00:06
In this question, we are reacting to butyne with each of the sets of reagents.
00:13
So in part a, we have one mole of hbr.
00:16
So we will add hydrogen to one carbon on the alkyne and bromine to the other.
00:23
And we'll have an alkyne left over.
00:25
Because this is symmetrical, we'll only get one unique product.
00:29
So we'll have bromine on one side and a hydrogen on the other.
00:34
And that will be our product for that.
00:37
And then for part b, now we have two moles of hbr, so we'll just do that same process over again twice, and we'll have two bromines on the same carbon, and that will become a ch2.
00:51
In the next one, we have one mole of br2.
00:55
So br2 will add two bromines, one bromine to each carbon, and it adds anti.
01:01
So we end up with a transalkene.
01:04
So we'll have bromine, bromine, and then we'll have our methylobin.
01:08
Groups here.
01:12
And then in part two, we have two moles of br2.
01:14
So that will just give us two bromines on each carbon.
01:17
So two bromines on this one and two bromines, oops, two bromines on this one as well.
01:28
And then the c -h here on the end...