Let's look at the formation of alcohols using a hydride reducing agent. 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 our ketone here. To refresh your memory about what a hydride is, that would be hydrogen with two electrons, which would give it a negative one formal charge. And so let's go ahead and show the transfer of a hydride from sodium borohydride to our carbonyl.
So remember our carbonyl is partially negative and partially positive, so this carbon right here wants electrons, and so it can get some electrons. from right here, right? So these two electrons are going to attack this carbon, pushing these electrons off onto the oxygen.
So let's go ahead and show what that's going to form. Alright, so we're gonna now have our carbon. bonded to a hydrogen, and then over here on the left we would have our oxygen, now with three lone pairs of electrons, so a negative one formal charge, and an R and an H. So let's follow those electrons.
So the electrons in blue right up here are gonna form this bond right here, and then this would be this hydrogen right here. So we've transferred a hydride, right, so a hydrogen and these two electrons to our carbonyl. Now the reason we need something like sodium borohydride is because you couldn't use something like sodium hydride by itself because hydride by itself is not a great nucleophile, it's a good base.
So the orbital is too small to interact well with the carbonyl carbon here, and so that's why we need a hydride transfer agent, something like sodium borohydride. So once we've added on our hydride, we could then protonate it. And you'll see different ways to do this in textbooks. One way would be to, in a second step, just add a protons and protonate your alkoxide to form your final product. And so that could be done in the workup.
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. So just depending on how you work up this reaction. So the important thing is the formation of your alcohol and by transferring a hydride. Right.
So a hydrogen right here and then two electrons, two of the carbon there. And so you're reducing your carbonyl. So if I go back over here to the left, so let's look at this aldehyde here or a ketone.
So I have a carbon with two bonds to oxygen. And by adding a hydride, now the carbon has only one bond to oxygen. So if you assign some oxidation states, you will see that this carbon here in red has been reduced, so the hydride reducing agents reduce the carbonyl to form an alcohol.
Let's look at an example using sodium borohydride. So over here we have a ketone, and we have sodium borohydride in methanol here. And so for our final products, we could go ahead and draw our ring is untouched, so we put that in here.
And then just think about adding a hydride, so a hydrogen and two electrons to our carbonyl carbon. So let's go ahead and draw. and draw this out here. So 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 workup.
to form an alcohol. So once again, let's go ahead and show the addition of that hydride. So this would be this hydrogen and these two electrons add on to form your alcohol.
And so here we're starting with a ketone, and we're ending with a secondary alcohol. So the carbon bonded to the OH is bonded to two other carbons. So that's formation of a secondary alcohol, reduction of a ketone to form a secondary alcohol.
Another hydride reducing agent is lithium aluminum. hydride, so let's look at this reaction here. So we have lithium aluminum hydride. Let's draw up the structure for this one.
So I would have aluminum, this time with four bonds to hydrogen like that, which would give aluminum a negative one formal charge, and so the lithium plus one ion would be there like that. And so once again, this is a hydride transfer agent. We're gonna think about transferring a hydride from lithium aluminum hydride to the carbonyl. And so let's try to be consistent.
here with colors, so I'm just gonna say these two electrons and this hydrogen are gonna come along, so we're going to attack here, and push these electrons off onto the oxygen. And this is definitely an oversimplified... mechanism, but once again, this is the easiest to think about.
So now, we would have our alkoxide over here on the left. So let me go ahead and draw in that negative 1 formal charge on our oxygen. And then over here on the right, we've added a hydride and we started with a hydrogen originally on our aldehydes.
Let me go ahead and show those. Alright, so the hydride that we added, I'm gonna say it's this right here, and these electrons, and then we started with a hydr... hydrogen originally on our aldehydes. I'm gonna say that's this hydrogen right here.
And so in the next step, we need to protonate our alkoxide. And so in the second step, you could add water or dilute acid or something like that to protonate your alkoxide, right? So 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. So we would form, let's see, this as our alcohol, so an OH. So let's go ahead and count our carbons here. just to make sure we did it right.
So this is carbon one, two, three, and four carbons to start with. And then we have carbon one, two, three, and four carbons to start with here. And so let's go ahead and show the hydride that we added. So we added this hydride onto here, and then there was already a hydrogen on that carbon starting out from our aldehyde, and so that's our product. And so going from an aldehyde over here on the left, and if we reduce...
our aldehyde, then we're going to form a primary alcohol. Alright, so 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. So 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 and show why.
So if you have lithium, aluminum, hydride, and water together, you get a violent and sometimes potentially dangerous reaction. So we would have our negatively charged aluminum here like that. And 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. And we're going to take this proton off of water.
And that's going to form hydrogen gas. So H2, which could be potentially dangerous. Sodium borohydride is not quite as reactive as lithium aluminum hydride and so you can use it in an alcohol but you can't use something a lot 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 reactivities. First let's talk about sodium borohydride. Sodium borohydride is So that would be the bond between boron and this hydrogen.
And then let's think about lithium aluminum hydride. So that would be aluminum bonded to this hydrogen here. So electronegativity differences. So boron has a value approximately 2, and aluminum has a value of approximately 1.5. So if you think about the electrons that we've been discussing for our hydride transfer, so these electrons right in here, boron is more electrodes.
negative than aluminum, so it wants those electrons in blue a little bit more. And since aluminum doesn't want those electrons as much, its value is not as great, therefore it's easier to give them away. So 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. And so therefore lithium aluminum hydride is more reactive and it will reduce more functional groups than sodium borohydride. will, and so let's look at what we have here for our starting material.
So we have this compound, and let's say we first do a reduction with sodium borohydride. Sodium borohydride is going to react with our aldehyde, so right here on our molecule, but it's not strong enough to reduce our ester over here, and so it's only going to react with this portion of the molecule. So let's go ahead and draw the final product if sodium borohydride is added.
So we're going to have our ring. it's going to reduce the aldehyde. And so if I think about the product, I could draw my OH here, and we started...
with one hydrogen on this carbon, and now we're going to have two. So let me go ahead and just go ahead and highlight those. I'm saying one of these is the one we already started with, and then one of these is the one that we added.
So let's say this one right here, and these electrons, this was our hydride transfer like that. And so we reduced our aldehyde to this alcohol. And the ester portion of the molecule remains untouched.
So let me go ahead and draw. And so the sodium borohydride is selective for this aldehyde here, so not strong enough to reduce the ester. Lithium aluminum hydride, however, will reduce both of these functional groups.
So let's go ahead and draw the product if that happens here. So let's go ahead and draw our ring. And then we're going to reduce the aldehyde to, once again, our alcohol.
So same way to think about it as before. So let's say this hydrogen was on there. there to start with, and then we transferred a hydride. We transferred hydrogen with two electrons like that.
And then we're also going to reduce our ester. So let me go ahead and draw the product there as well. So we're going to have an OH, and then we're going to have two hydrogens that were added on in this reduction actually.
So let me go ahead and show those. So this hydrogen right here and these electrons, this hydrogen right here and these electrons. So in this case, it reduced it. It added on two hydrides. and so we don't have time to 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.
And then of course in the second step we added a proton source here. So sodium borohydride under normal conditions will reduce aldehydes and ketones. Lithium aluminum hydride is much more reactive and it'll reduce things like aldehydes, ketones, esters, and carboxylic acids. And again, much more about that in later videos.