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Question: The below reaction is: \(C{H_3}C{H_2}Cl + KOH(aq) \to C{H_3}C{H_2} - OH + KCl\) A. nucleophilic ...

The below reaction is:
CH3CH2Cl+KOH(aq)CH3CH2OH+KClC{H_3}C{H_2}Cl + KOH(aq) \to C{H_3}C{H_2} - OH + KCl
A. nucleophilic substitution
B. electrophilic substitution
C. free- radical substitution
D. nucleophilic addition

Explanation

Solution

Generally, in the given reaction one group of the reactant is replacing the other group of the reactant to form the desired product. The chloride ions present in the ethyl chloride act as a nucleophile and the hydroxyl ions present in potassium hydroxide also act as a nucleophile.

Complete step by step answer:
In the given reaction between ethyl chloride and potassium hydroxide both the chloride ion and hydroxyl ion act as a nucleophile.
The reaction is shown below.
CH3CH2Cl+KOH(aq)CH3CH2OH+KClC{H_3}C{H_2}Cl + KOH(aq) \to C{H_3}C{H_2} - OH + KCl
In this reaction, ethyl chloride reacts with aqueous solution of potassium hydroxide gives ethanol and potassium chloride.
In the given reaction, the hydroxyl ion replaces the chloride ion to form the product.
In nucleophilic substitution reaction, nucleophile of one reactive species replaces the other nucleophile of other reactive species to form the product.
The group or atom which accepts the electrons and gets removed from the carbon is known as the leaving group and the molecule where the substitution reaction takes place is known as substrate. The leaving group is removed from the molecule as a neutral atom or an anion.
In nucleophilic substitution reactions, the reactivity strength of a nucleophile is known as nucleophilicity.
Thus, the reaction is a nucleophilic substitution reaction.
Therefore, the correct option is A.

Note: In nucleophilic substitution reaction, the stronger nucleophile of the compound displaces the weaker nucleophile from its compound to form the product. The nucleophilic substitution reaction is the same as the displacement reaction where the higher reactive metal displaces the lower reactive metal to form the product.