Primary alcohol?

Question

hi is this the correct equation for the reaction with Butan-1-ol with [o] acidified K2Cr2O7 ??

CH3CH2C2OH + [o] -> CH3C-C(=O)CH + H2O

thank you.

also what are the conditions that this experiemnt should be carried out in

Answer 1

The oxidation of alcohols goes through 2 states, first the aldehyde (which you have described, i think) and then a further step to the carboxylic acid.

Therefore the end product of the reaction will be CH3CH2C(=O)OH. It is rare that the oxidation reaction stops at the aldehyde CH3CH2C(=O)H

Answer 2

Butan-1-ol is oxidised in two stages.
First, butanal is formed.
CH3CH2CH2CH2OH + [O] ----> CH3CH2CH2CHO + H2O
Then butanoic acid is formed, especially if the reaction is performed under reflux.
CH3CH2CH2CHO + [O] -----> CH3CH2CH2COOH

The overall equation for the two-stage reaction is therefore
CH3CH2CH2CH2OH + 2[O] ----> CH3CH2CH2COOH + H2O

Several of the above answers have not got the correct number of C atoms.

Answer 3

Oxidation of butan-1-ol to form a carboxylic acid (butanoic acid)
Reagents: acidified K2Cr2O7
Condition: reflux
- CH3(CH2)2CH2OH + 2[O] -> CH3(CH2)2CO2H + H2O

What u have is the reaction of controlled oxidation of butan-1-ol to form a aldehyde (butanone)
Reagents: acidified K2Cr2O7
Condition: reflux with distillation
- CH3(CH)2CHOH + [O] -> CH3(CH2)2CHO -> H2O

Answer 4

No, it will be

CH3CH2CH2CH2OH + [O] ==> CH3CH2CH2CHO + H2O

propanaldehyde will be formed.
On prolonged exposure propanoic acid will be formed.

CH3CH2CH2CHO +[O] ==> CH3CH2CH2COOH

This reaction should be carried out in presence of Collin's reagent ie Cr2O3 + Pyridine.