What is the product formed when ethyl acetate | vedclass

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(B) Ethyl acetate $(CH_3COOC_2H_5)$ reacts with $CH_3MgI$ in two steps.Step $1$: Nucleophilic attack of $CH_3^-$ on the carbonyl carbon of ethyl aceta

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tert-butyl alcohol

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(B) Ethyl acetate $(CH_3COOC_2H_5)$ reacts with $CH_3MgI$ in two steps.Step $1$: Nucleophilic attack of $CH_3^-$ on the carbonyl carbon of ethyl acetate forms an unstable intermediate,which eliminates ethoxide to form acetone $(CH_3COCH_3)$.Step $2$: Since $CH_3MgI$ is in excess,it reacts with the formed acetone to produce a magnesium alkoxide intermediate,which upon hydrolysis yields tert-butyl alcohol ($2$-methylpropan-$2$-ol).Reaction: $CH_3COOC_2H_5 + CH_3MgI \rightarrow CH_3COCH_3 + Mg(OC_2H_5)I$.Then,$CH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3COMgI$.Finally,$(CH_3)_3COMgI + H_2O/H^+ \rightarrow (CH_3)_3COH + Mg(OH)I$.

What is the product formed when ethyl acetate reacts with an excess of $CH_3MgI$ followed by hydrolysis?

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