Provide an arrow-pushing mechanism, accounting also for the st...

Question

Provide an arrow-pushing mechanism, accounting also for the stereochemical outcome, of the first step (oxymercuration) of the three reactions in Figure 8.63.

Arrow-pushing mechanism for oxymercuration, showing stereochemical outcomes in three reactions with Hg(OAc)2 and NaBH4.

Accepted Answer

Welcome back, everyone show the mechanism and account for the stereo chemistry of the oxy mercury step of the following. The starting material is an alkene. It is 144 tri methyl cyclo he one in and in the first step of the reaction, we're given mercury to acetate and water to produce a mixture of two alcohols. Afterwards, we have our reduction step which allows us to remove mercury from each alcohol and produce the final alcohol, which does not have mercury in its structure anymore. So we want to think about the first step of the reaction and we want to propose a mechanism specifically, we're going to start with the given LK. So let's go ahead and redraw its structure. And now what we want to do is simply draw the structure of mercury to acetate. We're going to add mercury bonded to two acetates. And then let's go ahead and add the long pair on mercury that is present. So now that we have our structure, we are going to consider the mechanism, we know that mercury is a transition metal. So it is electro positives. And when it approaches the pi bond, the pi bond essentially attacks mercury, which attacks back using a Sloan pair to form the mercure iron. Of course, we have the loss of one of the acetate anions or specifically the O A group. Right now. From here, we are simply drawing our meum Ron. And here it is important for us to understand that we have two possibilities for the mercurial iron to form, we can add mercury above the ring or below the ring, right. So what we're going to do is simply remove the muscle group for now. And first of all, let's draw two wedges indicating that mercury is pointing up and then two dashes. In each case, let's add mercury bonded to the aceta in America. We will have a formal positive charge because now it is forming three bonds, the metal group will be pointing down in the first intermediate and it will be pointing up in the second one, right? Because if we have two wedges, the remaining group will be dashed. Well, then, so we have formed our intermediates. And what we want to recall from here is that nucleic water will attack the more substituted carbon because that carbon can form a more stable carbo cion. We locate our more substituted carbon. And now we are essentially showing the nuclear like attack stop, that would be our backside attack and we're going to open that ring. Now, let's go ahead and draw the two possibilities for our next intermediate. So what we're going to do now is just draw the first structure. And essentially, if we have a backside attack, we will have an aversion of configuration. Meaning if there was a, we now the oxygen atom will be on a dash. So simply what we're doing here is just drawing a dashed bond, then we're going to bond the water molecule. But of course, we're going to expand the two hydrogen atoms that are bonded to oxygen. Oxygen gets a formal positive charge. Then the metal group will be pointing up, right? We have an inversion of configuration and then on the adjacent carbon atom, we still have that wedge. Let's add mercury acetate, well done. And now for the second structure, we're going to observe a slightly different intermediate. But of course, we are still performing that backside attack. So what we're going to do here is simply draw a wedge, right? Because we have a backset attack and an inversion of configuration. So now water will be pointing up. And that means the metal group to the inversion of configuration will be pointing down on a dash. Similarly, we are adding mercury on a dashed bond. And from here, we simply want to think about the de protonation stuff for the derotation stuff. We can use the acetate and ion formed or we can use another molecule of water. So let's suppose that we go with another molecule of water, let's add the low pair of electrons onto oxygen. And now let's simply indicate that we are capturing one of the protons. And this is how we are forming the final alcohols. What we're going to do now is simply redraw the two structures. And we are just going to remove that extra proton as well as the formal positive charge on oxygen, right? Because now we are simply forming an alcohol group. Also, there is no necessity to show the oxygen hydrogen bond. We are simply going to add Noh group in each case. Well, then we have our mechanism ready. Let's go back and label everything that we have. So once again, in our mechanism, the very first step was the formation of the mercurial Ron. And as we said, we had two possibilities to form that ion either above the ring or below the ring. And then we were observing the nucleic attack by the water molecule that was a backside attack, meaning we had an aversion of configuration in each case at the electro center. And finally, after the def rotation step with water, we have managed to form the final alcohols. So let's go ahead and label the mechanism in full. Well, then that's our mechanism. Thank you for watching. And I hope to see you in the next video.

Provide an arrow-pushing mechanism, accounting also for the stereochemical outcome, of the first step (oxymercuration) of the three reactions in Figure 8.63.

Key Concepts

Arrow-Pushing Mechanism

Oxymercuration Reaction

Stereochemical Outcome

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