A quick chem Q on oxidation on alcohols!! HELP!!!

Hiiii guys, pls help me, I don’t understand the answer!

The question is: Write a half equation including electrons to show the oxidation of pentan-2-ol, CH3CH(OH)CH2CH2CH3, to pentan-2-one, CH3COCH2CH2CH3

What I got is CH3CH(OH)CH2CH2CH3 ➡️ CH3COCH2CH2CH3 + 2H+ +2e-

But the answer said it should be CH3CH(OH)CH2CH2CH3 + H2O ➡️ CH3COCH2CH2CH3 + 4H+ + 4e-

I don’t get why as there are 2 Os on the left of the equation and there’s only 1 O on the right!! Is that means that you don’t count the O in bracket or the answer is just wrong??

Thanks so much!!!

The oxygen in the H2O is the oxidising agent so you need that for the reaction to happen. Then you have 4 hydrogen ions which also results in 4 electrons

what is your level?

Because the answer is wrong, no wonder you're so confused.

H₂O shouldn't even be included since the number of oxygen atoms are already balanced.
Only the hydrogen atoms need balancing, but even in the answer this is wrong; it should only be 2H⁺ + 2e^-.
As a result, water does not act as an oxidising agent like someone mentioned. You don't include oxidising agents in an oxidation half reaction, since oxidising agents are their own species which are reduced during the reaction, and so they have their own reduction half equations. The oxidising agent would be something else, usualy acidified dichromate when oxidising alcohols, but the question doesn't ask about that.

Hence the correct answer is: CH₃CH(OH)CH₂CH₂CH₃ $\rightarrow$ CH₃COCH₂CH₂CH₃ + 2H⁺ + 2e^-

Uhhhh thanks so much!!! I have a chem exam tomorrow so quite stressed. I knew I was right!!!

Believe in yourself.
I wish you luck for tomorrow; the best thing you could do now though is get some rest