Ethyl phenyl ether is heated with excess HBr. The products are

A $C X_{6} H X_{6}$ and $CH X_{3} CH X_{2} - OBr$
B $C X_{6} H X_{5} - Br$ and $CH X_{3} CH X_{2} - OH$ , because bromide attacks the carbon of the stronger bond
C $C X_{6} H X_{5} - Br$ and $CH X_{3} CH X_{2} - Br$ , because both bonds break equally easily
D $C X_{6} H X_{5} - OH$ and $CH X_{3} CH X_{2} - Br$ , because an $s p^{2}$ aryl carbon cannot undergo SN2 or SN1 attack

Answer: D

Protonation of the oxygen makes either group a possible leaving group. Bromide can attack only the ethyl carbon (SN2), since backside attack at an aryl carbon is impossible and the phenyl cation is too unstable. The

C X_{6} H X_{5} - O

bond stays intact and phenol is released along with ethyl bromide. Phenol does not react further with HBr.

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Ethyl phenyl ether is heated with excess HBr. The products are

Answer: D

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