00:05
In this question, we are drawing the product of each of these compounds with excess hcl.
00:11
So starting with a, we know that if you have a terminal alkyne, the terminal carbon will get the hydrogen to make the more stable carbokadion in the middle here.
00:23
And then that will be where the cl attacks.
00:25
And if we have excess hcl, that will just happen twice for each of the pie bonds.
00:29
So our products will be this.
00:32
We'll have two chlorines on this carbon, and we'll have a ch3 on the end.
00:39
In part b, we have a symmetrical internal alkyne, which means that it is the same on this side and on this side.
00:47
So whichever carbon the chlorine ends up on first, that's where the second one will also go, because that does end up becoming a more stable carbocanion.
00:56
So we'll have, no matter which side it attacks first, it'll end up being the same product because it is symmetrical.
01:03
So that product will look like this.
01:05
3, ch2...