Draw the major products obtained from the reaction of one equi...

Question

Draw the major products obtained from the reaction of one equivalent of HBr with the following compounds. For each reaction, indicate the kinetic product and the thermodynamic product.

b. Chemical structure illustrating the reaction of HBr with a conjugated system, showing major products and their classifications.

Accepted Answer

Hey, everyone and welcome back, draw the products obtained in the reaction of the following compound with one equivalent of hyd hyen bromide and label them as the kinetic and the thermodynamic product do not include stereo isomers. Now, the structure that we're given is 32 L id in cyclopse and to predict the products, we essentially once think about the mechanism of that reaction. So first of all, we're going to think about one comma two and one comma four edition, right? Because we are using a dying, it has two double bonds and specifically, it's a conjugated dying. So that's why we're thinking about one comma two and one comma four addition, let's start with a terminal double bond and let's think about one com edition. So basically, we will have our protonation step, right. So according to the Marconi of hydrogen will be added to the less substituted carbon. And we're going to say that we are forming the following intermediate where we have our five member ring, we are going to draw one of our double bonds. And now because we're breaking the other eye bond, we said that according to the Marconi of, we want to form a more stable secondary carbo in this case, which of course can undergo resonance. So let's go ahead and immediately draw its resonance structure. And to do that because it's a conjugated dye, we're going to shift our pi electrons from the outer pi bond, right. And in that case, we are going to form a positive charge at a tertiary position. And now that we have two carbo cars formed, we are going to show the nucleic attack stop the bromite and ion will attack the first one. And then the second one. Now if you think about our carbon atoms, number one, number two, and then for the other one, that would be number one, number two, number three, number four, we can clearly see that the first product that we get and actually we can easily get it by adding our carbon bro bond, right? Because we have an attack. So let's draw our carbon roaming bond. Let's not worry about stereo chemistry just as indicated. So we got our one comma two product and now thinking about the one comma four product, right? Because we have our one comma four edition adding hygiene to position one and bromine to position four, we end up with the following product, we can simply draw our carbon roaming bond. There would be no stereo isomers form because the carbon form specifically the position that we're analyzing is not chiral, right. The ring is symmetrical with respect to the vertical line and that would be it. We got our two products. The first one is our one comma two edition and the second one is our one comma four edition, right, based on the positions labeled before. Now, we simply want to determine which one is the kinetic product and which one is the thermodynamic product. Let's remember that kinetic product is the one that forms foster and foster means that there must be an effect which allows us to explain why that reaction takes place faster. So when we think about the structure, we notice that the terminal double bond is basically closer to the molecule that it reacts with, right, the internal double bond is basically harder to approach. So one comma two addition here would correspond to the kinetic product and that's because of the proximity effect. OK. So we're going to use that term proximity effect. OK. So when we form our carbo ion, the bromite anion is much closer to the carbo. So that due to that proximity effect, we have our kinetic products, we can state that the one comma two product is our kinetic product. OK. And now thinking about the thermodynamic product, we have to remember that it corresponds to you the more stable product. OK? It might form lower, not necessarily, right. But the most important part is that it's more stable or the most stable product we can get. Now thinking about the products we have formed let's notice that the kinetic product, one comma two edition is a tetra substituted alkene. OK. So that's tetra substituted. And for the one comma four edition, we have 12, we out of substitutions, right? So that's try substituted. OK. Let's remember that alkin stability depends on how substituted Alkins are. The more substituted they are the more stable they get. So tetro substituted al would be more stable. And surprisingly, it tells us that not only the one comma two edition product is our kinetic product but also our thermodynamic. OK. Well done. So we got our two products, we are going to label them as our final answer. And in addition to that, we are going to conclude that the one comma two product is both kinetic and thermodynamic product of this reaction. Thank you for watching and I hope to see you in the next video.

Draw the major products obtained from the reaction of one equivalent of HBr with the following compounds. For each reaction, indicate the kinetic product and the thermodynamic product.
b.

Key Concepts

Hydrohalogenation

Kinetic vs. Thermodynamic Products

Carbocation Stability

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