Reduction of Aldehydes and Ketones - Video Tutorials & Practice Problems

Reduction Reactions Concept 1

Now, when it comes to the reduction of aldehydes and ketones recall, that reduction uses a reducing agent to add as many carbon hydrogen bonds as possible without breaking any carbon carbon bonds. Ok. So that's the whole idea of reduction. If we take a look here, we're only paying attention to what's within this shaded box. That's what's really the most important but stuff lies outside that box. So let's take a look. So if we're starting here over here on the right, we can say that the most oxidized form of carbon exists as carbon dioxide, carbon is making its maximum four bonds. All of them being to oxygen once I start the reduction process. So we're gonna reduce it some, we're gonna start to add as many carbon hydrogen bonds as possible. Carbon can only go up to four bonds. So to do that, we have to break one of these carbon oxygen bonds. So that's where this comes in carbon needs to keep making four bonds. So that's where this comes in oxygen. Two needs to make its ideal number of bonds, which is two. So we add a hydrogen to it as well. If we keep going. We enter the shaded region here. So here we keep adding carbon hydrogen bonds. So we'd have to get rid of this bond here. And that's how we wind up with this aldehyde. This alde high can undergo further reduction. So we're gonna add more carbon hydrogen bonds to it. Meaning we have to break this carbon oxygen bond here. So here goes, the h that I've added to the carbon oxygen still wants to keep making four bonds among two bonds. So he would have its hydrogen added to it as well. So we can see here that reducing an aldehyde has created an alcohol. If we wanted, we could continue further with strong reduction to help create an alkane. At the end, the alk being the last stop in terms of reduction process. But what we need to take from this is that we can reduce an aldehyde into an alcohol. And even if we had a key to, we could do the same because if we reduced that, we wanna add as many carbon hydrogen bonds as we want. So we'd again remove one of these carbon oxygen bonds. So we have our carbon with its two carbons, one on each side, it's newly gained hydrogen. And then the oxygen would also gain a hydrogen. So we can see here that we can reduce both aldehydes and ketones to produce alcohols. Ok. So that's the whole basis when it comes to reduce, reducing aldehydes and ketones doing so, changed them both into alcohols.

Reduction of Aldehydes and Ketones Example 1

Which of the following compounds could not be reduced. So if we take a look at our options, we have 22 dimethyl pentane, two methyl, one Pinel, three amyl, two hepton and then four Bromy Hepton acid. Now we don't need to draw all these the names themselves. Tell us the identities or functional groups. Pentane. This is an Alcaine. L means this is an aldehyde own means that this is a key toone and then OIC acid, pino acid that tells us this is a carbolic acid, right? So remember the whole line in terms of reduction, we start out at carbon dioxide as the most oxidized form of carbon. When we reduce that, we changed into a Carboy acid, it says what cannot be reduced. Well, we know that uh Carboy acid could be reduced if we wanted to, we could reduce it to a aldehyde. And we know that an aldehyde can be reduced to an alcohol. So this is out and this is out. We also know that a kone can be reduced to an alcohol as well, just like an aldehyde. The answer here is option A and Alcaine. Remember we said that the Alcaine is the end stop of our reduction line. Right? Once we get to that point, we've added as many hydrogens as we can to our carbon. So there's nothing else we can do. Remember, we cannot break carbon carbon bond. So even if that carbon is connected to another carbon, there's nothing we can do to sever that bond to add another hydrogen. So al cane represents the most reduced form. Therefore, it cannot undergo reduction. And therefore, option A is the answer.

Reduction of Aldehydes and Ketones Concept 2

Now, we can say here that aldehydes in ketones are reduced to primary and secondary alcohols respectively. Now, the reducing agent within these processes is H two. And the catalysts that help facilitate this reduction, we can have them as metals which are N I which is nickel, platinum or palladium. Now, the whole process, the whole result of this reduction is that the carbon oxygen gains an H and the carbon carbon also gains in. If we take a look here, we start out with an aldehyde. We're reducing it with H two A nickel. Remember reduction means we have to try to make as many carbon hydrogen bonds as we can without breaking any carbon carbon bonds to do this. We'd have to break one of the carbon oxygen bonds here. And when we break that bond, Carmen still needs to make its four bonds and oxygen would ideally want to still make its two. So this two basically gives itself up. One goes to the carbon, the other h goes to the oxygen. As a result of this, we see that we made a primary alcohol, a key toone. Well, a key Toone saying process, we're gonna say we use H two and remember we could use nickel platinum or palladium as our metal catalyst. And again, we're trying to get rid of one of these carbon oxygen bonds to make up for that lost carbon is going to bond to a hydrogen and oxygen is gonna bind to a, a hydrogen. What we make now is a secondary alcohol. So just remember when it comes to reduction, both aldehydes and ketones can be reduced. Aldehydes, typically create primary alcohols. Ketones. You um primarily make uh secondary alcohols. OK. So, they're gonna make secondary alcohols once they undergo reduction. All right. So, just keep that in mind.

Reduction of Aldehydes and Ketones Example 2

Determine the alcohol product formed in the following reaction. So here we have a kone and we're using our reducing agent of H two with our metal catalyst. Nickel. Remember, keytones get reduced to secondary alcohols. What's going to happen here is we're gonna cut one of these carbon oxygen bonds and to maintain four bonds for carbon and to maintain two bonds for oxygen, each one of them gains a hydrogen. So we'll have here is we'll have ch three, ch two still. This is still connected to this carbon which is still connected to this oxygen. And then we have stage two, stage three again. So this two, here, one of the HS will go to carbon and the other H will go to oxygen. We just created a secondary alcohol by reducing the ketone reactant.

Determine the alcohol product formed in the following reaction.

• A.
• B.
• C.
• D.

Determine which reactant should be used to produce the following alcohol.

• A.
• B.
• C.
• D.