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(D) The reaction of an alkyl halide with alcoholic $KCN$ is a nucleophilic substitution reaction ($S_N2$ mechanism) that produces an alkyl cyanide (ni

Vedclass · Accepted Answer

$n-$Propylchloride and alcoholic $KCN$

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(D) The reaction of an alkyl halide with alcoholic $KCN$ is a nucleophilic substitution reaction ($S_N2$ mechanism) that produces an alkyl cyanide (nitrile).To obtain butanenitrile $(CH_3CH_2CH_2CH_2CN)$,which contains $4$ carbon atoms in the chain plus the nitrile carbon,we need a starting alkyl halide with $4$ carbon atoms.$n-$Propylchloride $(CH_3CH_2CH_2Cl)$ has $3$ carbons,so it would yield butanenitrile $(CH_3CH_2CH_2CN)$ which is $3$ carbons in the chain plus the nitrile carbon,totaling $4$ carbons.Wait,butanenitrile is $CH_3CH_2CH_2CN$. This contains $4$ carbon atoms in total.Therefore,$n-$propylchloride $(CH_3CH_2CH_2Cl)$ reacts with $KCN$ to form $CH_3CH_2CH_2CN$ (butanenitrile).Reaction: $CH_3CH_2CH_2Cl + KCN 	ext{ (alc.)} ightarrow CH_3CH_2CH_2CN + KCl$.

Which of the following pairs of compounds on heating gives butanenitrile?

Similar Questions

Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?

Arrange the following compounds in order of decreasing rate of hydrolysis for $SN_1$ reaction:
$(I)$ $C_6H_5CH_2Br$
$(II)$ $p-CH_3-C_6H_4-CH_2Br$
$(III)$ $p-(CH_3)_2CH-C_6H_4-CH_2Br$
$(IV)$ $p-CH_3CH_2-C_6H_4-CH_2Br$

The decreasing order of reactivity of the compounds given below towards solvolysis under identical conditions is:

The product of the reaction is:

The correct reactivity order of the following compounds towards alkaline hydrolysis is:-
$I. \ p-CH_3O-C_6H_4-CH_2Cl$
$II. \ C_6H_5-CH_2Cl$
$III. \ p-CH_3-C_6H_4-CH_2Cl$
$IV. \ m-CH_3-C_6H_4-CH_2Cl$

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