When methylenecyclohexane is treated with a low concentration ...

Question

When methylenecyclohexane is treated with a low concentration of bromine under irradiation by a sunlamp, two substitution products are formed.

a. Propose structures for these two products. (b) Propose a mechanism to account for their formation.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle the following problem. Three metal butte one in gives to substitution products when treated with a low concentration of romaine under irradiation by a sun lamp proposed a mechanism to show the formation of these two products. As the problem suggests the starting material is three metal butte one in. So we are ready to do a four member chain with the double bond position number one and a metal group at position number three. And according to the problem, we're treating that Elkin with roaming in the presence of light. According to the problem, we are getting two products X and Y and we don't know their identities. However, we will propose a mechanism and we will understand what X and Y are now because we're using bro mean in the presence of life, we may recall that this is the free radical rumination. Recall that the first step of the free radical abomination mechanism is the initiation step. During the initiation step, the molecule of romaine is treated with light and we are observing the hom elliptic cleavage of the romaine romaine bond so that these electrons are UNP aired and we are producing two bromine radicals. So we may say that in the initiation stuff we are producing to most of romaine radicals. Now we may continue our mechanism with the propagation step and in the propagation step, this is where our starting material becomes important. We may simply draw it out once again and we understand that we want to find these hydrants that could be removed by the blooming radical. Recall that blooming this highly selective and generally it wants to attack the most substituted position in particular. We may find out which hae jin's could be removed and one of them is just right here, which would result in the most substituted product, right? Or the first radical that we are forming and it would be sufficiently stable for us. We want to make sure that the radical is performing is really stable And removing this hygiene with producers with a tertiary radical that is resonance stabilized. That's why we're choosing this hydrogen now the blooming radical form than the previous stop will attack so that we will have the harm elliptic cleavage of this carbon hydrogen bond in particular blooming would attack this electron from the carbon hydrogen bond would be used in the formation of the hydrogen booming bond and the remaining electron would be placed over here to form a new radical. So we may now draw the result on radical. We May one of all notice that there's HPR form plus that residents stabilized radical. So let's throw it out. So we have an unpaid election over here and we may notice that this radical is resonance stabilized just as we discussed before because if you look at these electrons in particular their pi electrons nearby. So we can essentially appeared and we can say that this electron might come here while one of the pi electrons which I'm here to form a new pi bonds and the remaining electron would be placed over here. So this tells us that the radical is resonance stabilized and we will draw the result in residence form. So now we have the position of the double bonded carbon number two. And then here we have our own paired electron. So these two are our residence forms. So we may include brackets to indicate the presence of resonance. And now this is where we are starting to think about the final propagation stop having formed these two radicals, we are going to react them with another molecule of romaine. So if you look at the first possibility, the bro mean the bromine molecule would be attacked by that radical and again, we would observe the harm a little cleavage right here. Right. Because one of the electrons would be used in the formation of the carbon bromine bonds and the other electron would be placed on roaming to form another radical of roaming. And similarly, we would observe a similar scenario for the second resident structure. If we think about the roaming roaming molecule, the radical attacks were forming a new carbon roaming bond and the remaining electron is used in the formation of a new radical. That's how we can now get our two products and the propagation stuff. So if we draw them, the first product would have blooming at this position and the second one would have blooming at this position where the um paired electron was. Now we have got our two products, we just have to keep in mind that there must be a termination stop. So let's include it immediately so that we don't forget that. Stop. And number three is our termination stop where we just want to terminate that reaction in determination step. You have at least three possibilities to combine any radicals that you see either to bromine radicals or two radicals of the intermediate that you have formed from your starting material or just roaming with your intermediate. But let's just make it simple and say suppose that we are taking to bromine radicals and now we're essentially forming that's starting blooming molecule. Right. We may say that determination step is done because we have combined two radicals to form a stable molecule. And now thinking back we may see that we have produced two products. The question is which one of them is the major one in which is the minor one. Recall that Elkins that are more substituted are more stable. And this elkin is tri substituted while this one is mono substituted, meaning this would be the major product since it's more substituted and therefore more stable. While this would be the minor one. And we can say this is our X in the equation and this is our why of course the order doesn't matter right, We just have two unknowns. We may simply say that these two are our final products. In addition to that, we clearly see our complete mechanism for this reaction. And we may conclude that the correct answer choice to this question is B. However, if you have any questions, don't have States, leave your comment down below in the comment section and thank you for watching.

When methylenecyclohexane is treated with a low concentration of bromine under irradiation by a sunlamp, two substitution products are formed.

a. Propose structures for these two products. (b) Propose a mechanism to account for their formation.

Key Concepts

Electrophilic Substitution Reactions

Radical Mechanism

Product Distribution

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