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Question

Column I Column II
A CH3CHBrCD3/on treatement with alc.KOH gives CH2=CHCD3 as a major product P E1 reaction
B PhCHBrCH3 reacts faster than PhCHBrCD3 Q E2 reaction
C PhCH2CH2F on treatement with C2H5OD/C2H5Ogives PhCD=CH2 as the major product R E1cB reaction
D PhCH2CH2Br and PhCD2CH2Brreact with same rate S First order reaction

A
AQ; BQ; CR, S; DP ,S
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B
AQ; BP, S; CR, S; DQ
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C
AP; BP, S; CR; DQ
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D
AP, S; BQ, S; CP, S; DR, S
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Solution

The correct option is A AQ; BQ; CR, S; DP ,S
AQ
BQ
CR,S
DP,S

In (A), the reactant, secondary alkyl halide reacts with alc. KOH. Alcoholic KOH is a strong base, it gives 1, 2 β elimination product. Thus, it is a E2 elimination reaction.
In (B), the reactant is a benzyl halide derivative. Benyl halide undergoes both E1 and E2 elimination reaction. Here, PhCHBrCH3 reacts faster than PhCHBrCD3 which means the deuterium is involved in rate determining step. Hence, the given reaction is E2 elimination.



In (D), the reactant is a benzyl halide derivative. Benyl halide undergoes both E1 and E2 elimination reaction. Here, PhCH2CH2Br and PhCD2CH2Br reacts with same rate which means the deuterium is not involved in rate determining step. Hence, the given reaction is E1 elimination.

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