2
$\begingroup$

I have been unable to find the reaction anywhere on the internet so I'm assuming the answer is no, although I would like an explanation as to why. I can't seem to find any reason why it would not work, provided the reaction is carried out in two steps. In the first step, H2N:- will act as a nucleophile, attacking the least sterically hindered carbon of the epoxide. The second step would of course be an acidic workup. As stated in the question, I would expect the same exact mechanism as the reaction of Grignard Reagents with epoxides (substituting sodium amide as the nucleophile) due to the high basicity of sodium amide and Grignard Reagents. Like I said, I'm not expecting the reaction to be viable. However, I would like to know exactly why not. Im assuming it's something to do with the Hydrogen atoms of the amide anion, though I could be wrong. Enlighten me.

$\endgroup$
2
$\begingroup$

It actually is possible but it works well only with simple epoxides so it is not wildly used. $NaNH_2$ is extremely basic (it can deprotonate alkynes) and too many functional groups can react with it: alkyl halides (elimination), aldehydes, ketones...

To work around that basicity, sodium azide $NaN_3$ is commonly used instead, followed by the reduction of the azido group into an amine.

For example, see this article.

$\endgroup$
  • $\begingroup$ Thank you! Since I was unable to find any info on this, I was beginning to think I was crazy for even considering it! And I was aware of the 1) NaN3 followed by 2) LiAlH4 method for forming primary amines, I was only considering this possibility as a possible more green (less chemicals used) and more efficient (less steps) method. However, with all the side reactions that could occur with the amide route, the better option is clearly the azide/hydride route. Id upvote you if I could but my rep is too low. $\endgroup$ – KeatonB Aug 17 '16 at 8:11
  • $\begingroup$ @Ki11akd0g Well maybe you are crazy but for another reason :-) For the reduction of azides, I usually preferred hydrogenation when possible, or the Staudinger reaction (triphenyhosphine with a little bit of water) when I had reducible moieties, though thriphenylphosphine oxide can be a pain to get rid of sometimes. $\endgroup$ – SteffX Aug 17 '16 at 8:19
  • $\begingroup$ Well from what I have heard, Lithium ions can also be a pain to get rid of as well. Guess you just have to pick between whatever the lesser of two evils is for a particular application X-) $\endgroup$ – KeatonB Aug 17 '16 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.