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  4. Arrange the following alkyl halides in decreasing order of the rate of β elimination reaction with alcoholic KOH: (i)<img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/chemistry/ce5fcaa76e029b601a0a2584195bf977-.png /> (ii) CH 3- CH 2- Br (iii) CH 3- CH 2- CH 2- Br

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Q. Arrange the following alkyl halides in decreasing order of the rate of $\beta$ elimination reaction with alcoholic $KOH$:
(i)image
(ii) $CH _{3}- CH _{2}- Br$
(iii) $CH _{3}- CH _{2}- CH _{2}- Br$

Hydrocarbons

Solution:

More stable alkene will be formed by $\beta$ - elimination, only when there are more number of $\beta$ -substituents. Since

(i) has two,

(ii) has no $\beta$ - methyl substitutive therefore; reactivity $\beta$ - elimination decrease in order:

(iii) has one $\beta$ - methyl substituent while

$( i )>( iii )>( ii )$