Identify A and B in the following reaction.

The correct answer is Option (1) → A: Benzal Chloride; B: Benzaldehyde

Chlorine reacts with toluene in two different ways. With a Lewis acid such as \( \ce{FeCl3} \) in the dark it substitutes on the ring. In sunlight (hν), without a catalyst, chlorine splits into free radicals, which attack the methyl side chain instead. Chlorination continues on the same carbon, and with enough chlorine two hydrogens of the methyl group are replaced, giving benzal chloride (A):

    \( \displaystyle \ce{C6H5CH3 + 2Cl2} \xrightarrow{h\nu} \ce{C6H5CHCl2 + 2HCl} \)

A carbon carrying two chlorine atoms is hydrolysed by water at 373 K to an unstable gem-diol, which loses water to give the aldehyde. This is a commercial method for making benzaldehyde (B):

    \( \displaystyle \ce{C6H5CHCl2 + 2H2O} \xrightarrow{\text{373 K}} [\ce{C6H5CH(OH)2}] + \ce{2HCl} \)

    \( \displaystyle [\ce{C6H5CH(OH)2}] \xrightarrow{-\ce{H2O}} \ce{C6H5CHO} \)

So A is benzal chloride and B is benzaldehyde.

Why the other options are incorrect

• Option (2) A: Benzyl Chloride; B: Benzyl alcohol — Incorrect. Benzyl chloride, \( \ce{C6H5CH2Cl} \), is only the first stage of side chain chlorination, and its hydrolysis gives benzyl alcohol, not an aldehyde. In the route with chlorine in light followed by water at 373 K, chlorination is carried on to the dichloro stage, which gives benzaldehyde.

• Option (3) A: o-Chlorotoluene; B: o-Cresol — Incorrect. Ring chlorination needs a Lewis acid such as \( \ce{FeCl3} \) in the dark, and an aryl chloride is not hydrolysed by water at 373 K; it needs NaOH at about 623 K and high pressure.

• Option (4) A: p-Chlorotoluene; B: p-Cresol — Incorrect. For the same reasons: light does not chlorinate the ring, and p-chlorotoluene is not hydrolysed to p-cresol by water at 373 K.

NCERT (2026-27 edition):
By Side Chain Chlorination Followed by Hydrolysis

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