Answer
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Hint: According to the statement of Anti-Markovnikov addition reaction, an addition reaction of a generic electrophile HX to an alkene or alkyne takes place where the hydrogen atom of the \[HX\] (hydrogen halide) becomes bonded to the carbon atom that had the least number of hydrogen atoms in the starting alkene or alkyne.
Complete step by step answer:
In Anti- Markovnikov addition reaction, the attack of a nucleophile takes place on that carbon atom of an unsaturated system (having double or triple bond) which has a more number of hydrogen atoms attached to it and the hydrogen atom gets attached to the adjacent carbon atom which has less number of hydrogen atoms attached to it.
The peroxide effect in anti- Markovnikov addition reaction is a reaction in which, with the help of organic peroxides present in the solution, we get a free radical chain reaction. The chain reaction is initiated by free radicals produced by an oxygen-oxygen bond in the organic peroxide breaking. There is a hemolytic cleavage of the $O - O$ bond. These free radicals extract a hydrogen atom from a hydrogen halide (mostly $H - Br$ ) molecule to produce halogen radicals.
So, the correct answer is Option c.
Note:
The change in the regioselectivity of the addition of hydrogen bromide to an alkene or alkyne in the presence of a peroxide is known as the peroxide effect or the Kharasch effect. In the absence of a peroxide, \[HBr\] adds to propene via an ionic mechanism (with a carbocation intermediate) to give 2-bromopropane. In that case, the Markovnikov's Rule is obeyed.
Complete step by step answer:
In Anti- Markovnikov addition reaction, the attack of a nucleophile takes place on that carbon atom of an unsaturated system (having double or triple bond) which has a more number of hydrogen atoms attached to it and the hydrogen atom gets attached to the adjacent carbon atom which has less number of hydrogen atoms attached to it.
The peroxide effect in anti- Markovnikov addition reaction is a reaction in which, with the help of organic peroxides present in the solution, we get a free radical chain reaction. The chain reaction is initiated by free radicals produced by an oxygen-oxygen bond in the organic peroxide breaking. There is a hemolytic cleavage of the $O - O$ bond. These free radicals extract a hydrogen atom from a hydrogen halide (mostly $H - Br$ ) molecule to produce halogen radicals.
So, the correct answer is Option c.
Note:
The change in the regioselectivity of the addition of hydrogen bromide to an alkene or alkyne in the presence of a peroxide is known as the peroxide effect or the Kharasch effect. In the absence of a peroxide, \[HBr\] adds to propene via an ionic mechanism (with a carbocation intermediate) to give 2-bromopropane. In that case, the Markovnikov's Rule is obeyed.
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