Question

Find the product of the following reaction:
$MeO-C{{H}_{3}}-Cl\xrightarrow{KCN}$
(A) $Cl-C{{H}_{3}}-CN$
(B) $MeO-C{{H}_{3}}-CN$
(C) $Me-O-C{{H}_{2}}-C{{H}_{2}}-CN$
(D) $CN-O-CN$

Answer 1

Hint : We know that in the given reaction, methyl chloride is a haloalkane and ethylamine is an amine. In order to convert a haloalkane to an amine, the haloalkane should be treated with ammonium. Here a primary haloalkane is being converted so the resultant would be a primary amine.

Complete Step By Step Answer:
Let us first write the molecular formulas of the given organic compounds. The molecular formula of methyl chloride is written as $MeO-C{{H}_{3}}-Cl.$ Here, we can see that methyl chloride contains only one carbon atom whereas ethylamine contains two carbon atoms. So our first step would be adding one carbon atom to the methyl chloride molecule. This can be done by treating methyl chloride with sodium cyanide i.e. $NaCN$ In this reaction the chlorine atom will be replaced or substituted by the cyanide ion. This reaction undergoes through the $S{{N}^{2}}.$ mechanism.
The product in this reaction is known as acetonitrile. Acetonitrile consists of a $C\equiv N$ bond and we need a $C{{H}_{2}}-N{{H}_{2}}$ bond to acquire ethylamine. This means that the cyanide molecule in acetonitrile should be added with hydrogens which is basically a reduction reaction.
$MeO-C{{H}_{3}}-Cl\xrightarrow{KCN}MeO-C{{H}_{3}}-CN+C{{l}^{-}}.$
Therefore, the correct answer is option B. i.e. the product of the following reaction is $MeO-C{{H}_{3}}-CN.$

Note :
Remember that it is to be noted that sodium in presence of ethanol is a strong reducing agent and hence it reduces the cyanide molecule directly to amide. If a milder reducing agent is used the cyanide molecule could potentially become an imide which is a $CH=NH$ molecule.