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Question: 2-Bromopentane is treated with a alcoholic KOH solution what will be the major product formed in thi...

2-Bromopentane is treated with a alcoholic KOH solution what will be the major product formed in this reaction and what is the type of elimination called?

A

Pent-1-ene, β\beta -Elimination

B

Pent-2-ene, β\beta -Elimination

C

Pent-2-ene, Nucleophilic substitution

D

Pent-2-ene, Nucleophilic substitution.

Answer

Pent-2-ene, β\beta -Elimination

Explanation

Solution

: CH3−CH2−CH2−CH∣Br−CH3→alc.KOH−HBRCH_{3} - CH_{2} - CH_{2} - \underset{Br}{\underset{|}{CH}} - CH_{3}\underset{- HBR}{\overset{\quad alc.KOH\quad}{\rightarrow}}

2-Bromopentane

CH3−CH2−CH=CH2−Pentene−CH3CH_{3} - CH_{2} - \underset{2 - Pentene}{CH = CH} - CH_{3}