00:01
For this question, we're given a series of alcohols, and we're asked how to prepare these alcohols using a greenyard reagent.
00:10
And so we're going to start with a.
00:12
So for a, we have two methyl, two propanol.
00:16
So if we draw the structure, we have propanol.
00:20
So that's going to be three carbons, two, meaning our alcohol is on the second carbon, and two methyl, meaning we have a methyl substituent.
00:32
Also on the second carbon.
00:34
Okay.
00:36
So we see now that we have a tertiary alcohol.
00:42
So to prepare using a greenyard reagent, we want to then start from a ketone.
00:50
So if we draw the ketone, we see that we can then sever this linkage and add methyl magnesium bromine.
01:07
And we also note that it if two or more alkyl groups are the same, that we can use an ester because an ester is going to add twice.
01:19
So we see that we do have more than one of the same alkyl group.
01:27
And so we could also imagine starting with an ester and then adding two equivalents of the following greenyard.
01:46
Okay, and we're going to draw a retro arrow.
01:51
Okay.
01:52
And so these are the ways that we could prepare a greenyard starting with alcohol in a.
01:59
So let's move on to b.
02:02
So in b, we have one methyl cyclohexanol.
02:07
So cyclohexan means six -membered ring.
02:13
All means alcohol.
02:15
And one methyl.
02:17
So we have a methyl substituent in the one position.
02:23
And so again, we see that we have a tertiary alcohol.
02:30
And so then we want to start with a ketone.
02:38
And in this case, we can add, we could imagine severing this bond and using methyl magnesium bromide.
02:52
So we can move on to part.
02:54
Part c now.
02:55
So for c, we see we have three methyl, three pentanyl.
03:00
So pentan, five member chain, three all.
03:10
So in alcohol at the three position and three methyl.
03:14
So methyl at the three position as well.
03:18
So again, just as in parts a and b, we see we have a tertiary alcohol...