Reduction

Concept: Concept: Reducing Agents

10m
Video Transcript

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 that those different reducing agents are.
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's 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, stuff like that, 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 I 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, in general, say reduction. But some types of reduction cause saturation.
Also, just another kind of tangent is that saturation and reduction are also caused by hydrogenation. 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 molecule. 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 in 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. Pretty much regardless of what the carbonyl looks like – today we're going to be reducing a lot of carbonyls. And 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.
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 five bonds, it wants to have four, I have to break a bond, so I 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. What I wind up doing is I wind up making 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.
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. That means that I don't need to say ten equivalents of my reducing agent in order for you to realize that it's going to react ten 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 hydrogen can react with, it will.
Let's go over our first reagent and this is actually a very important reagent in organic chemistry. We use it a lot in orgo one and in orgo two. This is lithium aluminum hydride, LiAlH4. It's also simplified, you could call it LAH. Why? Because it's lithium aluminum hydride. 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.
Let me show you guys what a typical reduction with LAH would look like. Well, here you'll 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.
You might be wondering, “Johnny, 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 more reduced than an alcohol. But another way you can think of it is that we're adding an equivalent of hydrogen because notice that before this reaction happened, I had one 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 one H here. Oops, I wanted to use a different color so you guys could see that. I'm going to use blue. One H here. And I added one H here. That's why I get an alcohol because I'm basically adding two hydrogens to both sides. 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? We're adding two H's, which this would be one equivalent of hydrogen.
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. 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 and alcohols don't attach in a ring. You can't have an alcohol with two bonds on both sides or that would be an ether. It wouldn't be an alcohol. We know that that bond is going to have to break.
Actually, what's going to happen is we're just going to get two alcohols on both sides. So I could count my carbons. I would say this is carbon one, two, three. I'm going to have a three-carbon chain. But then both of the things on both sides wind up becoming alcohols. So on number one – I was trying to use a different color again. On number one, I would get – this is number one – I would get an alcohol. Where basically this double bond O just got two H's added to it. It got an H at the bottom and an H at the O, like I just drew in the first example. Does that make sense? So nothing's changed there.
Now what's different is that for molecule or atom three, 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 OH as well. So you wind up getting actually when you have a cyclic ester like this, cyclic ester. 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 that this is an example of one. That's the firs 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. 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 one. What that means is that – 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. What that means is that when you encounter a carbonyl that is not an aldehyde or a ketone, nothing is going to happen because NaBH4 is not strong enough to reduce those reagents. We're only going to be able to reduce aldehydes and ketones.
Let's look at the product 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.
What is it not going to reduce? Let's go ahead and just list them out. It's not going to reduce ester. It's not going to reduce 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 be able to predict the products and tell what NaBH4 will react with and what it won't. Remember that LAH pretty much reacts with anything. Let's go ahead and do some practice problems. 

Concept: Practice: Intro

1m
Video Transcript

Alright guys, so now we're going to be doing some reduction practice problems just remember to keep in mind the difference between LAH and NaBH4, remember that one is strong and one is weak I'm going to test you on both of those concepts right now so go ahead and try to figure out the first problem.

Problem: What is the product of the reaction?

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Problem: What is the product of the reacion? 

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Problem: What is the product of the reaction?

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Reduction Additional Practice Problems

DRAW THE MOLECULAR STRUCTURE OF THE FOLLOWING REACTION PRODUCT(S).

Two major organic products (after workup) from the reaction of cis-4-methylcyclohexanol with  CrO3, followed by reaction with NaBH4

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Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw the answer in skeletal form.

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Which reagent(s) is/are capable of producing the following conversion?

A) LiAlH4 / Ether

B) NaBH4 / Ethanol

C) NaOH / H2O

D) CrCO3 / Ether

E) CH2PPH3

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The principal organic product of the following reaction is:

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Draw the organic products of the following reaction. Hint: There are  4 reactive sites of the molecule. 

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Identify the reactants then draw and identify the organic product of each reaction. 

 

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Provde the correct reagent(s) for the following reactions

a 1) xs CH 3MgBr    2) H2O

b 1) xs LAH           2) H  2O

c 1) xs Me 2CuLi    2) H 2O

d 1) NaOH             2) H  2O

e 1) H 2SO 4           2) H 2O

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Draw the organic product(s) for the following reaction. Indicate stereochemistry where appropriate. Assume an aqueous workup, when necessary. A reasonable answer may be “ No Reaction.”

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Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. Be sure to indicate the major product if more than one product is formed. Draw all answers in skeletal form. 

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One of the following reactions isn't a reduction reaction. Which one is not a reduction reaction?

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What product(s) is/are formed in the following reaction?

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Which product is formed from the following transformation?

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