Arrange the following alkyl halides in decreasing order of the rate of beta-elimination reaction with alcoholic KOH. (I) CH_(3)-underset(CH_(3))underset(|)overset(H)overset(|)(C)-CH_(3)Br (II) CH_(3)-CH_(2)-Br (III) 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 beta-elimination reaction with alcoholic KOH. (I) CH_(3)-underset(CH_(3))underset(\|)overset(H)overset(\|)(C)-CH_(3)Br (II) CH_(3)-CH_(2)-Br (III) CH_(3)-CH_(2)-CH_(2)-Br
Answer» IgtIIgtIII IIIgtIIgtI IIgtIIIgtI IgtIIIgtII Solution :`underset((I)" "("has 2 "beta-"substituents"))(CH_(3)-underset(CH_(3))underset(\|)overset(H)overset(\|)(.^(alpha)C)-.^(alpha)CH_(2)Br)""underset((II)" "("has no "beta-"SUBSTITUENT"))(overset(alpha)(C)H_(3)-overset(alpha)(C)H_(2)-Br) underset((III)" "("has 1 "beta-"substituent"))(CH_(3)-overset(beta)(C)H_(2)-overset(alpha)(C)H_(2)-Br)` More the number of `beta`-substituents (alkyl GROUPS), more STABLE alkene it will form on `beta`-elimination and more will be the reactivity. thus, the decreasing order of the RATE of `beta`-elimination reaction with alcoholic KOH is: IgtIIIgtII