Propose a mechanism to show the individual alkylations that fo...

Question

Propose a mechanism to show the individual alkylations that form this quaternary ammonium salt.

Accepted Answer

All right. Hi everyone. So for this question we have to suggest a plausible mechanism for the formation of the following ordinary ammonium salt. So in our starting material side here, we have an amine. Right? And we're reacting this amine with an alcohol. Hey light. So what I want to bring to your attention is that this amine has reacted with our alcohol highlight to generate a coa ordinary ocular ated ammonium salt. So um because our local highlight has a method group. Right? What I want you to notice is that going from start to finish this amine here has been inoculated, meaning that three local groups have been added to it and in particular three methyl groups from our local highlight have been added to our mean to generate a coa ordinary ammonium salt. So this is going to indicate that the reaction in question is an amine calculation right? Because the local group of Roko highlight is going to be added to haramain. And recall that um in a mean calculation, you're a mean is going to react with your alcohol. Hey light and that light is going to be um generally speaking it's going to be primary. Right? Because recall that a mean calculation proceeds via an sN two mechanism. So a primary Okay, oh hey light is going to be favored. So now because we know that it's going to proceed through an S and two mechanism, I'm going to go ahead and redraw our starting material on the bottom here because now we understand that that is going to function as a nuclear file. Right? And I'm also going to draw our local highlight on the side. So that's our actual highlight. And because we understand this is sn two are nitrogen here is going to behave as a nuclear file and it's going to attack the carbon of our actual hairline, therefore kicking out broening, so to speak. So on our first step um we are going to add our first methyl group. Two are mean, so this is our starting material. But this time we have an additional method group and our nitrogen is going to have a positive charge over here. So with that being said um because we have multiple method groups from our alcohol highlight, we understand that this occupation process is going to occur multiple times. So in order to for um for this process to happen multiple times we need um we need another lone pair of electrons on nitrogen. Right? So this is where our base here is going to come in, right because our base our h um CO three That's going to go ahead. And deep protein 8 1 of our hydrogen zone are mean to give an additional pair lone pair of electrons that is going to repeat the process. Right? So if I scroll down to give myself some space, we are going to have a secondary amine this time. No, I added an extra carbon in between. So now here is gonna be our lone pair instead of that hydrogen and this is gonna be our first methyl group. Right? So now the process is going to repeat again, another molecule of methyl bromide is going to react with our now secondary amine and the process is going to repeat. Right. So I'm going to go ahead and draw out a second molecule of methyl bromide and R. S. And two mechanism is going to repeat itself because the new lone pair of electrons on nitrogen is going to attack the electra filic carbon of the daylight, therefore generating another bromide ion. So we're going to have our second occupation happened and nitrogen is going to have a positive charge once again, this is gonna be our nitrogen here are the two methyl groups from methyl bromide and we're down to our last hydrogen and like I mentioned before, nitrogen is going to have a positive charge but we have to add one more alcohol group based on the structure of our co ordinary product. Right? We have to so far so we need to add one more. And in order for that to happen, our base here are H c 03 minus is going to d protein, eight are remaining hydrogen of r mean This is HM 03 and we are going to generate a lone pair of electrons on nitrogen by deep resonating it. So if I scroll down to give myself some more space, we are now um we now have a tertiary amine. So this is our lone pair and here are the method groups. So now in our last step, This is going to be our final SN two reaction. Because when we react an additional molecule of methyl bromide, Then another sn two reaction is going to take place. And once this third method group adds to our mean, we are going to end up with our co ordinary ammonium salt. So this is our nitrogen and now we have the three method groups that we have in our final product and so now we cannot deep throat in it. This molecule any further because we do not have any hydrogen is left. So this is gonna be our final product and that's going to have a positive charge and our counter ion is going to be bromide which we generated earlier. Right by our sn two mechanism. So this is going to be um your final answer. Right? So what essentially happens in an amine calculation is that for every hydrogen that you have in your I mean you are going to see an sN two mechanism, right? Because first your nuclear filic nitrogen attacks your Kyohei light. And to repeat that process, you are going to see the deep rotation of each of your hydrogen until finally you have a fully substituted nitrogen which cannot be deep rotated any further. So with that being said, um if you stuck around to the end, thank you very much for watching your answer here is going to be choice E and once again, thank you for watching and I hope you found this helpful.

Propose a mechanism to show the individual alkylations that form this quaternary ammonium salt.

Key Concepts

Alkylation Mechanism

Quaternary Ammonium Salts

Role of NaHCO3

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