Show how the following compounds can be synthesized from benze...

Question

Show how the following compounds can be synthesized from benzene:

a. N,N,N-trimethylanilinium iodide

Accepted Answer

Hey, everyone. And welcome back to another video show how the following compound can be prepared using the given starting material and any necessary regions, the starting material is tolu and we are producing N M M for tetra methyl benzene minium iodide, basically a quinary ammonium salt. So that if we think about to at the power position, we're adding nitrogen which is bonded to three metal groups and that it has a formal positive charge and the counter I which is the iodide anion. So let's think about the synthesis. We're going to draw a line. And specifically, we have to recall that because to corresponds to the benzene ring with the methyl group and since groups are Ortho and para directing. And in addition to that, because we need to add nitrogen at the power position. First of all, we're going to perform the nitration so that we add N 02. Let's recall that to perform nitration, we want to use nitric acid in the presence of sulfuric acid. So that what's happening here is that we are replacing the hydrogen atom at the power position with the nitro group. However, we are not really satisfied to see the nitro group, we want to get an A and specifically, first of all, we're going to convert our nitro group into the amino group. Let's recall that we can simply do that using hydrogenation. So we're going to use hydrogen gas in the presence of paladium over carbon. So catalytic hydrogenation will work for us and we will be able to produce amino group instead of nitro group. In this case, we are simply going to replace two oxygen atoms with two hydrogen atoms because we have performed our reduction. And finally, we have to think about our last step if we want to turn a primary amine into a quinary ammonium salt, right? So let's draw our final product and then let's think about the regions. We want to replace the two hydrogen atoms with two metal groups. And then we want to add one more methyl group which will give nitrogen its positive charge. And then we must have a counter I which is the iodide anion for that purpose, we are going to use methyl iodide. We want to make sure that it is given in access because we want to add three muscle groups. And then we want to use a base such as potassium carbonate. So that carbonate will act as a base. And let's think why this reaction works for us. So if we take our starting material in the last step, that's our primary. I mean, we are going to analyze how this reaction takes place. So the lone para nitrogen will essentially perform the attack at the electro Pyle carbon of methyl iodide followed by the loss of the leaving group. We are observing a simple essence reaction and then the stop, we're going to add our first metal group. So we can simply replace one of the hydrogen atoms with the metal group while one of the hydrogen atoms remains. However, for the purposes of the mechanism, we can simply show that both hydrogen atoms will be present. We are simply adding one metal group so far, so that nitrogen gets as positive charge. But this is where carbonate comes in, right? We said that we want to also use a base and carbonate specifically, which is C 032 minus with a Sloan pair, carbonate will essentially take one of the protons to restore the lone pair on nitrogen. OK. So that when nitrogen gets its lone pair, it can perform the same nucleic substitution reaction on another molecule of methyl iodide. And eventually, we can state that in several steps. When we repeat that process, we get our final product. But basically, to clearly visualize it, we can just redraw the product obtained after the de protonation stops. So you can understand it more clearly, we're removing one of the hydrogen atoms and now we are getting our loan pair on hydrogen, which will perform another S N two reaction so that we are going to replace another hydrogen with a muscle group, right. So let's admit the deep protonation step and immediately understand that we're getting another methyl group. And finally, the third one will be obtained after the final S N two reaction. And that's how we will get the quaternary ammonium sold, right? So, in each case, we're using methyl iodide in the presence of carbonate. And once again, we're going to use methyl iodide in the presence of carbonate. In the final step, we are simply getting our product so that we're going to get nitrogen with three metal groups. So that's how it gets its positive charge. And then the iodide and ion is basically our counter ion which forms during the loss of the leaving group well done. So we understand how we can get the questionary ammonium sold in our final step. And we can simply label our synthesis scheme. And because the problem was a bit extensive, we are going to review the steps to undergoes nitration so that we're adding our nitro group at the power position, even though some of them will be added to Ortho position, the major product will correspond to the par substituted product. Then using catalytic hydrogenation, we're forming our primary amine, which then undergoes several essence reactions in the presence of methyl iodide. And it's really important to have an excess amount of it in the presence of carbonate, which acts as a base. This allows us to replace both hydrogen atoms with metal groups and suits the presence of the lone pair on nitrogen. One additional metal group is added so that we have three metal groups total. And that's how we get our final quinary ammonium salt. Thank you for watching.

Show how the following compounds can be synthesized from benzene:
a. N,N,N-trimethylanilinium iodide

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Quaternization of Amines

Nucleophilicity and Electrophilicity

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