Reductive Amination

Concept: Concept: Reductive Amination

Video Transcript

In this video, we’re going to cook up a really important reaction called reductive amination. Recall back to your ketones and aldehydes section of the text that ketones and aldehydes when reacted with a primary amine in an acidic environment, what would you get? You would get the functional group called an imine. What was an imine? Remember that an imine is just a carbonyl carbon but instead of the O being attached to the C, you’re going to replace that O with an N. That N could be attached to up to one R groups. The reason that it can’t be two R group is because that would be called an enamine and the double bond would be in a different spot. We’re not going to be talking about enamines here. We're only talking about imines. This mechanism was a reversible reaction. Remember that you have your double-sided equilibrium arrows showing that you can go from the carbonyl to the imine and then back to the carbonyl. But regardless of which direction you’re going, you always had to pass through a really important intermediate. That intermediate was called the iminium cation. The iminium cation was an intermediate where your nitrogen had a positive charge on it. I’m not going to draw it right now but you'll see it in a little bit when I go through the mechanism.
Because this is a mechanism that I’ve already covered in another video, if you do want to brush up on the entire imine mechanism, feel free to just search in the Clutch search bar. Type in imine and this whole mechanism should pop up so you can refresh. What I'm going to do is since I’m assuming that you guys can go back to that video and watch if anything, I’ll be teaching the abridged mechanism here. Why this is important is because the imine mechanism is the first step of reductive amination. Let’s keep going. It turns out that instead of deprotonating at the end and making your imine, there's something else that you could do. You could react with a reducing agent called NaBH3CN. What you could do is in your first step, you could make an imine. This is your first step. Notice how I have a nitrogen like ammonia and acid. That’s what we would usually expect from an imine.
But what's different about this reaction is that in the second step, I would use a reducing agent. Remember, a reducing agent is one that adds hydrogen. So I’m just putting H in brackets just to generalize that this is a reducing agent. What we're going to find is that if you use the reducing agent in the second step, instead of just leaving it as an imine, what you're going to wind up getting is an amine. As you can see, this is actually a primary amine. It turns out that reductive amination is a really convenient way to make amines. Following the logic of these mechanisms, this is how it works. Your first step like we said is that imine reaction. What you do is you take your carbonyl. In this case, this would be an aldehyde and I would react it with an amine, an acid like we have here. What that would give you is this intermediate. This intermediate was called an iminium cation. This is the one I was referring to earlier. This is the last step before we make our imine. Usually what would happen is if we wanted to just make an imine, we would deprotonate. Usually we would use some of kind of base or a conjugate to deprotonate, take away the H and you would usually get the imine that looks like this. NH, carbon, H and we’d be done. This is called your imine.
But we're not going to do that because in this reaction, we want to get rid of this double bond altogether. We don't want the double bond there anymore because the double bond is an imine but we want an amine. We want one that has just the single bond. How can we add hydrogens to the double bond to make sure that it goes away? Basically, to reduce it. We can use my reducing agent. This is going to be basically sodium borohydride or NaBH4. Notice that it's almost the same as NaBH4 but it has a cyano group on it, CN. What the CN is going to do is it's an electron-withdrawing group, so it's going to make the sodium borohydride a little bit less strong of a reducing agent. This is actually going to be mildly reducing. I’m just going to write here mildly reducing. Since it’s mildly reducing, I don't have to worry about getting any kind of weird byproducts. All it’s going to do is it’s going to add hydrogens to the double bond. What we're going to do is in the last step of the mechanism, like I said I’m not going to go through the whole imine mechanism because this part of the mechanism, you can just watch the imine video if you want to know this part.
But I know that you’re probably interested in this part. You're wondering how does this happen. That's what we're going to go through now. You've got your sodium borohydride. I’m going to put your B, it’s got an H2CN. Let me just make sure that I have everything right. It looks good to me. By the way, because of the fact that this boron has four bonds on it, so it’s going to have a negative charge. When I wrote the sodium borohydride, I didn't write the negative charge. Where was that? Remember, there's a sodium around here. The Na+ is counterbalancing that so that’s why I didn’t write it as a negative charge. But once you ionize the sodium away, you let it go into a solution, you’re going to be left with this negative charge on the B. What's going to happen is that the electrons from my sodium borohydride with the CN group, this H can attack the carbonyl and push electrons up to the N. What we're going to basically do is we’re going to take one of these Hs. We’re going to take that red H. We’re going to attach it to the bottom of the iminium cation. Then I’m going to push the electrons up to the nitrogen. What that means is that now I’m going to have two new electrons here. That becomes the lone pair on the nitrogen and we have the extra red hydrogen that is now there. That red hydrogen came from basically inactivated version of my sodium borohydride because once again, I said that the CBN is an electron-withdrawing group so it kind of inactivates the reducing agent. That's really it. Now we have a primary amine.
For this next question, I want you guys to take this molecule. As you can see it's a ketone. Go ahead and react it with these two reagents. Try to predict what the product is going to be and then I'll go ahead and explain the whole reaction to you. So, go for it.