00:06
In this problem, we are naming each of the compound shown.
00:09
So starting with a, we have an al -kine and a bromine.
00:13
The al -kine will take precedence, so we will number from the side that makes the al -kine the lowest number.
00:18
So we'll go from here.
00:19
One, two, three, four, five, six.
00:22
We have a bromine on carbon five, so that will be five -bromo.
00:27
And then our al -kine is on carbon two.
00:30
And we have six carbon, so that's hexine for the al -kine.
00:36
For part b, again, we're going to number so that the alkyne is the lowest it can be, and we're going to look for the longest carbon chain, which is right here.
00:46
So we'll number from this side.
00:48
One, two, three, four, five, six, seven, eight.
00:53
We have a methyl group on carbon five.
00:56
So that's five.
00:58
Methyl.
01:00
Two, because that's where our alkyne is.
01:02
And then we have eight carbons, so that's octane.
01:08
For part c, again, we're going to number from the lowest point that the alkyne can be.
01:15
So that is 1, 2, 3, 4, 5, 6.
01:20
We have 2 methyl groups on carbon 5, so that's 2 to dye methyl, die to make sure that we know there's 2.
01:28
And then our alkyne is on carbon 2, and we have 6 carbon, so that's hexine.
01:36
On part d, again, we're going to number so let the alkyne.
01:39
Is the lowest number it can be.
01:41
So we'll start from this side.
01:43
One, two, three, four, five, six, seven...