Predict the major products of the following reactions, and giv...

Question

Predict the major products of the following reactions, and give the structures of any intermediates. Include stereochemistry where appropriate.

(p)

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve this problem which reads for the following reaction for the structure of the major product through the structures of intermediates, if there are any if applicable, include the cereal chemistry of the final product. If we examine that reaction in particular, we are starting with an elk. In the first reaction is the addition of mercury, mercury to acid eight, water and ethyl ether, followed by the addition of sodium boron, hydride and sodium hydroxide. So if we combine these two together, we may now recall that such a reaction is called oxy mur curation. The mercure ation in particular. The first step is our oxy Merck Eurasian steppe. And our second step is the dimmer curation steps. So we will try to draw the intermediates from reaction number one and the final products from reaction number two. If we start by drawing the owl king, we may remember the basic steps of the mechanism specifically. We have mercury to acetate. There's a lone pair on mercury and there are two acetate groups. In this step, we will notice that we are examining an electrolytic addition across the double bond. So the double bond attacks mercury, which expects to the presence of that lone pair, we will be forming a bridge. And simultaneously we have the loss of the leaving group, One of the two acid eights will act as a leaving group. So now we are forming our intermediate and now there are two ways to add mercury either above the ring or below the ring. Right? So we may indicate that by the presence of dashes and wedges, which will also change the way that this method group is pointing. So now if we are signing two wedges, there will be mercury with a formal positive charge because now it forms three bonds, there are still one at ST group left. This method group would start pointing down and nothing changes at this position. Another way is to add mercury below the bridge. So then our second structure would correspond to the following T natchez Mercury with a former positive charge and the estate car bonnet said Now the method group would be pointing out in this case we are nearly done, but now we need to add a water molecule. So now we wish to really recall one of the most important parts. The water molecule will attack a more substituted carbon atom. And if you compare the two carbon atoms, one and two of these positions, this is a more substituted one. So we will perform an attack here in particular because it's able to stabilize a positive charge better. So the water molecule will perform a backside attack right? We will have an inversion of configuration. So this water molecule would attack the carbon atom, we will open the ring and we will have this fragment as our leaving group. And if we look at the second pathway, there's exactly the same idea a backside attack at a more substituted carbon atom. Now we will draw our two intermediates dinner immediate from the first one. Because we had a backside attack, we will have an inversion of configuration. Nothing changes a disposition. There will still still be mercury bonded on a wedge. However, at this position we have inversion of configuration. So the metal group would start pointing up and the water molecule which later on gets deeper motivated by acid eight. So we may indicate that as well minus H plus the deportation step that would give us our final and remediate the water molecule would now be pointing down. But in this case that's a wage already because we have the protein ated it. So that's our first intermediate. We may label it because the problem clearly asks us to show intermediate and our second intermediate would be formed a very similar way. We are going to indicate the product and also deported nated. We may start by drawing the ring. Now again, nothing changes to this position, mary Carey would be pointing down similarly here, because this center does not participate in the reaction. And because what are performs a backside attack. The muscle group would be pointing down and the alcohol group would be pointing up. Now notice that it's important to understand that the relationship between the alcohol group and mercury, They're basically in anti arrangement. Right? In particular. Uh My apologies. Um the alcohol group is pointing up and Mercury is pointing down. Right, so they're in anti arrangement. And similarly here, the alcohol group is pointing down and mercury is pointing up. So that's the anti arrangement. These are our two intermediates. And we are nearly done. So we have our intermediate for the first reaction, the oximeter creation step. And we are pretty much ready to finalize our problem. In the next step, we are adding sodium world hide right in sodium hydroxide. So we may indicate that in here we have sodium borrow high tried and sodium hydroxide. Exactly same reaction goes here. And if we perform that reaction, we call that what it does, it essentially replaces your mercury with an additional hydrogen atom. Right in particular, we will have the cleavage of the carbon mercury bond and we will be adding a hydrogen atom. A disposition. And same goes here. So even though we have three different serial centers, let's label them, we will only end up with two total after the cleavage of the carbon mercury bond. All that we have to do is just add a hydrogen atom here. After removing mercury. This would be our two final products. We are ready to draw them. So, if we are adding a hydrogen atom here, since we already had one, there will be no stereo chemistry. And here we get our ring, the methyl group pointing away and the alcohol group pointing up. We may now label these as our final products. You may also notice that they are, they asked Siri immerse. That's because we have an inversion of configuration at one remaining Cairo center, right, and the other one is not even affected. So we are done with this problem. We see our two intermediates and two major products. They are formed in roughly equal quantities because we have a backside attack and two possibilities for that, and therefore the correct answer to this multiple choice question is D However, if you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching.

Predict the major products of the following reactions, and give the structures of any intermediates. Include stereochemistry where appropriate.
(p)

Key Concepts

Hydration of Alkenes

Reduction Reactions

Stereochemistry in Organic Reactions

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