Identify the product of the given reaction.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → p-Nitrobenzoic acid
A strong oxidising agent such as acidified sodium dichromate oxidises an alkyl group on a benzene ring to a –COOH group. The ring and its other substituents are not affected.
In p-nitrotoluene the side chain is \( \ce{-CH3} \); the \( \ce{-NO2} \) group stays where it is. The methyl carbon takes up three oxygen atoms, as the 3[O] in the scheme shows:
\( \displaystyle \ce{O2N-C6H4-CH3 + 3[O]} \xrightarrow{\ce{Na2Cr2O7}/\ce{H^+}} \ce{O2N-C6H4-COOH + H2O} \)
So the product is p-nitrobenzoic acid (4-nitrobenzoic acid).
Why the other options are incorrect
Option (1) p-Nitrobenzaldehyde — Incorrect. The aldehyde is only an intermediate; the strong oxidant takes it on to the acid. Stopping at \( \ce{-CHO} \) needs a mild, selective reagent such as chromyl chloride (Etard reaction).
Option (3) p-Nitrophenylacetic acid — Incorrect. \( \ce{O2N-C6H4-CH2COOH} \) has one carbon more than p-nitrotoluene. Oxidation cannot add a carbon atom; the carbon of the side chain itself becomes the carboxyl carbon.
NCERT (2026-27 edition):
8.7 Methods of Preparation of Carboxylic Acids
8.2.2 Preparation of Aldehydes – “(i) By oxidation of methylbenzene
Strong oxidising agents oxidise toluene and its derivatives to benzoic acids.”