
The addition of HBr to an alkene in the presence of peroxide is an example of:
A. Electrophilic addition reaction
B. Nucleophilic addition reaction
C. Free-radical addition reaction
D. Formation of carbocation as an intermediate
Answer
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Hint: When hydrogen bromide is added to alkenes in presence of peroxide, it reacts with anti-Markovnikov’s addition mechanism. Peroxides form free radicals that initiate the reaction and add bromine radical at terminal carbon.
Complete step-by-step answer:
Peroxides have a weak oxygen-oxygen bond which on heating results in homolytic fragmentation of this bond i.e. the bond breaks in order to leave one unpaired electron on each atom involved in the reaction. Strong sources of light such as floodlight or other source of light radiation which reaches into the near UV might also serve to weaken this bond.
Only a catalytic amount of peroxide is needed to get the reaction started, although one molar equivalent of HBr is essentially required to result in complete addition of HBr to the alkene.
This results in a highly reactive alkoxy radical which then abstracts hydrogen from H-Br, releasing a bromine radical. The bromine radical is the one that is added to the alkene from the molecule hydrogen bromide.
Preferably addition to an alkene tends to occur in such a way that the most stable free radical is formed, tertiary radical here in HBr. That’s the reason that bromine ends up on the least substituted carbon of the alkene. This tertiary radical then eliminates hydrogen from H-Br, liberating a bromine radical, and this way the cycle continues.

Hence, the correct option is (C).
Note: In absence of peroxide, if alkene reacts with hydrogen bromide, the attack of bromine is due to electrophilic addition reaction in which carbocation is formed as an intermediate. It is also called Markovnikov’s addition in which attack takes place on more substituted carbon.
Complete step-by-step answer:
Peroxides have a weak oxygen-oxygen bond which on heating results in homolytic fragmentation of this bond i.e. the bond breaks in order to leave one unpaired electron on each atom involved in the reaction. Strong sources of light such as floodlight or other source of light radiation which reaches into the near UV might also serve to weaken this bond.
Only a catalytic amount of peroxide is needed to get the reaction started, although one molar equivalent of HBr is essentially required to result in complete addition of HBr to the alkene.
This results in a highly reactive alkoxy radical which then abstracts hydrogen from H-Br, releasing a bromine radical. The bromine radical is the one that is added to the alkene from the molecule hydrogen bromide.
Preferably addition to an alkene tends to occur in such a way that the most stable free radical is formed, tertiary radical here in HBr. That’s the reason that bromine ends up on the least substituted carbon of the alkene. This tertiary radical then eliminates hydrogen from H-Br, liberating a bromine radical, and this way the cycle continues.

Hence, the correct option is (C).
Note: In absence of peroxide, if alkene reacts with hydrogen bromide, the attack of bromine is due to electrophilic addition reaction in which carbocation is formed as an intermediate. It is also called Markovnikov’s addition in which attack takes place on more substituted carbon.
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