Sodium triacetoxyborohydride, NaBH(OAc)3, is a mild
reducing age...

Question

Sodium triacetoxyborohydride, NaBH(OAc)3, is a mild reducing agent that reduces aldehydes much more quickly than ketones. It can be used to reduce aldehydes in the presence of ketones, such as in the following reaction: <
of reaction> (b) Propose a mechanism for the reduction of an aldehyde by sodium triacetoxyborohydride.

of reaction> (b) Propose a mechanism for the reduction of an aldehyde by sodium triacetoxyborohydride.

Accepted Answer

All right. Hi everyone. So for this question, it says that all the heights are reduced more rapidly than ketones by a mild reducing agent, sodium tries to Toxie borrow hydride or NABHOAC three. Even in the presence of the ketone, sodium tries to toxic borough hydride preferentially reduces Aldo hides. As shown in the following reaction. So here we have in our starting material on the left side we have a molecule that has both a ketone and all the hide functional groups. So when you react this with our sodium tries to toxic borrow hydride with acetic acid as a solvent, we end up with a final product that has the same ketone group. But instead of that, all the hide we started off with we have a primary alcohol instead. So our task is to propose the reaction mechanism underlying this selective reduction. So let's get started. Um I'm going to start off by actually drawing out our structure of our sodium. Try Toxie borrow hydrate. So to do that, I have a boron in the center with our three acid Toxie groups. Which reminds me a lot of acetic acid. So these are our three acid acid toxic groups and our hydride is simply a hydrogen atom connected to boron. So this is actually going to result in a negative charge for boron which is counterbalanced by that sodium that we have there. So now let's start with our actual mechanism. And in order to do that, I'm going to redraw our starting material at the bottom here. So this is our Aldo hide and this is our ketone. So now what's going to happen is that boron hates having a negative charge. So what's going to happen is that this hydride group is going to go ahead and attack that all the hide carbonnel because it's going to take those electrons connecting it to boron with itself. So boron is going to end up neutral in this case. So what's going to happen here is that we end up with the same hydrogen over here. But instead of a carbonnel, a C double bond, Oh we have an alc oxide instead because we have an o minus but the ketone stays the exact same. And now we have that hydride from our sodium tries to toxic borrow hydrate connected to that previously carbon no carbon and sodium is actually going to be a counter and ion counter ion. Sorry for our Alcock side instead because now it's the oxygen that has a negative charge, not boron. So now let's move on. Right? And in order to move on, I'm gonna scroll down to give myself a bit of space and now let's continue because we need a fully protein ated alcohol at the end of this reaction. But we have an all minus in our product. So in order to fix that, I'd like to remind you that we have acetic acid in solution. So we have acetic acid in solution which can fix our problem that we have for our Alcock side and I am So what's going to happen is that our own minus is going to take that hydrogen from acetic acid and therefore deep rotated. So we're going to end up with I should have drawn the arrow red. So let me just do that. There we go. And now we're done with the reaction. Now we've protein ated our alcohol. So let me just draw our now predominated alcohol are ketone is going to stay the exact same. And I'm going to actually redraw this to show that our hydrogen comes from the depot donation of acetic acid. So connected to this oxygen um is our carbon that has our hydrogen from our starting material. And another hydrogen from our sodium tries to toxic borough hydra. So that is it for the mechanism um This is our final product. It is a primary alcohol that we got by reducing an alga hide with sodium tristate, oxy borough hydrate. And that ketone was left completely alone. As you can see. Thanks for watching. And I hope this helped you

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