Leah here from leah4sci.com and in this video we'll look at carbonyl reduction using sodium borohydride. NABH4 is one of the two most common reducing agents you'll see with carbonyl compounds. The second one, lithium aluminum hydride will be discussed in the next video which you can find along with this entire Redox series, practice quiz and cheat sheet on my website leah4sci.com slash redox. Sodium borohydride has a sodium ion, a positive counter spectator ion that does not react and an anion of BH4.
If we write it out, the key here is the hydride ion which is an H with a lone pair of electrons and a negative charge. When you see the ending IDE, think negative. Chlorine as a negative is chloride, fluorine is fluoride and hydrogen is hydride.
It's a tiny atom with a negative charge making it a very strong base. The problem here is that with a carbonyl reduction, we're looking for a nucleophilic attack rather than a basic attack. So we need a carrier molecule that will slow it down a little and control the reaction. That's where having the BH3 comes in. Boron as an exception to the octet rule prefers to have just three attached electrons meaning a total of six and three bonds.
But in this molecule, we see boron with four attached valence electrons, a complete octet and therefore a formal charge of negative one. The negative charge is on boron but think of it as simply having a neutral BH3 attached to a reactive hydride and this will help you quickly make sense of the mechanism. If this idea is confusing, go back to my nucleophile versus base video which is linked in the description below.
When it comes to carbonyl compounds, sodium borohydride is a weak reducing agent. This makes it slower and safer but also limits what it will react with. When sodium borohydride reacts with a ketone, the product is going to be a secondary alcohol. The shortcut here is to simply break the pi bond so you have one bond left between carbon and oxygen. Add a hydrogen to the carbon atom, this is your reduction.
and also add a hydrogen to the oxygen atom so that it's not left negative. This gives us an extra hydrogen, extra hydrogen for a secondary alcohol. If an aldehyde reacts with sodium borohydride, we use the same trick but the product in this case will be a primary alcohol. We break the pi bond leaving a single bond between carbon and oxygen.
We add in a hydrogen atom and also add a hydrogen to oxygen so it's not left negative. This means the carbon has a single bond to oxygen, a blue hydrogen that we had initially from the aldehyde, a second hydrogen that was added for reduction and the hydrogen on the oxygen atom giving us a primary alcohol. Given that sodium borohydride only reacts with aldehydes and ketones, we can use it for selective reduction when the starting molecule has multiple carbonyls and we don't want to reduce them all. In this example, The starting molecule has two carbonyls.
The first is an aldehyde, the second is an ester. The sodium borohydride will only reduce the aldehyde but will not touch the ester. So we break the pi bond, add in those hydrogen atoms, redraw the skeleton for the product without the pi bond and then simply add in the hydrogen atoms that we want to show for that partial reduction.
You can also draw this product without showing any of the hydrogens. Because in line structure you don't expect to see all the hydrogen atoms. Before we look at the mechanism, let's review the resonance of a carbonyl to understand what makes this reaction happen. There's a pi bond between the carbonyl, carbon and oxygen. This pi bond can resonate up to the more electronegative oxygen atom to give me a resonance structure that has the two initial electrons and oxygen and the third pair which used to be a pi bond.
This gives oxygen. too many electrons and a formal charge of negative one but also leaves carbon with an incomplete octet and a formal charge of positive one. This separation of charge makes it unstable and unhappy so we're more likely to see the neutral form but the fact that this resonance can happen means that we can expect to see some version of the resonance hybrid with a partial negative on oxygen and a partial positive on carbon.
That partial positive is what makes carbonyl reactions happen. For this mechanism, we'll start with a simple ketone, acetone, which is simply propanone. The sodium ion is a spectator ion dissociated and dissolved in solution so we're only looking at the BH4-. Even though boron carries the negative charge, it's the electrons between boron and hydrogen that are most reactive. Boron doesn't want it, it wants to kick it away and the only way those electrons will leave is if they have something attracting them somewhere else.
In this case, that something else is a partially positive carbonyl carbon. which gives us the first step in our mechanism. The electrons between boron and hydrogen will break away from boron taking the hydrogen atom, the entire hydride with it to attack the carbonyl carbon.
This puts too many bonds in carbon so we kick out the pi bond pushing the electrons up onto the oxygen atom. To redraw the skeleton we have oxygen with two initial lone pairs of electrons, a third lone pair from the collapsing pi bond and a negative charge. And a single hydrogen atom attached to the former carbonyl carbon where the electrons forming that bond are the electrons that were previously bound to boron.
Realize that when the electrons break away, we separate it out. So we get BH3 wandering off into solution and hydride attached to the carbon. But now we're stuck with the negative oxygen, we want a neutral product.
For the final step, we'll use the solution itself to react. This reaction typically takes place at the same time. takes place in a polar protic solvent like alcohol or water.
If we show our solvent to be methanol, we have a CH3OH where the oxygen is partially negative pulling electrons away from hydrogen making that hydrogen atom partially positive. The negative oxygen on our molecule will reach for that partially positive hydrogen atom collapsing those electrons onto the oxygen. We'll show these electrons now as a bond between oxygen and hydrogen.
and the hydrogen it grabbed from the solvent for a final product that is a secondary alcohol. If you want to keep tabs on everything else, don't forget we have a positive sodium spectator that is now free to associate with the OCH3-for a sodium ethoxide side product. Be sure to join me in the next video where we look at lithium aluminum hydride and all the carbonyls that it can reduce. You can find this entire video series along with my redox practice quiz and cheat sheet. by visiting my website leah4sci.com slash redox.