Synthesize Novocaine from benzene and any other reagents of fo...

Question

Synthesize Novocaine from benzene and any other reagents of four carbons or fewer.

Accepted Answer

All right. Hi, everyone. So for this question, we have to show how to synthesize the following compound using benzene as your starting material and any other reagents of four or fewer carbon atoms. So if we go ahead and we compare the structure of benzene to that of our final product, we're going to notice that there are two substitutes, right? On the right side here, we have an and on the left side, we have an amine. So recall that these two groups can be added directly to our benzene ring or any aromatic system in general via electro phi aromatic substitution reactions. So that's what we're gonna be relying on for the sake of this problem. But um there's kind of an issue right, when we consider the fact that we cannot directly introduce an amine to an aromatic ring, but recall that instead, what we can do is introduce a nitro group to our benzene ring and then subsequently reduce it down to an amine. So that's what we're gonna do for that amine. So now let's go ahead and consider their positionings on the ring, right? And what I wanna bring to your attention is that our and am mean groups are in the paras position with respect to each other. But that's a little strange, right? Because recall that and nitro groups are what we consider to be electron withdrawing, right? And if they're electron withdrawing, then they're going to be um meta directing. What that means is that if we go ahead and install an ester or a nitro group directly onto the ring, any subsequent um additions to the ring are going to be in the metas position, not the paras position. So what that means is that we have to introduce an Ortho slash rector first and recall that all kill groups are um examples of orthopro directors, right? So we can go ahead and we can introduce something like a metal group to our ring and then direct our nitro group to the paras position. And that's gonna work out because recall that we can go ahead and oxidize um a methyl group to a carboxylic acid, which is exactly what we have here, one carbon from the benzene ring. So if we react our carboxylic acid with an alcohol, we can generate the ester that we have on the right side via eer application. So with that being said, let's go ahead and set up our substituent first. So like I mentioned before, we're going to set up first, our Ortho slash rector, which in this case is gonna be a met group. So recall that we can react um this bending ring via FRITO crafts cul. So I'm gonna take methyl react that with a OCL three as a catalyst. And we're going to have our Meho group on carbon one here. So now when I go ahead and I proceed with nitration of um of this ring system, then the nitro group is going to be added to the paras position. So I'm going to react our material here with a no three in the presence of H two. So four, because that's going to give me the nitration product for this ring. So now this is going to be our setup. And now, like I mentioned before, we can go ahead and oxidize the methyl group with KMNO four in the presence of heat. So our nitro group is gonna stay the same, but we're going to have a carboxylic acid on position one instead and I ran out of space there. So let me just put it back in frame. There we go. And like I mentioned before, our nitro group is going to stay untouched. So now we can go ahead and proceed with an ification reaction. But the question then becomes right, what's gonna be our alcohol? So now in order to figure out what our alcohol is going to look like, let's go back to our final product and let's notice here that our alcohol is going to have an amine separated by two carbons, right? So we're gonna have our, oh group at the very end with two carbons. And then this dimethyl amine that we have here. So recall that the alcohol for this ification step because we have both an amine and an oh group in the same molecule, we can react dimethyl amine in this case. So if I go ahead and I take dimethyl amine and I react dimethyl amine with a two carbon oxide. Recall that our dimethyl amine is going to open the um the oxide ring and generate an alcohol, right? So if I go ahead and I react these two compounds together, we're going to have our alcohol and now we can proceed with an ification reaction. So if I combine our carboxylic acid in our alcohol in the presence of H two, so four, we are going to have or Esther. So this is gonna be our benzene ring. Our nitro group is gonna stay untouched for now. But now we're going to have an. So these are two carbons and then are dimethyl meat and I ran out of space once again. So let me just move that over. So you can see it better, much better. So now we have our exactly like we have it in our final product, which means that our last step is simply going to be the reduction of our nitro group down to an amine. And recall that we can accomplish that by reacting our nitro group with elemental iron in the presence of hydrochloric acid. So at the very end, we're going to have NH two instead of no two. And of course, we're going to have our, just making sure I'm counting my carbons and there we go. So this is going to be your synthetic sequence. And I know this was a lot, but we are finished with the process. And your answer here is going to be choice D as in dog. So with that being said, um, if you stuck around to the end, thank you so much for watching and I hope you found this helpful.

Synthesize Novocaine from benzene and any other reagents of four carbons or fewer.

Key Concepts

Electrophilic Aromatic Substitution

Amine Functional Group

Carbon Chain Length and Reactivity

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