00:02
This is the answer to chapter three, problem number 31 from the smith organic chemistry textbook.
00:10
And in this problem, we're asked to draw four different isomers of c6h12o.
00:18
And they each need to have at least one different functional group.
00:23
And then we're asked to predict which of them will have the highest boiling point.
00:30
And so i don't think that we've learned hydrogen deficiency index yet.
00:38
But when we do, we could calculate the hdi here and see that whatever molecules we draw are going to have to have one double bond or one ring.
00:51
And so that's something to bear in mind.
00:54
And so the easiest way to get started, i think, is one, two, three, four, five, six.
01:07
So we can start here and put an aldehyde.
01:12
So this fulfills our formula, aldehyde.
01:20
Okay.
01:21
And so next, we can just move the location of that carbonyl.
01:28
To an interior carbon.
01:30
So, for example, this one, and that is going to be a ketone that fits our molecular formula.
01:38
So next, as i said, we have to think about other functional groups.
01:45
So other oxygen -containing functional groups would be an ether and an alcohol.
01:52
And so we can do the ether next.
01:57
And as i said, this is going to have to be a cyclic ether, because if you were to try to draw this as an open chain without a double bond in it anywhere, you would have one too many hydrogens.
02:14
And so drawing this cyclically is the easiest way.
02:18
And then actually we need one more carbon.
02:23
So this is going to be an ether...