Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH.

A.
B. CH3-CH2-Br
C. CH3-CH2-CH2-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 eliminates one molecule of HX to form an alkene. Hydrogen is eliminated from β-carbon atom, thus this reaction is know as β-elimination reaction. Stability of product that is alkene determines rate of reaction.
The 3° alkyl halide will give more substituted 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., 3°>2°>1° or A>C>B