Question

In the following reaction:
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\xrightarrow{{{\text{B}}{{\text{r}}_{\text{2}}}}}{\text{X}}\xrightarrow{{{\text{KCN}}}}{\text{Y}}$; ${\text{Y}}$ is:
A) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}}$
B) ${\text{NC}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CN}}$
C) ${\text{Br}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{CN}}$
D) ${\text{Br}} - {\text{CH}} = {\text{CHCN}}$

Answer 1

Hint:We are given a reaction sequence. The initial compound of the reaction sequence is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ is ethene which is an alkene. Ethene reacts with liquid bromine and an alkyl halide is formed which further reacts with potassium cyanide.


Complete solution:
We are given a reaction sequence as follows:
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\xrightarrow{{{\text{B}}{{\text{r}}_{\text{2}}}}}{\text{X}}\xrightarrow{{{\text{KCN}}}}{\text{Y}}$
The initial compound of the reaction sequence is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ is ethene which is an alkene. Ethene reacts with liquid bromine and an alkyl halide.
During the reaction, the double bond in ethene breaks and one bromine atom gets attached to each carbon atom. The reaction is as follows:
${\text{C}}{{\text{H}}_2} = {\text{C}}{{\text{H}}_2}\xrightarrow{{{\text{B}}{{\text{r}}_2}}}{\text{Br}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{Br}}$
When ethene reacts with liquid bromine 1,2-dibromoethane which is an alkyl halide is produced. 1,2-dibromoethane is also known as ethylene dibromide (EDB).
Thus, X is ${\text{Br}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{Br}}$ 1,2-dibromoethane.
1,2-dibromoethane further reacts with potassium cyanide. During the reaction, the cyano groups replace the two bromine atoms. This is a nucleophilic substitution reaction. The reaction is as follows:
${\text{Br}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{Br}}\xrightarrow{{{\text{KCN}}}}{\text{NC}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CN}}$
When 1,2-dibromoethane reacts with potassium cyanide 1,2-diaminoethane is produced.
Thus, is ${\text{NC}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CN}}$,1,2-diaminoethane.

Thus, the correct option is (B) compound.


Note:The reaction in which one nucleophile is replaced by another nucleophile is known as a nucleophilic substitution reaction. The group which takes the electron pair with it and gets displaced from the carbon is known as the leaving group. The leaving group leaves as an anion or as a neutral molecule leaving behind a carbonium ion.