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(D) The reaction involves the nucleophilic attack of $PhMgBr$ on the lactone (coumarin derivative). $1$. The first equivalent of $PhMgBr$ attacks the

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(D) The reaction involves the nucleophilic attack of $PhMgBr$ on the lactone (coumarin derivative). $1$. The first equivalent of $PhMgBr$ attacks the carbonyl carbon,leading to the ring opening of the lactone to form a phenoxide intermediate. $2$. The second equivalent of $PhMgBr$ attacks the newly formed ketone group to form a tertiary alcohol after aqueous workup. $3$. Upon treatment with $H_2SO_4$ and heating,the tertiary alcohol undergoes acid-catalyzed cyclization. The phenolic oxygen attacks the carbocation formed at the tertiary carbon,resulting in the formation of the cyclic product $(B)$. The oxygen atom labeled as $18$ remains in the ring throughout the process as shown in the mechanism.

Product $(B)$ in this reaction is

Similar Questions

In the following compounds,the ones that give positive iodoform test are:
$(I)$ $CH_3CH_2COCH_2CH_3$
$(II)$ $CH_3CH(OH)CH_3$
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$(IV)$ $CH_3CH_2COCH_3$
$(V)$ $PhCOPh$
$(VI)$ $PhCOCH_3$

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