00:01
All right.
00:01
This is where some of these more tricky questions are starting to appear.
00:07
How do you synthesize the following substances from benzene? so we're starting with, we have four compounds in total.
00:12
We're starting with metachloro -nitrobenzine.
00:16
As i read it out loud, i'm going to check to make sure i do the structure, right? so we have to start from benzene in all of these cases, right? so essentially you have to look at the substituents that are on your product and think about the order that you, have to add them in order to, one, make sure the reaction is actually possible.
00:36
The ring isn't deactivated enough that will prevent certain reactions that you've learned from happening.
00:41
But also make sure that you get the substituents to go in the correct place.
00:46
Right.
00:47
So in this case, we have to add a nitro group and a chloro group, and they have to be metered to each other.
00:52
So with two substituents, it's usually easier than when you have three or sometimes four substituents that you have to add because the relationship between them is going to guide you to which substituent you should add first.
01:07
So you're looking for a meta -director in this case because the first substituent is going to go on just fine, and then the second substituent is going to be guided by the first substituent.
01:17
So you know that in this case, the nitro group, being a strongly deactivating group, is going to guide the second substituent into the meta -position.
01:25
So we're going to solve this problem by first adding the nitro group using nitric and sulfuric acid, and then we're going to add the chlorine to the metaposition by cl2 and aluminum trichloride, which will put that solution, we'll put that chlorine right into metaposition for us.
01:43
So that was a pretty easy one.
01:45
Same thing going on here.
01:48
So i say easy.
01:49
Everyone's at different levels with this stuff, right? so i say easy because it's one of the more simple of these tricky problems.
01:56
For example, c and d in this question get a little tricky.
02:01
So in this case, you have the same kind of thing here, right? so you have these two groups, an ethyl group.
02:07
So you have metachloroethylbenzene.
02:09
We're starting from benzene again.
02:11
And in this case, it might not be completely apparent which one you have to add first, because if you add an ethyl group and if you add a chlorine group first, they're both going to be orthoparadirecting.
02:21
Right.
02:22
So regardless, we have to do a third.
02:26
Reaction.
02:27
So you have to do some chemistry at one of the, at one of the substituents, instead of on the benzene ring, you have to do chemistry with one of the substituents in order to convert it from an electron withdrawing group to an electron donating group.
02:41
So in order to get the relationship to be meta, the first two substituents, the first substituents going to go on fine, right? the second substituent is going to be guided by the first substituents.
02:51
So if you add the eiffel group first, it's going to favor orthopara for the chlorine.
02:56
But if you, instead of adding the ethyl group, let's say you add an acyl group that has two carbons.
03:03
So you know that that acyl group, for example, like acetyl chloride here, two carbons, it's an acyl group with aluminum chloride.
03:17
That's going to give you right there.
03:26
So that'll give you a two carbon acyl group.
03:30
So now you can add the chlorine.
03:32
That's what carbonials attached to benzene rings in the benzoposition are electrolytic drawing groups.
03:37
Now you could add the chlorine by chlorination to get that to go into the metaposition.
03:49
And then the last thing you do is you have to get rid of that carbonyl.
03:53
So you just learned about hydrogenation with palladium on carbon, or just poitium.
04:01
But you've learned about just palladium.
04:03
Now we're introducing this.
04:05
Palladium on carbon is just the surface that allows you to reduce carbonyles, benzylic carbonyles, to methylene groups, to hydrocarbons.
04:16
So that'll give you, that let's go from that carbonyl, then i'll make it disappear, leaves his water, and then you could have a methylene group there.
04:25
So that's how you get to metachloro -ethylbenzene.
04:29
Part c here is where it gets a little tricky, right? so four chloro, one nitro, two propyl benzene.
04:37
So you have to kind of think about how you're going to get these substituents on here, starting from benzene.
04:46
So you have a relationship of para between the nitro group and the chloro group that are, the nitro group is ortho to the propal group, and then the chloro group is meta to the propor group.
05:02
So if you add the probal group first, you'll be able to get the nitro group on there.
05:07
But then the nitro group is a stronger director than the probal group.
05:12
So whatever substituent you add after you add, the nitro group is probably going to go meta.
05:17
Right.
05:17
So if you start adding stronger activator or deactivators onto the ring, they're going to guide where the substituents go.
05:26
So also keep in mind that you have to add an alkyl group.
05:29
So if you're going to do any kind of fridyl crafts, the ring can't have an electron withdrawing group.
05:34
It can't be a deactivated ring.
05:35
Otherwise, the frito crafts isn't going to work...