Alcohols, Phenols and Ethers: JEE Advanced PYQ, 2007–2026

Answer: (A)

Excess sodium replaces the hydrogens of both hydroxyl groups, giving NaOCH2CH2ONa. Each O⁻ is a nucleophile that attacks CH3I in an SN2 step (Williamson synthesis), forming an O–CH3 bond and releasing I⁻. Two such substitutions give 1,2-dimethoxyethane, CH3OCH2CH2OCH3, with 2 mol CH3I consumed per mole of diol.

Answer: (C)

PCC is a mild oxidant: it converts primary alcohols to aldehydes (without going on to acids) and secondary alcohols to ketones. Tertiary alcohols have no H on the carbinol carbon and are not oxidised. In the triol, the CH2OH becomes CHO (-2 H), the CH(OH) becomes C=O (-2 H), and the C(CH3)(OH) stays. The formula changes from C6H14O3 to C6H10O3, so the molar mass is 134 - 4 = 130 g/mol. Oxidation of the tertiary OH (128) is impossible, oxidation of only one group (132) is incomplete, and oxidation to the acid (146) would need a stronger oxidant.

Answer: (A)

Step 1: Phenol is acidic enough to react with NaOH, giving sodium phenoxide (C6H5O⁻ Na⁺). Step 2: The phenoxide ion is a nucleophile and reacts with CH3Br by a Williamson-type SN2 reaction to give anisole, C6H5OCH3. Step 3: HI protonates the ether oxygen. I⁻ then attacks the less hindered methyl carbon by SN2. It cannot attack the sp² phenyl carbon, because the C–O bond there has partial double-bond character from resonance and the ring blocks backside attack. The products are phenol and CH3I. Phenol does not give iodobenzene with HI, for the same reason.

Answer: (D)

Acid-catalysed hydration adds OH to the more substituted carbon (Markovnikov, through the more stable carbocation). Hydroboration-oxidation adds OH to the less substituted carbon (anti-Markovnikov). We need an alkene whose more substituted carbon is tertiary-forming and whose less substituted carbon is a secondary carbon. In 2-methylbut-2-ene, (CH3)2C=CH-CH3, Markovnikov OH goes on C2 (bearing two methyls) to give the tertiary alcohol 2-methylbutan-2-ol. Anti-Markovnikov OH goes on C3, giving (CH3)2CH-CH(OH)-CH3, 3-methylbutan-2-ol, a secondary alcohol. 2-Methylbut-1-ene and 3-methylbut-1-ene give primary alcohols by hydroboration, and pent-2-ene gives secondary alcohols by both routes.

Answer: (A)

In the Williamson synthesis the alkoxide attacks an alkyl halide by SN2. SN2 works for methyl and primary halides. Secondary halides partly eliminate, and tertiary halides give almost only alkene because the alkoxide acts as a base. To make an ether with a tertiary alkyl group, the tertiary group must be on the alkoxide and the halide must be methyl or primary. So sodium tert-butoxide with CH3I gives (CH3)3C–O–CH3. The reverse combination gives 2-methylpropene. tert-Butyl alcohol has no leaving group. Aqueous NaOH cannot generate tert-butoxide in useful amounts.

Answer: (A)

Phenol is strongly activated, and in a polar solvent such as water the OH group is ionised enough to bring the ring to full ortho/para bromination. Excess bromine water gives a white precipitate of 2,4,6-tribromophenol, C6H3Br3O. Its molar mass is 72 + 3 + 240 + 16 = 331 g mol⁻¹. Moles of phenol = 4.7 / 94 = 0.05 mol, and the reaction is 1:1. Theoretical mass = 0.05 × 331 = 16.55 g. At 80% efficiency, mass = 0.8 × 16.55 = 13.24 g.

Answer: (A)

The Williamson synthesis needs a good nucleophile, the alkoxide or phenoxide ion. An aqueous NaOH solution can generate it only if the conjugate acid is acidic enough. Phenol is acidic because its conjugate base, phenoxide, has resonance stabilisation with the negative charge shared over the ring. Ethanol is a much weaker acid (similar to water), and its ethoxide is destabilised by the +I effect of the ethyl group. NaOH therefore leaves it mostly undissociated, so a much stronger base source, sodium metal, is used for alcohols.

Answer: (A)

Principle 1: a phenoxide ion is a good nucleophile. It reacts with a primary alkyl halide by SN2 (Williamson synthesis), so PhONa + C₂H₅Br gives PhOC₂H₅ (ethoxybenzene) + NaBr. Principle 2: ethers are cleaved by HI. In an alkyl aryl ether the aryl C–O bond is stronger because the oxygen lone pair is delocalised into the ring, giving partial double-bond character, and an sp² carbon does not undergo SN2. So I⁻ attacks the alkyl carbon, and PhOC₂H₅ + HI gives PhOH + C₂H₅I. The phenol cannot be converted to iodobenzene by HI. With excess HI any alcohol formed from the alkyl side is converted to the alkyl iodide, so the final products are phenol and iodoethane.

Answer: (A)

Solubility in NaOH but not in NaHCO3 shows the compound is weakly acidic, a phenol rather than a carboxylic acid, which rules out benzyl alcohol. Bromine water substitutes at the free positions ortho and para to –OH. The product C7H5Br3O means three H atoms of C7H8O have been replaced by Br. Counting free ortho/para positions relative to –OH gives two for o-cresol, two for p-cresol and three for m-cresol, so the compound is m-cresol.

Answer: (A)

A base converts an alcohol or phenol into its conjugate base only to the extent that the alcohol/phenol is a stronger acid than the conjugate acid of the base. The conjugate acid of OH⁻ is water (pKa about 15.7). Phenol (pKa about 10) is more acidic than water because its conjugate base is stabilised by delocalising the negative charge into the ring, so NaOH forms phenoxide essentially completely. Ethanol (pKa about 16) is no more acidic than water, so NaOH gives only a trace of ethoxide, and a much stronger base such as Na or NaH is needed. Phenol is weaker than carbonic acid, so the option based on NaHCO3 is wrong.