The major product 'P' formed in the following reaction is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1-Chloro-1-phenylethane, C6H5CH(Cl)CH3
Chlorine in light (hν) and without a Lewis acid reacts by a free radical mechanism, so it attacks the side chain, not the ring, just as toluene gives benzyl chloride on side-chain chlorination. The hydrogen removed is the one on the benzylic carbon, because the benzylic radical \( \ce{C6H5C} \)·\( \ce{HCH3} \) is stabilised by resonance with the ring:
\( \ce{C6H5CH2CH3 + Cl2} \xrightarrow{\ce{h\nu}} \ce{C6H5CH(Cl)CH3 + HCl} \)
P (\( \ce{C8H9Cl} \)) is 1-chloro-1-phenylethane.
Why the other options are incorrect
Option (2) 2-chloro-1-phenylethane, C6H5CH2CH2Cl — Incorrect. Substitution at the terminal \( \ce{CH3} \) goes through a primary radical, which is far less stable than the benzylic radical; this is only a minor product.
Option (3) 1-chloro-2-ethylbenzene — Incorrect. Ring chlorination needs a Lewis acid catalyst (\( \ce{FeCl3} \)) in the dark, not light.
Option (4) 1-chloro-4-ethylbenzene — Incorrect. Again a ring substitution product, formed with \( \ce{Cl2} \)/\( \ce{FeCl3} \), not with \( \ce{Cl2} \)/hν.