Tetraethylammonium bromide

Transcription

In the last video, we saw that amines are actually reasonably decent nucleophiles. And, in particular, we saw some ethylamine attack some bromoethane in an Sn2 reaction. And right after that happened, we had some diethylammonium because the nitrogen had given away an electron to this carbon right over here. So it had a positive charge. And so you had diethylammonium bromide. The bromine had left, taking an electron. It then had a negative charge. And then we said that that could actually be in equilibrium, where another ethylamine goes and takes a hydrogen from the diethylammonium and turns that into diethylamine. So we essentially had a primary amine turn into, or I guess you could say, converted into a secondary amine. Now, the one thing that might have popped into your brain near the end of that video is, hey, isn't this also a good nucleophile? And you say look, it's got this lone pair. It's more electronegative than the things that it's bonded with. Maybe this too could do a nucleophile attack or maybe some type of an Sn2 reaction. And you would be correct. So you could imagine, if you have some more bromoethane hanging around-- let me draw some more bromoethane. So let me draw it like this. So this is our bromo group. And then let's say we have a hydrogen above like that. That's kind of upside down from the last one we did, but same idea. Then we have another hydrogen behind, just like that. And this is obviously a carbon. You could imagine that now you have this diethylamine. It can have an Sn2 reaction with the bromoethane. So before it was just the ethylamine. Now it's the diethylamine that's acting as a nucleophile. So this electron right here in yellow could be given to this carbon right at the same time that this bromine leaves and takes that electron. And the bromine already had one, two, three, four, five, six, seven valence electrons. And this is really almost identical to what happened in the previous video, except now we are starting with a secondary amine, and we're going to see it's going to turn into a tertiary amine, as opposed to the last one, where we started with a primary amine and turned into a secondary amine. So what's this going to look like after we've done the attack? So the bromine will have left. It's now the bromide anion. So now we have the bromide anion. It has its seven original valence electrons and now it took one more electron from the carbon. It now has a negative charge. And now this ethane group right here-- so this carbon right here is this carbon right over here-- is now going to be attached-- let me draw it this way. It's now going to be attached to this nitrogen. So this yellow electron is now attached to that carbon, and it's bonded to that green electron, which on the nitrogen or that the nitrogen has. And then this nitrogen has two other ethyl groups attached to it, one, two, and then it also has this hydrogen over here. And because this nitrogen gave away an electron, it now has a positive charge. And I didn't draw these two hydrogens here on this carbon; they're implicit. I could draw them. One hydrogen, one hydrogen, and then another hydrogen. But just to keep things simple, I'm going to leave it off, just so it becomes very clear that we now have triethyl. Let me make this clear. So we have one, two three ethyl groups. So now this is triethyl and it's a positively charged nitrogen. It's bonded to four things. And essentially, it has given away an electron, so this is triethylammonium. And if you view it as being an ionic compound with the bromide, then you would call this triethylammonium bromide, which this would be the salt version, the ionic compound right there. But we have our triethylammonium, and then you could imagine, we had that ethylamine floating around before. If you still have enough of that ethylamine floating around-- so let me draw some of that ethylamine. Ethylamine looks like this. It has two hydrogens. It has a lone pair. It has a slightly partially negative charge right here. Nitrogen is very electronegative. It's the most basic of the neutral functional groups that we've studied. It might want to capture this hydrogen proton, so it captures it by giving it its electron. And then that electron that was associated with that hydrogen atom is now given back to this nitrogen. And now, once again, I'll draw this as being in equilibrium. I'll do this in equilibrium because one could swipe the hydrogen from the other. But now this guy over here becomes ethylammonium. So I'll draw-- so this is what he was before. And now he's bonded to this hydrogen. So now he's bonded to this green hydrogen over here. Let me make this green just here. So that's ethylammonium. And we saw this in the last video: ethylammonium. And then this guy right over here, and, of course, you have the bromide floating around, but this guy over here now is just a nitrogen bonded to three ethyl groups: one, two, three. It lost its hydrogen, gained an electron, so it's now neutral again. And so this guy is triethylamine. And since he gained that electron, he had an electron here, an electron there, so this is the electron that he already had and that magenta electron he gets back, and he has a lone pair. So you might say, hey Sal, but couldn't this guy, couldn't triethylamine act as a nucleophile? And you would be right. So the triethylamine, if you still have some of that bromoethane hanging around-- so let me draw some bromoethane. And obviously, there's some hydrogen here. There is some hydrogen here. If you have some more of that bromoethane hanging around, this guy could then attack that bromoethane and then you would be left with-- what would that look like? Let me do it in that same color. So in that green color, you have the nitrogen. It has its three original ethane groups or ethyl groups that's attached to it. It has this electron. But it's now bonded to this carbon over here. And let me be very clear. This was an Sn2 reaction. So right when it attacks that carbon, this bromine, and we saw this multiple times, can take that other electron from the carbon, and so this becomes another bromide anion. I'm going a little bit faster than I've done in the past because we've seen this many, many, many, many times. And now this guy is attached to this ethyl group. So this ethyl group I could just draw like this. He gave this carbon, this magenta electron there, and so we have this new bond. And I could also draw the hydrogens. We have this hydrogen and this hydrogen. We could draw them off like this, hydrogen and hydrogen, but that makes the drawing a little bit more confusing. But what just happened? This guy gave away an electron. Let me make it clear. He gave away an electron so he has now a positive charge. You have the bromine turned into bromide. It took an electron. But what is this now? This is tetraethylammonium. So we have four. It's a quaternary ammonium. So this is tetra, tetra for four: one, two, three, four. Tetraethylammonium. Ammonium, instead of amine, because it has that positive charge. The nitrogen has four bonds. It's given away an electron. Tetraethylammonium, and then if you view it as a salt with the bromide anion, it's tetraethylammonium bromide. So the whole point of this video and the last video is-- in the last video, I just wanted to show you that an amine, it's reasonably basic. It can act as a nucleophile, and show you how it would react, show you the mechanism. But what I wanted to show you in this video is that that reaction just keeps going. And so when you start with the original ingredients, when you started with the ethylamine and the bromoethane, you might say, oh, this is pretty simple. You're just going to end up with a little bit of diethylamine. But the reality is you're going to end up with all of this stuff. It's going to be all mixed up in every different way. You're going to end up with some diethylamine in there. You're going to have some ethylamine in there. You're going to have some ammonium in there. You'll have some triethylammonium in there. You'll have some ethylammonium. in there. You're going to have a little bit of everything. Oh, let me be clear here. I just want to make sure I didn't misname any of this stuff. No. Yes, I think I named it right. Sometimes I get confused with the ammoniums and amines. But the whole point of this is that this reaction can keep going and this Sn2 reaction can keep occurring in a bunch of different ways.