Acetaminophen is the active ingredient in Tylenol. How can ace...

Question

Acetaminophen is the active ingredient in Tylenol. How can acetominophen be synthesized from benzene?

Chemical structure of acetaminophen, featuring a benzene ring with an amine and hydroxyl group.

Accepted Answer

Hey, everyone. The compound below is involved in the synthesis of proxy meta cane show how to synthesize the compound from pheno. The compound that we are referring to is three nitro, four Propox benzo acid. We want to synthesize it from Phey. Let's think about the possible synthesis. We're going to redraw a funeral and we understand that there are three substitutions that we want to add to our ring. There's a carboxylic acid, an ether and a nitro group. So before we think about our sentences, let's think about the sequence of the addition of those substitutions because Nitrol group is meta erecting, we understand that the only substitution that is me to it is the carboxylic acid group. So in instead of adding nitro at the very beginning, we're actually going to add it at the end because we have an ether group which is Ortho part directing. So that's good for us. And we have our carboxylic acid group which is meta directer. So they're both directing at the metas position where we have our nitro group. So the nitro group will be added as the final one. Now because carboxylic acid is meta erecting and Ether is Ortho power directing, we're going to add ether at the very beginning so that we get our power substituted product as the major product, right? So that leaves us with the addition of the acid in our second step. So let's try and see how that works for us. Let's think about the formation of an ether and to form an ether, we can simply use the idea of the Williamson ether synthesis, right. So our goal is to first of all deproteinate pheno and that can be achieved with the usage of a strong base such as sodium hydride and the hydro and I will deproteinate pal to form P oxide. When oxide is formed, we can perform an essence or reaction. We need a primary alky hali, right, which has three carbons. So let's draw three carbon chain and let's add a good living group such as bromine. And during that essence reaction, we're going to use the pin oxide anion as the nuclei which will attack the electrolytic carbon, followed by the loss of the leaving group that will essentially give us the required ether structure. So now we're going to replace hydrogen with a prop group. So let's add 123 carbon atoms well done. Now, let's think about our next step. So now if you want to form a carboxylic acid, one of the easiest way is I'm sorry. One of the easiest ways is to essentially add a muscle group at the power position. Let's recall that we can simply use the Friedel Craft calculation, right? If we want to add a methyl group, we're going to use chloral methane. So that would be C H three C L and we want to use a Luis acid catalyst, aluminum chloride. So we are performing the Friedel Crafts calculation in which specifically let's read the structure more clearly. So once again, we have our prop group and now let's remember that during the Friedel Crafts calculation because we have and activating substi that works for us, we are going to form or and power products but power substituted product will be the major one, right, due to a small and not really significant ster hindrance effect. But Ortho power is slightly minor product and parra product is the major one. So we're adding our muscle group at the power position. And now that we have our metal group, we can perform the side chain oxidation during the side chain oxidation. If we have a metal group bond to the benzene wing, we can easily convert it into carboxylic acid using two steps. So the first step is the usage of potassium permanganate, which is a strong oxidizing agent as well as basic conditions and heat. And the second step because we end up with a carboxylate and ion during that reaction. Since we have basic conditions, we want to use acidic conditions to form an actual carboxylic acid group. Meaning in the stop, we obtain the required carboxylic acid. So let's draw the benzene ring, the ether group. So we have oxygen purple group on it said, and now instead of the metal group, we are going to get our coo H group, right, our carboxylic acid group. And from here, our last step is to add the nitro group, as we said before, we are already in an ideal spot because the carboxy group is metal directing and the ether group is or power directing. So they're both directing at the same position, meaning we can use nitration nitric acid in combination with sulfuric acid will allow us to replace hydrogen with the nitro group. So you can simply add the nitro group and we understand that we managed to get our final product. So let's go ahead and add the nitro group that corresponds to our final synthesis scheme since we got the required product. But to make it for us simpler to think about the final answer of the problem, we can just summarize all of the steps at once. So let's review what we did. We essentially started with pal and we managed to form our farm product. So let's read to our final product that would be our proxy group, right? So we're going to add our ether our nitro group as well as our carbolic acid group. And we had several steps. So first of all, we started with the addition of sodium hydride to form P oxide. Second, we added prophyl bromide to perform an essence two reaction in which we had substitution and we formed our ether. So that was the Williamson ether synthesis. Number three, we performed the Friedel Crafts calculation so that we managed to add the metal group at the or and power positions. But power was the major product. Number four, we performed the side chain oxidation using potassium permanganate basic conditions and heat. Number five, for the resultant carboxylate and Ron, we had to use acidic conditions to we to essentially form the carboxylic acid group. Number six, the final step was the nitration so that we added our Nitrol group and those will be our steps to perform their required sentences. Thank you for watching and I hope to see you in the next video.

Acetaminophen is the active ingredient in Tylenol. How can acetominophen be synthesized from benzene?

Key Concepts

Electrophilic Aromatic Substitution

Nitration and Reduction

Acetylation

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