift can occur to reach a more stable cation before product forms — so a 2° alcohol can give a product from a 3° cation. Always check for a shift in any SN1/E1 alcohol reaction.

Quick check: When (R)-butan-2-ol reacts with HBr under SN2 conditions, what is the stereochemical outcome?
Answer: This is a secondary alcohol. Under SN2-favoring conditions you get inversion → (S)-2-bromobutane; under SN1-favoring conditions (excess HBr, heat) you get a racemic mixture.

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Protection of Alcohols

Know the logic

In multi-step synthesis a reactive –OH is temporarily masked as an inert derivative, then deprotected later. A good protecting group installs without touching the rest of the molecule and removes selectively (orthogonally).

Most common groups to recognize:

• Silyl ethers (TMS, TBS): made with TMSCl (or TBSCl) + base; stable to base/neutral conditions; removed by fluoride ($F^-$, e.g. TBAF) or dilute acid. The driving force is the very strong Si–F bond.

$\ce{ROH + TMSCl ->[\text{base}] RO-TMS ->[\text{TBAF or H3O+}] ROH}$

• Acetals (for 1,2-/1,3-diols): diol + aldehyde/ketone under acid → cyclic acetal (e.g. acetonide); stable to base, hydrolyzed by aqueous acid. (Sugars are often protected this way.)

Optional

THP ethers (from dihydropyran) are another cyclic-acetal protecting group, cleaved by dilute acid.

Quick check: You protect a free –OH as a TMS ether, reduce an ester elsewhere, then need the –OH back. How do you remove the TMS group?
Answer: TBAF (fluoride) or dilute aqueous acid — fluoride attacks silicon, breaking the O–Si bond.

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Preparation of Mesylates and Tosylates

Know the logic

Instead of protonating –OH (acidic, risks SN1 rearrangement), convert it to a sulfonate ester — a superb leaving group — without breaking the C–O bond, so stereochemistry stays predictable.

• Mesylate (–OMs): alcohol + MsCl ($\ce{CH3SO2Cl}$) + base ($\ce{Et3N}$).
• Tosylate (–OTs): alcohol + TsCl + pyridine/$\ce{Et3N}$.

$\ce{ROH + CH3SO2Cl ->[\ce{Et3N}] RO-SO2CH3 + HCl}$

The –OMs/–OTs leaving ability comes from charge delocalized over three sulfonate oxygens.

Stereochemistry — The Critical Point

Must know

Mesylation/tosylation breaks only the O–H bond, so configuration is retained at carbon. The leaving group is then displaced:

• Subsequent SN2 → inversion at carbon → net inversion overall.
• Subsequent SN1 → racemization.

Worked example: (R)-2-butanol + MsCl/Et$_3$N gives (R)-mesylate (C–O intact, config retained); then NaBr does SN2 with inversion → (S)-2-bromobutane. Net: (R)-alcohol → (S)-bromide.

Quick check: If that mesylate instead undergoes SN1 (polar protic, heat, no strong nucleophile) with $\ce{Br^-}$, what's the outcome?
Answer: Ionizes to a planar carbocation → $\ce{Br^-}$ attacks either face → racemic 2-bromobutane.

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Common Confusions & Tricks

1. 3° alcohols don't oxidize. A chromium reagent on a tertiary alcohol = "no reaction" (no C–H on the carbinol carbon). The most common wrong answer.

2. PCC stops at aldehyde; aqueous Cr(VI) goes to carboxylic acid. PCC is anhydrous, so the aldehyde can't hydrate and over-oxidize. Aqueous oxidants (Jones, $\ce{K2Cr2O7}/\ce{H2SO4}$, $\ce{KMnO4}$) take 1° all the way.

3. Alcohols are WEAKER acids than water, not stronger. ROH pKa ~16–18 > water 15.7. But phenol (~10) is much more acidic. Alkyl groups destabilize the alkoxide; resonance stabilizes phenoxide.

4. Mesylation/tosylation retains configuration; the subsequent SN2 inverts it. The sulfonylation step does not invert (C–O intact); only the substitution step does. Net of Ms/Ts + SN2 = inversion.

5. Fluoride removes silyl groups. Counterintuitive, but the strong Si–F bond is the driving force.

6. 1° favors SN2; 3° favors SN1. With HBr: 1° = SN2 (inversion); 3° = SN1 (racemization possible). 2° is the ambiguous middle — the MCAT asks which factors push it each way.

7. PCC = partial oxidation (aldehyde); Jones/KMnO₄ = complete (carboxylic acid).

8. Dehydration favors elimination at higher temperature. High temp + strong acid → alkene (E1); lower temp favors substitution.

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Key Equations

Equation/Expression	Variables & Usage
$\text{p}K_a(\text{ROH}) \approx 16\text{–}18$	Acidity of aliphatic alcohols; less acidic with more alkyl substitution
$\text{p}K_a(\text{PhOH}) \approx 10$	Phenol; far more acidic (resonance-stabilized phenoxide)
$\ce{R-CH2OH ->[\text{PCC}] R-CHO}$	Primary → aldehyde (PCC, anhydrous; stops here)
$\ce{R-CH2OH ->[\text{Jones or } \ce{KMnO4}] R-COOH}$	Primary → carboxylic acid (aqueous/strong oxidant)
$\ce{R2CHOH ->[\text{[O]}] R2C=O}$	Secondary → ketone (stops at ketone)
$\ce{R3COH ->[\text{[O]}] \text{N.R.}}$	Tertiary: no oxidation
$\ce{ROH + HX -> RX + H2O}$	Alcohol → alkyl halide; 3°/2° SN1, 1° SN2; HI > HBr > HCl
$\ce{ROH + CH3SO2Cl ->[\ce{Et3N}] RO-SO2CH3}$	Mesylation; retains config at C; makes excellent leaving group
$\ce{ROH + TsCl ->[\text{pyridine}] RO-SO2Ar}$	Tosylation; same principle
$\ce{ROH + TMSCl ->[\text{base}] RO-TMS ->[\ce{F^-}] ROH}$	Silyl ether protection/deprotection; removed by fluoride
$\ce{ROH + (CH3)2C=O ->[\ce{H+}] \text{acetonide}}$	Cyclic acetal protection of 1,2-/1,3-diols; cleaved by aqueous acid