How many monochlorination products can be obtained from the ra...

Question

How many monochlorination products can be obtained from the radical chlorination of methylcyclohexane? Disregard stereoisomers.

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next question. How many products are formed from the radical chlorination of methyl cyclops in the parentheses, it says ignore stereo isomers A four B six C eight or D 12. So let's think about the reaction that's taking place. Here. We start with methyl cyclops. So we have a five carbon ring that's our cyclops and it has a single methyl group on it. And our reaction is radical chlorination. So we have light or heat activated formation of a radical halogen. So we have a, a di Halide molecule, light or heat breaks it apart and we have this radical and that starts a chain reaction where in the end, a chlorine atom will replace one of the hydrogen atoms of an alkane. Now, if we were asked about the major product, we wanted to know in what proportion these products would exist that would depend on the stability of the radical intermediate that's formed. When our reaction starts, our di Halide will be broken apart, generating a halogen radical. That radical will pull off a hydrogen from our alk leaving behind a free radical leaving a single electron on one of the carbons. So depending on what type of carbon that is, it will be more or less stable. So a tertiary carbon radical will be more stable than secondary, more stable than primary, more stable than a meth radical. So if we were asked what the major product is, we would look for the most substituted carbon and have the chlorine there. But we're not asked for the major product, we're asked how many products. So that means we want to see, want to see all possible products. So they'll all be generated just in different amounts. So for all possible products, we want to see how many different types of hydrogens do we have on our molecule. So how many different types of carbons do we have is our answer. Well, we have this terminal carbon on the methyl group. Then I've put a blue dot there. So we have three hydrogens on that. And then we have a red dot on the two carbons that are adjacent to the carbon with our methyl group, which I actually forgot here. Apologize about that. So I'll put a green dot on the carbon that has the methyl group attached. So that's a tertiary carbon. So we've got our blue methyl group carbon that's primary our green tertiary carbon that has the methyl group substituent, then red for the two carbons adjacent to that green carbon. They are, you know adjacent to the carbon with a methyl group. And we have one more different type of carbon. If we go a little farther away from our methyl group, I'll note that in a purple dot Actually, that looks a little dark. Let's do it with a yellow dot So a nice bright yellow dot for our fourth type of carbon, which is two carbons away from our carbon with a methyl substituent. So since we have four types of carbons that we could remove a hydrogen from, we will have four possible pro possible products. So we'll start with our blue carbon. Let's put on our metal group and at a chlorine on to that methyl group carbon, we'll highlight that in light blue. So it doesn't cover it up showing that that was that blue carbon, that's one type of product. And then our green carbon, our tertiary carbon, we can put a chlorine on there. So there's our molecule and then the same carbon that has a methyl group there, we add a chlorine on that one. So we're going to go ahead and highlight that in green. So that's product number two. Now, those are singular carbons. There's only one carbon like that on our molecule. But as we go around our ring, our ring being symmetrical, we're going to have the two red carbons. So we're placing a hydrogen in either red position will result in the same molecule. So let's add our chlorine onto one of those adjacent carbons. And we'll highlight that in red to show that, that matches where the red carbon was. And then finally, our fourth type of carbon which we labeled in yellow. So if we attach the chlorine two carbons away from the methyl group, and again, both yellow carbons will give the same molecule. So our question was how many products are formed? The answer is we form four different products. So that would be choice a four. You can see that if we wanted to, uh if, if we had listed a product for every single carbon, we'd end up with two more products, we'd end up with six. But that would mean that we ignore the fact that four of our carbons are, are in two symmetrical pairs and then Joyce D 12. Well, there are 12 total hydrogen atoms in this molecule, two on four of the carbons in our ring, one on our tertiary carbon and then three on the methyl carbon. But again, for instance, all the hydrogen on the methyl carbon, replacing any one of them will give the same molecule given that we were told to ignore stereo isomers. Well, in the case of metal, that would all be symmetrical. But uh that's why choice 12 is wrong because replacing many of these hydrogens are give the same product if you replace one of them. Now, if you labeled one of them as a deuterium or something that would be different. But that's why choice is wrong. And choice C eight also wrong because it is too many hydrogens, too many products, ignoring some of our identical molecules that are generated. So our answer here is the smallest possible number choice, a four possible products based on the four different types of carbons. And thus the four different types of hydrogens that could be replaced by a chlorine. So again, we have the product where the chlorine is added to our methyl carbon, the chlorine is added to our tertiary carbon that has the methyl group substituent as well where the chlorine is added to the carbon adjacent to the methyl group and where the chlorine is added two carbons away from the carbon with the methyl group substituent. See you in the next video.

How many monochlorination products can be obtained from the radical chlorination of methylcyclohexane? Disregard stereoisomers.

Key Concepts

Radical Chlorination

Monochlorination Products

Methylcyclohexane Structure

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