Question

What product do you get when you react an alkyne with NaNH3 and CH3Br?

Answer 1

For simplicity's sake, please just use CH3-C(triplebond)C-CH3

Answer 2

Are you sure that isn't supposed to be NaNH2? If that is the case, and I believe it is, you will see that there is a possibility of an acid-base reaction between the ammonia and acetylene.

The pKa of ammonia is about 36 and the pKa of acetylene is about 26. All that means is the lower they are, the more likely they are going to be deprotonated when in solution with the other.

But wait, I'm looking at the pKa of ammonia and what you have is sodium amide. But I've done it correctly, you see right now it's a base and what you should do is imagine what the acid-base reaction would be. You'd get ammonia and sodium acetylide, right? This is a reaction in equilibrium, so it will continue to flop the hydrogens (and sodiums) back and forth, but the pKa's tell us the most comfortable state is when acetylene is deprotonated in this equilibrium. For the sake of understanding why we compare pKa's, I hope this helps. In the real world you will never have 100% purity in your reactions, but I've digressed too much!

So now that you have this deprotonated acetylene, also called acetylide anion, now what? Well, we essentially have a carbon that can act as a base and become acetylene again, or it can act as a nucleophile on the bromomethane. Both can, and do happen. But the fun thing here is that Bromine is a very good leaving group since it is so electronegative, it doesn't mind. There's a really polar bond between methane and bromine, so bromine will gladly leave when the electrons on the acetylide's deprotonated carbon are nearby. This forms a carbon-carbon bond, and what do we know about these? They're incredibly non-polar and stable. So the likelihood of reversing this reaction with bromine coming in and somehow breaking a carbon-carbon bond with the acetelide as a leaving group is practically impossible! Now you won't have any of the reverse reaction going on like we did with the acid-base part of it.

There are a few more possibilities and things to think about here, but overall I think that covers it fairly well. Once you get it, it's really not so bad and you can reason through a great deal of reactions with nearly no knowledge as long as you know a few pKa's and electronegativities.