4
$\begingroup$

An exam question involved the following reaction -

enter image description here + enter image description here forms enter image description here

Both reactants (2-methylbutanoic acid and 2-methylbutanamine) are optically active. The question used pure R-enantiomer of the acid, and a racemic mixture of the amine, amine in excess.

However, it also suggested that the R isomer of acid will not react equally well with both enantiomers of the amine. I could not understand why, since the reaction does not involve any stereocentre. My best (and probably silly) guess would be that one configuration would encounter greater steric hindrance during attack of the nitrogen atom on carbon, so that product will be formed less. However, I could not find any resource to support this, nor did I find this very likely as the tertiary carbon atoms are separated by 4 bonds - so steric hindrance probably won't have a large effect.

Could anyone please confirm if the configuration does indeed affect the reactivity, and if yes, how and why?

$\endgroup$
10
  • 1
    $\begingroup$ The effect might be very small. but it's definitely nonzero. $\endgroup$ Jul 14, 2021 at 10:21
  • $\begingroup$ @orthocresol But what is the reason for that effect? Is it steric hindrance, as I proposed? $\endgroup$
    – TRC
    Jul 14, 2021 at 10:22
  • $\begingroup$ What are the reaction conditions? $\endgroup$
    – Waylander
    Jul 14, 2021 at 10:23
  • 5
    $\begingroup$ The bottom line is that it will be different, simply because the products and transition states are diastereomeric. Thus, there’s no way it can be exactly the same. Having a model or a rationale to understand which one is faster is nice, but secondary: you must first accept that they are different. $\endgroup$ Jul 14, 2021 at 13:47
  • 1
    $\begingroup$ If you got your answer, feel free to self-answer it :) $\endgroup$ Jul 15, 2021 at 7:54

1 Answer 1

1
$\begingroup$

Thanks to @orthocresol for helping me via the comments. Here is a reasonable explanation for it -

The final product comprises of two different diastereomers, of configurations RR and RS. Diastereomers, unlike enantiomers, have some inherently different properties, other than optical rotation.

Consequently, the reaction that forms two different diastereomers will proceed through intermediates and transition states that will also have different properties. Since they differ in aspects other than optical rotation, their products will be formed in different amounts. Just because the products are different, the amounts formed are also bound to be different - they essentially become two different reactions. Even if the difference is very slight.

$\endgroup$

Your Answer

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

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