Arrange the following alkyl halides in decreasing order of the rate of -elimination reaction with alcoholic KOH. (A) H_(3)C-underset(CH_(3))underset(|)overset(H)overset(|)(C)-CH_(2)Br (B) CH_(3)-CH_(2)-Br (C) CH_(3)-CH_(2)-CH_(2)-Br

Arrange the following alkyl halides in decreasing order of the rate of

1.	Arrange the following alkyl halides in decreasing order of the rate of -elimination reaction with alcoholic KOH. (A) H_(3)C-underset(CH_(3))underset(\|)overset(H)overset(\|)(C)-CH_(2)Br (B) CH_(3)-CH_(2)-Br (C) CH_(3)-CH_(2)-CH_(2)-Br
Answer» `A gt B gt C` `C gt B gt A` `B gt C gt A` `A gt C gt B` SOLUTION :Alkyl halides on heating with alc. KOH eliminated ONE molecule halogen acid and to form alkane. Hydrogen is eliminated from BETA carbon atom. Nature of alkyl groups determines rate of reaction. Order of reactivity is `3^(@) gt 2^(@) gt 1^(@)` or `A gt C gt B`