a. Identify two alkenes that react with HBr to form 1-bromo-1-...

Question

a. Identify two alkenes that react with HBr to form 1-bromo-1-methylcyclohexane without undergoing a carbocation rearrangement.

b. Would both alkenes form the same alkyl halide if DBr were used instead of HBr? (D is an isotope of H, so D⁺ reacts like H⁺.)

Accepted Answer

Hey, everyone in today's video, we're going to solve the following problem. A determine which two alkins would react with hyen bromide to produce one bromo, one methyl cyclopse without undergoing Carbocaine intermediate rearrangement. And part B, if we use deuterium bromite instead of hydrogen bromide, would there be the same alkyl halite products? Now, let's start with part A and we want to form 1 bromo, methyl cyclo pense. So let's think about that product. We need to try a five member ring and we understand that at position one, we have two substations, one of them is a group and another one is a bromo substituent. So we got our product, we want to think about the two Alkins that would produce this product upon hydro halogen. And to do that, we can simply think about the loss of the lele because when we are thinking about hydrogenation, the final step is the nucleic attack by the bromide. And so it makes sense to think in reverse and we are just going to think about the loss of the leaving group and we will analyze the carbo formed. So we need to draw a five membered ring a metal group at position one. And at that same position, we have our positive charge at carbon. So we understand that we have our tertiary carbo intermediate. And we also have to recall that this cardboard iron is formed upon the addition of hydrogen to less substituted carbon. So now we can easily think about it. Basically, we're going to label our carbon that B R C positive charge in blue. OK. Because that carbon must have been double bonded, then there are only two adjacent carbon atoms. Let's label them in green. There is one more which we can label in black. But basically, it's equivalent to the one above it because we have a line of symmetry passing through our ring. So we only have two unique carbon atoms where hygiene could have been added during the protonation stopped. So let's see that let's just remove these hygienes and let's see what sort of alkins we we could get. So for the green pathway, if we essentially remove that hygiene and form a double bond, we get the following our five membered ring, we will add our metal group and our double bond. So that's one methyl cyclo pan, one in. And similarly, if we think about removing our hygiene from the metal group, we will get our double bond at a different position. So let's show the resultant keen and blue in particular, we will have a five membered ring and then we will have a double bond and that structure is methylene cyclopse. So we got our two keys. Now, let's understand once again why they do work for us. It's basically because during the protonation step, when each of the two given kes reacts with hydro brom acid, we are adding hydrogen to less substituted position according to the more cognitive rule to form a more stable tertiary carbo ion. So if you think about the green structure, if we add hydrogen to a less substituted position, we're getting our tertiary carbo. And similarly, if we think about methylene cyclopse, that's exactly the same thing. We are adding hygiene to our CH- 2 group. We are getting our methyl group and a positive charge at a tertiary position. And then the Bromy Line attacks to give us our final product. So we can label our two kings as our final answer. For part A, we can label it as part A and now we are ready to move on to part B. If we use deuterium bromide instead of hydrogen bromide, would there be the same Alky Halite products? Now that we got our Alkins, we can just take a look at the reaction starting with the 1st 11 methyl cyclo one. We're just going to redraw it. We have to remember that the first step is the protonation step. The ethereum is an isotope of hydrogen. So the reaction sequence will be pretty much similar to the one that we would observe with high cham. In the first step, the pipe bomb attacks it, it bonds to deuterium just as to hydrogen at a less substituted position. The more cognitive rule still holds right, because chemical properties of hydrogen and deuterium are pretty much similar. So we are going to draw our intermediate that will be our tertiary carbo carrion. And now the ethereum bonds to the adjacent position. So we have the, when the car car is formed, the bromite and iron at and we get our final product. So we got our final product by simply adding bromine to the carbon atom that has our metal group. So that's how we got our first product. Now, if we think about methylene cyclopean, what about the product formed with it? We have our five member, we have our double bond according to the mechanism. We are going to add the ethereum to a less substituted position. So if we draw the reaction arrows, there will be an attack by the pi on the ethereum, followed by the formation of the bromide and ion. And here we get exactly same tertiary carbo ion just as expected. However, we have to be extra careful because now the siege two group would become CH 2D, right? We are basically adding the ethereum to that Sieh two group. So let's add it. And then the roommate and iron attacks the carbo as we have shown it before and we are ready to draw our final product, we have our five membered ring at position one, we have bromine and then we have our C H two D group, right? So now the question was, do we form the same kill hay lights? The answer is no, basically, if we just look at the structures, they are not the same, right? The original structure, one bromo, one methyl cyclopse is not equivalent to the two structures drawn. And it's basically due to the fact that instead of adding hydrogen, we're adding deuterium, which changes the identity of the product. Indeed, the two products are even different, right? Because we have the uteri bond to two different carbon atoms. While in the previous structure, we only got one product. So for part B, the answer is no, we are getting two different products which are not equivalent to the product in part A. So that'll be it. We have labeled the two products for part B and we can conclude our findings. So for part A, we got two products. One of them, my apologies, we got, we got two Alkins as our starting materials. Those were one methyl cyclo, one in and methylene cyclopse. For part B. We can say that no two different alkyl halite will be formed if we use the uteri bromite instead of hydro bromide. And that would correspond to answer choice A and the multiple choice question. Thank you for watching. And I will see you in the next video.

a. Identify two alkenes that react with HBr to form 1-bromo-1-methylcyclohexane without undergoing a carbocation rearrangement.
b. Would both alkenes form the same alkyl halide if DBr were used instead of HBr? (D is an isotope of H, so D⁺ reacts like H⁺.)

Key Concepts

Alkene Reactivity

Carbocation Stability

Isotope Effects in Reactions

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