4-Nitrobenzoic acid is to be converted into 4-nitrobenzyl alcohol, so that the –COOH group is reduced but the nitro group is left unchanged. Which of the following reagents is best suited for this conversion?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Diborane (B2H6), followed by H3O+
Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride or, better, by diborane. Diborane does not easily reduce groups such as ester, nitro and halo, so it reduces the –COOH group and leaves the –NO2 group as it is:
\( \displaystyle \ce{O2N-C6H4-COOH} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{B2H6}} \ce{O2N-C6H4-CH2OH} \)
Why the other options are incorrect
Option (1) Sodium borohydride (NaBH4) — Incorrect. NaBH4 reduces aldehydes and ketones but does not reduce the carboxyl group, so the acid stays unchanged.
Option (2) Tin and hydrochloric acid — Incorrect. Sn and HCl reduce the nitro group to an amino group, giving 4-aminobenzoic acid; the –COOH group is not reduced.
Option (3) Zinc amalgam and concentrated hydrochloric acid — Incorrect. This is the Clemmensen reduction, which reduces the carbonyl group of aldehydes and ketones to a –CH2– group; it is not used to turn –COOH into –CH2OH.
NCERT (2026-27 edition):
8.9.3 Reactions Involving –COOH Group – “Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride or better with diborane. Diborane does not easily reduce functional groups such as ester, nitro, halo, etc. Sodium borohydride does not reduce the carboxyl group.”