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Question:

enter image description here

Answer given: D

My answer: C

My reasoning - I outlined the following steps in the reaction:

  1. Nucleophilic attack of pyrrolidine on the carbonyl, leading to the carbon atom now having one $\ce{-OH}$ group and attached to the nitrogen of pyrrolidine.

  2. Heat is mentioned, so I assumed it is for dehydration. Dehydration, as far as I know, should result in the most stable double bond i.e. the one with maximum alkyl substituents. So I formed an enamine with double bond on the right side.

  3. Nucleophilic attack of enamine's $\ce{C=C}$ on the leftmost carbon atom of $\ce{H2C=CH-CN}$, which is Michael addition.

  4. $\ce{H3O^+}$ to provide proton to the carbanion.

These steps lead to the formation of C. For D to be formed, step 2 must be altered, and the less substituted (and less stable) double bond must be formed. Why does that happen?

I also can see that the nucleophile $\ce{C=C}$ is sterically hindered in step 3. That will lower the yield of step 3. However, that steric hindrance or comfort in step 3 should not affect the formation of $\ce{C=C}$ in previous step 2.

I am also open to the fact that the answer key given might be wrong. But in case the key is right and I am wrong, I would like to know why my step 2 is wrong.

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  • $\begingroup$ I think the answer key is probably wrong (judging from the other options it seems like you really had to decide between the 2 enols really. $\alpha-H$ calculation makes it clear that we should prefer C. What is this cis-trans it is mentioning in opt. B and D? I don't see an stereogenic center.. $\endgroup$ Jul 12 at 8:41
  • $\begingroup$ @TRC I can see 2 pathways for this reaction, one through additon reaction(in first step) other through removal of acidic H and my prof had taught, that in such condtions carry out acid base first as this reaction is faster. I also doubt whether acc to your mech dehydration would not happen, I might be wrong about this though. $\endgroup$
    – Rishi
    Jul 12 at 9:05
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    $\begingroup$ 1 last point, this is a really bad question, the h3o+ at the end will hydrolyse the cyanide. Usually if you have to question the author's intention more than think about the question it is not a good qn. :( $\endgroup$ Jul 12 at 9:16
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    $\begingroup$ @Waylander Yes, the enamine does get hydrolyzed when $\ce{H3O^+}$ is added - that's why the options again depict cyclohexanone. As for the concentration and temperature, that's another fault of this exam - they give us only half the experimental information and expect us to know what products will be formed. $\endgroup$
    – TRC
    Jul 12 at 10:27
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    $\begingroup$ @Waylander I agree, I still think it should be mentioned though. $\endgroup$ Jul 13 at 6:37
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According to this site here citing Carey & Sundberg and Morrison & Boyd

"In case of 2-methyl cyclohexanone, this planarity would cause steric clashes between the methyl group and the pyrrolidine hydrogen atoms. Due to this, formation of the less stable 6-position carbanion is preferred and hence forces reactions to proceed at this position."

enter image description here

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    $\begingroup$ In turn, "site here" cites Carey & Sundberg and Morrison & Boyd. Referencing primary sources (i.e. these textbooks) is preferred. $\endgroup$
    – andselisk
    Jul 12 at 10:43
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    $\begingroup$ The answer is D as @Waylander has noted. Here is the original Stork paper: pubs.acs.org/doi/10.1021/ja00885a021 $\endgroup$
    – user55119
    Jul 12 at 16:19
  • $\begingroup$ @Waylander I think that while this is a problem in normal situations, I would rather take heat as an assumption that the reaction is under thermodynamic control and only final product matters.. $\endgroup$ Jul 13 at 6:49

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