The light-initiated reaction of 2,3-dimethylbut-2-ene with N-brom...

Question

The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products: b. The bromination of cyclohexene using NBS gives only one major product, as shown on the previous page. Explain why there is no second product from an allylic shift.

Accepted Answer

Hey everyone and welcome back. The reaction of one comment to the Method Cycle vaccine with NBS under ultraviolet light produces a mixture of two products. However, the reaction of cycle pension with NBS only produces one product despite the generation of a resonance stabilized radical, explain why this reaction does not yield a second product. Now, one of the ideas to answer this problem is to think about the free radical mechanism of NBS in the presence of ultraviolet light. So what we can see here, we can just draw the structure of NBS. That would be our five member ring where we have nitrogen, Adam bonded T brimming. We are going to draw that shortly. So that's our rain nitrogen is bonded to booming. And we also have these carbonell groups. So you can think of this as a my group pride. So to Carbondale groups have been added and now because there's generally an impurity of H B R N B S can easily react with H B R T, produce roaming even a small amount or a trace amount of grooming is sufficient to initiate the reaction. So we are good, we can now drool the remaining product before we proceed. Okay. So now roaming that has been generated can initiate the reaction. So as we know, we have our initiation step where we have that harm elliptic cleavage of that booming booming bond, right? We need to use light and these two electrons are unfair to give us two roaming radicals. Now the initiation stop has been performed, right? We got our two blooming radicals and now we can think of the chain propagation stop or basically propagation in the propagation step. This is where cyclops unseen becomes important and we control cyclops unseen, we can add a double bond in this position. And what we have to keep in mind because we have a reaction with N B S, we are only interested in a lilic positions. So because this is where our double bond is the adjacent to carbon atoms here and here they are considered to be a little carbons and we are only going to look at the hydrogen atoms bonded to them. But if you look at the line of symmetry, these two Olympic positions are same, right? You can just visualize there's this diagonal and with respect to that diagonal, the two sides of the ring are symmetrical. So we are only going to look at one of them supposed to expand this hydrogen for our mechanism. What happens is that one of the bromine radicals from the initiation stop this bro mean radical attacks, right. So we have an attack. And the electrons here are unfair to form a new hydrogen booming bond while the remaining electron from the original bond is placed onto carbon. And that's how we form our radical here. So let's draw it. We have our double bond and the position of the unpaid electron is adjacent it. So we have our radical just here. Now, we have formed our radical and yes, it undergoes residence for sure because there's an adjacent double bond. And if we show the formation of a new radical due to resonance, we can simply say that these this electron can essentially be transferred over here, breaking the pi bond, one of the electrons will be transferred over here to form a new pi bonds. We are pairing these two electrons while the remaining electron from the pi bond will be transferred onto the adjacent carbon atom. And this is how we will get our resonance structure. So now let's draw it out. And now you may notice that you have a double bond, a disposition, right? And the electron is over here. But now if you look at these two residents forms, aren't they equivalent? Suppose that you rotated right? You will get exactly the same or specifically, you may just flip it if you rotate it slightly and then flip it, you will get exactly the same structure, the ian paired electron and the position of the double bond there exactly same relative to each other. In each structure. So you may now notice that if you think about the reaction of each radical with the molecule of grooming, you will get two equivalent products. So let's see how that looks like if we take the first one for complete clarity, right? Instead of saying that they're exactly the same, we will just draw cheap possible products. So the molecule of roaming comes in the radical attacks, we are unfair ng these electrons and this is how we get our first product. Let's draw it a bit more neatly. Okay. So that's our first product. And now if you think about this radical instead, once again, the radical attacks, we are unp airing these electrons breaking the bond to produce a new radical and we are getting another product. Now blooming is attached over here and we have our double one just add this position. And now if you look at these two structures aren't the same, you can simply rotate this structure. So if we just rotated a bit, we will see the following. Let's suppose that we rotate that rotate. Now suppose that your rotated clockwise and then you will have grooming over here. You want to align these blooming atoms and you will have the double bond at this position. So they look very similar. But now imagine that the last transformation you perform here is just flipping it with respect to that diagonal. If you flip it, you will keep roaming at the same position, but the double bond will appear here. So that implies these two structures are exactly the same. And this is the answer why we do not have two different products. The answer is because cycle up ensign psycho pension, my apologies because we're looking at cycle a benzene has two equivalent Elik positions has to equivalent were same, right, same a la leche physicians because there's that line of symmetry, as we said before. Of course, we went above and beyond. And we looked at all the possibilities including the mechanism. But it's just sufficient to understand that if you want to have a shortcut because it has that line of symmetry and the two Olympic positions are equivalent to each other. That means there would only be one possible product, right. And therefore we may conclude that the correct answer choice to this multiple choice question is the the two Olympic radicals form from cyclops benzene are equivalent. And that would be it if you have any questions, don't hesitate to leave your comment down below in the comment section. And thank you for watching.

The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products: b. The bromination of cyclohexene using NBS gives only one major product, as shown on the previous page. Explain why there is no second product from an allylic shift.

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