Question

$C{H_3}Br + KCN(alc.) \to X\xrightarrow{{Na + {C_2}{H_5}OH}}Y$
What is Y in the series?

Answer 1

Hint: Alkyl halide in the presence of nucleophile undergoes nucleophilic substitution reaction. Here $CN^-$ acts as a nucleophile replacing the $Br^-$ of alkyl group which results in the formation of alkane nitrile. Alkane Nitrile in the presence of $(Na^{+}, {C_2}{H_5}OH$) undergoes reduction which leads to the formation of amines.

Complete step by step answer:
We know alkyl halide undergoes substitution reactions in the presence of nucleophiles. Here Bromomethane In the presence of KCN undergoes nucleophilic substitution reaction and yields the compound X ,i.e methanenitrile.

$C{H_3}Br + KCN(alc.) \to C{H_3}CN + KBr$

Now we know nitrile undergoes reduction in the presence of $(Na^{+}, {C_2}{H_5}OH$), leading to the formation of ethyl amine.
Here sodium in the presence of alcohol acts as a good reducing agent reducing the nitrile into the respective amine.

$C{H_3}Br + KCN(alc.) \to C{H_3}CN + KBr\xrightarrow{{Na + {C_2}{H_5}OH}}C{H_3}C{H_2}N{H_2}$

Hence in the given series of reaction first nucleophilic substitution reaction taken place which convert Bromomethane into methanenitrile. Further methanenitrile in the presence of strong reducing agent ,i.e (Na+C2H5OH) undergoes reduction which yields ethylamine as a product.
Therefore the compound Y is ethylamine.
Here it should be noted that alkyl halide in the presence of alcoholic solution of KCN undergoes SN2 reaction. SN2 stands for nucleophilic substitution bimolecular reaction. In SN2 reaction the nucleophile attack on the alkyl halide and the removal of leaving group here Br- occurs in a single step. The SN2 reaction is preferable in primary halide groups due to less steric hindrance.

Note:
Here if the tertiary haloalkane is used in place of primary haloalkane, then the substitution reaction does not occur as in place of substitution reaction it undergoes elimination reaction which leads to the formation of alkene.