Question

Please include how to figure out the half reactions, step by step. I.e., I dont just want the answer, I want to be able to understand it. Please and thanks so much!

Reaction of borneol with HOCI under acidic conditions - balance the following 2 half reactions under acidic condition. Write the overall reaction based on the balanced 2 half reactions Он CI НОС Overall reaction:

Answer 1

LLLLLLLLLLLLLLL LLLLLLL LLLLLL LLLLLLL LLLLLLLLLLLL LLLLLLL LLLLLLLLLLLL LLLLLLL LLLLLL LLLLLL LLLLLLL LLLLLLLLLLLLL LLLLLL LLLLLLLLLLLLLL HOCI the hypochlorus acid gives proton: This is actual mechanism how reaction procceds: H-O - 0-0 + H+ CI H₃ C CH3 H₃C CH3 H₃ C CH3 H₃cy CH₃ Н.С HAC OM - a H₃C H H₃C . HH Theretical half reactions: H+ + HOCI -------> H2O + CI- and seconf half reaction: HACCHE HC CH3 + H2O + CI + H3O+ + HCI нс Т. ОН

Overall reaction is: H₃C- CH₂ H₃ C CH₃ + 2H3O+ + CI- YH CI H3C OH As per the mechanism: HOCI protonates the alcohol hydroxyl group making that hydroxyl group as a good leaving group. Further, the hypochlorite ion attacks as a nulceophile to substitute the protonated hydroxyl group. Thus the new C-OCI bond is formed. In the next step, the proton attached to carbon bearing OCI group is very acidic in nature and hence it is abstracted by water molecule to form CEO and at the same time 0-CI bond is also broken. This leads to formation of CEO leaving behind H3O+ and Cl- ions.