Balancing the Oxidation of 2-Butanol with CrO3

[Soaring Crane]

Homework Statement

I am trying to figure out the complete reaction equation for the oxidation of a secondary alcohol, 2-butanol, with CrO3 (and H2SO4).

Homework Equations

See below.

The Attempt at a Solution

In this oxidation, the alcohol is turned into a ketone, not a carboxylic acid, correct?

Therefore, the simple synthesis setup would be:

CH3CH2CH(OH)CH3 + CrO3 + H2SO4 -> CH3CH2C(=O)CH3

Now for the redox reaction with the chromium ions:

(Please note that I am slightly rusty on balancing redox reactions, and I really don’t know if this is remotely valid.)

CrO3 -> Cr(+3)
CH3CH2CH(OH)CH3 -> CH3CH2C(=O)CH3 ??

3 e- + CrO3 + 6 H(+) -> Cr(+3) + 3H2O
CH3CH2CH(OH)CH3 -> CH3CH2C(=O)CH3 + H(+) + 1 e-

3 e- + CrO3 + 6 H(+) -> Cr(+3) + 3H2O
3*(CH3CH2CH(OH)CH3 -> CH3CH2C(=O)CH3 + H(+) + 1 e-)


3 e- + CrO3 + 6 H(+) -> Cr(+3) + 3H2O
3 CH3CH2CH(OH)CH3 -> 3 CH3CH2C(=O)CH3 + 3H(+) + 3 e-
CrO3 + 6 H(+) + 3 CH3CH2CH(OH)CH3 -> Cr(+3) + 3H2O + 3 CH3CH2C(=O)CH3 + 3H(+)

CrO3 + 3 H(+) + 3 CH3CH2CH(OH)CH3 -> Cr(+3) + 3H2O + 3 CH3CH2C(=O)CH3 ?

Thank you.

 

[chaoseverlasting]

Yeah. Looks right.

 

[Soaring Crane]

I just came across this equation for the oxidation in a book.

2CrO3 + 6 H(+) + 3 CH3CH2CH(OH)CH3 -> 2Cr(+3) + 6H2O + 3 CH3CH2C(=O)CH3

Why is it that the bold coefficients are two times more than those in my equation?

Did I do something wrong?

Thanks again.

 

[chemisttree]

3 e- + CrO3 + 6 H(+) -> Cr(+3) + 3H2O
CH3CH2CH(OH)CH3 -> CH3CH2C(=O)CH3 + H(+) + 1 e-

Rethink this... pay attention to the hydrogens I have bolded.