# Reactivity of NBS

The given correct answer is (A). According to me there would be 2 enantiomeric pairs.

My doubt is that it does not react with NBS but it does react with Cl/hv though both have the same initiation step.

My guess is that this may be because Br is present in very small amount i.e. as an impurity so this may have something to do with it.

• Look here for the final product. It doesn't have active hydrogen (allylic or benzylic) to react with NBS. C and D are also correct. But it would produce more than one enantiomeric pairs when reacted with $\ce{Cl2}/h\nu$ and therefore A is the correct answer. May 16 '21 at 7:21

Initially, I wasn't going to give an answer to this question but suggested the choice A is the correct answer in my comment. However, since Random Quark hasn't give a satisfactory answer (yet I appreciate the explanation), it prompt me to give my thoughts as an answer.

As I pointed out in my comment to the question, the reaction given by OP is an example of Mozingo Reduction:

Thus, the final product is methylcyclopentane, which does not contain any acidic hydrogen because there are no carbonyl or double bond on the structure. Also, although the starting compound, (2S)-2-methylcyclopentanone is an optically active compound, it would loose its chirality due to reduction (gaining plane of symmetry).

1. As shown in the above diagram (insert in the bottom box), methylcyclopentane will get chlorinated in the presence of $$\ce{Cl2}/h\nu$$ at 1, 2, and 3-positions, and on the methyl group as well (only 2- and 3-chloro products are shown because they are the compounds in concern). It is also possible to get di-chlorinated, tri-chlorinated, and etc. based on the conditions used. Thus, as shown in the scheme above there would be more than one enantiomer pair by this reaction. Therefore, the statement A is incorrect.
2. As pointed out above, the final product does not contain any acidic hydrogen because there are no carbonyl or double bond on the structure. Therefore it cannot react with the presence of NBS with or without initiator. Therefore, the statement B is correct (and methylcyclopentane is optically inactive as well).
3. Two optically active alkenes (they probably mean by (3R)-3-methylcyclopentene and (3S)-3-methylcyclopentene) would give the final product (methylcyclopentane) when reduced by $$\ce{H2}$$ $$(\pu{1 mol})$$. Therefore, the statement C is also correct.
4. Finally, the reaction with $$\ce{H2/Raney Ni}$$ is a redox reaction. Thus, the statement D is correct as well.

Hence, the only incorrect statement is A and consequently, it is the correct choice as given.

Firstly, about (A), you are correct that two enantiomeric pairs are formed. You have misread the question, it says the INCORRECT option.

Then coming to why Cl2/hv reacts but NBS doesn't give its usual bromine substitution. As you already know, these reactions are free radical reactions that need a free radical initiator like light, peroxide, heat, etc. Cl2/hv has the presence of sunlight that is able to do the homolytic cleavage of the Cl-Cl bond.

This is Mozingo's Reduction, and product formed would be methylcyclopentane. The reason $$\ce{Cl}$$ free radical reacts with this compound, but NBS doesn't is because $$\ce{Br}$$ free radical is formed in low concentration in NBS, so it only reacts to form stable free radical, namely, allylic and benzylic free radicals only. Also, you are right, $$2$$ enantiomeric pairs would be formed, the question is asking for $$\mathbf {incorrect}$$ statements.