I am using fluorescamine to determine the degree of polymer conjugation to primary amines in a protein. Although I've found literature on the general reaction scheme, it's pretty vague.
Does anyone happen to know the exact mechanism of fluorescamine reaction with primary amines (i.e., flow of electrons, intermediates, bond formation) that leads to the final product?
I've included a link to the general reaction.
http://what-when-how.com/molecular-biology/fluorescamine-molecular-biology/
The amine attacks the "acetalic" quaternary carbon opening the five-membered non-lactonic ring to yield a sort of vinylogous lactam; the lactonic ring is also opened to yield an "hydroxy-acid"
Hi William, I'm also vexed by the lack of a detailed reaction mechanism between fluorescamine with primary amines.
The best that I can gather is that the primary amine is reacting with the aldehyde form of the spiro lactone in fluorescamine.
This is my best guess, See the attached image: 1,4 addition of amine to beta carbon, subsequent ellimination that generates an hemiketal and the amine trans to the phenyl ring, tautomerization to diketone, and finally 1,2 addition of the amine to the ketone and proton transfer to generate product.
I don’t think that the attack goes at acetalic quaternary C. There are 2 centers in fluorescamine able to react with amines. However, I cannot imagine the way to the desired product, i.e. 2-hydroxypyrrolinone, via the amine attack at the ester center. However, it is possible to write down the reaction sequence via the initial reaction of amine with alpha,beta-unsaturated ketone fragment (i.e. at dihydrofuranone ring). Roscoe is right here. He proposed a good scheme. But, I can propose a modified variant. The first step is also a usual nucleophilic conjugate addition (1,4-addition) when amine attacks beta-position of unsaturated carbonyl compound. However, a carbanion is first formed, the negative charge provided by Nu- species is delocalized in –CH-C=O fragment. I think that in this intermediatre structure a carbanion can undergo a fast transformation releasing the carboxylate anion at the ester fragment. RHH substituent is still nucleophilic and attacks C atom from which COO group has just leaved. This is a cyclization via a simple nucleophilic substitution, and OH becomes a leaving group. Something like that. I think the mechanism is described somewhere a long time ago.
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