
In the presence of peroxide, HCI and HI do not give anti-Markovnikov’s addition to alkenes because:
A. one of the steps is endothermic in HCI and HI
B. both HCI and HI are strong acids
C. HCI is oxidizing and the HI is reducing
D. all the steps are exothermic in HCI and HI
Answer
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Hint: Anti-Markovnikov rule states that when there is an addition of a hydrogen halide to an asymmetric alkene, the acidic hydrogen (H) is attached to the carbon which has more alkyl substituents, and the other or the negative group is attached to the carbon which have more hydrogen substituents.
Complete step by step answer:
When we add hydrogen halides in the absence of peroxides, hydrogen halides will add according to the Markovnikov rule, which states when there is an addition of a hydrogen halide to an asymmetric alkene, the acidic hydrogen (H) is attached to the carbon which has more hydrogen substituents, and the other or the negative group is attached to the carbon which have more alkyl substituents.
But according to the question the case is in the presence of organic peroxides, when we add hydrogen bromide with peroxides the bromides will follow radical mediated reactions. And breaking of the peroxide O-O linkage will induce a Br radical which will proceed by adding to the less substituted or in other words towards the more hydrogen substituents side of the alkene (Anti - Markovnikov).
For example: we have propene as a substrate and HBr is added in presence of peroxides the reaction goes via Anti-Markovnikov addition as the reaction is exothermic.
$C{H_3}CH = C{H_2}\xrightarrow[{Peroxide}]{{HBr}}C{H_3}C{H_2}C{H_2}Br$
But, HI and HCl do not follow this same procedure as the addition of Cl and I radicals to the alkene in an Anti-Markovnikov way is an endothermic reaction and therefore it is unfavourable.
Hence, option A is the correct answer.
Note: This is not the case in the absence of peroxides as all hydrogen halides proceed through Markovnikov’s addition. Markanikov’s addition state that In an addition reaction of a protic acid HX (HCl,HBr etc.) to an alkene or alkyne, the hydrogen atom of HX becomes bonded to the carbon atom that had the greatest number of hydrogen atoms in the starting alkene or alkyne.
Complete step by step answer:
When we add hydrogen halides in the absence of peroxides, hydrogen halides will add according to the Markovnikov rule, which states when there is an addition of a hydrogen halide to an asymmetric alkene, the acidic hydrogen (H) is attached to the carbon which has more hydrogen substituents, and the other or the negative group is attached to the carbon which have more alkyl substituents.
But according to the question the case is in the presence of organic peroxides, when we add hydrogen bromide with peroxides the bromides will follow radical mediated reactions. And breaking of the peroxide O-O linkage will induce a Br radical which will proceed by adding to the less substituted or in other words towards the more hydrogen substituents side of the alkene (Anti - Markovnikov).
For example: we have propene as a substrate and HBr is added in presence of peroxides the reaction goes via Anti-Markovnikov addition as the reaction is exothermic.
$C{H_3}CH = C{H_2}\xrightarrow[{Peroxide}]{{HBr}}C{H_3}C{H_2}C{H_2}Br$
But, HI and HCl do not follow this same procedure as the addition of Cl and I radicals to the alkene in an Anti-Markovnikov way is an endothermic reaction and therefore it is unfavourable.
Hence, option A is the correct answer.
Note: This is not the case in the absence of peroxides as all hydrogen halides proceed through Markovnikov’s addition. Markanikov’s addition state that In an addition reaction of a protic acid HX (HCl,HBr etc.) to an alkene or alkyne, the hydrogen atom of HX becomes bonded to the carbon atom that had the greatest number of hydrogen atoms in the starting alkene or alkyne.
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