Synthesize from benzene. (Hint: All of these require ...

Question

Synthesize from benzene. (Hint: All of these require diazonium ions.)

(c) 2-methyl-5-hydroxybenzoic acid

Accepted Answer

All right. Hi everyone. So for this question we have to show how to synthesize the following compound. Starting from Benzene hint, the synthesis requires di ammonium ions. So, before we start talking about our final product, let's recall really quickly that the dye Azzoni um ion can be used as an intermediate for the addition of various substitue ints on an aromatic ring. But two specific examples are crab oxalic acids and hydroxy soul groups. So with that being said, let's go ahead and draw out our final product so that we can figure out what's different compared to Benzie. Right, so two ethyl five hydroxy benzoate acid is going to be a carb oxalic acid on our benzene. Right. With an ethyl group on carbon to or in the Ortho position and also a hydroxy group on carbon five. So that's going to be in the meta position relative to our carb oxalic acid. So, there's a lot going on here. Right? Because we do need to add three substitutes to benzene, but we can break it down into a couple of steps. So, like the question mentions, we do need dyas ammonium ions as an intermediate to add both our crop oxalic acid and a wage groups to our ring. Right, so the first thing we can do here is if we start with benzene then our synthesis can start off with our frito crafts calculation to go ahead and generate an ethyl benzene derivative from our benzene. Right, Because that doesn't require any intermediates, we can go ahead and simply react our benzene ring with a two carbon alcohol highlight, such as um ethyl iodide in the presence of A L. C L three as a catalyst. Right? So when we add those two things together, we're going to introduce our ethyl group right off the bat. Right? But now um now let's go ahead and start with the addition of our crab oxalic acid. Right? And what I want to bring to your attention is that relative to the ethyl group? The crab oxalic acid is in the Ortho position, not the peril. So recall really quickly that ethyl groups and alcohol groups in general are what we consider to be Ortho or para directing, meaning that um new substitutes are going to be added to the Ortho or para positions. But recall that the para position is generally favored due to less starik hindrance. Right? So because we have our carb oxalic acid um in the Ortho position and not the para position. That means we need to block the para position right to prevent that addition from occurring. So recall that if we go ahead and react our ethyl benzene with S. 03 in the presence of sulfuric acid, we're going to generate a um cell phone oh group in the para position which is going to subsequently um it's gonna subsequently block the para position. Right? So this is going to be our protecting group and our group is still gonna be on carbon one here um relative to this ring. So now like I mentioned before, this is going to be our protecting group. So now we can go ahead and um react this molecule with our protecting group installed via nitrate in right, because in order to use dye Azzoni um a di ammonium ion, I should say as an intermediate for the addition of a crab oxalic acid. We must introduce nitrogen to the ring and we introduced nitrogen to the ring um via nitrate in reaction of course. So recall that that is going to be with H and 03 in the presence of sulfuric acid. So we're going to do this and that's going to add um that's going to add our nitro group to the Ortho position and it's a little bit out of frame. So let me just move it before I draw the rest. That's better. So now, like I mentioned before, our nitro group is going to be on the para position on the Ortho position because the para position is physically blocked. Alright, so now that um now that the nitro group has been installed we can go ahead and remove are protecting group, right? And we can simply remove the protecting group by adding water in the presence of heat to our system. So if we go ahead and introduce water in the presence of heat, that's an H. I swear then we're going to end up with our ethyl benzene that has a nitro group um in the Ortho position and nothing else. Right, so this is gonna be our ethyl group and here's our nitro. So now that that's done, recall that um yes we need a nitro group to introduce a di ammonium ion um to a given molecule. But recall that we must first reduce that nitro group um two and a mean before we can go ahead and generate a di ammonium ion. So we can go ahead and reduce this nitro group by reacting it with elemental iron in the presence of hcl. So that's going to reduce our nitro group down to the primary aromatic amine that we're going to require to generate a di Azzoni. Um I'll so now with that being said, let's go ahead and generate our di Azzoni um ion. Right? And we're going to do that by reacting this primary aromatic amine that we've just reduced from a night trial or a nitro group. And we're going to react this with sodium nitrate in the presence of hcl. So this is going to be our brand new di Azzoni um ion. Now instead of that primary aromatic amine, we're going to have an end to plus with chloride as a counter ion that stems from hcl. So now that that's there, we can't forget about our group of course. So now our dyas ammonium ion is ready to be converted into um into a carb oxalic acid. But recall that um from the diocese oni um ion. We can go ahead and install um a nitro group and then hydrolyzed hydrolyzed the night trial to give a carb oxalic acid. So if I scroll down to give myself some space the first thing we're going to do is introduce um the first thing we're gonna do is introduce your night trial right? So we're gonna react this with your copper cyanide. And we're going to have our di Azzoni um group be replaced by a night trial. So this is your ethnic group. And now we're going to have a night trial. But like I mentioned before we can undergo hydraulic sis of that night trial to generate a carb oxalic acid instead. So if I react this with acid in the presence of heat that is going to hide relies our night trial down to a crawl oxalic acid. Yeah and now we have a crab oxalic acid in the Ortho position relative to um relative to the ethyl group. But the fight is not over right because we still need um the O. H. Group that's in the meta position relative to the carb oxalic acid and pear a relative to our ethyl group. So like we were talking about in the beginning of this question we can use a dye Azzoni um ion once again to um to get to that O. H. Group once again. So what we need to do is we need to introduce another nitrogen to the correct position so that we can go ahead and generate our dia sodium ion and then convert or die Azzoni um to um to an O. H. Group. Okay so it's gonna be a lot more steps but at least we know what we're doing right so we can go ahead and we can um proceed with another round of nitrate in right to introduce that second nitrogen and therefore that second guy Azzoni. Um Right so we have another equivalent of H. N. 03 in H. Two S. 04. And here's what we're going to have. So this is our ethyl group, this is our crab oxalic acid. And now we are going to have once again a nitro group. So now our nitro group um can be converted to a Die Azzoni. Um But first right we must reduce it down to a primary aromatic amine. So this nitro group is going to be reduced with elemental iron in the presence of hydrochloric acid. Mhm. So this is now your primary aromatic mean. And now for our Die Azzoni. Um Right. Mhm. I know this is a lot but we're almost done guys I swear. So now we have a second Die Azzoni um we have N two plus and cl minus as our counter ion. Yeah and now our very last step is going to be replacement of the dye Azzoni um with an O. A troop. And recall that we can do that by reacting are Di Azzoni um here um with water in the presence of heat. So now yeah, we're going to have our hydroxy group in addition to our ethyl group and our carb oxalic acid. So at long last right, we made it to um the end of this problem. Now I know this was a lot of steps but we made it to the end, right? And it's worth mentioning here um just before we end the video that first of all, your answer is going to be choice. E but notice how your your ethyl group, your fo group is an Ortho para director and your crop oxalic acid is a meta director. So our new O. H. Group that we just installed, it is meta to our meta director and it is pera to our Ortho para director. Mhm. So that is why we needed to introduce a substitue ints in the order in which we did so right to make sure that um that the positionings are going to be matching up that they add up. So with that being said, if you stuck around to the end, thank you so much for watching. And I do hope you found this helpful

Synthesize from benzene. (Hint: All of these require diazonium ions.)
(c) 2-methyl-5-hydroxybenzoic acid

Key Concepts

Benzene and its Derivatives

Diazonium Ions

Functional Group Transformation

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