1
$\begingroup$

It is well known that peptides are synthesized by a so called "condensation" reaction between an amine and a carboxylic acid group to form the final amide moiety (because it releases water). In its general form, it resembles the picture below.

Depicted here

However, what is the actual chemical mechanism behind this reaction? My thoughts are that the amine can deprotonate the carboxylic acid and form a carboxylate - which is a very bad electrophile - so how would it form the amide from nucleophilic attack? I am aware that in the body, it is possible that this reaction is mediated by biochemical catalysts but I am still not sure of the overall mechanism. If this reaction does indeed involve biochemical catalysts, is it possible to complete the synthesis without them?

$\endgroup$
1
  • 2
    $\begingroup$ Getting a carboxylic acid to react directly with an amine is very difficult, that's why the body uses a huge ribosome to facilitate it. In a lab these couplings are also very complex, everything is protected apart from your target carb. acid and amine, which are then joined with coupling agents like DCC or HATU. Either way you need to activate your carboxylic acid for anything to happen. $\endgroup$ – Jabbamanga Apr 1 at 21:13
0
$\begingroup$

Here is a proposed mechanism in a 2005 paper. One thing to point out is that this is a proposed mechanism but it is all I could find:

enter image description here

link to pdf: https://www.pnas.org/content/pnas/102/35/12395.full.pdf

Typically, these reactions involve the nitrogen attacking the partially positive carbonyl carbon as a form of nucleophilic substitution. However, OH is a bad leaving group so often, reagents such as DCC can be used to make the OH into a good leaving group before reacting. This is why it is easier to make nylon using a diacid chloride instead of a diacid.

$\endgroup$

Your Answer

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

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