Question

Electrolysis of aq. $C{H}_{3}COONa$ gives $\left(B\right)$ at anode. $\left(B\right)$ when heated with $C{l}_2$ at ${400}^{0}C$ gives $\left(C\right)$. $\left(C\right)$ when reacts with excess of alcoholic $N{H}_3$ gives $\left(D\right)$. The final product $\left(D\right)$ is:

• A. $(C_2{H}_5{)}_{3}N$
• B. $(C_2{H}_5{)}_4{N}^{+}C{l}^{-}$
• C. $(C_2{H}_5{)}_{2}NH$
• D. $C_2{H}_{5}N{H}_2$

Answer 1

Answer: (B)

$(C_2{H}_5{)}_4{N}^{+}C{l}^{-}$

The final product $\left(D\right)$ is tetraethyl ammonium chloride $(C_2{H}_5{)}_4{N}^{+}C{l}^{-}$.

$aq.C{H}_{3}COONa\stackrel{electrolysis}{\to }\underset{Batanode}{C_2{H}_6}\underset{Heat}{\overset{C{l}_2,{400}^{0}C}{\to }}\underset{\left(C\right)}{C_2{H}_{5}Cl}\underset{excess}{\overset{alc.N{H}_3}{\to }}\underset{\left(D\right)}{(C_2{H}_5{)}_4{N}^{+}C{l}^{-}}$

The electrolysis of aq. sodium acetate gives ethane at anode. On heating with chlorine at ${400}^{o}C$, ethyl chloride is obtained. Treatment with alc. ammonia (excess) gives tetraethyl ammonium chloride.