What is the product of the following reaction?

The correct answer is Option (1) → m-Nitrobenzaldehyde

In electrophilic substitution of a substituted benzene, the group already on the ring decides where the new group enters:

• Electron donating groups (–OH, –\( \ce{NH2} \), –\( \ce{CH3} \)) activate the ring and direct the electrophile to the ortho and para positions.

• Electron withdrawing groups (–CHO, –COOH, –\( \ce{NO2} \), –CN) deactivate the ring and direct the electrophile to the meta position, because they remove electron density mostly from the ortho and para positions.

The –CHO group of benzaldehyde withdraws electrons by both the −I and −R effects, so it is meta directing. With the nitrating mixture (conc. \( \ce{HNO3} \) + conc. \( \ce{H2SO4} \)), the \( \ce{NO2^{+}} \) ion attacks the meta position. The reaction is kept at a low temperature (273-283 K) so that the easily oxidised –CHO group is not oxidised:

    \( \displaystyle \ce{C6H5CHO + HNO3} \xrightarrow[273\text{-}283\ \mathrm{K}]{\text{conc. }\ce{H2SO4}} \ce{m-O2N-C6H4-CHO + H2O} \)

So the product is m-nitrobenzaldehyde.

Why the other options are incorrect

• Option (2) p-Nitrobenzaldehyde — Incorrect. The para position is electron-poor because of the −R effect of –CHO, so para substitution is not favoured.

• Option (3) m-sulphonatobenzaldehyde — Incorrect. In the nitrating mixture, conc. \( \ce{H2SO4} \) protonates \( \ce{HNO3} \) and produces the electrophile \( \ce{NO2^{+}} \), so a nitro group enters the ring. Sulphonation needs fuming sulphuric acid (oleum).

• Option (4) Benzoic acid — Incorrect. The low temperature (273-283 K) keeps the –CHO group from being oxidised, and the reagent acts as a nitrating mixture, so the product carries an –\( \ce{NO2} \) group.

NCERT (2026-27 edition):
Electrophilic Substitution Reaction

Note: The same/similar question has been asked in different years by NTA as per the details given below: