Radical chlorination, a reaction studied in Chapters 5 and 11,...

Question

Radical chlorination, a reaction studied in Chapters 5 and 11, replaces a hydrogen with a chlorine. Assuming there is no regioselectivity (all hydrogens can react equally), how many different chloroalkanes are possible upon reaction of each alkane with one equivalent of Cl₂
(a)
(b)
(c)
(d)

(a)

(b)

(c) Line drawing of a zigzag hydrocarbon chain, representing a generic alkane structure for chemical analysis.

(d) Structural formula of a chloroalkane showing a carbon chain with a chlorine substituent.

Accepted Answer

Welcome back, everyone. Chlorine is used in radical chlorination to replace hydrogen in an L cane. If all hydrogens respond identically and there is no reach of selectivity. How many distinct chloral alkanes can be produced when each alkane reacts with one equivalent of chlorine, we're given four structures ABC and D and four answer choices. Let's think about this problem. And let's recall that in chlorination, we want to, first of all understand how many non equivalent hydrogens we have in each structure. So for part A, we have pens, it's a five member rank. And to identify those non equi equivalent hydrogens, we first of all have to locate planes of symmetry. We understand that we have one of them, which is a vertical plane of symmetry and we can have the next one diagonally, which also tells us that one side of the ring is equivalent to the other side and so on, right. So essentially, we can have multiple of them passing through each of the five carbon atoms. OK. So let's draw them for simplicity. Instead of drawing them, we can just realize that there is a free rotation of that rank in each of our carbon atoms is secondary, right? We only have one type of carbons. So that's a secondary carbon and all of them are equivalent because we can have that pre rotation. So we have only one type of non equivalent hydrogens, right. So let's try that we only have a one type. In this case, if we have have one type of non equivalent hydrogens, because all of those hydrogens are the same, they're all bonded to the same type of carbons. It tells us that there's only one type of hydrogen to be removed. Therefore, we can have one possible product and we can eliminate the answer choices C and D, right? Because A can only have one product moving on to B let's think about the given structure. And let's label types of carbons. We have our primary carbons. There are three metal groups bonded to a tertiary carbon, right? So essentially, it immediately tells to us that there are two types. How do we know that? Well, essentially we have a central tertiary position, there's one hydrogen available to us. And then we have three metal groups. They are all equivalent because they're bonded to the same central carbon. Therefore, we can have two products here, which is consistent with the remaining answer. Choices B can have two products. Now moving on to C let's draw pense and let's immediately identify the vertical plane of symmetry passing through carbon number three. Now along that plane of symmetry we first will have two primary carbon atoms than two secondary carbon atoms. And the central one is an additional second re carbonatum. So if we count the number of pairs, we have one pair of primary carbons, one pair of secondary carbons and one secondary carbon that is unique. So we have three types which gives us three possible products. OK. So far it's still option A or B for our answer. So we just want to determine the number of products for D let's draw the given structure and let's understand any possible symmetry. And immediately we can tell that there is a diagonal plane of symmetry. So let's label non equivalent carbons. First of all, it's really important for us to notice that there are four equivalent metal groups because they're bonded to the same carbon atom. Right? Specifically, we notice that there is a tertiary carbon atom that has two equivalent metal groups. There are primary, the reason why they are equivalent is because they're exactly the same metal groups bonded to the same carbon atom. And the reason why we have four of them being all equivalent to each other is due to the diagonal plane of symmetry. So the four metal groups will be equivalent, giving us one type of carbon and therefore one type of hydrogen to be replaced with chlorine. And in addition to that, we have an additional pair of tertiary carbon atoms that are equivalent based on the diagonal. So Now, we have two types and two possible products. Now, if we look at the answer choices, this gives us option B as the right answer, right? Because D is four for option A and we have just said that there are actually two types. Therefore, the correct answer to this problem would correspond to choice B A gives us one product B two, C, three and D two. That would be it for today. And thank you for watching.

Radical chlorination, a reaction studied in Chapters 5 and 11, replaces a hydrogen with a chlorine. Assuming there is no regioselectivity (all hydrogens can react equally), how many different chloroalkanes are possible upon reaction of each alkane with one equivalent of Cl₂
(a)
(b)
(c)
(d)

Key Concepts

Radical Chlorination

Regioselectivity

Chloroalkanes

Watch next

Radical chlorination, a reaction studied in Chapters 5 and 11, replaces a hydrogen with a chlorine. Assuming there is no regioselectivity (all hydrogens can react equally), how many different chloroalkanes are possible upon reaction of each alkane with one equivalent of Cl₂(a) (b) (c) (d)

Key Concepts

Radical Chlorination

Regioselectivity

Chloroalkanes

Watch next

Radical chlorination, a reaction studied in Chapters 5 and 11, replaces a hydrogen with a chlorine. Assuming there is no regioselectivity (all hydrogens can react equally), how many different chloroalkanes are possible upon reaction of each alkane with one equivalent of Cl₂
(a)
(b)
(c)
(d)