Question

KOH (alcoholic) + CH3 C(CH3)2 CH2Br (1), The reactants in the condition given will under
go:

A. Nucleophilic substitution reaction.

B. Elimination reaction.

C. Nucleophilic addition,

D. None of the above.

While I know typically elemination will occur but here there is no hydrogen with beta carbon.

so is it substitution or none?
​

Answer 1

Explanation:

R−Cl+KOH(aq)→R−OH+KCl

The ionization of aqueous KOH produces hydroxide ions which are strong nucleophiles. Hence, alkyl chlorides undergo substitution to form alcohol.

R−CH

2

−CH

2

−Cl+KOH(alc)→R−CH=CH

2

+KCl+H

2

O

Alcoholic KOH solution gives alkoxide ion which is a strong base. It abstracts β hydrogen atom of alkyl chloride. A molecule of HCl is eliminated and an alkene is formed.

Note: The basicity of hydroxide ion is much lower than the basicity of alkoxide ion as hydroxide ion is significantly hydrated in aqueous solution.

Hence, hydroxide ion cannot abstract β hydrogen atom of alkyl chloride