1. Match the following:

\begin{array}{} & \textbf{Column I}&&\textbf{Column II} \\ \text{(A)} & \ce{CH3-CHBr-CD3} \text{ on treatment with alc. KOH gives}& \text{(P)}& \text{E1 reaction}\\ &\ce{CH2=CH-CD3} \text{ as a major product} \\ \text{(B)} & \ce{Ph-CHBr-CH3} \text{ reacts faster than } \ce{Ph-CHBr-CD3} & \text{(Q)} & \text{E2 reaction}\\ \text{(C)} & \ce{Ph-CH2-CH2Br} \text{ on treatment with } \ce{C2H5OD/C2H5O-}& \text{(R)}& \text{E1cb}\\ &\text{gives } \ce{Ph-CD=CH2} \text{ as the major product.} && \text{reaction}\\ \text{(D)} & \ce{Ph-CH2-CH2Br} \text{ and } \ce{Ph-CD2-CH2Br} \text{ react with the }& \text{(S)} & \text{First Order}\\ &\text{same rate.} && \text{reaction} \end{array}

The answer given in the exam solutions is

A − Q; B − Q; C − R,S; D − P,S

Question source: IIT-JEE 2006 (memory based)

My answer: A - Q ; B - Q ; C - Q ; D - Q

I am giving reasons for my answers and please correct me for the same.

B: elimination reaction not possible for weak nucleophiles so assuming strong nucleophile.

C: elimination reaction not possible (1$^0$ carbon and good leaving group) for weak nucleophiles so assuming strong nucleophiles

D: elimination reaction not possible (1$^0$ carbon and good leaving group) for weak nucleophiles so assuming strong nucleophiles. In my opinion C and D have same answers because statements doesn't have impact on answer.

EDIT: (For TRC's answer) For C: I agree that I had initially ignored the fact that an $\alpha\ce{-H}$ has been replaced by $\ce{D}$. But that doesn't imply E1cb mechanism. If E1cb occurs than also there will be no replacement of $\alpha\ce{-H}$ by $\ce{D}$ as shown below. Also $\ce{Br}$ is good leaving group than fluorine. enter image description here

Lets say intermediate is formed by E1cb and reaction shown in image takes place (but it is weird and absurd thought). But than also E1cb and E2 will compete with winner being E2 according to conditions given in problem. Also if carbanion forms in equilibrium with (P), concentration of (P) will be much less and can't influence major product.

enter image description here

  • $\begingroup$ Your arrows are not standard. The arrows should be from oxygen's lone pair onto the hydrogen, and bond electrons moving down to the carbon. $\endgroup$
    – Babu
    Commented Aug 11, 2021 at 16:29
  • $\begingroup$ I understand. I say the end calculations must be presented in line with standard rules, otherwise , this question may confuse future readers who may not know how to correctly draw arrows. This is further compounded by the fact that most coachings do not emphasize on mechanistic study to my knowledge. $\endgroup$
    – Babu
    Commented Aug 11, 2021 at 16:34
  • 1
    $\begingroup$ @Buraian I personally want to update all my posts but after my exam finishes. Anyway you may edit it and I may assure that I will approve it. $\endgroup$
    – Jay
    Commented Aug 11, 2021 at 16:35
  • $\begingroup$ chemistry.stackexchange.com/questions/155806/… $\endgroup$
    – Babu
    Commented Aug 11, 2021 at 16:50

1 Answer 1


$\ce{C-H}$ bond and $\ce{C-D}$ bonds are not the same in energy. $\ce{C-D}$ bond is stronger. So breaking $\ce{C-D}$ bond is more difficult and slower than breaking $\ce{C-H}$ bond.

I hope you know the mechanisms of $\mathrm{E1}$ and $\mathrm{E2}$ reactions. In $\mathrm{E1}$ reaction, a carbocation is formed first, which is the rate-determining step (rds). Since rds does not involve breaking of $\ce{C-H}$ (or $\ce{C-D}$) bond, hence rate is same, regardless of whether $\alpha$ hydrogen is $\ce{H}$ or $\ce{D}$. So roughly equal amounts of both products are formed.

In $\mathrm{E1}$ reaction, the rds involves breaking of $\ce{C-H}$ bond. Since $\ce{C-H}$ bond is easier to break than $\ce{C-D}$ bond, so major product is the one where $\ce{C-H}$ bond is broken during elimination.

Now let's look at the options one by one -

A) Since the product with broken $\ce{C-H}$ is major, it means it is $\mathrm{E2}$.

B) Same reasoning as above.

C) Here we see an $\ce{\alpha-H}$ being replaced by $\ce{D}$, meaning a carbanion is formed as intermediate due to very strong base ethoxide. So it is following $\mathrm{E1cB}$ mechanism.

D) They have said that rate is same for both $\ce{\alpha-H}$ and $\ce{\alpha-D}$, meaning $\mathrm{E1}$ mechanism is followed. $\mathrm{E1}$ is first order.

Disclaimer: I have been told earlier that the concept of rds is inaccurate by a very experienced user here. I agree with him. This answer is from a purely theoretical point of view, and strongly in the context of JEE only, where rds is useful and prevalent.

  • 1
    $\begingroup$ @Jay Out of the given options in column 2, the only mechanism with the possibility of H replaced by D is E1cb. You will agree that since the question asks for it the reaction must be following some mechanism. The H-to-D replacement rules out the possibility of E1 and E2. The only choice you have is E1cb. I'm sure you must be familiar of how frequently option-elimination is used in JEE MCQ's to get the correct answer ;-) $\endgroup$
    – TRC
    Commented Jun 19, 2021 at 9:40
  • 1
    $\begingroup$ @Jay also, during E1cb the species with $\alpha-H$ and $\alpha-D$ are in equilibrium with the intermediate carbanion. Since C-D bond is stronger, it is favourable to form C-D bond from the intermediate carbanion and hence the concentration of species with $\alpha-D$ becomes higher, so major product is the alkene with H removed (eliminated), and D retained. $\endgroup$
    – TRC
    Commented Jun 19, 2021 at 9:44
  • 1
    $\begingroup$ @ShoubhikRMaiti If in the above question by me, you can't see the linked question, it's because that question got closed and deleted. In that question, I had given an answer, but Martin, a moderator here, pointed out that rds was inaccurate and downvoted my answer, without any further explanation (If you can give an answer to my question, that I have originally linked, it would be most helpful!) $\endgroup$
    – TRC
    Commented Jun 20, 2021 at 4:56
  • 2
    $\begingroup$ I saw your question, and I think your argument with rds is completely right. There is nothing more for me to answer. $\endgroup$
    – S R Maiti
    Commented Jun 20, 2021 at 9:02
  • 1
    $\begingroup$ @ShoubhikRMaiti Thank you so much for the clarification! $\endgroup$
    – TRC
    Commented Jun 20, 2021 at 14:36

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