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enter image description here

The solutions manual said that enamine B with its roughly sp2 methyl group can force the nitrogen out of the plane and this messes around with the conjugation.

How about the fact there is a significant degree of resonance donation by the nitrogen, which places a significant partial negative charge on carbon. In enamine A, this carbon is secondary. In enamine B, this carbon is tertiary. Extra inductive donation from the methyl group destabilizes the product.

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    $\begingroup$ This is a complex question. We are dealing with very small energy differences (1-2 kcal/m) and a number of factors each play a subtle role. Understanding kinetic vs. thermodynamic control and the energy differences between ground states and transition states are probably the two keys. If isomer distribution was simply based on the effect that steric factors have on conjugation, then why would 2,5-dimethylpyrrolidine produce 90% of isomer B? If you really want to understand this then see here and $\endgroup$ – ron Dec 7 '14 at 16:09
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    $\begingroup$ here. Like most things, when you dig into something, often it is not simple. $\endgroup$ – ron Dec 7 '14 at 16:11
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Conjugation effects are not very important in this case. The main reason is steric hindrance. If the more substituted alkene is formed, then both the methyl group and the nitrogen ring will be on the same plane, and repulsions will destabilize the molecule.

In the case of A, the methyl group is more free to move in a pseudo-chair conformation of the cyclohexene ring.

But there are exceptions, for example:

enter image description here

In the second case the more substituted alkene is formed, because is more stable and there is essentially no difference in regard to steric hindrance.

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  • $\begingroup$ The steric hindrance exemplified in Enamine B is an A-1,3-strain: en.wikipedia.org/wiki/Allylic_strain $\endgroup$ – jerepierre Dec 7 '14 at 16:01
  • $\begingroup$ Yep, in this case the A-1,3 strain is lower than the hypothetical interaction with nitrogen substituents, so trans olefin is formed. $\endgroup$ – Altered State Dec 7 '14 at 16:10
  • $\begingroup$ @AlteredState would you say that the inductive and hyperconjugative effect has a role or not? $\endgroup$ – Dissenter Dec 8 '14 at 15:28
  • $\begingroup$ When it comes to evaluate thermodynamic stability of a more or less substituted alkene, hyperconjugation always has a role. If the destablization caused by the interaction between the amine ring and the sp2 fixed substituent is more important than the hyperconjugative stabilization effect of the more substituted alkene being formed, then you obtain the less substitued enamine, like the example you proposed. $\endgroup$ – Altered State Dec 8 '14 at 16:29
  • $\begingroup$ @AlteredState what about inductive effects? Why do you only mention hyperconjugative effects? $\endgroup$ – Dissenter Dec 9 '14 at 16:04

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