The major product formed in the reaction is

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(A) The starting material is a $\gamma$-lactone ($5$-membered cyclic ester) with a methyl group at the $\alpha$-position.When treated with excess $PhM

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(A) The starting material is a $\gamma$-lactone ($5$-membered cyclic ester) with a methyl group at the $\alpha$-position.When treated with excess $PhMgBr$ (a Grignard reagent),the following steps occur:$1$. The Grignard reagent attacks the carbonyl carbon of the ester,opening the ring to form a ketone intermediate.$2$. Since $PhMgBr$ is in excess,it further attacks the newly formed ketone to produce a tertiary alcohol.$3$. Upon acidic workup $(H_3O^+)$,the alkoxide is protonated to form the final diol product.The ring-opening of the lactone results in a chain where the original carbonyl carbon becomes a tertiary alcohol with two phenyl groups,and the original oxygen of the ester becomes a secondary alcohol at the other end of the chain.Thus,the product is a $1,4-$diol: $Ph_2C(OH)-CH_2-CH_2-CH(OH)-CH_3$.

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