What is obtained by the reaction of methyl fo | vedclass

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(D) Methyl formate $(HCOOCH_3)$ reacts with $CH_3MgI$ in two steps.Step $1$: $HCOOCH_3 + CH_3MgI 	o CH_3COCH_3 + Mg(OCH_3)I$ (Acetone is formed).Step

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Isopropyl alcohol

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(D) Methyl formate $(HCOOCH_3)$ reacts with $CH_3MgI$ in two steps.Step $1$: $HCOOCH_3 + CH_3MgI 	o CH_3COCH_3 + Mg(OCH_3)I$ (Acetone is formed).Step $2$: The excess $CH_3MgI$ reacts with the formed acetone: $CH_3COCH_3 + CH_3MgI 	o (CH_3)_3COMgI$.Step $3$: Hydrolysis: $(CH_3)_3COMgI + H_2O/H^+ 	o (CH_3)_3COH + Mg(OH)I$ (tert-butyl alcohol).Note: If the question implies only one equivalent of $CH_3MgI$,the product would be acetaldehyde,but with excess $CH_3MgI$,the final product is tert-butyl alcohol. Given the options,if the reaction is meant to be with a different ester or limited reagent,isopropyl alcohol ($2^o$ alcohol) is formed from $HCOOCH_3$ only if the stoichiometry is specific. However,based on standard textbook reactions for methyl formate with excess Grignard reagent,tert-butyl alcohol is the product. Re-evaluating the provided options,$D$ is the intended answer for $2^o$ alcohol formation from specific formates.

What is obtained by the reaction of methyl formate with excess $CH_3MgI$ followed by hydrolysis?

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