Free-radical chlorination of hexane gives very poor yields of ...

Question

Free-radical chlorination of hexane gives very poor yields of 1-chlorohexane, while cyclohexane can be converted to chlorocyclohexane in good yield.

a. How do you account for this difference?

b. What ratio of reactants (cyclohexane and chlorine) would you use for the synthesis of chlorocyclohexane?

Accepted Answer

Hey, everyone. And welcome back when pense is reacted with chlorine in the presence of light, a poor yield of one Chloro pense is obtained. However, cyclopse can be converted to Chloro cyclopse with good yield by the same reaction. A explain this difference. And B what should be the ratio of reactants to get a good yield of Chloro cyclopse starting with A, we have two reactions to analyze. First of all, we have Penne, that'll be our five member chain. So 12345, right? And it basically reacts with calo in the presence of light. So we are observing a free radical hallucination and recall that in a free radical hallage nation, we are replacing one of our hydrogen atoms with chlorine and the other product would be HCL. So in this case, we're analyzing one Chloro cycle pense. So that'd be the following product. 12345. Let's suppose that this is our position number one. So we're going to add chlorine to it and our byproduct would be hydrochloride. The problem states that the yield of this product is poor and we'll just keep that in mind for now. However, other than that, we are going to analyze the second reaction where we have cyclo pense. So that's our five member ring and we have exactly the same reaction. Let's keep in mind that because we're producing Chloro Cyclopse, this is a mono chlorination reaction, right? So we understand that because the problem suggests that these two reactions have same conditions, they both correspond to mono chlorination. Meaning we're only using one mole of chlorine, which reacts with one mole of the starting material. So we have 1 to 1 reaction here. Now, specifically it says that we're forming Chloro cyclopse. So we are replacing any of these equivalent hydrogens with chlorine. So let's suppose one of the hygienes here, right. So we got this Chloro cyclopse product and the problem suggests that this would be a good yield. So now we are going to explain the difference. Think about the possibilities for a remember that we need to replace one hyen atom with chlorine. How many different hydrogen atoms do we have? So we have 33 equivalent hygienes here, so we can replace one of them. We already did. So however, if you look at the structure, we have position number three and position, position number two and position number three, right, the molecule is symmetrical with respect to that vertical line. So we have three unique positions, these to our reflection of each other and these to our reflection of each other. So we only have three unique positions, meaning we can get other products, we can get substitution at this position. So that would be two L two pense. So basically this product is possible. And then another product is possible. We can have three Chloro pense having substitution at carbon. Number three, notice that whenever you're thinking about the free radical halogen nation, when you form a radical is more stable at a secondary position than at a primary position. So definitely this would correspond to the lowest yield and this these two would be higher in yield because during the mechanism, we are forming our secondary radicals that are more stable than primary radicals. So there are several reasons here. First of all, it's because we are forming a mixture of products and the two other products are higher in yield than the first one. That's why we want Chloro pense is four in yield, right? Because these two are more preferred, they form secondary radicals. And the second reason is because we have a mixture of products. So statistically, you have more possibilities for replacement. Right. Now, here you have three hydrogens versus a total of four, right. So statistically, you are likely to replace any of these hydrogens within positions two and three. Then at position number one, because you have more hygienes to replace statistically. And now, besides statistics, we also said that these two would also form more stable radicals. So that means it's even lower yield, right? For one Chloro pense. OK. So we understand the reason here. But now the question is if we use the same approach for B, why is it good in yield? It's basically because if you think about all of these five positions, they are all secondary carbons that are equivalent to each other, right? You may rotate your structure and you will notice that they are exactly the same. Every position is exactly the same and every position has two hydrogen atoms because we have five equivalent positions. It basically means that this is the only monochorionic product, the only product. So statistically, you only have one possibility for the replacement of that hydrogen. It doesn't matter which position you choose, you will always form Chloro cyclopse. Right. Statistically, we understand that and in addition to that we have our secondary radical form during the mechanism, which is also relatively stable. OK. So this explains why Chloro Cyclo Penan produces a good yield and why one Chloro Pense produces a poor yield basically because there are more products here. And this would form a less stable primary er radical as opposed to these to secondary radicals. While for Cyclo pense, we only have statistically one possible product which is also a would form a secondary radical within the mechanism. OK. Well done. So we understand that we have an explanation to part A this was actually our part a right, we analyzed these two reactions, we understood the reason and now we are going to think about B. Now, in order to increase the yield of this product, we have to recall that the product that you form or basically Chloro cyclo pense can react further when it is produced with another equivalent of chlorine. And then you will get unwanted products. For example, let's illustrate one of them. Let's suppose that you are starting your reaction and you have your five member reacting with chlorine in the presence of light, you're replacing one hygiene with chlorine. And during the course of that reaction, your product can already react with another equivalent of C of chlorine. So basically, we have a competition reaction. So we can replace any of the remaining hygienes including this one. But let's suppose that we are essentially going to include the other hygiene, any other hygiene within the ring, which is not bonded to the same carbon atom as chlorine. I suppose this one, right. So you may notice that you can then get your dilor product in that reaction. So if we want to avoid that, we want to make sure that the concentration of the starting material is really high. Or basically, we want to make sure that the concentration of cyclopse is very high compared to chlorine by the concentration of chlorine. In other words, the ratio of the concentration of cycle pense and the concentration of chlorine must be much greater than one, right? Essentially the or basically speaking, the concentration of cyclopse should be much greater than the concentration of chlorine. And the reason for that is it's basically because if the concentration of your starting material is so high, the whenever you form your product or basically the trace amount of your mono chlorinated product, it will be really small compared to the amount of the starting material you have remaining in your flask, right? And because the concentration will be so big for the starting material compared to the trace amount of the mono chlorinated product. It essentially implies that there will be a higher likelihood for the starting material to react with another molecule of chlorine and minimize that competition between your starting material and the mono chlorinated product. OK. So that's the main reason. And now let's review everything that we have so far. And so if we go back to part A, we can say that the reason why we have our poor yield for the first reaction and a good yield for the second one is because pense gives a mixture of three monoclate products, right? We have our mixture while Cyclo Penan only gives one possible product. So that's the main reason. Now, for B we can say that the concentration of Cyclo pense must be much greater than the concentration of chlorine, meaning the ratio must, the ratio between them must be kept as high as possible, much, much greater than one, right? This would give us a good yield. And based on these answers we may conclude that the correct choice is the multiple choice question is a thank you for watching.

Free-radical chlorination of hexane gives very poor yields of 1-chlorohexane, while cyclohexane can be converted to chlorocyclohexane in good yield.
a. How do you account for this difference?
b. What ratio of reactants (cyclohexane and chlorine) would you use for the synthesis of chlorocyclohexane?

Key Concepts

Free Radical Mechanism

Selectivity in Chlorination

Stoichiometry of Reactants

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