Question

Which compound in each of the following pairs will react faster in $S{N}^2$ reaction with $\stackrel{⊝}{O}H$?
(A) MeBr (I) and MeI (II)
(B) $M{e}_{3}C-Cl$ (III) and MeCI (IV)

• A. (A)-I, (B)- III
• B. (A)-I, (B)- IV
• C. (A)-II, (B)- III
• D. (A)-II, (B)- IV

Answer 1

Answer: (D)

(A)-II, (B)- IV

Reason for A- (Il), Iodine is a better leaving group than Br−Br−
Reason for B- (IV) will show less steric hindrance than (III) so in the $1^o$ halide the reaction will be faster.