On boiling with concentrated hydrobromic acid | vedclass

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(A) The reaction of phenyl ethyl ether $(C_6H_5-O-C_2H_5)$ with concentrated hydrobromic acid $(HBr)$ involves the cleavage of the $C-O$ bond. Since t

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Phenol and ethyl bromide

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(A) The reaction of phenyl ethyl ether $(C_6H_5-O-C_2H_5)$ with concentrated hydrobromic acid $(HBr)$ involves the cleavage of the $C-O$ bond. Since the $C-O$ bond between the phenyl group and oxygen has partial double bond character due to resonance,it is stronger and does not break easily. Therefore,the cleavage occurs at the $O-C_2H_5$ bond,resulting in the formation of phenol $(C_6H_5OH)$ and ethyl bromide $(C_2H_5Br)$. The reaction is: $C_6H_5OC_2H_5 + HBr \rightarrow C_6H_5OH + C_2H_5Br$.

On boiling with concentrated hydrobromic acid,phenyl ethyl ether will yield:

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