Question

The hybridization state of carbon atoms in the product formed by the reaction of ethyl chloride with aqueous KOH is:

• A. $sp$
• B. $s{p}^2$
• C. $s{p}^3$
• D. $s{p}^{3}d$

Answer 1

Answer: (C)

The function of aqueous KOH is to convert halogen to hydroxy group. So, ethyl chloride reacts with aq. KOH to give ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ as: $\begin{array}{c}{C}_2{H}_5\\ \underset{Aq.}{O\stackrel{+}{H}}\end{array}\to \underset{\text{Ethyl}\text{alcohol}}{C_2{H}_{5}OH}+KCl$ In ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ all the four valencies of carbon are single bonded. So, hybridization of carbon atom in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $\text{s}{\text{p}}^3.$