Predict the major products of the following reactions.
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Question

Predict the major products of the following reactions.

b. trans-hex-3-ene + peroxyacetic acid (CH₃CO₃H) in water

c. 1-methylcyclohexene + MMPP in ethanol

Accepted Answer

Hey everyone and welcome back to this video. We are going to tackle the following problem, examine the following reactions and predicts the major products here. We have two reactions A and B. And both of them involve Elkins. This is a trans AL king and this one is a sizzle king. Right? Because the two superstitions within the trans version are on opposite sides of the double bond and for the cis one they're on the same side of the double bond. Now these Elkins that'll go to different reactions. Reaction number one for A tells us that we are adding parasitic asset. And because there is a C. 03 age group, we immediately recall that this is a proxy asset, meaning whenever Al Kane's react with proxy assets, they produce ip oxides. So we will produce any box. And in the first step we may just label that that will form any box side followed by water addition of water will essentially imply that we will have the ring opening of any box hide. So that would be the hydro Eliza's reaction right now for B. If we look at our AL king that says this falcon and we're using mm PP. That's magnesium mona proxy tallied And in particular it's a weaker acid than a parasitic asset. But essentially it forms an epoxy as well but we will not have any ring opening. We're essentially not going to perform hide releases so that we may just stick with our epoxy it and that'll be our final product. So now that we have performed our analysis, let's see how that looks like if we start with reaction A and if we withdraw our al king, we understand that in the first stop, we are forming an ip oxide. Right? We are going to use parasitic acid to form an ip oxide. Parasitic acid would be CH three C 03 H. So we may use that shorthand notation for these three oxygen atoms. And we may simply say that if we think about drawing that structure, oxygen will be added across a double bond. So we're breaking the pi bond and we're just adding oxygen to these two carbon atoms in the ring. Right? So that's how we keep oxide. And we also understand that the two method groups are pointing in opposite directions. One of them is pointing up. The other one is pointing down over here, right, what else we have? We have hydrogen atom with this position. So that would be pointing down and there's another hydrogen over here pointing out now, essentially we will need to protein eight that oxygen there is a source of protons, right? We have a parasitic assets so there is a source of acidic protons and after we protein it that this would be our good leaving group. So the water molecule could actually attack any of the two positions because they're equally substituted. But recall that if hydra laissus is taking place in acidic conditions were generally attacking a more substituted position. But because they're equal a substitute, that we may just choose to attack one of the two positions. So let's suppose here and that'll be our backside attack. Right? We also need to make sure that this part is protein ated to make sure that it's a good leaving group. This is taking this reaction is taking place in acidic conditions once again. Right. So when the water molecule attacks any of the two positions, because they're equally substituted, we may essentially have the loss of the leaving group in the next step. And we will form one of our products. The other product will be formed during the attack of the other position, which will just be, it's in Aunt Emma. Right. So we may just draw one of our products and let's do that. Our product that we get here would look like this. We may start by drawing the central carbon carbon bond during the loss of the leaving group. Nothing changes regarding the configuration over here. So we may just say that the wages pointing up on a solid line, there's a hydrogen atom pointing up and a muscle group pointing down. However, here you may wish to recall that this is a backside attack, meaning a wage is being added from the bottom side and it should be anti to this alcohol group. So the other alcohol group will be over here and now regarding these bonds. We are going to see muscle pointing up, hydrogen pointing down. So this would be our one comma two dial. Right? This is one of our products that we get. Plus it's in Antrim are just as we said, there might be an executive position as well. Right? Because they're equally substituted. So we are just going to swap our priorities in particular. We may just swap two hydrants. We may just add hydrogen here on a dash here on a watch and just change these two methyl groups to the opposite once so that we get. It's an anti however, it's just sufficient to show one of the products that's perfectly fine and just say the other product is it's an anti that's fine. So we may essentially label this is our final product now for B. If we look at B as we previously said, mm PP magnesium monopoly proxy Tallit forms an epoxy without any ring opening because it's a weaker asset. We don't have any acidic hydraulics. Is the solvent here is try chloral methane. So we're just stopping at the dip oxide formation part. And we may simply say that if we start with an alkaline, there will be double bond in addition of mm P. P in try chloral methane essentially just forms that peop oxide. So we will form our ip oxide and that would be it. Right. That's our final product. We're not performing any ring opening here because there are no there's no acidic hydrolyzed this, right? And we may say that now looking at parts A and B. our intermediate in two cases was exactly the same. We formed an ip oxide in a However, we needed to open the ring of ip oxide because we used water. So we had acidic hydrolyzed sis. Right during the protein nation stop, we formed a good leaving group and there was a backside attack by the water molecule, which formed one comma two, dial and it's in and hammer. However, look at reaction be mm P P is a weaker asset and there's no water present that would help us perform that ring opening so we get our ip oxide as the final product instead of it being an intermediate. That's why these two are our final products. And we may now conclude that the correct choice to this multiple choice question is A However, if you have any questions, don't have States, leave a comment down below in the comment section and thank you for watching

Predict the major products of the following reactions.
b. trans-hex-3-ene + peroxyacetic acid (CH₃CO₃H) in water
c. 1-methylcyclohexene + MMPP in ethanol

Key Concepts

Electrophilic Addition Reactions

Peroxyacetic Acid as an Oxidizing Agent

MMPP and its Role in Organic Synthesis

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