Alcohols and Phenols
Alcohols (R–OH) and phenols (Ar–OH) both carry a hydroxyl group, yet their reactivity differs sharply because of the aromatic ring in phenol. The AAMC MCAT content outline expects you to compare the two classes, rank their acidity and explain it with resonance, and know how alcohols are prepared (hydration, reduction of carbonyls) and oxidized (1° → aldehyde → acid; 2° → ketone; 3° resists).
Difference between Alcohol and Phenol
Both classes contain a hydroxyl group, but in alcohols the –OH is attached to an sp3 carbon of an aliphatic chain, whereas in phenol the –OH is bonded directly to an sp2 carbon of a benzene ring. This single structural difference governs every chemical contrast between them.
- Alcohol
- A compound in which –OH is bonded to a saturated sp3 carbon. General formula R–OH. Example: CH3CH2OH (ethanol).
- Phenol
- A compound in which –OH is bonded directly to a benzene ring. General formula Ar–OH. Example: C6H5OH (carbolic acid).
Side-by-side comparison
| Property | Alcohol (R–OH) | Phenol (Ar–OH) |
|---|---|---|
| C–OH carbon | sp3 (aliphatic) | sp2 (aromatic ring) |
| Example | CH3CH2OH (ethanol) | C6H5OH (carbolic acid) |
| Acidity (pKa) | ~16–18 (weaker than water) | ~10 (stronger than water, weaker than COOH) |
| Reaction with NaOH | No reaction | Forms sodium phenoxide (salt) |
| Reaction with NaHCO3 | No reaction | No reaction (too weak to liberate CO2) |
| Reaction with Na metal | Yes — gives RONa + H2↑ | Yes — gives ArONa + H2↑ |
| FeCl3 test | No color | Violet / purple complex |
| Lucas test (ZnCl2/HCl) | Distinguishes 1° / 2° / 3° alcohols | Not applicable |
| Esterification with RCOOH | Easy | Slower (less nucleophilic O) |
| Conjugate base stability | Alkoxide RO− — no resonance | Phenoxide — stabilized by resonance over o, p ring carbons |
| Aromatic ring substitution | N/A | Highly reactive at o, p positions (ring activator) |
Nomenclature, Structure and Reactivity of Alcohols
Alcohols are classified by the number of carbons attached to the carbon bearing the –OH group. The chemistry of an alcohol is dominated by two reactive sites: the polar O–H bond and the C–O bond.
IUPAC nomenclature
- Choose the longest carbon chain that contains the –OH group.
- Replace the parent alkane suffix –e with –ol (e.g. methane → methanol).
- Number the chain to give the carbon bearing the –OH the lowest locant.
- Examples: CH3OH = methanol, CH3CH2OH = ethanol, (CH3)2CHOH = propan-2-ol, (CH3)3COH = 2-methylpropan-2-ol.
Classification
- Primary (1°)
- –OH on a carbon bonded to one other carbon. Example: ethanol CH3CH2OH.
- Secondary (2°)
- –OH on a carbon bonded to two other carbons. Example: propan-2-ol (CH3)2CHOH.
- Tertiary (3°)
- –OH on a carbon bonded to three other carbons. Example: 2-methylpropan-2-ol (CH3)3COH.
Physical properties
Alcohols form intermolecular hydrogen bonds through the O–H group, so their boiling points are markedly higher than those of comparable alkanes or ethers of similar molar mass. Lower alcohols (C1–C3) are completely miscible with water; solubility falls off as the hydrocarbon tail lengthens.
Preparation of alcohols
Acid-catalyzed addition of water across a C=C follows Markovnikov's rule — –OH goes to the more substituted carbon: CH2=CH2 + H2O ⟶{H+} CH3CH2OH. Oxymercuration–demercuration gives the Markovnikov alcohol with no carbocation rearrangement; hydroboration–oxidation (BH3, then H2O2/OH−) gives the anti-Markovnikov alcohol via syn addition.
Aldehydes → 1° alcohols and ketones → 2° alcohols with NaBH4 or LiAlH4; carboxylic acids and esters are reduced to 1° alcohols only by the stronger LiAlH4. Grignard reagents (R′MgX) add to formaldehyde → 1° alcohol, to other aldehydes → 2° alcohol, and to ketones → 3° alcohol after aqueous workup.
Reactions of alcohols
2 R–OH + 2 Na → 2 R–ONa + H2↑. Sodium displaces the hydroxylic hydrogen to give a sodium alkoxide and hydrogen gas. The reaction is slower than that of water.
R–OH + HX → R–X + H2O. Reactivity order of HX: HI > HBr > HCl. Reactivity order of alcohols: 3° > 2° > 1° (because tertiary carbocations are more stable in the SN1 pathway).
Reagent: anhydrous ZnCl2 in concentrated HCl.
3° alcohol → turbidity immediately.
2° alcohol → turbidity in 5–10 minutes.
1° alcohol → no turbidity at room temperature.
R–CH2–CH2–OH ⟶{conc. H2SO4, 170°C} R–CH=CH2 + H2O. Ease of dehydration: 3° > 2° > 1°. Mechanism: E1 (3°, 2°) or E2 (1°). Follows Zaitsev's rule — the more substituted (more stable) alkene is the major product.
1° alcohol ⟶{[O], KMnO4 or K2Cr2O7} aldehyde ⟶{[O]} carboxylic acid. PCC (a milder, anhydrous oxidant) stops a 1° alcohol at the aldehyde.
2° alcohol ⟶{[O]} ketone (no further oxidation under mild conditions).
3° alcohol — not oxidized under normal conditions, because the carbon bearing the –OH has no C–H bond to break.
R–OH + R′COOH ⟶{conc. H2SO4} R′COOR + H2O. Reversible; driven forward by removing water.
R–ONa + R′–X → R–O–R′ + NaX. The sodium alkoxide attacks the alkyl halide via SN2; works best with 1° halides — tertiary halides give elimination instead.
Nomenclature, Structure and Reactivity of Phenols
In phenol, C6H5OH, the –OH is bonded directly to the aromatic ring. The ring activates strongly toward electrophilic aromatic substitution and, through resonance, stabilizes the conjugate base — the two features that dominate its MCAT chemistry.
IUPAC nomenclature
- The parent is "phenol" (retained IUPAC name) or "benzenol".
- Substituents are numbered such that –OH is C-1.
- Common examples: 2-methylphenol (o-cresol), 4-nitrophenol, 2,4,6-trinitrophenol (picric acid), 1,2-dihydroxybenzene (catechol), 1,3-dihydroxybenzene (resorcinol), 1,4-dihydroxybenzene (hydroquinone).
Structure and bonding
The oxygen lone pair is donated into the π system, increasing electron density at the ortho and para positions. This makes phenol an activating, ortho/para-directing substrate in electrophilic aromatic substitution and explains its higher acidity relative to alcohols.
Acidity of phenol
Phenol pKa ≈ 10 vs ethanol pKa ≈ 16. Reason: the phenoxide negative charge is delocalized onto the ring (partial C–O double-bond character), which the alkoxide cannot do. Electron-withdrawing groups (–NO2, –CN, –X) at the ortho/para positions further stabilize the anion and increase acidity — nitrophenols are more acidic than phenol, and 2,4,6-trinitrophenol (picric acid, pKa ~ 0.4) is nearly as strong as a mineral acid.
Reactions of phenol
C6H5OH + NaOH → C6H5ONa + H2O. Phenol is acidic enough to dissolve in dilute NaOH — alcohols are not. (Distinguishing test.)
Phenols give a deep violet/purple color with neutral aqueous FeCl3. Alcohols give no color. Diagnostic test for a phenolic –OH.
C6H5OH + 3 Br2 (aq) → 2,4,6-tribromophenol (white precipitate) + 3 HBr. No catalyst needed — the ring is so activated that all three ortho/para positions are attacked at once.
Dilute HNO3 at low temperature gives a mixture of o- and p-nitrophenol. Concentrated HNO3 with H2SO4 drives full substitution to picric acid (2,4,6-trinitrophenol).
Sodium phenoxide (C6H5ONa) + R–X → C6H5–O–R (an alkyl aryl ether) + NaX. The phenoxide oxygen is the nucleophile in this SN2 step — phenols cannot be the alkyl halide partner because aryl halides do not undergo SN2.
Worked MCQs
Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — that's where the deeper concept lives.
Q1. The Lucas reagent (anhydrous ZnCl2 in concentrated HCl) reacts immediately with which of the following?
2-methylpropan-2-ol is a tertiary alcohol; it forms a stable 3° carbocation that captures Cl− at once, producing an immediate cloudiness. Secondary alcohols take 5–10 minutes; primary alcohols give no turbidity at room temperature.
Q2. Phenol is more acidic than ethanol primarily because:
When phenol loses a proton, the resulting negative charge is delocalized over the ortho and para ring carbons. This resonance stabilization lowers the energy of the phenoxide ion, shifting the ionization equilibrium far to the right relative to ethanol, whose ethoxide has no such delocalization.
Q3. Which alcohol cannot be oxidized by acidified K2Cr2O7 under normal conditions?
Oxidation of an alcohol removes the hydrogen on the carbon bearing the –OH. A tertiary alcohol has no such C–H bond, so it cannot be oxidized to a carbonyl. Primary alcohols oxidize to aldehydes then carboxylic acids and secondary alcohols to ketones; tertiary alcohols resist oxidation unless very harsh conditions break the C–C skeleton.
Q4. Phenol on reaction with excess bromine water gives:
The ring in phenol is so strongly activated by the –OH that no Lewis-acid catalyst is required. All three ortho/para positions are brominated simultaneously, precipitating white 2,4,6-tribromophenol from solution.
Q5. The Williamson ether synthesis works best with:
Williamson synthesis proceeds via SN2: the alkoxide is a strong nucleophile/base and 1° halides have the least steric hindrance. With 3° halides, the alkoxide acts as a base instead, giving alkene by E2.
Quick Recap
- Alcohol: –OH on sp3 C; phenol: –OH on sp2 C of an aromatic ring.
- Acidity: carboxylic acid > phenol > water > alcohol.
- Lucas test: 3° instant, 2° in minutes, 1° no reaction.
- Oxidation: 1° → aldehyde → acid; 2° → ketone; 3° → no reaction.
- Williamson ether synthesis prefers 1° halides (SN2).
- Preparation of alcohols: hydration of alkenes (Markovnikov; anti-Markovnikov via hydroboration) and reduction of carbonyls (NaBH4/LiAlH4, Grignard).
- Phenol gives violet color with FeCl3; dissolves in NaOH (alcohols don't); ring is activating/ortho-para directing (e.g. + 3 Br2 → 2,4,6-tribromophenol).