00:06
This question asks us to draw the products of each of these reactions with one mole of propine.
00:13
So i've drawn them all out here.
00:15
So part a, we have one propine with one mole of hbr.
00:20
So if we have hbr, that means that we're going to have the pie bond attack the h and kick off the bromine.
00:30
And then we'll make a carbocation, which needs to be the most stable carbocation we can make.
00:34
So that'll be the one on this carbon rate.
00:37
Here and that's where our bromine will attach.
00:39
So we'll end up with ch3, cbr, and then we'll still have that double bond there.
00:47
In part b, we have two moles of hbr, so the only difference here is we're going to do the same process twice.
00:53
So now we'll have ch3 and we'll have two bromines on that carbon and a ch3 on the end.
01:01
Now we've used both pie bonds.
01:04
In part c, we have one mole of br2.
01:08
So br2 adds anti, which will give us a trans alken.
01:13
So we'll have an alken.
01:15
We'll have br's opposite each other.
01:17
And then we'll have an h on one side and the methyl group on the other.
01:23
And then in part d, we now have two moles of bromine.
01:26
So now we'll have two bromines on each carbon.
01:28
So we'll have ch3.
01:30
And then this will have two bromines.
01:33
And this will have two bromines.
01:37
And there's an h on the end.
01:38
End.
01:40
In part e, we have a, again, a terminal alkyne with acid catalytic hydration with the mercuric ion to help us out...