# What is the general mechanism for a lactam ring reacting with water and an acid?

For example, let us say we have 2-Pyrrolidone and react it with water and $\ce{HCl}$. I do know that the product is a salt, $\ce{HO(C=O)CH2CH2CH2NH3Cl}$.

I've tried coming up with a mechanism but fell short. I do know that at the first step we must have protonation of the carbonyl-oxygen which gives us a resonance stabilized carbocation intermediate. Nitrogen can better handle a positive charge.

For the second step, we use the addition of water to the carbonyl-carbon followed by a proton transfer from the oxygen to the nitrogen. But when we do this we get $\ce{NH2-}$ which is not favorable. This is the step that I'm stuck on. After this step it would be easy to see the collapse of the tertiary carbocation intermediate.

• You're right that -NH2 is a poor leaving group, even worse than another common "bad" leaving group, -OH. What do we do to an alcohol to get the hydroxyl to leave? – electronpusher Jun 12 '17 at 3:50
• We protonate it – Navy_Colors Jun 12 '17 at 3:54
• Exactly, the simplest and most common way to activate leaving groups. Could that work with -NH2? – electronpusher Jun 12 '17 at 3:55
• @electronpusher has nailed it. The hydrolysis mechanism is quite similar to the mechanism of ester hydrolysis. Does that help you? Btw, if you get your answer, do consider posting it as an answer here, that would help future visitors to learn :) – Pritt Balagopal Jun 12 '17 at 4:08
• I have the solution. I can't post it here because I'm using the ship's computer (a military computer) and we don't have USB ports, so I can't transfer an image from my mobile phone. There is also no webcam utility for screenshots – Navy_Colors Jun 12 '17 at 4:15