# Product of primary amine and acid anhydride

How does an acid anhydride react with a primary amine? I thought it'll be some sort of condensation reaction, so I made the product by removing a water molecule and got a cyclic product. However, the product given is an open chain molecule (major).

How does the reaction occur? How does the ring open? Could someone please give a detailed mechanism?

P. S. What is the general reaction for primary amines and acid anhydrides?

It should be very clear that a primary amine is a nucleophile and an acid anhydride an electrophile. In the series of carbonyl and carboxyl derivatives, an acid anhydride is classified as quite reactive: more so than aldehydes or ketones, much more than esters or amides but less than acid chlorides. Thus, we should all agree that the first step is the nucleophilic attack on the $\ce{C=O}$ π system via a tetrahedral intermediate to give N-methyl succinic monoamide.

Scheme 1: Mechanism of the monoamide formation.

With that, we can consider whether further reactions will take place. Where we originally had an anime and an acid anhydride, we now have a carboxylate (the $\ce{C=O}$ bond that wasn’t attacked) and a secondary amide. The secondary amide still has a lone pair we might consider to be nucleophilic — but that lone pair is taking part in amide resonance and is thus not available for nucleophilic attack (at least not well). And to remove the amide’s proton, we would need a strong base which we don't have. Thus, the amide side of the equation is far less reactive.

What about the carboxylate? Thus highly symmetric structure is also very unreactive towards nucleophilic attack in the same way as an amide. Due to the negatively charged oxygen donating one of its lone pairs to the π system, the carbon atom is much less electrophilic than in an acid anhydride. Thus, in this case also there is no desire to react.

Since we have two functional groups of low reactivity, the reaction ends here. However, with enough heat (supplying activation energy), it is possible for the sluggish intramolecular reaction between the the amide and the carboxylate to occur. In principle, the mechanism is the same except that you must transfer a few protons and remove water.

Scheme 2: potential onwards reaction to succinimide under harsh conditions.

• How do you explain the simultaneous formation of N-methylsuccinamic acid? – arya_stark Dec 17 '17 at 18:00
• @user28968 Well, with enough heat, some of the amide may attack the carboxylate anyway, but it is going to be a slow process. – Jan Dec 17 '17 at 18:01
• That's particularly the reaction I need the exact mechanism for. I'll be very grateful if you could help me with the same – arya_stark Dec 17 '17 at 18:02
• @user28968 Wait … maybe I’be got my nomenclature wrong but the major product according to your question is the one whose formation my answer describes. I thought that was the one you needed? (In any case I can’t draw until tomorrow.) – Jan Dec 17 '17 at 18:05