Propanenitrile on reaction with ethylmagnesiu | vedclass

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Question

(D) The reaction of propanenitrile $(CH_3CH_2CN)$ with ethylmagnesium iodide $(C_2H_5MgI)$ in the presence of dry ether proceeds as follows:$1$. Nucle

Vedclass · Accepted Answer

$3-$Pentanone

Vedclass · Answer

(D) The reaction of propanenitrile $(CH_3CH_2CN)$ with ethylmagnesium iodide $(C_2H_5MgI)$ in the presence of dry ether proceeds as follows:$1$. Nucleophilic attack of the ethyl group from the Grignard reagent on the carbon atom of the nitrile group leads to the formation of an imine magnesium complex: $CH_3CH_2CN + C_2H_5MgI \rightarrow CH_3CH_2C(C_2H_5)=NMgI$.$2$. Acid hydrolysis $(H_3O^+)$ of this imine complex results in the formation of a ketone and ammonia: $CH_3CH_2C(C_2H_5)=NMgI + H_2O/H^+ \rightarrow CH_3CH_2COCH_2CH_3 + NH_3 + Mg(OH)I$.$3$. The product formed is $CH_3CH_2COCH_2CH_3$,which is $3-$pentanone (pentan-$3-$one).Therefore,the correct option is $D$.

Propanenitrile on reaction with ethylmagnesium iodide in the presence of dry ether gives an imine complex. This imine complex on acid hydrolysis forms:

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