The product of the following reaction is /are

The correct answer is Option (2) → 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 (1) p-nitrobenzaldehyde. — Incorrect. The para position is electron-poor because of the −R effect of –CHO, so para substitution is not favoured.

• Option (3) o-nitrobenzaldehyde. — Incorrect. The ortho position is also electron-poor for the same reason.

• Option (4) p-nitrobenzaldehyde and o-nitrobenzaldehyde. — Incorrect. An ortho and para mixture is formed by activating groups, not by the deactivating –CHO 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: