Alcohols, phenols and ethers for NEET

This chapter rewards anyone who keeps asking "where is the lone pair, and where does the charge go?" Phenol is acidic because its anion spreads the negative charge into the ring. The same ring, pushed full of electrons by –OH, takes three bromines at once from bromine water. Ethers split with HI along the path that suits SN1 or SN2.

The 2026 NMC syllabus names a few items directly: identifying primary, secondary and tertiary alcohols, the mechanism of dehydration, the acidic nature of phenol, its halogenation, nitration and sulphonation, the Reimer–Tiemann reaction, and the structure of ethers. Expect questions on acidity order, the product of a named reaction, which test separates two alcohols, and which halide–alkoxide pair makes a given ether.

Structure

Ethers are bent and polar. They cannot hydrogen-bond with each other, so their boiling points are close to those of alkanes of similar mass and far below alcohols. They can accept hydrogen bonds from water, so small ethers are about as soluble as alcohols of similar size.

Preparing alcohols and phenols

Alcohols:

• Acid-catalysed hydration of alkenes follows Markovnikov's rule: propene gives propan-2-ol.
• Hydroboration–oxidation (B₂H₆, then H₂O₂/OH⁻) gives the anti-Markovnikov alcohol: propene gives propan-1-ol.
• Reduction of aldehydes and ketones (H₂/Pd, NaBH₄ or LiAlH₄) gives 1° and 2° alcohols. LiAlH₄ also reduces carboxylic acids to 1° alcohols.
• Grignard reagents: methanal gives a 1° alcohol, any other aldehyde a 2° alcohol, and a ketone a 3° alcohol, after hydrolysis.

Phenol: from chlorobenzene (NaOH at 623 K and 300 atm, then acid), from benzenesulphonic acid (fused NaOH, then acid), from benzene diazonium chloride (warm water), and from cumene: air oxidation gives cumene hydroperoxide, which dilute acid splits into phenol and acetone.

Identifying 1°, 2° and 3° alcohols

Lucas turbidity is the insoluble alkyl chloride separating out. It forms by SN1, so it tracks carbocation stability: 3° fastest. PCC (pyridinium chlorochromate) stops the oxidation of a 1° alcohol at the aldehyde.

Mechanism of dehydration

Ethanol with concentrated H₂SO₄ at 443 K gives ethene. The steps:

1. Protonation (fast): the –OH oxygen takes H⁺ to give the protonated alcohol, CH₃CH₂–OH₂⁺. Water is now a good leaving group.
2. Carbocation formation (slow, rate-determining): water leaves to give CH₃CH₂⁺.
3. Loss of a proton (fast): H⁺ leaves from the β-carbon and the C=C bond forms: CH₂=CH₂.

Because step 2 decides the rate, ease of dehydration follows carbocation stability: 3° > 2° > 1°. NCERT's conditions show this: 1° needs conc. H₂SO₄ at 443 K, 2° needs 85% H₃PO₄ at 440 K, and 3° needs only 20% H₃PO₄ at 358 K. Where two alkenes are possible, the more substituted one is the major product.

Acidity: alcohols and phenols

Alcohols react with sodium to give alkoxides and H₂, but they are weaker acids than water. Among alcohols, acidity falls from 1° to 3°, because alkyl groups (+I) push electron density onto the oxygen.

Phenol (pKa about 10) is far more acidic than ethanol (about 15.9). In the phenoxide ion the negative charge is delocalised over the ring by resonance, which stabilises it; the alkoxide ion has no such help. Phenol dissolves in NaOH but does not release CO₂ from NaHCO₃. Carboxylic acids do, which is how the two are told apart.

Substituents change phenol's acidity:

Electron-withdrawing –NO₂ stabilises the phenoxide, most strongly from the ortho and para positions, where it takes part in resonance. Electron-donating –CH₃ destabilises it, so cresols are slightly weaker than phenol. The order: picric acid > p-nitrophenol > o-nitrophenol > m-nitrophenol > phenol > cresols.

Electrophilic substitution in phenol

–OH strongly activates the ring and directs to ortho and para.

The o- and p-nitrophenols are separated by steam distillation. The ortho isomer has an intramolecular hydrogen bond, so it is steam-volatile. The para isomer hydrogen-bonds between molecules and stays behind.

Kolbe's reaction: sodium phenoxide is heated with CO₂ under pressure, then acidified, to give salicylic acid (2-hydroxybenzoic acid). The phenoxide ion is even more reactive than phenol towards electrophiles, so it attacks weakly electrophilic CO₂.

Reimer–Tiemann reaction: phenol with chloroform and aqueous NaOH at about 340 K gives a substituted benzal chloride that is hydrolysed to salicylaldehyde (2-hydroxybenzaldehyde). The electrophile is dichlorocarbene, :CCl₂, formed from chloroform by the base.

Also know: phenol distilled with zinc dust gives benzene, and oxidation with chromic acid gives benzoquinone.

Ethers

Williamson synthesis is an SN2 reaction of an alkoxide with an alkyl halide: R–ONa + R′–X → R–O–R′ + NaX. It works well only when the halide is primary. With a tertiary halide, the alkoxide acts as a base and elimination wins.

Cleavage by HI. Ethers are split by hot concentrated HI:

• A methyl alkyl ether gives methyl iodide and the alcohol, because I⁻ attacks the less hindered methyl carbon (SN2).
• A tert-butyl methyl ether gives tert-butyl iodide and methanol, because the tertiary carbocation forms (SN1).
• Anisole gives phenol and methyl iodide. The O–aryl bond has partial double-bond character and is not broken.

Anisole (methoxybenzene) undergoes ortho/para substitution. Bromination in ethanoic acid gives mainly p-bromoanisole; Friedel–Crafts and nitration also go ortho and para.

Worked NEET-style questions

Example 1: choosing the Williamson pair

Which combination makes 2-methoxy-2-methylpropane, (CH₃)₃C–O–CH₃?

• Route A: (CH₃)₃C–Br + CH₃ONa. The halide is tertiary, so CH₃O⁻ removes a β-hydrogen and gives 2-methylpropene. Wrong.
• Route B: CH₃–Br + (CH₃)₃C–ONa. The halide is methyl, ideal for SN2. Route B is correct.

Example 2: Grignard route to a 3° alcohol

Which carbonyl compound and Grignard reagent give 2-methylbutan-2-ol, (CH₃)₂C(OH)CH₂CH₃?

• The OH carbon carries CH₃, CH₃ and C₂H₅, so it came from a ketone.
• Propanone + C₂H₅MgBr, or butanone + CH₃MgBr, followed by hydrolysis.

Example 3: acidity order

Arrange: ethanol, phenol, water, 4-nitrophenol.

• 4-Nitrophenol (7.1) > phenol (10) > water (15.7) > ethanol (15.9).
• 4-nitrophenol > phenol > water > ethanol.

Practice MCQs

1. Which alcohol gives turbidity with Lucas reagent immediately? (a) butan-1-ol (b) butan-2-ol (c) 2-methylpropan-2-ol (d) methanol
2. The rate-determining step in acid-catalysed dehydration of an alcohol is: (a) protonation of –OH (b) formation of the carbocation (c) loss of H⁺ (d) attack of HSO₄⁻
3. The strongest acid among these is: (a) phenol (b) o-cresol (c) m-nitrophenol (d) p-nitrophenol
4. Phenol with bromine water gives: (a) o-bromophenol (b) p-bromophenol (c) 2,4,6-tribromophenol (d) bromobenzene
5. The Reimer–Tiemann reaction on phenol gives: (a) salicylic acid (b) salicylaldehyde (c) benzaldehyde (d) picric acid
6. Anisole with hot HI gives: (a) iodobenzene and methanol (b) phenol and methyl iodide (c) benzene and methanol (d) iodobenzene and methyl iodide
7. o-Nitrophenol is separated from p-nitrophenol by steam distillation because the ortho isomer has: (a) a higher molar mass (b) intramolecular hydrogen bonding (c) intermolecular hydrogen bonding (d) ionic bonding
8. Heating 2-methylpropan-2-ol with copper at 573 K gives: (a) 2-methylpropanal (b) propanone (c) 2-methylpropene (d) no reaction

Answers

1. (c) A tertiary alcohol forms the stable 3° carbocation fastest.
2. (b) Loss of water to form the carbocation is slow.
3. (d) –NO₂ at para stabilises the phenoxide by resonance and induction.
4. (c) The strongly activated ring is brominated at all three o/p positions.
5. (b) The –CHO enters ortho to –OH.
6. (b) The O–aryl bond survives.
7. (b) Intramolecular H-bonding keeps the molecules separate and volatile.
8. (c) A 3° alcohol has no hydrogen on the carbon carrying –OH, so it cannot be dehydrogenated; it dehydrates instead.

What to do next

• Write the three-step dehydration mechanism from memory, then redo it for a 2° alcohol and name the major alkene.
• Learn the pKa table in order; say the reason for each position aloud.
• Practise five Williamson pairs, rejecting the route with a 3° halide.
• Revise SN1 and SN2 in haloalkanes and haloarenes, then move to aldehydes, ketones and carboxylic acids.