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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: shook0 on February 02, 2022, 02:05:44 PM

Title: Cyclohexanol (C6H11OH) + K2Cr2O7
Post by: shook0 on February 02, 2022, 02:05:44 PM

Hi, I'm supposed to set up a reaction between Cyclohexanol (I'm German, so I'm not sure if this spelling is correct in English) and K2Cr2O7. I thought it's a classical oxidation, however, the oxidation numbers don't change if I see it correctly. Can anybody help me please?

 C6H11OH + K2Cr2O7 :rarrow:  ?

Title: Re: Cyclohexanol (C6H11OH) + K2Cr2O7
Post by: Babcock_Hall on February 02, 2022, 02:17:02 PM

Can you show us what your oxidation numbers are?

Title: Re: Cyclohexanol (C6H11OH) + K2Cr2O7
Post by: shook0 on February 02, 2022, 03:23:14 PM

C6H11OH = C: +II; H +I; O: -II
The same applies to C6H11OOH

Title: Re: Cyclohexanol (C6H11OH) + K2Cr2O7
Post by: Babcock_Hall on February 02, 2022, 03:58:31 PM

Can you think of a different organic product?

Title: Re: Cyclohexanol (C6H11OH) + K2Cr2O7
Post by: Borek on February 02, 2022, 04:38:06 PM

Quote from: shook0 on February 02, 2022, 03:23:14 PM

C6H11OH = C: +II; H +I; O: -II
The same applies to C6H11OOH

No, ROOH contains an oxygen-oxygen bond, so these O don't have oxidation numbers of -2.

Not that it is an expected product of an alcohol oxidation.