An exam question involved the following reaction -
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?