How does a primary amine react with phthalic anhydride to form an imide?

[Horseb0x]

I'm trying to figure out the mechanism for the reaction between phthalic anhydride and a primary amine, yielding an imide. Can't find it anywhere.

 

[sjb-2812]

What nucleophiles can you see in your pot? What electrophiles?

 

[Horseb0x]

The amino group is a nucleophile but I'm not too sure about phthalic anhydrides electrophilicity. The carbonyl carbons of the anhydride would be electrophilic enough I suppose. Heres my guess. The amino group attacks one of the carbonyl carbons of the anhydride which breaks the C=O pi bond and forms a tetrahedral intermediate. The quaternary ammonium salt then ejects a proton which protonates the alkoxy O.

The rest I don't understand. Someone drew a mechanism for me and the next step she drew was the carbonyl pi bond reforming and yielding a C=OH+ group. First time I've ever come across a C=OH+ group. Is that step correct?

 

[sjb-2812]

The amino group is a nucleophile but I'm not too sure about phthalic anhydrides electrophilicity. The carbonyl carbons of the anhydride would be electrophilic enough I suppose. Heres my guess. The amino group attacks one of the carbonyl carbons of the anhydride which breaks the C=O pi bond and forms a tetrahedral intermediate. The quaternary ammonium salt then ejects a proton which protonates the alkoxy O

Instead of ejection of the proton at the tetrahedral intermediate stage, how about the C=O pi bond reforms, kicking out the alkoxy oxygen as carboxylate (reasonably good LG), then proton transfer to a secondary amide / carboxylic acid hybrid. Whilst amides aren't usually that basic, perhaps the proximity of the amide to the acid means you can repeat the nucleophilic attack on the other carbonyl?