00:06
In this problem, we are asked to draw all 14 alkynes that have the formula c7h12.
00:12
So the way i like to approach this type of problem is we'll start out by drawing all of the straight chain versions of the alkyne, and then we'll start adding substituents as methyl groups, ethelgroups, whatever, instead of straight chains.
00:26
So to start off, we'll have straight chain with the alkyne on the end.
00:32
So that's one, two, three, four, five, six, seven carbon.
00:36
There.
00:37
Then we'll put it on number two.
00:39
So one, two, three, four, five, six, seven carbons.
00:45
And then we'll put it on number three.
00:46
So one, two, three, four, five, six, seven carbons there.
00:53
So that's all of our unique straight chain alkanes.
00:58
Or sorry, alkynes.
01:00
So now we will add a methyl group.
01:02
So we'll do a six carbon chain and we'll add a methyl group.
01:04
So if we start with it on the end, one, two, three, four, five, six, we can have a methyl group here.
01:14
Or one, two, three, four, five, six.
01:18
We can have a method group here.
01:20
Or one, two, three, four, five, oops, one, two, three, four, five, six.
01:27
We can also put the methyl group down there.
01:31
And then we can do the methyl groups with the al -qaeda in the second carbon.
01:37
So if we start here, that's one, two, three, four, five, six.
01:43
We can have it here.
01:44
Or one, two, three, four, five, six.
01:47
We can also put it up there.
01:50
So now we're at one, two, three, four, five, six, seven, eight.
01:53
We could also do a methyl group where we have it on the third carbon.
01:59
So one, two, three, four, five, six.
02:03
We can put a methyl group there.
02:06
Then we could also do two different methyl groups.
02:10
So again, starting with when we have the alkyne on the first carbon.
02:17
So we'll do a five carbon chain now.
02:22
So one, two, three, four, five.
02:27
And then we can do two methyl groups.
02:28
So we can put them both on that carbon.
02:32
One, two, three, four, five.
02:34
They can be both on this carbon.
02:38
Or they can be an opposite carbon.
02:39
So one, two, three, four, five.
02:42
We can have one here and one here.
02:44
So that is one, three, four, five, six, seven, eight, nine, ten, eleven, twelve.
02:49
So there's two more.
02:51
So we can do an ethel group instead of a methyl group where we still have it on that first carbon.
02:57
So one, two, three, four, five...