Ethyl acetate xrightarrow[{(2),HOH}]{{(1),C{H | vedclass

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(A) Ethyl acetate $(CH_3COOC_2H_5)$ reacts with excess Grignard reagent $(CH_3MgBr)$ in two steps.Step $1$: The first equivalent of $CH_3MgBr$ attacks

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$2-$methylpropan$-2-$ol

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(A) Ethyl acetate $(CH_3COOC_2H_5)$ reacts with excess Grignard reagent $(CH_3MgBr)$ in two steps.Step $1$: The first equivalent of $CH_3MgBr$ attacks the carbonyl carbon of ethyl acetate to form a ketone intermediate,acetone $(CH_3COCH_3)$,and releases ethanol $(C_2H_5OH)$.Step $2$: Since $CH_3MgBr$ is in excess,the second equivalent of $CH_3MgBr$ attacks the carbonyl carbon of the newly formed acetone to form an alkoxide intermediate,magnesium salt of tert-butyl alcohol.Step $3$: Upon hydrolysis $(HOH)$,the alkoxide is protonated to yield the final product,$2$-methylpropan-$2$-ol ($tert$-butyl alcohol).The reaction is: $CH_3COOC_2H_5 + 2CH_3MgBr$ $\rightarrow (CH_3)_3COMgBr + C_2H_5OMgBr$ $\xrightarrow{H_2O} (CH_3)_3COH + C_2H_5OH + Mg(OH)Br$.

Ethyl acetate $\xrightarrow[{(2)\,HOH}]{{(1)\,C{H_3}MgBr\,\left[ {Excess} \right]}}P$
The product $P$ will be

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Ethyl acetate $\xrightarrow[{(2)\,HOH}]{{(1)\,C{H_3}MgBr\,\left[ {Excess} \right]}}P$The product $P$ will be