Question

Which of the following on heating with aqueous KOH, produces acetaldehyde?
(A) $C{{H}_{3}}COCl$
(B) $C{{H}_{3}}C{{H}_{2}}Cl$
(C) $C{{H}_{2}}ClC{{H}_{2}}Cl$
(D) $C{{H}_{3}}CHC{{l}_{2}}$

Answer 1

Hint: Acetaldehyde is the aldehyde group consisting of two carbons. So, the reactant should also contain only two carbon to give acetaldehyde as the product.

Complete step by step solution:
The reactant reacting with aqueous KOH gives the nucleophilic substitution of the halide group to give the aldehyde substituted group as the product. Now, let us see about the nucleophilic substitution reaction of alkyl halides with KOH (aq.). Aqueous KOH is the alkaline salt. It dissociates to give hydroxide ion which further acts as a strong nucleophile. Hydroxide ion produced replaces the halogen atom in the alkyl halide. Thus, alcohol molecules are formed. This reaction is called the nucleophilic substitution reaction. Generally,
$R-X+KOH\to R-OH+KX$
Illustration- Now, our given data consists of formation of acetaldehyde,
For (A)- Reaction of acetyl chloride with aqueous KOH gives acetic acid as follows,
$C{{H}_{3}}COCl+KOH\to C{{H}_{3}}COOH+KCl$
Thus, $C{{H}_{3}}COCl$ on reaction with aq. KOH cannot form the required product.
For (B)- Reaction of chloro ethane or ethyl chloride with aqueous KOH gives ethanol or ethyl alcohol as follows,
$C{{H}_{3}}C{{H}_{2}}Cl+KOH\to C{{H}_{3}}C{{H}_{2}}OH+KCl$
Thus, $C{{H}_{3}}C{{H}_{2}}Cl$ on reaction with aq. KOH cannot form the required product.
For (C)- Reaction of 1, 2 dichloroethane with aq. KOH gives ethylene glycol (simply glycol) or ethane 1, 2 diol as follows,
$C{{H}_{2}}ClC{{H}_{2}}Cl+2KOH\to C{{H}_{2}}OHC{{H}_{2}}OH+2KCl$
Thus, $C{{H}_{2}}ClC{{H}_{2}}Cl$ on reaction with aq. KOH cannot form the required product.
For (D)- Reaction of 1, 1 dichloroethane with aq. KOH gives acetaldehyde as the final product as follows,
$C{{H}_{3}}CHC{{l}_{2}}+2KOH\to C{{H}_{3}}CHO+2KCl+{{H}_{2}}O$
Thus, $C{{H}_{3}}CHC{{l}_{2}}$ on reaction with aq. KOH forms the required product by following mechanism;
-The $C{{H}_{3}}CHC{{l}_{2}}$ on reaction with aq. KOH replaces its two chloride ions with hydroxide ions.
-The hydroxide ions form bonds with the same carbon and thus, the bond length is small in this case.
-Thus, due to the bulky nature of hydroxide ions, formation of water molecules takes place.
Hence, acetaldehyde is formed.

Therefore, option (D) is correct.

Note: We have two carbon atoms in every case but the formation of the desired product depends upon the substitution pattern of KOH.