Concept: Concept: General Reaction4m
In these next few videos, I’m going to be discussing amine alkylation. By the word amine, I'm not talking about an angry alkylation. Obviously, I'm talking about the functional group amines. Now that we’ve settled that one, let's go ahead and get into the reaction.
A huge part of this section of your text, the amines portion of your text is going to be trying to figure out how to synthesize primary amines. Primary amines are useful for lots of different applications. We're always trying to figure out what's the best way to make a primary amine of my choice. The go-to reaction that we would possibly think of is why don't we use an alkyl halide because if you look at an amine, it's got a very nucleophilic lone pair. There isn't a full formal charge on this molecule at all. I'm just using the NU negative as my symbol for a nucleophile. I’ve got an electrophile which is the carbon that is attached to my halide. This opens up for a reaction that we learned a long time ago. It’s one of those mechanism you’re never allowed to forget – backside attack or SN2. Theoretically, we can make primary amines through the reaction of an amine and an alkyl halide. Let’s go ahead and just talk about this for a second. I'm not going to draw the whole mechanism but we’re just going to draw an abridged one here. You would have the lone pair doing a backside attack on my electrophilic carbon. I would kick out my leaving group. What you would wind up getting just from this very first reaction is you’d have an NH3 that's now attached to a three-carbon chain. I’m just going to write 1, 2, 3. One of the problems with that is that you're going to have a positive charge because now you have an extra bond to that nitrogen that wasn't there before. That's why we always run this reaction in a basic environment, something like NaOH. The OH negative is going to serve to deprotonate the product. So then hopefully by the end, we're actually getting our primary amine. I’ll just draw that one more time, NH2 with our three-carbon chain.
Everything seems pretty straightforward. It seems like this should be our go-to reaction for making primary amines because it’s so easy to use. I could just literally throw out any alkyl halide in there as long as it was primary or secondary or methyl and I could get an SN2. Remember that SN2s are tricky. They don’t work on tertiary alkyl halides because the leaving group is too backed up. It's too difficult to access. But we've got another problem on top of that. It's not just that SN2 has its limitations. One of the biggest problems of this reaction is that we still have a nucleophilic lone pair at the end of the reaction. Notice that your product is just as nucleophilic as it was at the beginning. What we get is a possibility of multiple alkylations. What we’re going to find is that the predominant product in these reactions is going to be a polyalkylated amine which kind of sucks if you're trying to make a primary amine. There is a way to counteract that which is to use excessive amounts of amine. If you can drown the solution with tons of your amine, and then have very little of your alkyl halide, that reduces the chances of getting multiple R groups on the same nitrogen. But it’s still quite a limiting reaction and one that we're not going to find very synthetically useful. We're going to rely on other methods of making primary amines for this reason. Let's go ahead and look at the next video. In the next video, I’m going to be explaining the entire polyalkylation mechanism of an amine.
Concept: Concept: Mechanism4m
Let's explore, what this reaction would look like in a situation where there isn't hexa amine. Alright, so let's go ahead and just bring down that Amine that we were using before we were using basically NH3, right? And, let's go ahead and react that with the same alkyl halide we were using before. So, let's use a propyl chloride, okay? So, anyway. So, what we get in the first step is an SN2 mechanism running this in the basic environment. So, what we're going to say is that you're going to get an H, 2, H. Now, with an R group and a positive charge, your OH would then deprotonate and what you're going to get as a result of the first step is you're going to get NH2 with your propane group and you're also going to get I guess water, right? And your Cl minus as a leaving group, I'm going to start omitting those as we move forward because it's not really that important, I just care about the organic product. So, now we've got this primary amine wonderful, we start off with basically ammonia and now we have a primary amine. Now, if this reaction stopped here that would be awesome, the problem is going to be that's not going to stop here, it's going to keep reacting. So, we're going to wind up getting is this reacting with my second equivalent of alkyl halide. So, I then take my lone pair, I react, I kick out the chlorine.
Now, I've got something like this, I have to start drawing a little smaller to fit all this in, I'm going to get a three carbon chain on one side, I'm going to get another three carbon chain on the other, two H's, positive, this gets deprotonated cuz I'm in a basic environment and now I've got a secondary amine. So, guess what happens guys? it can react again. So, anyway, this reaction keeps on going. So, I'm just going to write arrow, arrow, arrow, imagine what winds up happening towards the end, what you're going to wind up getting is a nitrogen that actually has four propyl groups on it, because it completely substituted, every single hydrogen was deprotonated and every single bond is now an alkyl group, the problem with that is that that's not what I wanted, I mean, if you wanted an ammonium derivative, great, this is the reaction for you, but this isn't very useful for me, if what I'm trying to get is a primary a primary amine, this is very far from the mark, okay? So, just letting me know, this is the reason that we're usually going to stay away from the inoculation and we're going to use some more interesting unique ways of making primary amines that are more controlled, that are in a more controlled setting, okay? So, that being said, I do want to show you guys an example where this reaction may be helpful. So, go ahead and do this practice problem, this just so you know heads up is related to your basically aromaticity section. So, if you haven't gone through your aromatic reactions yet you're going to be a little lost here, you can always return to this after but just keep in mind, this is going to draw from, what we learned in your aromatic section of your text, okay? So, go ahead and try to figure out what two reagents would be needed to make this transformation and then I'll show you guys.
Concept: Example: Propose a Synthesis4m
So, we've got two transformations that we need to keep track of here, one, I need to turn my tertiary amine, okay? Which would be referred to as an aniline derivative, right? Because, this is an and aniline with two R groups, so it's an aniline derivative, somehow I need to turn my tertiary amine into a quaternary amine, you might have a better idea of how to do that now that we just went through this reaction, maybe giving you some ideas, when have you seen a quaternary amine before being made, another transformation that it needs to make is I need to add a chlorine to the meta position of the benzene, because it's meta to the original group that was there. So, we've got basically two reactions that need to happen, one of them is going to have to be an amine alkylation because amine alkylations are really surefire way to get quaternary amines, once you have a quaternary you can't keep alkylating. So, it's the endpoint. So, that's actually gonna be the perfect reaction for this. Now, over here to add the chlorine, that's going to be an EAS halogenation or an EAS chlorination. Now, the question is, does it matter which order these are in? or kind I just put them in any order? the order matters guys because right now this is an o.p, director, okay? aniline derivatives are o.p. directors, they're one of the strongest ones that there is, whereas notice my quaternary amine has what kind of charge? positive, guys remember what a positively charged substituent is going to do, it's going to be a deactivator, it's going to be a withdrawing group, there's now an electron withdrawing group, it's actually one of the strongest electron withdrawing groups because it has full positive charge right there and it's going to be a meta director. So, what that shows you guys is that amine alkylation is actually a sequence group, okay? And again, this is all drawing from your aromaticity section but just letting you guys know that it is a sequence group because this group will, can alter the sequence of the reagents for synthesis. So, in this case since I want meta activity, I want a meta directed product, I'm going to go ahead and do my alkylation first, the specific R group I'm adding is just the CH3. So, I could just use a metal alkyl halide. So, I'm going to use CH3Cl, okay? Now, normally we would run in amine operation in base. So, I would say CH3Cl and NaOH but there's not going to be anything to deprotonate here, because I'm literally just adding the fourth R group on. So, that's all I need, I don't need base here, in fact, if I put base it's not going to do anything, there's nothing to deprotonate, okay? You would only use base if you were starting from a primary or secondary or methyl, not methyl but ammonia, ammonia or primary amine or secondary amine, but if you're starting from a tertiary like this then there's really going to be nothing to deprotonate. Now, the second reaction is going to be the EAS halogenation. So, you guys remember the reagents for that, you got this. So, Cl2 and FeCl3 as my lewis acid catalyst, so that is going to make my chlorine atom in the meta position because now I have a strong meta director and we're good, okay? So, hope that made sense guys, this is really the only synthetically useful application of amine alkylation, we're not going to use it a lot for synthesis because of the complications that we just discussed with poly alkylation. Alright, so let's move on to the next reaction.
Show how you would use direct alkylation to synthesize the following compounds.
(a) benzyltrimethylammonium iodide
Propose a mechanism to show the individual alkylations that form this quaternary ammonium salt.
Answer the following questions about the Alkylation of Amines by circling T (True) or F (False).
1. The lone pair of electrons on a primary amine cannot react in a nucleophilic manner. T or F
2. The substitution of the alkyl halide used in polyalkylation will have an effect on the yield of the product. T or F
3. With excess base and alkyl halide a polyalkylation does not result in polyalkylated amines. T or F
4. One way to prevent over alkylation is to run the reaction with excess amine. T or F
Indicate the order in which the intermediates would appear during the conversion of 1 into 2.
1) 1 → I → IV → II → III → 2
2) 1 → V → I → II → 2
3) 1 → IV → I → III → II → 2
4) 1 → I → III → II → 2
5) 1 → V → IV → II → III → 2
Treatment of ammonia ( 1) with a large excess of bromoethane ( 2) will form
Indicate the order in which the intermediates would appear during the conversion of 1 into 2.
a) 1 → I→IV → II→ III→ 2
b) 1 → V → I→ II → 2
c) 1 → IV → I → III→ II → 2
d) 1 → I→ III → II → 2
e) 1 → V → IV → II → III → 2