X is:
X is:
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The Correct Option is C
Solution and Explanation
The H of HCl adds to the terminal \(\mathrm{CH_2}\) (Markovnikov), giving the secondary carbocation \(\mathrm{C_6H_5CH_2\overset{+}{C}HCH_3}\).
A 1,2-hydride shift from the adjacent benzylic \(\mathrm{CH_2}\) gives the benzylic carbocation \(\mathrm{C_6H_5\overset{+}{C}HCH_2CH_3}\), which is stabilised by resonance with the ring.
\(\mathrm{Cl^-}\) attacks this carbocation, giving \(\mathrm{C_6H_5CH(Cl)CH_2CH_3}\) (1-chloro-1-phenylpropane).
This is the structure in option (3), with Cl on the carbon attached to the ring and an ethyl group on the other side.
Answer: (3) \(\mathrm{C_6H_5CHClCH_2CH_3}\)
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