
Which of the following acids adds to propene in the presence of peroxide to give anti-Markownikoff's product
A. \[HF\]
B. \[HCl\]
C. \[HBr\]
D. \[HI\]
Answer
162.9k+ views
Hint: According to the anti-Markovnikov rule, when a hydrogen halide is added to an asymmetric alkene, the acidic hydrogen is attached to the carbon atom that has the most alkyl substituents, and the other or negative group is associated to the carbon atom that has the most hydrogen substituents.
Complete step-by-step answer:In order to know that the Markovnikov rule states that when a hydrogen halide is added to an asymmetric alkene, the acidic hydrogen is attached to the carbon with the most hydrogen substituents and the other or the negative group is attached to the carbon with the most alkyl substituents. This rule applies when there are no peroxides present.
However, the situation in the question is that when organic peroxides are present, adding hydrogen bromide will cause radical-mediated reactions to occur with the bromides. The breaking of the peroxide \[O - O\] linkage will also produce a \[Br\] radical, which will proceed by adding to the side of the alkene that has more hydrogen substituents than the less substituted side (Anti - Markovnikov).
As an illustration, when \[HBr\] is added to propene in the presence of peroxides, the reaction proceeds by Anti-Markovnikov addition since it is exothermic.
\[C{H_3} - CH = C{H_2} + HBr\]
The peroxide effect is not observed in addition of HCl and HI. As, the H–Cl bond being stronger than H–Br bond, is not cleaved by the free radical, on the other hand the H–I bond is weaker and iodine free radicals combine to form iodine molecules but not add to the double bond. The HF is very reactive, so we don’t use this for peroxide reaction.
Option ‘C’ is correct
Note: All hydrogen halides go through Markovnikov's addition in the absence of peroxides but only HBr undergoes anti Markovnikov's addition in the presence of peroxide. According to Markanikov's addition that is the carbon atom that previously housed the most hydrogen atoms receives the hydrogen atom when a negative part is added to another C-atom of an alkene or alkyne.
Complete step-by-step answer:In order to know that the Markovnikov rule states that when a hydrogen halide is added to an asymmetric alkene, the acidic hydrogen is attached to the carbon with the most hydrogen substituents and the other or the negative group is attached to the carbon with the most alkyl substituents. This rule applies when there are no peroxides present.
However, the situation in the question is that when organic peroxides are present, adding hydrogen bromide will cause radical-mediated reactions to occur with the bromides. The breaking of the peroxide \[O - O\] linkage will also produce a \[Br\] radical, which will proceed by adding to the side of the alkene that has more hydrogen substituents than the less substituted side (Anti - Markovnikov).
As an illustration, when \[HBr\] is added to propene in the presence of peroxides, the reaction proceeds by Anti-Markovnikov addition since it is exothermic.
\[C{H_3} - CH = C{H_2} + HBr\]
The peroxide effect is not observed in addition of HCl and HI. As, the H–Cl bond being stronger than H–Br bond, is not cleaved by the free radical, on the other hand the H–I bond is weaker and iodine free radicals combine to form iodine molecules but not add to the double bond. The HF is very reactive, so we don’t use this for peroxide reaction.
Option ‘C’ is correct
Note: All hydrogen halides go through Markovnikov's addition in the absence of peroxides but only HBr undergoes anti Markovnikov's addition in the presence of peroxide. According to Markanikov's addition that is the carbon atom that previously housed the most hydrogen atoms receives the hydrogen atom when a negative part is added to another C-atom of an alkene or alkyne.
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