00:07
In this question, we are assigning systematic names to each of these compounds.
00:12
So starting with a, we're going to number this along the longest carbon chain so that either one of the pie bonds ends up the lowest number that it can be.
00:22
So that we'll start over here.
00:24
One, two, three, four, five, six, seven.
00:28
So this is a seven carbon chain.
00:31
We have methyl groups at three and six.
00:34
So three, six, die, and then we have our, so seven carbon chain, so that's hept.
00:45
And then we have an al -keen at number two, so two -een, and we have an al -kind at number four, so four, ein.
00:54
And then we also need to acknowledge the stereochemistry of this double bond right here.
00:59
So to do that, we'll decide if it's e or z.
01:02
And to do that, we will look at each side and figure out what the most important thing on each side is.
01:08
So over here, it's the methyl group, and on the other side it is this whole side of the molecule.
01:14
And because those are both below the double bond, that makes this z.
01:18
So this will be 2z.
01:22
Oops, this is a 2, not a 3.
01:26
Oh, no, i was right before.
01:29
2z, 3 ,6, thy methyl hept 2in, 4.
01:33
For b, again, we want a number, so let the alkyne is the lowest it can be, that's the only really big thing we've got going on here.
01:41
So we'll start from this side.
01:42
One, two, three, four, five, six, seven, eight.
01:48
We do have a couple of substituents.
01:51
We have a methyl group and a tert -butal group.
01:55
So on five, we have tert -butal.
01:59
And on two, we have methyl.
02:03
And then our al -kine is on number three...