Answer
Verified
114.3k+ views
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key
Hybridization of Atomic Orbitals Important Concepts and Tips for JEE
Atomic Structure: Complete Explanation for JEE Main 2025
Trending doubts
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Collision - Important Concepts and Tips for JEE
Ideal and Non-Ideal Solutions Raoult's Law - JEE
Current Loop as Magnetic Dipole and Its Derivation for JEE
JEE Main 2023 January 30 Shift 2 Question Paper with Answer Keys & Solutions