Question

Select the reagent that will yield the greater amount of substitution on reaction with $C{H_3} - C{H_2} - Br$:
(A) $C{H_3}C{H_2}OK$ in dimethyl sulfoxide (DMSO)
(B) ${(C{H_3})_3}COK$ in dimethyl sulfoxide (DMSO)
(C) Both A and B will give comparable amounts of substitution
(D) Neither A nor B will give any amount of substitution

Answer 1

As we know that alkyl halides undergo nucleophilic substitution reaction because the electrophilic alkyl halide can form a new bond with the nucleophile resulting in replacing of halogen present at the alpha carbon and the beta hydrogen can be replaced to yield a greater amount of products like alkene.

Complete Step by step answer: We know that alkyl halides undergo nucleophilic substitution reaction because alkyl halide being electrophile have the ability to form a new bond with the nucleophile thereby replacing the halogen atom present at the alpha carbon also the beta hydrogen can be replaced to yield a greater amount of product that is alkene.
We also know that Dimethyl sulfoxide is used as solvent in nucleophilic substitution reactions and the potassium ethoxide which possesses an ethoxide ion acts as a nucleophile and the potassium tertiary butoxide possesses a tertiary nucleophile. Remember that the nucleophilicity is inversely proportional to size of the nucleophile.
As we can see the nucleophile in potassium ethoxide $(C{H_3}C{H_2}{O^ - }{K^ + })$ is less sterically hindered and stronger as compared to potassium tert-butoxide ${(C{H_3})_3}C{O^ - }{K^ + }$ which is sterically hindered and bulky nucleophile therefore weak, so potassium ethoxide can easily attack the beta-hydrogen in bromoethane and substitution of this beta hydrogen results in ethene formation.
Therefore, from the above explanation we can say that potassium ethoxide in dimethyl sulfoxide will yield the greater amount of the product.

Hence the correct answer is (A).

Note: Always remember that the nucleophilicity is inversely proportional to the size of nucleophile, less bulky the nucleophile, lesser will be its hindrance and stronger will be the nucleophile whereas if bulky is the nucleophile, more is the steric hindrance and weak is the nucleophile.