00:05
In this problem, we are naming compounds.
00:08
So for a, we have both an alken and an alcohol here.
00:13
The alcohol needs to be the lowest number we can make it because that is the highest priority thing on this compound.
00:20
So if we go from this side, it would be one, two, three, four.
00:23
On this side it also be four.
00:25
So we're going to go from this side because that makes both the alken and the alcohol the lowest that they can be.
00:30
So one, two, three, four, five, six, seven.
00:35
So we have seven carbons on here.
00:41
So that is hept.
00:43
And then the al -keed will name first.
00:46
That'll be two.
00:47
It's on second carbon, e 'en.
00:50
And then the alcohol is on carbon four.
00:52
So for all.
00:53
Hept 2 -ein, 4 -all.
00:55
You could also say 2 -hept -ein -for -all...