0
$\begingroup$

If we consider this reaction: Compound 1 + Compound 2 --> Compound 3

I understand that if a reaction is under kinetic control, the less stable (e.g. a mono-substituted diene) is formed and if the reaction is under thermodynamic control the more stable stable (disubstituted diene) is formed.

However, if we wanted to go from the kinetic product to the thermodynamic product, would this reaction be considered as being under kinetic control? In other words, are the reactants under kinetic/thermodynamic control, or is it the products?

I think this is more of a terminology question.

Improve this question
$\endgroup$
1
  • $\begingroup$ You are considering a reaction Compound A --> Compound B --> Compound C or two alternatives, A <--> B compared to A --> C. Whatever way B --> C goes, there is no point in asking wether that's thermodynamic or kinetic, because it's both. And you don't address reversibility in your question text, why is it in the heading? $\endgroup$
    – Karl
    Jul 26, 2018 at 4:22

1 Answer 1

Reset to default
1
$\begingroup$

A product formed under kinetic conditions may or not be reversible under the conditions of the reaction. Here is an example of one that is reversible under the reaction conditions. Kinetic control is determined by the difference in activation energy (ΔEa) of the two pathways. The thermodynamic equilibrium is determined by ΔΔGo. Acetoacetic ester (AAE 1) is a mixture of keto and enol forms. Bromination of AAE in an inert solvent rapidly gives α-bromoAAE 2 in addition to HBr. This is the product that is formed faster having a lower activation energy. This reaction is also reversible, initiated by HBr (microscopic reversibility). Eventually, the more stable γ-bromoAAE 3 is formed from AAE 1 by enolization toward C4 (γ-position) of AAE under the influence of HBr.

enter image description here

Improve this answer
$\endgroup$
5
  • $\begingroup$ I think I got i! But, if with the addition of HBr, the more stable (thermodynamic) product was not formed from the less stable product (less stable --> intermediate --> more stable), would this reaction (or lack of) be considered to be under Kinetic Control? $\endgroup$
    – HSB
    Jul 26, 2018 at 15:23
  • $\begingroup$ There is no HBr added. It is produced as a by-product of the bromination. In this regard, the rxn. is also auto-catalytic. For a non-reversible kinetic/thermodynamic process, see the deprotonation of 2-methylcyclohexanone at en.wikipedia.org/wiki/… $\endgroup$
    – user55119
    Jul 26, 2018 at 18:42
  • $\begingroup$ If you run the reaction for a short period of time or at low temperature and monitor the progress of the reaction to optimize bromide 2, then the rxn. is kinetically-controlled. If on the other hand, you wait longer and/or operate at room temperature, then it will be thermodynamically-controlled. $\endgroup$
    – user55119
    Jul 26, 2018 at 19:01
  • $\begingroup$ @user55119 What if the products was higher in energy than the reatants? Would it make sense then to speak about "kinetic" or "thermodynamic" control? $\endgroup$
    – ado sar
    Nov 8, 2020 at 18:28
  • $\begingroup$ @ado sar: If the products are higher in energy, then the reaction is endothermic and moves to the left. The higher peak is rate limiting. $\endgroup$
    – user55119
    Nov 8, 2020 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.