Question

Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Answer 1

Hint: It is found that addition of HBr to propene yields 2-bromopropane in accordance with Markovnikov’s rule. Whereas, in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane in accordance with anti-Markovnikov’s rule.

Complete Step by step solution:
- Firstly let’s discuss about formation of 2-bromopropane:
Addition of HBr to propene yields. This reaction is in accordance with Markovnikov’s rule. We can see that it is basically an ionic electrophilic addition reaction in which the electrophile $({{H}^{+}})$ adds to form a more stable ${{2}^{\circ }}$ carbocation. And in the second reaction, it is found that the nucleophile $\left( B{{r}^{-}} \right)$ fastly attacks carbocation to form 2-bromopropane. We can see the mechanism:
\[H-Br\rightleftarrows {{H}^{+}}+B{{r}^{-}}\]
\[C{{H}_{3}}C{{H}^{+}}C{{H}_{3}}+B{{r}^{-}}\to C{{H}_{3}}CHBrC{{H}_{3}}\]
- Let's discuss about formation of 1-bromopropane:
In the presence of benzoyl peroxide, it is found that the same reaction yields 1-bromopropane. This reaction is in accordance with anti-Markovnikov’s rule. We can see that this reaction involves a free radical mechanism. It is found that, as a result of the action of benzoyl peroxide on HBr or we can say by the addition of HBr to propene in the presence of benzoyl peroxide, Br-free radical is obtained. We can see the reaction:
\[2{{C}_{6}}{{H}_{5}}CO{{O}^{\bullet }}\to 2{{C}_{6}}^{\bullet }{{H}_{5}}+2C{{O}_{2}}\]
\[{{C}_{6}}^{\bullet }{{H}_{5}}+H-Br\to {{C}_{6}}{{H}_{6}}+B{{r}^{\bullet }}\]
The Br-free radical further adds to propene to form the more stable ${{2}^{\circ }}$free radical:
 \[C{{H}_{3}}CH=C{{H}_{2}}+Br\xrightarrow{slow}C{{H}_{3}}C{{H}^{\bullet }}C{{H}_{2}}Br\]
Now, free radical that is obtained rapidly abstracts a hydrogen atom from HBr to form 1-bromopropane:
\[C{{H}_{3}}C{{H}^{\bullet }}C{{H}_{2}}Br+HBr\xrightarrow{fast}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br+B{{r}^{\bullet }}\]

Note: - We can see here that the main difference in Markovnikov’s rule and anti-Markovnikov’s rule is Markovnikov’s rule proceeds through an ionic mechanism and involves the formation of intermediate carbocation.
- Whereas, anti-Markovnikov’s rule proceeds through free radical mechanisms and involves the formation of primary and secondary free radicals.