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The addition of HI in the presence of peroxide does not show anti-Markovnikov behavior becauseA. the HI band is too strong not to be broken homolyticallyB. HI is reducing agentC. I free radical so formed readily combine with each other to give `I_(2)` moleculeD. I comine with H to give back HI |
Answer» Correct Answer - C H-I bond is weak (71/Kcal/mole) breaks homolytcically 1 free radical, it so formed combine each other produces `I_(2)` molecule. |
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