00:01
We're at problem 36.
00:02
Okay, we're given four molecules here.
00:04
Part a is your bromobutane.
00:07
Part b is your aline.
00:11
Part c, we have two hydroxypropinoic acid.
00:15
Okay, so hopefully you see why that is.
00:18
We have three carbons, right? one, two, three.
00:22
And this makes it the propanoic acid, and we have the two hydroxy group.
00:29
And finally, we have our three methyl hexane.
00:33
So 3 methyl hexane, you have 6 carbons, and you have that methyl group on carbon 3.
00:40
All right.
00:41
So this problem is asking for us to draw these molecules.
00:45
So part a, we want an s in nantrumor, okay, part b we want r.
00:51
Part c, we want r, and part d we want s.
00:55
So the way we're going to do that is we're going to take these groups.
01:03
So these groups hanging off of the molecules.
01:07
We're going to take those and draw them either going into the board or out of the board in order to create that inanchomer that we're looking for.
01:17
So part a here we want an s inantrumor.
01:21
So usually this, not usually always.
01:27
So bromine in this case would be your priority number one.
01:31
This is your priority number two and this is your priority number three okay so if you do that you're going to have this clockwise direction i'll just put it off to the side you're going to have this clockwise direction which will give you an r in angiomer which is not what you want so in order to to flip that we want the hydrogen remember there's a hydrogen that is with the bromine here you want this hydrogen to be going out of the board okay, so once it is going out of the board and it is priority four, this rnannchomer is going to turn into s...