Reaction of ethyl formate with excess of CH_3 | vedclass

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(D) The reaction of ethyl formate $(HCOOC_2H_5)$ with excess $CH_3MgI$ proceeds in two steps. First,the nucleophilic attack of $CH_3^-$ on the carbony

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Isopropyl alcohol

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(D) The reaction of ethyl formate $(HCOOC_2H_5)$ with excess $CH_3MgI$ proceeds in two steps. First,the nucleophilic attack of $CH_3^-$ on the carbonyl carbon of ethyl formate leads to the formation of an intermediate,which eliminates an ethoxide ion to form acetaldehyde $(CH_3CHO)$. Since $CH_3MgI$ is in excess,it further reacts with the formed acetaldehyde to produce an alkoxide intermediate. Finally,upon hydrolysis,this intermediate yields a secondary alcohol. The overall reaction is: $HCOOC_2H_5 + 2CH_3MgI \xrightarrow{H_3O^+} (CH_3)_2CHOH + Mg(OH)I + C_2H_5OH$. The product formed is isopropyl alcohol $(CH_3CH(OH)CH_3)$.

Reaction of ethyl formate with excess of $CH_3MgI$ followed by hydrolysis gives

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