Concept: Forcing Ortho substitution5m
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.
Example: Multistep Synthesis3m
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.
Problem: Beginning from Benzene, synthesize the following compound.6m
Predict the product for the following reaction.
Propose a synthetic pathway from the indicated starting material to the designated product
Complete the following reaction by drawing the structure of the principal major product. Indicate relative stereochemistry where necessary. If there is no reaction, write NR.