I'm trying to figure out the following reaction mechanism

There's a reference paper by Yanagi and Akiyoshi(1959) showing the hydroxyethylation of imides resulting in the following

I have two questions :

(1) Why would the hydroxylethylation occur in the 1 position instead of the imide at the 3 position? I would imagine that this reaction occurs for the more acidic amine, but I have no idea why there doesn't seem to be any literature about that reaction.

(2) Why would this reaction produce a HydroxyEthyl and not a urethane, as shown here :

Any insights would be appreciated, thanks!

Yanagi and Akiyoshi : http://pubs.acs.org/doi/abs/10.1021/jo01090a601

Reaction in red (from a patent application): http://www.huntsman.com/performance_products/Media%20Library/a_MC348531CFA3EA9A2E040EBCD2B6B7B06/Products_MC348531D0B9FA9A2E040EBCD2B6B7B06/Carbonates_MC348531D1109A9A2E040EBCD2B6B7B06/files/jeffsol_alkylene_carbonates_synthesis_of_hydroxy_alkyl_urethanes.pdf


2 Answers 2


The second part of the question is easy to answer: the imide attacks at C3 of propylene carbonate instead of C1 (the carbonyl carbon) because carbonates are stabilized (double ester-resonance), whereas carboxylate is a so-so leaving group.

The initial product from nuclephilic attack at C3 is a carbonate half-ester, which subsequently loses CO2 to form the observed N-hydroxyethyl imide.


I would expect that the two N-H groups would be similar acidity. The conjugate base from deprotonation at N-1 is resonance stabilized through both carbonyls, through the carbonyl at C-2 directly and through the carbonyl at C-4 via the C-5/6 double bond.


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