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: 2,3-dimethylbut-2-ene

NBS, hv —>

+

a. Give a mechanism for this reaction, showing how the two products arise as a consequence of the resonance-stabilized intermediate.

Accepted Answer

Hey, everyone. And welcome back in today's video, we are going to solve another problem. The reaction of 12 dimethyl cycle heine with N BS under ultraviolet light produces a mixture of two products. The reaction is given below, draw a mechanism for this reaction that accounts for the formation of these products. Now, first of all, let's keep in mind that because this reaction involves an N BS in the presence of light that implies a free radical Hallen nation. And in particular, bro, because N BS is a source of bromine and therefore, the mechanism that we are going to discuss here will correspond to free radical, bro. This will be the basis for our thought process. Let's begin before we look at the mechanism. Let's just understand how bro is formed for that purpose. We're going to draw N BS that would be our five member ring with nitrogen bonded to carbon chain. We have our carbon groups adjacent to nitrogen and then imagine is bone, bro. Now, usually N BS has a trace amount of hydrogen bromide impurities. So we can have that hydrogen bromine exchange, right? And we are going to form bromine that way. So let's draw the result on products our five membered ring now nitrogen contains hydrogen. We ha we still have these carbonyl groups. And for the other one bromine bonds to bromine to form Romine molecule, which can now participate in our first step, which is called the initiation step. Remember that whenever we are starting a free radical allegation mechanism, the first step is the initiation step in the initiation stop. We are observing that Bromine Bromine Holi bond cleavage, which essentially implies that we are going to apply ultraviolet light. In this case, we can label this as H new. That's a general representation of light which is an energy of the photon. And we are going to illustrate the homo bond cleavage. The bromine bromine electrons are essentially unpaired. One electron from the original pair is placed onto each bromine to form two bromine radicals. So we are getting two bromine radicals, each bromine radical as the name implies, contains an unpaired electron. Now that we have initiated our reaction because we have formed our radical. We are ready to think about the next step with the, which is the propagation step in the propagation step. This is where the actual starting material comes in and we are ready to redraw it. So that would be our six number ring, our double bond sea metal groups. Now, before we begin, we have to recall that bromine is going to replace one of the hydrogen atoms with an SP three hybridized carbon atom. But because we have plenty of them, we have to keep in mind that bromine is very selective in this reaction because bromine is selective, it essentially wants to replace the hydrogen at a position that would be the, that would correspond to a more stable or the most stable radical. And such position is generally an ay position. Whenever we are considering double bonds for bromine, it always prefers that allylic position. So the ay position is essentially the position for the carbon atom that is adjacent to the double bond. Notice how this compound has a horizontal line of symmetry. And due to that line of symmetry, the two parts be above and below the line are equivalent to each other, meaning it just sufficient to look at the top position. And we noticed that there are two carbon atoms that are a lilic. One of them is over here in blue and another one is to the left in blue. There are two aly positions because these two carbon atoms are bonded to a double bonded carbon atom. And these two are equivalent to the top ones. Now, the question is which of the two aly positions are we going to choose? And the answer is the secondary one, this is a primary position. So it would form a less stable primary resonance stabilized radical. But the second one would form a secondary resonance stabilized radical. So that would be more favorable for us in terms of stability because bromine selects the most stable possible radical to react with. So that means we are going to expand this carbon hydrogen bond because this is the hydrogen that we are going to remove. And now the booming radical formed in the initiation stuff comes in attacks. And this is where we are forming our hyen bromine bond. Because carbon hydrogen bond is cleaved, we're using one of the two electrons here to bond hydrogen to bromine. And the remaining electron is placed onto adjacent carbon atoms to form a new radical. So let's look at the products formed here. The first product that we're seeing is hydrogen bromide. And the second one is our newly formed radical from our starting material. So we can control that radical. We may notice that we still have these metal groups, our double bond. And now there is an unpaired electron over here at this position for our carbon atom. Now, when we form this radical, we have to think about resonance stabilization because as we said, this is the el elect position, there's an adjacent pi bond. So notice how we can essentially pair this electron with one of the electrons from the pi bond to form a pyon at a different position. But that means the remaining electron from this, this pyon would be placed onto that adjacent carbon atom. Me, this is how we are getting another form of that radical because it's resonant stabilized. So let's draw resonance arrows for this one, we are going to include our re arrow. And now the result on radical, we may simply draw it in a different color, let's suppose green. This result on radical now has a double bond at disposition, right? And now two metal groups. And we noticed that there there is an unpaired electron at this position, right, we can essentially circle it. Now, there are two radicals formed, we may label them them as number one in red and number two in green. And now each of these two radicals can participate in the propagation step. And this is how we get a mixture of products. So let's start with the red one. Let's label it as number one for clarity, let's redraw it. This is our unpaired electron pi bond and our metal groups. And now in this reaction, we are going to use a molecule of bromine because radicals in the propagation staff react with molecules, the radical essentially attacks. We are un pairing these bromine bromine electrons, one of them is used with this electron to form carbon bromine bond and then the other electron forms a new bromine radical. So we can now to our products, the first product corresponds to one of our products in the original reaction actually. So let's draw it out. We notice that we are adding bromine at this position, our double bond, our two metal groups. And of course, we are forming, bro mean a radical. Now, if we think about the green one, we can label it as number two, let's draw our six member ring, our double bond, two metal groups and an unpaired electron. So now, once again, a molecule of bromine comes in and reacts with that radical. So we are breaking that bromine bromine bond, there's another homo lithic cleavage and we are forming our carbon bromine bond followed by the formation of a new bromine radical. Let's indicate our products. The first product will correspond to one of the two products formed in the main reaction. So we can draw it out. That would be our double bond our metal group. And now at the adjacent position, we have a methyl group and a bromo substituent and the other one is mean a radical, right, as we saw before. So these are indeed our final products. Because if you saw the reaction scheme, it contained these two products, meaning our mechanism is so far consistent with the reaction, we simply want to finish this reaction because we got our final products and to finish it, we need to include the termination step. Step. Number three is termination for the termination. Stop. Our goal is to combine two radicals into a molecule that has no unpaired electrons to make it a a more stable the structure. So we can pair these two radicals, for example, or two bromine radicals or bromine with any of these two radicals to get a more stable structure. For simplicity, let's just think that the two bromine radicals here react to form a stable bromine molecule that will be easy for us to draw. And because there's no actual requirement to draw anything else, we can draw one of the probably four possibilities, right. So that's, that will be easy for us to terminate the reaction. So one bromine radical reacts with another bromine radical to form bromine diatomic bromine. So we can just add an arrow and illustrate that BR two diatomic bromine is formed after the termination that we understand that we are done, we got our products, we saw all of our steps for this mechanism. And we may simply conclude that based on this mechanism, the correct answer choice, this multiple choice question. Yes. D however, if you have any questions, don't hesitate to leave a 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: 2,3-dimethylbut-2-ene NBS, hv —> + a. Give a mechanism for this reaction, showing how the two products arise as a ­consequence of the resonance-stabilized intermediate.

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The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products: 2,3-dimethylbut-2-ene
NBS, hv —>
+
a. Give a mechanism for this reaction, showing how the two products arise as a consequence of the resonance-stabilized intermediate.