00:01
This is the answer to chapter three, problem number 41 from the smith organic chemistry textbook.
00:09
And in this problem, we're given four fairly complex molecules, and we're asked to predict the water solubility of each.
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And so in order to do that, remember this rule that the book suggests that i've written up the top of the screen here, which is for every five carbons in a molecule.
00:31
It needs to have one polar functional group capable of hydrogen bonding in order to be water soluble.
00:40
So looking at a, we can see right off the bat that 1, 2, 3, 4, 5, 6, 7.
00:48
There are 7 carbons in this molecule and 1, 2, 3, 4, 5, 6 polar functional groups.
00:56
So this molecule is very polar and it is water soluble.
01:02
So that's caffeine, by the way.
01:06
So looking at this steroid molecule that we have for b, so we can see that there's an ether here, the och3.
01:18
So this is an ether.
01:19
This is obviously an alcohol.
01:22
And that's it.
01:23
Two polar functional groups that could h bond and many more than 10 carbons.
01:31
We have three, six -membered rings and a five -membered ring, and then some branches and stuff.
01:38
So, you know, well over 10 carbons.
01:42
And so the answer to b is going to be not water -soluble.
01:50
Because remember, in order to be water -soluble, it needs to have one polar functional group per five carbons.
01:57
And that's not the case in b...