Hey everyone. In organic 2, I just finished a chapter on carboxylic acid derivatives. My professor gave us some practice problems to look at that he said would help us on the next test. One of these problems is similar to the intramolecular synthesis of a lactam, which my book seriously only has one paragraph on that is only a description of lactams in general. I was wondering if anyone could give me some info on how to do these? I've attached one of the sample problems, but I'm clueless how to go about doing it?
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it looks like the Nitrogen is acting like a nucleophile and is attacking the double bond O to give the ring, but i don't remember how the OH is formed...maybe by protonation? the double bond O being protonated by an H+ and forming the OH and then when the Nitrogen does it's nucleophilic attack, the other OH bond dissociates as water and you form the ring...
hopefully that might help. i don't quite remember. it's been a year since i've taken ochem. 🙂
hopefully that might help. i don't quite remember. it's been a year since i've taken ochem. 🙂
Haha, it's great the things you learn in organic that you will never really need again.
You mean Lactone (cyclic ester)
The product that you are showing is not a lactam but a cyclic hemiaminal. Hemiaminals are normally formed from amine and aldehydes. Your reactants that you have, the amine and carboxylic acid, would normally form a lactam. A lactam is a cyclic amide. So in order to get to the product from your attached picture from the reactants of the picture you would first have to reduce the carboxylic acid to an aldehyde and then have the amine (with the help of a proton source; the proton will activate the aldehyde) act as a nucleophile and react with the aldehyde (an electrophile) to give you that cyclic hemiaminal. I hope this helps.
Swenis said:
The best way to do this reaction is to first convert to an acyl halide then let the reaction proceed. This way you end up with an excellent leaving group and the secondary amine can react more readily. I would suggest using SoCl2/TEA. I believe that in acidic conditions you will end up with the compound that you see but if you end up with a carbonyl then you can always reduce.
