Draw the products of the following reactions, including their ...

Question

Draw the products of the following reactions, including their configurations:

Chemical reaction diagram illustrating hydrogenation and addition reactions with reactants and products labeled.

Accepted Answer

Hey, everyone. And welcome back in today's video, we are going to solve another problem, determine the major products of the reactions given below and include stereo chemistry. If applicable, the starting material is an kem. It's one metal cycle hex one E and we have two reactions reaction. A addition of bromine and the presence of water and reaction B addition of bromine and the presence of methanol. So let's remember that those reactions and specifically reaction A is a halo hydron reaction in which we are starting with our six member ring. So let's try it out and essentially we're forming out bro bridge. We can illustrate that by the attack by the pi electrons on the bromine molecule. So we are going to expand our bromine bromine bond and then bromine that has a lone pair attacks back to form the bromo ion. Now, we are ready to draw the bum formed and we have to understand that there are two possibilities to add bromine either above the ring or below it. So specifically, we're going to draw out our six membered ring in one of the structures. We're going to draw a bromine pointing up on two wedges Rome will have a formal positive charge. And then for the other structure, we just have to understand that we are going to get bromine below the ring, right? Because we can add it from the top piece or the bottom pa. So that means bromine will be pointing down on two dashes. Now, this intermediate form is really important for us because reactions A and B start with addition of bromine. Now for reaction A, we are just going to use nuclear file, which is water. And we have to remember because water is a weak nuclear file. In particular, it doesn't have any charge, it will attack the more substituted carbon in a backside attack manner. So we are attacking our carbon followed by the loss of the living group. So we have two possibilities. We have drawn one of, of the attack steps and then the other attack will take place at a different form of our mon bridge, right where we have drawn to the ashes. In each case, we have our backset attack, meaning we will have an inversion of configuration at the position. So we are going to say that there will be two products after the de protonation stop. So let's indicate that we are dealing with our deep protonation stop. And then we are ready to draw our two products for reaction A. So the first one will have a six member ring. Now we understand that we will have an inversion of configuration. So that water will actually be pointing down because Bromine was pointing up. And specifically, we will have our alcohol group after the derotation step. So we're just going to state that there's an a wage group pointing up and there's a metal group, actually O H will be pointing down, right. So let's fix that. That's the metal group which is pointing up and then the O age group is pointing down. Well, nothing really changes at the adjacent position and bromine is winding up. So we have our N C arrangement between our alcohol group and bromine just as expected within halo hydron reactions. And the other product will essentially have bromine winding down, right? Because it's a living group. So nothing changes regarding configuration at the adjacent position, we have an inversion of configuration. So the alcohol group will be pointing up on a wedge and then the muscle group will be pointing down on a dash. Well done. So we got our two products. And now for reaction B, there's no necessity to think that much about the mechanism. We understand that we are still using bromine with the same starting with cereal. So we will essentially have a very similar scenario in which the bro will now be attacked by methanol instead of water. So that's the whole difference. So we are simply going to change our nu phyle to methanol. Now we are going to add C H 30 H that's our nuclear file. OK. And now in each case, the lone pair and oxygen similarly to what we saw in, in water will attack the more substituted position once again, because it's a neutral, weak nuclear file. And we have, we will have equivalent products. So we can just use our template for the previous reaction to draw them out. And of course, we will need to deproteinate oxygen to get our alcoy group. In this case, that will be metox. So instead of our alcohol group, we will have a Mathy group. In each case, O C H three and O C H three. Well done. So we have obtained our four products, two for reaction A and two for reaction B. We essentially used our halo hyd transformation mechanism in which we understood that each reaction will proceed via the formation of the bro bridge. There are two possibilities for that to happen, either to add it from the top phase or the bottom phase. And then each of them will be attacked by our nu phyle, either water or methanol. After the deep protonation step, we are forming a pair of in an tumors. In each case, in which we have the anti arrangement between our alcohol group and bromine or alcoy group and bromine. We have labeled the four products in green boxes and we are ready to conclude that video. Thank you for watching and I hope to see you in the next One.

Draw the products of the following reactions, including their configurations:

Key Concepts

Halohydrin Formation

Regioselectivity

Stereochemistry

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