[siriipha]

I need to write an balanced equation to the oxidation of 2-octanol to 2-octanone. In the lab we used Ca(OCl)2 as the oxidizing agent, and water. The reaction happens in a aciditc solution as CH3COOH and CH3CN was also in the mixture, before it got extracted away and washed with Dichloromethane, NaHCO3 and water, then Dichloromethan was evaporated away from the solution and pure 2-octanone was obtained from a fractional-destillation.

I draw the 2-octanol and 2-octanon molecule and found out that they oxidized with +3e-, which gives me:

Ox: C8H17OH -----> C8H16O + 3e- + 2H+
Red: Ca(OCl)2 + ? ----> Ca2+ + 2OCl-
Balanced: ?

I cant see how Ca(OCl)2 get reduced in the reaction?? I wrote the oxidation numbers and Cl went from +1 to +1.5 in 2OCl-? 

[Borek]

How did you got 1.5 for chlorine in OCl^-?

[siriipha]

How did you got 1.5 for chlorine in OCl^-?

I calculated with 2OCl-. But I figured out I just need to use ClO- as the oxidation agent.
the balanced equation should be:

ox: C8H17OH --> C8H16O + 2e + 2H+
red: ClO- + 2e- + 2H+ --> Cl- + H2O
Balance: C8H17OH + ClO- --> C8H16O + Cl- + H2O

if I´m not wrong.

[Borek]

Looks correct now.