4
$\begingroup$

It is known that, with regard to the elimination reaction, the Zaitsev product is the thermodynamically more stable product, which is also often the more substituted alkene while the Hofmann product is the thermodynamically less stable product and is usually the less substituted alkene. I have always thought that the substituents refer to alkyl groups. This is because the basis of the thermodynamic stability lies in hyperconjugation with $\ce {C-H}$ and $\ce {C-C}$ bonds.

However, I recently encountered a test question in which we were supposed to identify the Zaitsev product, from the two given product alkenes, based on the groups attached to the alkene $\ce {C=C}$ bond. The alkene double bonds both had two alkyl groups attached to them. The difference between them is that one had an additional $\ce {-OH}$ substituent while for the other double bond, the two other substituents were just hydrogen atoms. On this basis, the teacher said that the alkene with the $\ce {-OH}$ substituent was more substituted and hence, it was the Zaitsev product while the other alkene was the Hofmann product. I would like to ask if this identification was made correctly. Can the $\ce {-OH}$ group be considered to have increased the degree of substitution of an alkene and hence, allow it to be identified as the "Zaitsev product"?

Of course, it is known that tautomerisation would occur for such an alkene product as it is an enol. However, let us ignore this fact in this case.

$\endgroup$
4
  • $\begingroup$ I think I would need a sketch of the products … $\endgroup$
    – Jan
    Oct 10 '19 at 14:34
  • $\begingroup$ To disregard tautomerization, perhaps the case of a halogen substitutent should be considered instead of the OH? $\endgroup$ Oct 10 '19 at 15:17
  • $\begingroup$ As to your first paragraph, to say that the Zaitsev product is the thermodynamic product does not imply that it was formed under thermodynamically controlled conditions. The ratio of Zaitsev/Hofmann (Z/H) products in the hydroxide elimination of 2-bromo-2-methylbutane is not necessarily the same as the Z/H ratio in the acid-catalyzed equilibration of the two isomers. $\endgroup$
    – user55119
    Oct 10 '19 at 15:43
  • $\begingroup$ @electronpusher Yes, we could also consider a halogen substituent. $\endgroup$ Oct 11 '19 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.