Question

In the reaction sequence ${{C}_{2}}{{H}_{5}}Cl+KCN\xrightarrow{{{C}_{2}}{{H}_{5}}OH}X$, the molecular formula of X is:
A.${{C}_{2}}{{H}_{5}}CN$
B.${{C}_{2}}{{H}_{5}}NC$
C.${{C}_{2}}{{H}_{5}}OH$
D.${{C}_{2}}{{H}_{8}}O$

Answer 1

Hint: Haloallkanes undergo nucleophilic substitution (SN) reactions, to obtain a compound in which the halo group is replaced by the nucleophile. Nucleophilic reactions are of two types. SN-1 and SN-2 reactions. They differ in taking place in two and one steps respectively.

Complete answer: a reaction sequence is given in which chloro ethane reacts with hydrogen cyanide, to form a product. The reaction happens in the presence of ethanol. This reaction is a type of nucleophilic substitution reaction 2, or ${{S}_{N}}^{2}$ reaction. It happens in one step and the leaving group is replaced by a nucleophile.
In this reaction the leaving group is chloride, Cl, which is replaced by cyanide, CN, also a product which is formed with potassium chloride. The reaction is as follows:
${{C}_{2}}{{H}_{5}}Cl+KCN\xrightarrow{{{C}_{2}}{{H}_{5}}OH}{{C}_{2}}{{H}_{5}}CN+KCl$
Therefore, the product formed is ethanenitrile, or ethyl cyanide. Hence , the molecular formula of X, ethyl cyanide is, .${{C}_{2}}{{H}_{5}}CN$, so option A is correct.

Additional information: CN is considered as an ambident nucleophile, this means that CN can attack the carbocation from any site. It can attach as either CN or NC. Here the nucleophile, CN is attacking from the carbon side, to from ethyl cyanide. $N{{O}_{2}}$also acts as an ambident nucleophile.

Note: When the CN attacks from the other side, then isocyanide is formed. The CN nucleophile attacks from the other side, when the reaction takes place with AgCN. The CN attaches from the carbocation in the form of NC and the product is isocyanide or alkyl carbyl amine.