Arrange the following alkyl halides in decreasing order of the rate of $\mathrm{\beta }$-elimination reaction with alcoholic KOH.

A.

B. ${\mathrm{CH}}_{3}-{\mathrm{CH}}_{2}-\mathrm{Br}$

C. ${\mathrm{CH}}_{3}-{\mathrm{CH}}_{2}-{\mathrm{CH}}_{2}-\mathrm{Br}$

1. A>B>C

2. C>B>A

3. B>C>A

4. A>C>B

Hint: Stability of alkene decide the rate of reaction

Alkyl halides on heating with alcoholic potash elminates one molecule of HX to form an alkene. Hydrogen is eliminated from $\mathrm{\beta }$-carbon atom, thus this reaction is know as $\mathrm{\beta }$-elimination reaction. Stability of product that is alkene determines rate of reaction.

The $3°$ alkyl halide will give more subtituted alkene and it is most stable alkene. The $1°$ alkyl halide give less substituted alkene has it is least reactive. The order is as follows:

i.e.,