2-Bromobutane on heating with alcoholic KOH f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The reaction of $2-$bromobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).According to $Saytzeff's$ rule,the m

Vedclass · Accepted Answer

$80\% \beta-$Butylene $+ 20\% \alpha-$Butylene

Vedclass · Answer

(D) The reaction of $2-$bromobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).According to $Saytzeff's$ rule,the major product is the more substituted alkene.$2-$Bromobutane $(CH_3-CHBr-CH_2-CH_3)$ undergoes elimination to form $2-$butene ($\beta-$butylene) and $1-$butene ($\alpha-$butylene).$2-$Butene is more stable due to hyperconjugation and steric factors,making it the major product $(80\%)$.$1-$Butene is the minor product $(20\%)$.Therefore,the correct composition is $80\% \beta-$butylene and $20\% \alpha-$butylene.

$2-$Bromobutane on heating with alcoholic $KOH$ forms

Similar Questions

Heating methyl chloride with zinc in a closed tube produces...............

What are $X$ and $Y$ in the following reaction sequence?

Acetone is mixed with bleaching powder to give

Which of the following undergoes $S_{N^{1}}$ reaction?

Which one of the following forms propane nitrile as the major product?

Vedclass Test Series

Exam Paper Generator

Online Exam Module