To reduce a molecule means to increase the hydrogen content of that molecule. In order for reduction to take place, you need reducing agents. So in this topic, we're going to explore what those different reducing agents are. All right. So as we just said, the definition of a reducing agent is anything that's going to be used to reduce or add hydrogens to a molecule. Now categorically, there are a few different ones we can use. These reagents are all going to add hydrogens across pi bonds because what we're doing here is we're going to be taking double bonds, triple bonds, and such, and we're going to be adding hydrogens to them. Unsaturated hydrocarbons and carbonyls are able to be reduced through these agents. Remember that unsaturated means that they don't have the total amount of hydrogens possible. So as we add hydrogen, we're actually saturating. So just so you guys know, reduction is a form of saturation. Those are two different words that can be used almost interchangeably. We'll usually say reduction, but some types of reduction cause saturation. Also, just another tangent is that saturation and reduction are also caused by hydrogenation. So if you've ever heard of any reaction that is a hydrogenation, that's a reduction because it's adding hydrogen and it's increasing the saturation of the molecules. These are all words that you should really group together in your mind because we're going to be using them a lot when we're discussing this topic. I just want to show you guys the general mechanism of what's going on. Now it's going to depend particularly on the reagents, but basically, both reagents that we're going to learn about today are both simply a source of H- ion, which is what we called a hydride ion. They're both a source of hydride. And pretty much regardless of what the carbonyl looks like, today we're going to be reducing a lot of carbonyls. Pretty much regardless of what it looks like, the hydride is pretty much always going to attack in the same way, which is a mechanism called nucleophilic addition. Now in this topic, I don't need you to be a master of nucleophilic addition, but just kind of humor me as I go through the first few steps just so you guys can kind of get an idea of what's going on. My H- attacks the carbonyl carbon. There's a positive charge on that carbon because of the dipole pulling away from it, which makes the H- attracted to it. When I make that bond, carbon doesn't like to have 5 bonds. It wants to have 4. I have to move electrons up to the O-. What I wind up getting is an extra H at the bottom. That's the one that just attacked. An O- at the top and that O- winds up protonating. So what I wind up doing is an alcohol out of my original carbonyl.
That's what the reagents we're going to learn about today do. They all do something like this. All right. So let's go ahead and read a little bit more about reduction. Reducing agents are going to add to all of the pi bonds present. So that means that I don't need to say 10 equivalents of my reducing agent in order for you to realize that it's going to react 10 times if it can. We're just going to blast away as many carbonyls as possible in this reaction even if I don't put how many equivalents of my reagent I have. That's why it says multiple equivalents of hydrogen will react if possible. Just think anything that hydrogen will react with, can react with, it will.
Let's go over our first reagent. This is actually a very important reagent in organic chemistry. We use it a lot in organic chemistry 1 and 2. This is Lithium Aluminum Hydride, LiAlH4. It's also simplified. You could call it LAH. Why? Because it's lithium aluminum hydride. So LAH is like the strongest reducing agent we know of. It blasts pretty much everything. So it's going to be able to reduce any carbonyl compound into an alcohol. So let me show you guys what a typical reduction with LAH would look like. Well, here you notice that we have an aldehyde. Is an aldehyde a type of carbonyl? Yes. So what I'm going to do is I'm going to wind up getting something that looks like this. The carbon stays intact, but now instead of that being a double bond O, it turns into an alcohol. Now you might be wondering, well, how would I know that that would happen? Well, first of all, you're always going to go to an alcohol. You're never going to go to less more reduced than an alcohol. But another way that you can think of it is that we're adding an equivalent of hydrogen because notice that before this reaction happened, I had 1 hydrogen at the bottom. That hydrogen is still there. But now what happened after the reaction is over is that I added one equivalent of hydrogen. What does that mean? It means I added 1 H here and I added 1 H here. So that's why I get an alcohol because I'm basically adding 2 hydrogens to both sides. And remember, that has to do with the mechanism how one H attacks from the bottom and one H comes from the protonation step. Does that make sense? Cool. So we're adding 2 H's which is this would be one equivalent of hydrogen.
So let's look at this next one. How about if our functional group is in a ring? That's a little bit more tricky. This is an ester, right? Is an ester a type of carbonyl? Yes. So in this case, we're going to wind up reducing both sides. We're going to chop this bond right there in the middle because we know that this eventually has to become an alcohol. Alcohols don't attach in a ring. You can't have an alcohol with 2 bonds on both sides or that would be an ether. It wouldn't be an alcohol. So we know that that bond is going to have to break. And actually, what's going to happen is that we're just going to get 2 alcohols on both sides. So I could count my carbons. I would say this is carbon 1, 2, 3. I'm going to have a 3 carbon chain. But then both of the things on both sides wind up becoming alcohols. So on number 1, I would get this is number 1, 2, and 3. I would get an alcohol where basically this double bond O just got 2 Hs added to it. It got an H at the bottom and an H at the O as I just drew in the first example. Does that make sense? So nothing's changed there. Now what's different is that for molecule atom 3, it has an O coming off of it. That O is just going to get an extra proton. So this one is going to become an alcohol as well. So you wind up getting actually when you have a cyclic ester like this, you're actually going to wind up getting what we call a diol. Now do I care that you remember that detail specifically? No. But if you have learned about diols before, you know this is an example of one. So that's the first thing. LAH, strongest reducing agent we know, it's going to basically transform every carbonyl. Let's move on to our next reducing agent.
We also have something called a weak reducing agent. Now a weak reducing agent is going to be very similar except it has more limitations. It's only going to be able to add one equivalent of oxygen as opposed to more than 1. So what that means is that and that's the technical term, but what I really care that you know is that it's only going to be able to reduce aldehydes and ketones. Remember, CHO is an aldehyde. So what that means is that when you encounter a carbonyl that is not an aldehyde or a ketone, nothing's going to happen because NaBH4 is not strong enough to reduce those reagents. So we're only going to be able to reduce aldehydes and ketones. So let's look at the products for the first one. Would NaBH4 react with that aldehyde? Yes. In fact, we're going to get the same exact alcohol that we would have gotten over here in the first example. Same exact alcohol. Will it reduce an ester? No, it will not. It is not strong enough. So, what is it not going to reduce? Let's go ahead and just list them out. It's not going to reduce an ester. It's not going to reduce a carboxylic acid. It's not going to reduce an amide. These are carbonyls that are too highly oxidized for the molecule to reduce. So NaBH4 is not going to be able to touch those, but it will be able to react with carbonyls such as ketones and aldehydes. So the first reaction works. The second one doesn't. Overall, reduction isn't that hard. You don't even really need to know the mechanism most of the time. You just need to predict the products and tell what NaBH4 will react with and what it won't. Remember that LAH pretty much reacts with anything. So let's go ahead and do some practice problems.
