- Formation of hydrates
- Formation of hemiacetals and hemiketals
- Acid and base catalyzed formation of hydrates and hemiacetals
- Formation of acetals
- Acetals as protecting groups and thioacetals
- Formation of imines and enamines
- Formation of oximes and hydrazones
- Addition of carbon nucleophiles to aldehydes and ketones
- Formation of alcohols using hydride reducing agents
- Oxidation of aldehydes using Tollens' reagent
Formation of alcohols using hydride reducing agents
Reduction of aldehydes, ketones, and esters using sodium borohydride and lithium aluminum hydride. Created by Jay.
Want to join the conversation?
- At11:20what's the excess product when LiAlH4 reduced the Ester group?(3 votes)
- Greetings Sir,
I couldn't find the video in which Cannizaro Reaction is explained.
Thanks in anticipation.(3 votes)
- I don’t think there is a KA video on the Cannizzaro reaction.
Here are the links to two Youtube videos.
- can any methyl shifts happen here to make the alcohol more stable?(2 votes)
- Rearrangements like methyl shift and hydride shift occur only when a reaction involves carbocation intermediate (a +vely charged on a carbon), in order to make a more stable carbocation( stability order of carbo cations: 3'>2'>1'). In this reaction there is no intermediate involving a carbocation, so NO rearrangements occur.(5 votes)
- At2:58can the alcohol R''OH give of a proton? wouldn't water act as a stronger acid?(2 votes)
- I believe that R"OH and H₂O are alternatives for the solvent – so you usually wouldn't have both present.(2 votes)
- But why the carbonyl ester is harder to reduce ?(1 vote)
- I believe the carbonyl in an ester is harder to reduce because esters (like carboxylic acids) are resonance stabilized.(2 votes)
- Why does the reduction of a ketone using NaBH4 create a racemic mixture?(1 vote)
- The aldehyde (or ketone) that we start with is a planar molecule. Consequently attack by the NaBH4 from 'above' or 'below' is equally easy. Those two directions will result in the two different mirror image stereoisomers.(2 votes)
- how can we convert 2 methyl cyclo hexanone to 2 methyl 4 pentenoic acid using several mechanism means it been converted to alcohol first.(1 vote)
- the hydrides of group 3-A are unstable except boron which form (B2H6)?why about aluminium which also form (Al2H6)?(1 vote)
- At1:40, I thought good bases were also good nucleophiles because bases like to donate electron pairs?(1 vote)
- Why is ester more difficult to reduce than a ketone or aldehyde that LiAlH4, and not NaBH4, is able to reduce it?(1 vote)
Voiceover: Let's look at the formation of alcohol using a hydride reducing agent and we'll start with sodium borohydride. If I wanted to draw the dot structure for this, it would have boron with four hydrogens. And that would give boron a negative one formal charge. Then we'd also have our sodium cation here. You could think about the hydride reducing agents as hydride transfer agents. We're going to transfer a hydride to our aldehyde or to our ketone here. To refresh your memory about what a hydride is, that would be hydrogen with two electrons, which give it a negative one formal charge. Let's go ahead and show the transfer of a hydride from sodium borohydride to our carbonyl. Remember our carbonyl is partially negative and partially positive. So this carbon, right here, wants electrons. It can get some electrons from right here. These two electrons are going to attack this carbon, pushing these electrons off onto the oxygens. Let's go ahead and show what that's going to form. We're going to now have our carbon bonded to a hydrogen, and then over here on the left, we'd have our oxygen now with three lone pairs of electrons. a negative one to formal charge and an R and an H. Let's follow those electrons. The electrons in blue, right up here, are going to form this bond right here. Then this would be this hydrogen right here. You transferred a hydride, a hydrogen and these two electrons to our carbonyl. Now the reason we need someone like sodium borohydride is because you couldn't use something like sodium hydride by itself, because hydride by itself is not not a great nucleophile, it's a good base. The orbital is too small to interact well with the neo carbon here. That's why we need hydride transfer agent, something like sodium borohydride. Once we've added on our hydride, we can then protonate this. You'll see different ways to do this in textbooks. One way would be to in a second step just add protons and protonate your alkoxide to form your final product. That can be done in the work-up or sometimes you'll see textbooks just list sodium borohydride with either an alcohol, like methanol or ethanol or water, and you could protonate your alkoxide that way too. Just depending on how you work-up this reaction. The important thing is the formation of your alcohol and by transferring a hydride. A hydrogen right here, and then two electrons, two with the carbon there. Your reducing your carbonyl. If I go back over here to the left... Let's look at this aldehyde here or a ketone. I have a carbon with two bonds to oxygen and by adding a hydride, now the carbon has only one bond to oxygen. If you sign some oxidation states you will see that this carbon here in red has been reduced. The hydride reducing agents reduce the carbonyl to form an alcohol. Let's look at an example using sodium borohydride. Over here we have a ketone and we have sodium borohydride in methanol here. For our final product, we could go ahead and draw our ring is untouched, so we put that in here. Then just think about adding a hydride. A hydrogen and two electrons to our carbonyl carbons. Let's go ahead and draw this out here. We're going to add hydrogen and two electrons to our carbonyl carbon, we're going to form an alkoxide, and then we're going to protonate the alkoxide in the work-up to form an alcohol. Once again let's go ahead and show the addition of that hydride. This would be this hydrogen and these two electrons add on to form your alcohol. Here we're starting with the ketone and we're ending with a secondary alcohol. The carbon bonded to the OH, is bonded to two other carbons. That's formation of a secondary alcohol, reduction of a ketone to form a secondary alcohol. Another hydride reducing agent is lithium aluminum hydrides. Let's look at this reaction here. We have lithium aluminum hydride. Let's draw up the structure for this one. I would have aluminum, this time with four bonds to hydrogen like that. Which would give aluminum a negative one formal charge. Lithium plus one ion would be there like that. Once again this is a hydride transfer agent. We're going to thing about transferring a hydride from lithium aluminum hydride to the carbonyl. Let's try to be consistent here with colors. These two electrons and this hydrogen are going to come along and attack here and push these electrons off onto the oxygen. This is definitely an oversimplified mechanism. But once again, it's just the easiest to think about. Now, we would have our alkoxide over here to the left, let me go ahead in draw in that. Negative one formal charge on our oxygen. And over here on the right, we've added a hydride. We started with a hydrogen originally on our aldehydes, let me go ahead and show those. The hydride that we added, I'm going to say this right here, these electrons. And then we started with hydrogen originally with our aldehydes, I'm going to say this hydrogen right here. In the next step we need to protonate our alkoxide. In the second step you could add water or dilute acid or something like that to protonate your alkoxide. It grabs a proton from here, pushes these electrons off, and then you form your alcohol products. Let me go ahead and draw that in here. We would form, let's see, this as our alcohol, an OH. Let's go ahead and count our carbons here just to make sure we did it right. This is carbon one, two, three, and four carbons to start with. Then we have carbon one, two, three, and four carbons to start with here. Let's go ahead and show the hydride that we added. We added this hydride on to here and then there was already a hydrogen on that carbon starting out from our aldehyde. That's our product. Going from an aldehyde over here on the left and if we reduce our aldehydes then we're going to form a primary alcohol. Let's go ahead and analyze this. This is a primary alcohol, the carbon that's bonded to our OH is bonded to one other carbon, a reduction of an aldehyde using lithium aluminum hydride would give us a primary alcohol as our target. Now for lithium aluminum hydride, you definitely have to show these two different steps. Lithium aluminum hydride reacts violently with water and let's go ahead a show why. If you have lithium aluminum hydride and water together you get a violent and sometimes potentially dangerous reaction. We would have our negatively charged aluminum here like that. If you mix these in at the same time, pretty much what's going to happen is you're going to transfer a hydride. You're going to get this and this. We're going to take this proton off of water and that's going to form hydrogen gas. H2 which could be potentially dangerous. Sodium borohydride is not quite as reactive as lithium aluminum hydride so you can use it in an alcohol. But you can't use something like lithium aluminum hydride, you have to make sure everything is completely dry when you're doing that. Let's talk a little bit more about why lithium aluminum hydride is more reactive than sodium borohydride. Let's look at this next picture here and let's discuss the different reactivates. First let's talk about sodium borohydride. That would be the bond between boron and this hydrogen. Then let's think about lithium aluminum hydride. That would be aluminum bonded to this hydrogen here. Electronegativity differences. Boron has a value of approximately two and aluminum has a value of approximately one point five. If you think about the electrons that we've been discussing for our hydride transfer, these electrons right in here, boron is more electronegative than aluminum. It wants those electrons, in blue, a little bit more. Since aluminum doesn't want those electrons as much, it's value is not as great, therefore it's easier to give them away. It's easier for these electrons to be given away if it's bonded to aluminum. That's what makes it more reactive, aluminum is more willing to transfer the hydride either to water or to a carbonyl. Therefore lithium aluminum hydride is more reactive and it will reduce more functional groups than sodium borohydride will. Let's look at what we have here for our starting material. We have this compound. Let's say we first do a reduction with sodium borohydride. Sodium borohydride is going to react with our aldehydes right here on our molecule. But it's not strong enough to reduce our ester over here, so it's only going to react with this portion of the molecules. Let's go ahead and draw the final product if sodium borohydride is added. We're going to have our ring. It's going to reduce the aldehydes, if I think about the products I can draw OH here. We started with one hydrogen on this carbon now we're going to have two. Let me go ahead and then go highlight those. One of these is the one we already started with, and then one of these is the one that we added. Let's say this one right here in these electrons, this is our hydride transfer like that. We reduced our aldehydes to this alcohol. The ester portion of the molecule remains untouched. Let me go ahead and draw that. The sodium borohydride is selective for this aldehyde here. I'm not strong enough to reduce the ester. Lithium aluminum hydride, however, will reduce both of these functional groups. Let's go ahead and draw the product if that happens here. Let's go ahead and draw our ring, and then we're going to reduce the aldehyde to once again our alcohol. Same way to think about it as before. Let's say this hydrogen was on there to start with, and then we transferred a hydride. We transferred a hydrogen with two electrons like that. Then we're also going to reduce our ester. Let me go ahead and draw that product there as well. We're going to have an OH, and then we're going to have two hydrogens that were added on in this reduction actually. Let me go ahead and show those. This hydrogen right here and these electrons, this hydrogen right here and these electrons. In this case it reduced it, it added on two hydrides. We don't all the time talk about why in this video, but in a later video we'll talk about the mechanism for reducing an ester using lithium aluminum hydride. Then of course in the second step we added a proton source here. Sodium borohydride under normal conditions will reduce aldehydes and ketones. Lithium aluminum hydrid is much more reactive and it will reduce things like aldehydes, ketones, esters, and carboxylic acids. And again, much more about that in later videos.