When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunla...

Question

When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunlamp is shone on the mixture, three products result.

(a) Give the structures of these three products. 
(b) Propose a mechanism that accounts for the formation of these three products

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve another problem. The reaction of methylene cycle hexen with NBS in presence of sunlight produces a mixture of two products through the structures of these two products and write a mechanism for the reaction. Now let's think about this problem. First of all, we have to understand that the starting material is methylene cycle hexen meaning we are going to draw our six member drink and we are going to attach a methylene group to it. So that would essentially imply in addition of a ch two group, two hour cyclo hexane. According to the problem, we're treating this compound with NBS in the presence of sunlight. We're basically light right? We can essentially right age new for that, which implies that there is light in this reaction and we are producing two products. There will be two unknowns. One of them will be our structure acts and the other unknown would be structure wise. We are just going to add these to present that there will be two products for this reaction. Now, we are going to use this later when we determine the identities of X and Y based on the mechanism. But now let's recall all of the steps needed for this reaction overall. First of all, this is a free radical hallucination. Right? NBS would be a source of roaming atoms for our reaction and we are going to perform a free radical halogen nation. Let's recall the structure of NBS. We are going to draw a five member ring where one of the atoms within the ring is a nitrogen atom. So we can essentially draw that structure as follows. Okay, and now we have to add to carbonell groups, two adjacent carbon atoms with respect to nitrogen. So we are going to add to carbondale groups and then there is bromine attached to nitrogen. Now because NBS usually has a trace amount of hydro bro make acid as impurities. We are going to show how we are producing bromine molecule. So NBS reacts with the impurity of hydrogen bromide and we are going to form our substitution product. In this case we are going to exchange hydrogen with roaming and this is how we are getting our two products and one of them will be used in our free radical hell origination stop. So now we have nitrogen bonded to hydrogen the original carbonell groups and now here is our booming molecule. Now even a small amount of bromine is sufficient to initiate our reaction. And we may recall that the first step in the free radical halogen nation is the initiation stop. So we are now ready to think about our mechanism. So the first step or the initiation stop involves the bromine molecule previously formed. We are going to expand that blooming blooming bond and we may recall that here is where we are using light because light promotes the hom elliptic cleavage of that roaming roaming bond essentially. The harm political leverage implies that we're using these electrons from the booming booming bond and we are uncaring them by forming two bromine radicals with one uncared electron on each roaming. So that gives us two bro mean radicals. Okay, so we see our initiation step clearly and we are ready to move on. We may recall that the next step is the propagation step. So let's write down propagation and in the propagation step, this is where our original structure comes in. Let's draw a methylene cyclo hexane. So that would be the structure that we drew previously at the beginning. And we need to think about the position that will be reactive. First of all, let's remember that we need to make sure that we're using an a hydrogen atom from an sp three hybridized carbon atoms. So there will be 12345. Of course. These two positions are equivalent. These two are equivalent as well and this one is another position where we have an sp three carbon atoms. So we have a line of symmetry. So we have three possible positions for our three A radical halogen nation. And we have to keep in mind that roaming is selective in this reaction. It essentially wants to be added to a position that forms the most stable radical. Right, in other words blooming is added to analytic position because it will be residents stabilized. We have two positions which are equivalent to each other because we have a line of symmetry because luminous added to analytic position. We are going to expand that carbon hydrogen bond and we understand that the broom in radical form than the previous stop essentially a tax. So the electron so the broom in radical attacks. And we are going to form a new hydrogen bomb in bond by breaking this carbon hydrogen bond and a new pairing the two electrons meaning the remaining electron would be placed onto carbon. So this is how we are getting our next intermediates. You may notice that one of the products is H P. R. So that be hydrogen bromide formed because we are pairing these two electrons in the bond and now we are forming a radical from our substrate. So we are ready to draw that uncared electron within methylene cycle Hexen notice how we are forming an electron at this position. So this is our newly formed radical. And it is really important to understand that it is resonant stabilized because it's a little right. As we previously said, this is the reason why romaine prefers to attack that position because it's a lilic and its resonance stabilized. You may easily notice how to draw the residence form by breaking this pi bond. And essentially we are going to unfair these electrons from here. So the first electron comes in to create a pi bond at this position and the other electron comes from the pi bond. So we are creating a pipe on at a different position and the remaining electron from that same pi bon is placed at the top carbon over here. So we are forming another radical in or basically just drawing the residence form of the previous one. Right. So now we can essentially add an era where we're presenting the residents form. And let's draw the result on the form. So, first of all, we would have a single bond at the top because we have broken the pi bond. Right? And this is where our radical is formed. And now the position of the double bond. The first we have a double bond formed over here. Okay, so these are our residence forms and we understand that the formation of the two products correspond to these two radical intermediates. So let's think about the next propagation stop, let's suppose that the first structure would be our structure number one. Right. This is our first radical form. And the second structure will be our structure number two. We are going to use a red color and a green color to indicate how these radicals react further. So starting with the Red one. Number 1, this radical, let's redraw it. It's still the propagation step. Right. We are not done with the reaction. So the radical formed for number one would react with another molecule of bro means we're going to draw that romain romain bond. And here we are going to show how the reaction takes place where essentially have an attack. Once again, we're pairing an electron from carbon and one of the original bonding electrons from booming and this allows us to form that carbon roaming bond and also a roaming radical over here. Right, we are going to unfair one of the electrons and place it on roaming. So here we are going to show our product formed. Let's see that. So the product form here would be Our six member ring with a double bond at the Great Top. Let's make it a bit clear to make sure that our drawing is consistent and I'll notice how we are creating this carbon bromine bonds at this position. Right? So that would be blooming attached to carbon. And then the other product is the blooming radical. So this is how we get our first product. Alright, and now let's think about the second product in green. So number two, if we redraw that radical we have our double bond and now once again another molecule of romaine comes in, we can expand that roaming roaming bond, we have an attack. One of the electrons from the roaming roaming bond comes in to create that carbon roaming bond and the other electron forms the booming radical. This is how we get our next product. So now notice that we are creating ch to be our group. Let's start with our ring. So that would be our six member ring, our double bond. And now here we have carbon with two high jin's bonded ce brahmins. So that's our second product and the other product is another bro mean radical. Now because we have formed our two products, we are ready to analyze the termination. Stop, right. We are done with propagation termination essentially implies that you have multiple possibilities to combine two radicals into a neutral molecule. You may choose any of them, you can choose to booming radicals combining into a booming molecule. You can choose this radical combining with that one. Apparently there were a bulky but it's still theoretically possible. Right? You can combine booming radical with the first radical in and you can combine booming radical with the green radical. So as you can see there are multiple visibilities for that to occur. For simplicity, we are not restricted. So we can just say that the reaction is terminated when she blooming radicals combined throw me in a radical combines with another booming radical to form a neutral roaming molecule. Of course the radicals are also neutral but they contain unpaid electrons. Right, so we are just forming a more stable molecule with no Owen paired electrons. This is our termination stuff. Most importantly we have to understand that these are our two products discussed in the problem. And now that we clearly see those products, we are going to draw them in the reaction scheme. We have analyzed our mechanism just as required by the problem and we are just going to include our final products for this reaction at the very top for our X and Y. Labels. So looking at the word top, the first structure X corresponds to. So if we redraw it, that'll be our double bond, our six member ring. And then we are essentially adding blooming at this position. That's our X. The first unknown that we had and now thinking about the other one, we have to remember that we obtained a six member ring, then we had a double one at a different position and then ch two bromo substitution. We may erase X and Y. Right, Because we have determined our unknowns this those would be our two products. We can just say this first one plus the second one, and we may label them once again as our final products for this reaction. Just as the problem stated, we indeed have two products. We have seen the mechanism from start to finish and we are ready to conclude that the correct answer to this multiple choice question is C. However, if you have any questions, don't hesitate to leave a comment down below in the comment section and thank you for watching

When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunlamp is shone on the mixture, three products result. (a) Give the structures of these three products. (b) Propose a mechanism that accounts for the formation of these three products

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