Blocking Groups - Sulfonic Acid

Concept: Concept: Forcing Ortho substitution

Video Transcript

In this video, we're going to discuss a synthetic strategy for benzene reactions that’s called blocking. There's only one reversible EAS reaction and that's sulfonation. Remember that sulfonation, you use
concentrated sulfuric acid. It produces SO3 and that reacts, makes a sulfonic acid group. But remember that we could very easily remove the sulfonic acid with heat and acid or even steam. That would be enough to take off the sulfonic acid group.
That's going to make it really special. It gives a real synthetic advantage to us. This is why. Imagine that you're trying to get an ortho substituent. Remember that we talked about ortho, para directors favor. They have ortho and para substitution. But really they’re usually going to favor the para position. It’s difficult to get things to add ortho. When you're trying to get high yields, it's not going to get a 15% yield of ortho or 20% yield of ortho.
Imagine if you’re trying to get an 80 or 90% percent yield of ortho. How do you do that? The only way to do that is to block the para position. That's the advantage of using sulfonation because sulfonation is actually that’s one of the biggest things it’s used for. It’s used to block the para position so that you can force ortho substitution.
Remember, ortho substitution is usually not highly favored because of steric hindrance. But if you can block that para position, then you’re going to force ortho substitution. It’s brilliant. This is called a blocking group because it's not actually found in the final product. In the end, you're not going to know that the sulfonic acid was ever there because you’re going to take it off. It just was there to block the second EAS reactions.
Let's go ahead and look at this. Let's say that you're trying to get some kind of substituent in the ortho position and that’s it. All you want is ortho products. You want ortho substitution. You better not just react the electrophile straight away with the toluene. In this case, we have toluene. Why? Because you’re going to get so much para that it’s going to be almost worthless. You’re going to get very little ortho. Then you have to separate it and you get very little yield. It sucks. But what you could do is you can do a sulfonation first. A sulfonation reaction, concentrated sulfuric acid and heat, that's going to make SO3. What that's going to do is it’s going to make a sulfonic acid here in the para position. Why is it going to be para? Because we know that the para position is favored. In fact, this is going to be over 90%. It’s going to be para; very, very high yields.
Now what we do is we can do our EAS reaction. Now you do your second reaction. This is our first reaction. This is your second reaction. You do an EAS. What happens? Notice what types of directors do you have. CH3 is ortho,para-director. It's going to direct either to here, to here or to here. Are you guys following so far? You have the ortho positions and the para.
Can the EAS reagent actually add para? No. Notice we also have a meta-director. Notice that my sulfonic acid is a meta-director because you have a very, very strong partial positive here. In fact, this is one of our strong deactivators. This is going to be what type of director? Meta. What positions does it direct to? It's going to direct to here and to here. See what's going on? We’re actually getting both groups directing to the same spots that are ortho.
Now when I react my EAS, it’s going to add. The electrophile is going to add specifically here or the other one but they're both the same. After my electrophile has added, then I can use my dilute acid to desulfonate. What I wind up getting is just the hydrogen here. In the end, it’s called a blocking group because it’s not found in the final product. It was there simply to block the para position and allow me to produce ortho substitution.
That's the whole point behind this topic. Let's go ahead and do a practice problem to really reinforce this. Predict the product of the following multistep synthesis. Do all three steps and then I'll show you guys how it works.

Concept: Example: Multistep Synthesis

Video Transcript

So guys in the first step I'm going to sulphonate. So that sulphonation reaction is going to produce a solphonic acid group. So it's going to be para because like I told you if you get very high yields of the para product because para's favoured over ortho. Now we're going to going to nitrate. Concentrated H N O 3 is a nitration reaction so now what we're going to do is we're going to get nitration and where's it going to add? Well guys it has to add here. So it has to add here because the O H is still an ortho para director, it's still going to direct ortho para but it can't add para anymore because there's something there already so then you have to add ortho and then as you guys know dilute acid, heat is going to desolphonate and I'm going to end up with my final product that looks like this. O H and N O 2.

So I get my ortho substitution. So I know there might be a few of you out there wondering Johnny, if you did the nitration straight on phenol wouldn't the major product be ortho anyway because I actually talked about how this is one of the exceptions how you can get hydrogen bonding so it would actually favour ortho. You're exactly right and I'm glad you've been paying such close attention but the problem is it still wouldn't be that great it would be like sixty percent if you just did it by itself so if all you did was you did concentrated H N O 3, what you would get is a phenol with a nitro group in about sixty percent yield but if you do this three step pathway it's not much harder you're going to get a ninety percent plus yield, see what I mean? So even in a situation where it would have been favoured to go ortho it's still better to use a blocking group so you can get a higher yield using, blocking that para position. Does that makes more sense now? Awesome guys, great example. So let's move on to the next topic.

Concept: Practice 1: Retrosynthesis

Video Transcript

Hey guys, let's take a look at the following practice question. So here it says beginning from benzene synthesise the following compound. So here we look, it says we're making this compound here which is 1, 2 disubstituted and here I'm saying this is the only isomer which means I cannot make any side products. Now here we have an alkyl group, alkyl groups are ortho para directors and here since it's an ortho para director it would want the chlorine to go here, here or here. Now it would have preferred the chlorine to go to the para position. Now chlorine itself is also an ortho para director. Now corn is an ortho, para director so it would have preferred it to go here, here or here but it would have wanted it to go para. So something's weird about this problem. Those both are ortho, para directors but the groups next to them are ortho. For some reason we got the ortho product instead of the more stable para product. So what you need to realize here is anytime you have a 1, 2 disubstituted benzene then a blocking group could have been used.

So when I say blocking group I mean that the para position was blocked somehow by a group so that the next incoming group had no choice but to go ortho thus giving us this 1, 2 disubstituted product then later the blocking group was taken off. Now the blocking group that we're talking about here is your S O 3 H group, your solphonic acid group which we add through solphonation. So we know we're going to have you solphonation at some point here and then later take it off. Now here we have to decide who goes on first, should the alkyl group go on first or should the chlorine go on first? Now aklyl has one imitation they have is that when meta directors are directly attached to benzene they don't work so this is true for alkylation and acylation. Now chlorine is not a meta group but it is weakly deactivating so it's better safe than be sorry, it's better to be safe than sorry. So here we're going to say let's add the alkyl group first and then add the chlorine later because even though it's not a metadirector it's still weakly deactivating so there's a chance that it could interfere with my alkylation step. So we're going to start out with benzene and since we need room guys let me take myself out of the image. So we're going to start out with benzene, we want to add a two carbon chain to benzene. Now remember the chlorine is going to leave to attach to this aluminum, it's okay that that carbon becomes a carbocation because it will not rearrange because the chain is only two carbons long. So there it goes right there remember it's an ortho, para director so it wants the incoming group to go either ortho or para. Now para would be the more stable position because the two groups would be far away from each other. Now at this step we'll do sulphonation. So we do S O 3 over H 2 S O 4 or you can see a concentrated H 2 S O 4, that's also good. Both are solphonation. Now this alkyl group is still an ortho, para director but S O 3 H is a metadirector. Now here who gets to decide where the next group goes? Remember ortho, para always beats meta directors in terms of the sighting where the next group goes. So we're going to do chlorination to add that C L, the ortho, para director wants it to go ortho since the para position's already taken out so there goes our products so far.

Lastly we just need to get rid of that blocking group, the S O 3 H group so just do H 3 O plus steam and heat and that will do desolphonation, it will remove that S O 3 H group and just like that we've isolated our product. Another reason that we want to do alkylation before chlorination is because, think about it, if we added the chlorine first that's great the next step I would have done is I would have added the sulphonic acid group, S O 3 H. Now S O 3 H is a full meta group since it's a meta director I can't do alkylation later because once that meta group gets added on to benzene it would stop alkylation from occurring at all so it was wise for us to first do alkylation, get that alkyl there on safely and then do sulphonation. The alkyl group's already on benzene so it doesn't matter is a metagroup comes in later we've already got it on benzene, then chlorination, then desolphonation. That was the key to this question. 1, 2 disubstituted benzenes are usually formed if we have a blocking group blocking the para position so that the next goes to ortho and then do desolphonation at the end.