Question

The hybridization state of carbon atoms in the product formed by the reactions of ethyl chloride with aqueous potassium hydroxide is
(A) $sp$
(B) $s{p^2}$
(C) $s{p^3}$
(D) $s{p^3}d$

Answer 1

Hint: Ethyl chloride has chemical formula ${C_2}{H_5}Cl$ and it reacts with aqueous $KOH$ by ${S_N}^1$ mechanism that is nucleophilic substitution of first order. It is also called as solvolysis because it is the solvent which acts as a nucleophile. The solvent used must have a high value of dielectric constant.

Complete step-by-step answer: Nucleophilic substitution takes place in two steps
-Formation of carbocation
-Attack of nucleophile
The reactant undergoing nucleophilic substitution has a good leaving group. As the leaving group leaves the reactant, there is a formation of an intermediate that is a carbocation.
The nucleophile approaches carbocation from front direction as well as back side leading to formation of racemic products.
Thus, in ethyl chloride, $Cl$ is a good leaving group and leaves to form ${C_2}{H_5}^ + $. Then the nucleophile $O{H^ - }$ attacks the carbocation to form ${C_2}{H_5}OH$.
the nucleophilic substitution of ethyl chloride with aqueous potassium hydroxide is represented by the following reaction
 ${C_2}{H_5}Cl + KOH \to {C_2}{H_5}OH + KCl$

There are all single bonds in ethyl alcohol.
Hence, the hybridization of carbon is $s{p^3}$ as (there are four single bonds attached to the carbon atom) that will undergo mixing.

Option ‘C’ is correct

Note: If the alcoholic $KOH$ was used in place of aqueous $KOH$ then instead of substitution reaction elimination reaction takes place. In case of elimination reaction, $s{p^2}$ hybridized carbons will be there in the product.