Question

Which one of the following compounds will give in the presence of peroxide a product different from that obtained in the absence of peroxide.
A. 1-butene,$HCl$
B. 1-butene, $HBr$
C. 2-butene, $HCl$
D. 2-butene, $HBr$

Answer 1

The addition of halogen halides in alkene gives alkyl halide in general. When the alkene is symmetrical we get only one theoretical product whereas in case of unsymmetrical alkenes two different rules are available, i.e.
Markovnikov’s rule
Antimarkovnikov’s rule

Complete step by step answer:
As per the above question, we have 2-butene which is a symmetrical alkene involved, so we will get only one theoretical product. When $HCl$ reacts with 2-butene as in Option C, the formation of 2-chloro butane is observed. The chemical reaction for the above is given as:

Similarly, in the case of option D i.e. during the addition of $HBr$ in 2- butene, there is also the formation of one product i.e. 2-Bromo butane. The chemical reaction for the same is given as:

Options A and B involve the addition of unsymmetrical alkenes, if the reaction proceeds in the presence of peroxide then antimarkovnikov’s rule applies else in absence of peroxide markovnikov’s rule apply. Option A involves the addition of $HCl$, so antimarkovnikov’s rule is not applicable. So according to markovnikov’s rule the product of the reaction is given as:

In the case of option B, i.e. during the adaddition of $HBr$ in 1-butene, both markovnikov’s and anti markovnikov’s rule apply. When the reaction proceeds are per markovnikov’s, i.e. in absence of peroxide we obtain 2-Bromo butane. The reaction involved is shown below:

And in the presence of peroxide, i.e as per anti markovnikov’s, the product obtained is 1-Bromo butane. The reaction involved is shown below:

Hence we can say in case option B, the same product is obtained in the presence of peroxide as well as in the absence of peroxide.

So, the correct answer is Option B.

Note: Antimarkovnikov’ s rule is also known as the peroxide effect and is applicable only in case of $HBr$ because both the steps are exothermic but in the case of $HCl$ and $HF$ the second step involving the reaction of carbon, radicals are endothermic whereas in the case of $HI$ the first step involving the addition of iodine radicals to an alkene is endothermic.