# Formation of enamine for unsymmetrical cyclohexanone derivatives

Question:

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.

• 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.. Jul 12, 2021 at 8:41
• @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. Jul 12, 2021 at 9:05
• 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. :( Jul 12, 2021 at 9:16
• @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.
– TRC
Jul 12, 2021 at 10:27
• @Waylander I agree, I still think it should be mentioned though. Jul 13, 2021 at 6:37