00:01
So this question here asks us to go through various synthesis.
00:05
And the first one is going from three hexines.
00:07
So one, two, three, carbons.
00:09
And at the third one, there's a triple bond.
00:11
So one, two, three, four carbons, five, six carbons, two, three hexanone.
00:19
So one, two, third carbon, we have a carbon double bonded to oxygen.
00:25
And so this is just testing your knowledge of knowing that there is an acid -catalyzed hydration that goes from an alpine, a triple bond to a ketone.
00:33
That goes to the more substitute position, but doesn't matter here because both sides will result in the same product.
00:40
And it needs to be catalyzed by hgso4, and you need to have a strong acid.
00:45
So generally, h2s .o4 and water will do.
00:50
The next question asks us to go from benzene to metabromoacetifanone, and that's a of complicated one to pronounce there.
01:03
But what we have is we have the phenone part tells us that we have a we have a fennel group basically means meaning we have this the benzene from before the oh and the aceto group often basically refers to this is where acetone comes from we have a c double bond into o and then one branch on here over here and another branch over here and so we have on one side we're told we have this group the phenone aceto and then um meta to it we have so this is if this is the one branch ortho meta para meta to it we have a bromo group and so um since it's not uh specified here we we're the cedo uh prefix sort of means that we have, normally we're expecting to have another carbon here.
02:10
And so this is what we're aiming for.
02:12
And so this is going to be electrophilic aromatic substitution.
02:16
And we have two different branches to add.
02:19
We have to add the bromine and this ketone acetyl branch.
02:25
And so because bromine and other halites are orthoparasacting, it's not going to give this regio selectivity and so but acetyl groups are meta directing and so that means we want to add the acetyl group first and so different books will write this in different ways but we always need is we need the portion that we're trying to add right here plus a leaving group and some acid some some lewis acid to catalyze this so alcl3 this will give us this product and then to add it meta, so right here or right here, we will add, we'll have bromine br2 with another catalyst f .e .b .r.
03:20
3.
03:20
So that's the second question.
03:22
The third question asks us to go from bromobenzene.
03:27
So that's this.
03:31
It's phenone.
03:32
And so that is basically half of our product from last time right there.
03:43
And so the way we can do this is we can, seeing the bromine here gives us a little clue as to one of the steps we can do.
03:55
And also seeing that we have six carbons over here and we have eight carbons over here...