(A) In elimination by E2 mechanism, C−D bond being stronger than C−H bond, elimination occurs on side of
CH3 group
(B) Since
Ph–CHBr–CH3 reacts faster than
Ph–CHBr–CD3, C–H or C–D bond cleavage is involved in the rate determining step. This is possible if the reaction follows E2 mechanism.
(C) This reaction proceeds as follows:
Since the rate depends on the concentration of
PhCH2CH2Br only, the reaction follows first-order kinetics and the mechanism involves formation of conjugate base. So it proceeds through E1cb mechanism.
(D) Since the rate is same for
PhCH2CH2Br and
PhCD2−CH2Br, it implies that C-D or C-H bond is not involved in the rate determining step. The reaction occurs in two steps. In the first step, conjugate base of the reactant is formed and the next step is the rate-determining step which involves the loss of a bromide ion from the carbanion. The rate of the reaction depends only on the concentration of one reactant so, it follows first-order kinetics and occurs by E1 mechanism.