The compound which can form intramolecular hydrogen bonds is

The correct answer is Option (2) → Salicylaldehyde

• A hydrogen bond forms between an H atom attached to a strongly electronegative atom (here the H of an \( \ce{-OH} \) group) and a lone pair on another electronegative atom (here the O of a \( \ce{C=O} \) group). When both groups belong to different molecules, the bond is intermolecular. When both groups are in the same molecule and close enough to reach each other, the bond is intramolecular.

• On a benzene ring, the \( \ce{-OH} \) and \( \ce{-CHO} \) groups can reach each other only when they are on neighbouring carbons (ortho position). The H of \( \ce{-OH} \) then bonds to the O of \( \ce{-CHO} \), closing a stable six-membered ring.

Salicylaldehyde is 2-hydroxybenzaldehyde (o-hydroxybenzaldehyde), \( \ce{HOC6H4CHO} \) with the two groups ortho. So it forms an intramolecular hydrogen bond. o-Nitrophenol has the same arrangement, where the O of the ortho \( \ce{-NO2} \) group takes the place of the O of \( \ce{-CHO} \).

Why the other options are incorrect

Option (1) m-Hydroxybenzaldehyde — Incorrect.

• The \( \ce{-OH} \) and \( \ce{-CHO} \) groups are at positions 1 and 3, too far apart to bond within the molecule.

• Its \( \ce{-OH} \) group forms hydrogen bonds with \( \ce{-CHO} \) groups of other molecules (intermolecular).

Option (3) Benzaldehyde — Incorrect. Benzaldehyde, \( \ce{C6H5CHO} \), has no O–H bond, so it has no hydrogen to donate. It cannot form a hydrogen bond with itself, either within one molecule or between its molecules.

Option (4) p-Hydroxybenzaldehyde — Incorrect. The two groups are at opposite ends of the ring (positions 1 and 4), so only intermolecular hydrogen bonds are possible. This associates the molecules, as in p-nitrophenol.

NCERT (2026-27 edition):
Chemical Reactions