Using 1,2-dimethylcyclohexene as your starting material, show how...

Question

Using 1,2-dimethylcyclohexene as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.) If a chiral product is shown, assume that it is part of a racemic mixture.
a.

a.

Accepted Answer

Welcome back everyone into this video. We are going to solve the following problem. Show how you can synthesize three bromo two metal butte one in using two metal butte one in as a starting material. If the product contains a stereo center, assume that the final product is part of a Euro systemic mixture. We may begin our problem by drawing our reagents and products in particular. What we have here is that our starting material is to metal butte one in. So we may draw a four member chain. B one in essentially implies that we have a double bond at position number one and position number two contains a metal group. So that's two metal but one in And according to the problem, we wish to produce a product named three bromo two metal butte one in. Again, we are going to draw a butte one in or member chain with a double bond. At position number one, Position number two still contains that metal group and now position number three Hasbro ming. If you compare the two structures, you may notice that the main difference between them is that position number three, which had two hydrogen atoms now contains hydrogen and blooming. So we are simply replacing one of the hydrogen atoms with broom in. Right? This should tell us a lot because first of all that we are adding that blooming atom or simply speaking, we're replacing one of the hydrogen atoms at the olympic position. In other words, this position is a little bit because it's adjacent to the double bond. Right? So because this is a little hell origination, we simply have to recall that the set of regions that were using that NBS in the presence of light. And let's see why that works for us. In particular. If we start with our structure the world first stop is that the booming radical that has been formed and the initiation stop attacks the Elik hydrogen in particular. We have two hydrogen atoms. This carbon has a little position number one. So that's position number one and that's position number two. Right. And they both have hydrogen atoms. So this is what we have so far roaming radical attacks and we are forming the HBR bond followed by the formation of that radical. Right at position number one. So our first possibility is to form this radical over here with an unpaid electron. However, if roaming attacks the other position, we may draw it over here, we are forming the HPR bond and then the position of the radical will be different. It's now at position number two. And we may draw that radical as follows. So that would be the original metal group. That double bond, the empire, the electron is now over here. Right. And these are our t possible radicals. They are also residents stabilized. But if you notice it doesn't really matter whether we have a double bond at this position or that position. Since they are equivalent. Now this is where these radicals react with the booming molecule. And if we think about the first reaction, right? The radical attacks, we are forming our carbon bromine bonds and another radical of grooming while here we performed the same step at a different position. So the radical attacks, we're forming the carbon roaming bond and we are losing the brahmin radical into our mixture. And now if we look at these two reactions, let's label them as reaction number one and reaction number two. Okay, so if we look at reaction number one, the product for this reaction would have rooming attached to this position. So that would be CH two B. R. And now for the 2nd 1 we would attach grooming at the position where we wanted to go. Right? So we are going to draw our structure. And now instead of having that compared electron, we are forming the carbon booming bond. So we actually using this set of regions which as you recall, is used whenever we have an olympic rumination and that's what we're doing here. We want to make sure that we retain the position of the double bond. We simply want to replace one of the hydrogen atoms at the olympic position with grooming. And for these purposes we're using NBS in the presence of heat or light. So we may simply say that, okay, this is our correct answer. And this is our set of regents, right? But it's very important to keep in mind that we are forming a mixture of products to products are formed product number one and product number two. So the product that we desire to obtain is just one of the two products that we're actually forming in this reaction. However, for the purposes of this problem, we are only interested in that product, so we are done, and we may state that the correct answer to this multiple choice question is C. However, if you have any questions, please leave your comment down below. Thank you.

Watch next