$[\mathrm{P}]$ on treatment with $\mathrm{Br}_{2} / \mathrm{FeBr}_{3}$ in $\mathrm{CCl}_{4}$ produced a single isomer $\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{Br}$ while heating $[\mathrm{P}]$ with sodalime gave toluene. The compound $[\mathrm{P}]$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Hydrocarbons

[P]

on treatment with

Br_{2} / FeBr_{3}

CCl_{4}

produced a single isomer

C_{8} H_{7} O_{2} Br

while heating

[P]

with sodalime gave toluene. The compound

[P]

is:

(A)(see structure)
(B)(see structure)
(C)(see structure)
✓ Correct
(D)(see structure)

Answer: (C)

Solution

Heating [P] with sodalime removes a carboxyl group and leaves toluene, so [P] is a methylbenzoic acid. Bromination with

Br X_{2} / FeBr X_{3}

gives only a single mono-bromo isomer, which is possible only when the methyl and carboxyl groups are para to one another: the methyl group (ortho/para directing) and the carboxyl group (meta directing) then both point to the same single position. Thus [P] is para-toluic acid, 4-methylbenzoic acid.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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