To synthesize m-ethylbenzenesulfonic acid, a student attempted th...

Question

To synthesize m-ethylbenzenesulfonic acid, a student attempted the Friedel–Crafts alkylation of benzenesulfonic acid with bromoethane. Do you predict that this reaction was successful? If not, propose an alternative synthesis.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem. A student tried to produce three propelled benzene cell phonic asset from benzene cell phonic asset and one chloral propane through the Friedel crafts calculation, is this a feasible reaction to get the desired compound? If not suggest an alternative synthesis. Now to begin this problem, We are just going to think about the students approach. Now the starting material is three propel apologies. That would be benzene, sulfur tannic acid. Right, so we are going to throw our benzene ring. We are going to add our cell phonic asset group, S 03 H and according to the students decision the product desired would be three pro pill Benzene cell phonic assets. So we're going to add our cell phonic asset group at position number one. If we count positions counterclockwise, that'd be 123. So here we are going to add our proposal substitution and the reaction used was one chloral propane, that would be three member chain with chlorine at position number one because it's a Friedel crafts calculation, we also need to use aluminum chloride as our catalyst. Now, essentially, you may immediately notice that there are several issues. First of all in this reaction, we need to generate an electro file. So let's look at the first issue with this problem, Number one is the generation of an electro file. You may recall that if you are starting with an electro file, in this particular case, we are first of all using one chloral propane. Alright, and the lone pair on chlorine will be essentially incorporated in the free orbital of aluminum and we are going to form our first complex. So that would be Our three member chain. Now, chlorine, it's bonded to aluminum chloride. Chlorine would have a formal positive charge and aluminum would have a formal negative charge. And now we are generating our actual electro file. We have the loss of this fragment and we noticed that we are forming an unstable primary carbo cast iron. Right? So we always have to check whether or not hide right or metal shift as possible. And indeed you may notice that there is this adjacent hygiene which can shift two position number one. So we have one comma two, that hydride shift and this allows us to get a more stable secondary carb. Okay, Ryan for our reaction. So this one is a primary one and this one is is a secondary one. A secondary carbo Karen would actually be our electro file. So that's our electro file. So in the product we would actually get an isopropyl substitute in this reaction. Right? So this is reason number one. The reason number one why this is not a feasible reaction is because we're getting an ayso propel substitute print instead of a straight propel chain. So if you visualize that you can essentially imagine that the actual substitution here would look as follows. It would be an isopropyl substitute print which essentially looks like this rather than a straight chain or a normal chain propel. So we are going to eliminate that and we understand that this is the first reason. Now, however, there is a more important reason regarding the deactivation effect. Let's crawl down. No two. The second reason why this reaction is not feasible, as we already understood is because of the presence of our cell phonic asset group. You may recall that groups such as cell phonic asset group or the nitro group, they are strongly deactivating. Now, this is the second reason why this reaction is not feasible essentially, you may simply recall that strongly deactivating means decreasing the electron density within the ring. Okay, now, if we are decreasing the electron density within the ring, a lower electron density would essentially have a hard time attacking the electro file. You may recall that the pi bond would attack an electro file to get our substitution product eventually. Right, so there if there is a low electron density because the symphonic asset group is strongly deactivating and therefore decreasing the electron density. A decrease in electron density would essentially not allow us to achieve such a high yield of the product and the yield would be negligible e small. So it would not be feasible reaction for us. So that's the second reason why that's method is not feasible. So now, essentially if this method is not feasible, we are going to propose another method for this synthesis because there are two issues with this reaction. We are going to fix both of them. So, first of all for this reaction, let's suppose that we are starting with a benzene ring and instead of forming that the activating substitute print, which is our cell phonic acid group, we are going to change the root of our reaction. And in the first step we are going to add a proposal chain. Okay, now, because we said previously that we cannot use one chloral propane because we would get an isopropyl substitute wint. We are going to change and calculation to isolation so that means we are going to use a carbon compound. We need to make sure that we're using our chlorine bond is a carbon in Alright. And essentially we need to make sure that we're using three carbon atoms because we need to have a three member chain. So we have 12 and that would be the third one. And of course because this is the Friedel crafts isolation, we're using aluminum chloride as our catalyst. Now you may recall that were similar forming that carbon carbon bond, we're replacing chlorine with a bond between the carbon group and our ring. So we may get our first product that way, that will be our carbonell group. Two more carbon atoms. Now the question is, do we have to reduce it immediately? The answer is no. So we're not yet getting our proposal chain. Remember if we get a proposal chain, it's actually Ortho and para directing. We actually want to have a made a product at position number three. So we're going to keep the carbon eel group which is perfectly fine. It's directing at the Ortho. My apologies at the main opposition. Right? So we will have our substitution product at either this position or that position. They're both equivalent. So that means we're not rushing and we're not yet reducing it to a pro ball chain. Instead, we are going to add our symphonic acid group And to do that. You may simply recall that the cell phone ation reaction is reversible. So we will keep that in mind and we will draw reversible arrows for this. Stop. We need to regions fuming sulfuric acid. So we may label this as sulfur cry oxide in the presence of sulfuric acid or basically fuming still very Cassidy and we understand that because Carbonell group is made a directing. We would add it at this position exactly where we need to have it. So that would be our six member ring our Carbonell group. And now at the main opposition or position number three with respect to carbon ill, that would be our S. 03 age group. And now we are done right because we need to get an eye. So pro my apologies to get a normal pro bill chain. We just need to reduce it and you may simply recall that there's an easy way to do that which is called the Clemmensen reduction Clemmensen reduction essentially implies that you're going to use a specific set of regions. You may recall that you want to use zinc mercury amalgam. So that would be saying mercury as well as and a clear solution of hydrochloric acid. So these regions imply the Clemmensen reduction which essentially allows you to reduce their carbon ill into an actual l kill chain. So we are just going to get rid of that ceo double bond. Right? And we are going to add two hydrogen atoms here instead. So this is how we will get our final product. Let's draw the final product. That would be Our six member drain At position # three We have our cell phone and cast the group actually. Now this will be our position number one, right? Because this symphonic acid group has a higher priority. So this is position number one. And at position number three we have our propel changes as required after the Clemens center reduction stopped. So this is our synthesis scheme which would correspond to a feasible reaction to form the desired product. And we may recall that the original scheme was not feasible because of the deactivated system of reactant. Remember that symphonic acid group is strongly deactivating and Friedel crafts reaction fails in such a case. And the second reason was that one chloral propane just wouldn't work for us. See to the fact that we would form an isopropyl substitute parent instead of a straight propel chain, and that said, we may simply conclude that the correct answer choice. This multiple choice question is C, However, if you have any questions, don't have seats, leave your comment down below in the comments section and thank you for watching.

To synthesize m-ethylbenzenesulfonic acid, a student attempted the Friedel–Crafts alkylation of benzenesulfonic acid with bromoethane. Do you predict that this reaction was successful? If not, propose an alternative synthesis.

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