Now we've got nothing left to do than to just go through the mechanism. And this is the part that's a little bit interesting. So let's go ahead and get started. Something that I told you guys that's very unique about Boron early in the semester, actually in the very first chapter, I talked about the way boron looks. Okay? And something that's unique about boron is that it has an empty p orbital. Since it has that empty p orbital, that's going to make it really good at doing what? Do you guys remember? It's going to be an amazing electron pair acceptor. In fact, BH3 is an extremely strong Lewis acid. Acid meant that I don't donate protons. Lewis means that I accept electrons. This is a really good electron pair acceptor. So when my double bond sees that empty p orbital, it's going to be like hey, I want to give my electrons to that p orbital. Okay? But my boron is going to have 2 different choices. Either it can go down here and basically make a bond to that carbon or it can go down here to the more substituted position and make a bond to that carbon. Okay? And it turns out that the one that it's going to prefer is going to be the one that is the least sterically hindered or the one that is the easiest to approach. So what that means is that my boron is actually going to choose to orient its p orbital right underneath the least substituted carbon.
What that means is that I'm going to get a transition state that looks like this, where basically I've got a partial bond to the b and my BH2 is there. Then I've got my H over here and I'm going to have a partial bond to my h. So what's basically going to happen, let me just show you the mechanism really, really quick, is going to be a cyclization reaction. So my BH3 goes like this. This is my BH3 and I have a double bond here. Right? And that double bond says okay, I'm going to give my electrons to that empty orbital and then this single bond says I'm going to give my electrons to this bond right here. So what that does is it's going to make a transition state that looks like this, where now I have my methyl group there and I'm going to have a partial bond to B, partial bond to H, partial bond. So all of these bonds are being broken and created at the same time. Okay? So that's what my transition state looks like.
Now does that make sense kind of how the double bond donates its electrons to the and then the donates its electrons to the bond that is breaking on the more substituted side. Okay? So that's why I get a transition state that looks like this. Now what you're going to notice is that the BH2 and the H are on the same side of the ring. They're cis. And the reason is because since it's making a ring, a ring can either be on the top or can be on the bottom. But it can't be trans. It can't be like one of them is at the top and one of them is at the bottom. That wouldn't make sense. Okay? So what that means is that that's why we get syn addition with hydroboration because of this 4 membered intermediate.
Now let's go to the oxidation step. So basically what happens after the transition state is that these bonds fully form. So what that means is that this bond fully forms and this bond fully forms, giving me just a single giving me just a single bond to BH2 on one side and a single bond to H on the other. Does that make sense? So basically the transition state just showed when all the bonds were being broken and made at the same time. Now my oxidation state just showed when all the were being broken and made at the same time. Now my oxidation step is going to have the final hydroboration done at the end. Okay? So now what we're going to do is we're going to do the oxidation step. And it turns out that for this step, just like oxymerc, you don't need to know the mechanism. The reason is because the mechanism is really, really long. It goes through what's called a tri triboroester and it's just a very, very long mechanism that professors don't require you to draw the whole thing. Okay? So all you're going to need to know is that you're going to use an oxidizing agent, H2O2 to turn this into an alcohol. Okay? And the base is going to help as well. Alright? So what that means is that at the end, I'm going to get something that looks like this. I'm going to get an alcohol in the least substituted position, and I'm going to get an H that assists to that. And then this is where my methyl group would go. Okay? And if I were to draw this out in an actual planar structure, what you would see is that it's going to look like the one that we had up above, where basically what we have is an o towards the back, an H towards the back, this has to do with the syn addition. Okay? And if those are in the back, that means that my methyl group must be going towards the front and that means that this H must be going towards the front as well. Okay? And if you look at this and if you look at this product up here that I drew, they're the same thing. Okay? So basically, what I was just drawing was the entire mechanism of the top general reaction. Does that make sense? Now keep in mind that this could have happened with any source of boron. It didn't just have to be BH3. The only difference would have been that I just have a different looking group in my intermediate or in my transition state. I'm going to have a slightly different looking group. But the boron is still going to have that p orbital that coordinates with the double bond. Alright? So I hope this mechanism wasn't too confusing, but it is supposed to be a little bit hard. This is one of the trickier mechanisms that we deal with in this chapter. So I hope that you guys didn't get too freaked out by it. Let's go ahead and move on.
