In the reaction shown below,the products are:

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(D) The reaction of anisole $(C_6H_5OCH_3)$ with $HBr$ involves the protonation of the ether oxygen atom by the acid.This is followed by the nucleophi

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Phenol and $CH_3Br$

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(D) The reaction of anisole $(C_6H_5OCH_3)$ with $HBr$ involves the protonation of the ether oxygen atom by the acid.This is followed by the nucleophilic attack of the bromide ion $(Br^-)$ on the less sterically hindered alkyl group.Since the phenyl group is attached to the oxygen,the $C-O$ bond between the phenyl ring and oxygen has partial double bond character due to resonance,making it stronger and harder to break.Therefore,the $Br^-$ ion attacks the methyl group $(CH_3)$,resulting in the formation of phenol $(C_6H_5OH)$ and methyl bromide $(CH_3Br)$.

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