Alcohols, Phenols and Ethers: JEE Main PYQ, 2020–2025

Answer: (D)

In the Williamson synthesis an alkoxide ion displaces halide from an alkyl halide. The ethoxide ion attacks the primary carbon of 1-bromopropane and bromide leaves, giving CH₃CH₂–O–CH₂CH₂CH₃, 1-ethoxypropane. A primary halide is the best substrate because substitution then wins over elimination.

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)

Williamson synthesis works best with primary alkyl halides, since the alkoxide attacks by SN2. With a tertiary halide, the strongly basic alkoxide removes a β-hydrogen instead, causing E2 elimination. The major product is 2-methylpropene, with NaBr and ethanol as by-products.

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 Williamson synthesis, an alkoxide (nucleophile) attacks an alkyl halide (electrophile) by SN2. The halide should be primary or methyl so that backside attack is not blocked and elimination is avoided. Alkoxides can be bulky, such as tert-butoxide, because the steric hindrance matters mainly at the electrophilic carbon. Therefore the tert-butyl group goes on the oxygen as the alkoxide, and the methyl group goes on the halide.

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)

Acid-catalysed hydration follows Markovnikov’s rule. H⁺ adds to the terminal carbon to give the more stable secondary carbocation, CH₃–CH⁺–CH₃, which water attacks, giving CH₃CH(OH)CH₃.