Question

[P] on treatment with \(\mathrm{Br_2/FeBr_3}\) in \(\mathrm{CCl_4}\) produced a single isomer \(\mathrm{C_8H_7O_2Br}\) while heating [P] with sodalime gave toluene. The compound [P] is :

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Sodalime decarboxylation to toluene fits phenylacetic acid and all three toluic acids; a single ring-brominated isomer requires both substituents to direct to the same two equivalent positions, which happens only in p-toluic acid.

Answer 1

The Correct Option is C

The acid is \(\mathrm{C_8H_8O_2}\); replacing one H by Br gives \(\mathrm{C_8H_7O_2Br}\). Sodalime (\(\mathrm{NaOH/CaO}\), heat) removes \(\mathrm{CO_2}\): phenylacetic acid gives toluene, and o-, m- and p-methylbenzoic acid also give toluene. So the clue is the bromination.
Phenylacetic acid: \(\mathrm{CH_2COOH}\) is an o,p-director, so ortho and para bromo products (a mixture).
o-Toluic acid (COOH at C1, CH₃ at C2): both groups direct to C3 and C5, which are different positions, so two isomers.
m-Toluic acid (COOH at C1, CH₃ at C3): CH₃ activates C2, C4 and C6, none of them equivalent, so several isomers.
p-Toluic acid (COOH at C1, CH₃ at C4): COOH (meta director) and CH₃ (ortho director, para blocked) both direct to C3 and C5, which are equivalent by symmetry. Only one monobromo product forms: 3-bromo-4-methylbenzoic acid.
So [P] is p-methylbenzoic acid.
Answer: (3) p-methylbenzoic acid (4-CH₃C₆H₄COOH)