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In general, more substituted alkenes are more thermodynamically stable. However, in certain cases such as enamine alkylation, the less substituted double bond is favored. How do we know that which is favored during a particular reaction? Can we have a generalized rule for it?

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closed as too broad by Mithoron, Todd Minehardt, A.K., Mathew Mahindaratne, Waylander Feb 10 at 13:26

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to ChemSE. Try to be more specific in phrasing your question. Specific examples would help. Your question is too broad as it stands. $\endgroup$ – user55119 Feb 8 at 3:19
  • $\begingroup$ Thanks for your comment.This question is intended to be general.I mean to say I would like to understand in what contexts more substituted 'ene' is favored and also what is the driving force for the formation of less substituted 'enes' although more substituted 'enes' are stable. $\endgroup$ – Science123 Feb 8 at 3:43
  • $\begingroup$ There is no single rule; chemistry can sometimes tend to be a collection of various rules which must be applied at different times. Learning chemistry entails learning when to use which rule. $\endgroup$ – orthocresol Feb 9 at 0:22
  • $\begingroup$ Arguably, it's also unclear; I don't know what 'enamine alkylation' is referring to. The answer currently refers to relative stabilities of the enamines, which is unrelated to their alkylation. It would be helpful if you could clarify (and also come up with a title that describes the context). $\endgroup$ – orthocresol Feb 9 at 0:28
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I'll address your concerns about alkene stability in general and enamines in particular. It is useful to examine the acyclic permutations of C6H12. The heat of formation of these isomers is a good measure of their relative stability. As the double bond is increasingly substituted with alkyl groups, the heat of formation becomes more negative, i.e., more stable. Tetrasubstitution is better than trisubstitution which is better than disubstitution. Clearly, monosubstitution is the least desirable case. The stabilization by alkyl groups in this case is due to the hyperconjugation of the bonding σ-CH orbital and the anti-bonding π*orbital. The more opportunities for this interaction, the more the stabilization. Heats of formation are available at the NIST site.

The pyrrolidine enamine of 2-methylcyclohexanone prefers the trisubstituted over the tetrasubstituted position. Because there is a strong resonance contribution (planarity) in the enamine, the tetrasubstituted isomer has strong steric interactions between the methyl group and the methylene adjacent to the nitrogen in the pyrrolidine ring. The equilibrium is not spontaneous but it is catalyzed by water or traces of acid. In the trisubstituted isomer, the methyl group adopts an axial position to avoid an A1,3 interaction if the methyl group were equatorial.

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1) International Institute of Standards and Technology, Dept. of Commerce, USA.
2) F. Johnson and S. K. Malhotra, J. Am. Chem. Soc., 1965, 87, 5492.

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I'm not sure where the alkene is during enamine alkylation. I can only see an alkene in the enamine starting material, so I will try to explain the alkene formation selectivity for enamine formation.

The less substituted alkene is favored for enamine formation because in the enamine, the N atom pushes electron towards the alkene by resonance. A less substituted carbon atom accommodates the negative charge better than a more substituted carbon atom, so a less substituted alkene is more stable in this case.

For other reactions such as alcohol dehydration, the more substituted alkene is formed because the intermediate has a carbocation character. The more stable intermediate is the more substituted intermediate, thus giving the more substituted alkene as the major product.

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  • $\begingroup$ Thanks for answering.So, in general, if there is a probability for negative charge to accommodate through resonance, a less substituted one is favored and vice versa? $\endgroup$ – Science123 Feb 8 at 4:18
  • $\begingroup$ It all depends on the mechanism of the specific reaction. No such general statements can be made. $\endgroup$ – mck Feb 8 at 6:46

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