Oxidation of an alcohol

There exist two oxidations of an alcohol .

Is oxidation is known as complete.
Is it is spared.

Oxidation supplements of an alcohol

To oxidize an alcohol completely, one burns it using a flame. At the end of the reaction, all alcohol reacted and the formed products are carbon dioxide (CO₂) and the steam (H₂O).

During an oxidation the carbonaceous chain of alcohol supplements is modified.

Here the equation of the general reaction and an example:

$\ rm {C_nH_ {2n+1} OH + \ frac {3n+1} {2} O_2 \ xrightarrow {flame} nCO_2 + (n+1) H_2O}$

$\ rm {C_4H_9OH + \ frac {13} {2} O_2 \ xrightarrow {flame} 4CO_2 + 5:00 _2O}$

Controlled oxydation of an alcohol

By controlled oxydation, one can obtain from an alcohol:

if it is a primary education alcohol, a Aldéhyde
if it is a secondary alcohol, a Cétone
if it is a tertiary alcohol, it does not have there oxidation

To carry out a controlled oxydation, it is necessary to make react alcohol with a Oxydant like the ion MnO₄^- permanganate or PCC. It occurs then a reaction of Oxydo-réduction.

During a controlled oxydation only the group characteristic hydroxyl of alcohol is concerned.

Controlled oxydation of a primary education alcohol

example of the propan-1-ol

Half-equations of oxydoreduction: it must there as many have electrons for oxidant than for the reducer (from where the multiplications)

×2 {MnO₄^- + 5e^- + 8:00 ⁺ = Mn²⁺ + 4:00 ₂O}
×5 {CH₃-CH₂-CH₂-OH = CH₃-CH₂-CHO + 2e^- + 2:00 ⁺}

equation of the reaction: one combines the two half-equations by removing the electrons

acid 5CH ₃-CH₂-CH₂-OH + 2MnO₄^- + 6:00 ⁺ $\ xrightarrow {}$ 5CH ₃-CH₂-CHO + 2Mn²⁺ + 8:00 ₂O

The acid medium allows the reaction. Generally, one uses SO₄^2- to acidify the solution.
Moreover, when the oxidant is in excess, there is formation of a Carboxylic acid

Controlled oxydation of a secondary alcohol

example of the propan-2-ol

Half-equations of oxydoreduction: it must there as many have electrons for oxidant than for the reducer (from where the multiplications)

×2 {MnO₄^- + 5e^- + 8:00 ⁺ = Mn²⁺ + 4:00 ₂O}
×5 {CH₃-CHOH-CH₃ = CH₃-CO-CH₃ + 2e^- + 2:00 ⁺}

equation of the reaction: one combines the two half-equations by removing the electrons

acid 5CH ₃-CHOH-CH₃ + 2MnO₄^- + 6:00 ⁺ $\ xrightarrow {}$ 5CH ₃-CO-CH₃ + 2Mn²⁺ + 8:00 ₂O

The acid medium allows the reaction. Generally, one uses an aqueous solution of sulphuric acid to acidify the medium réactionnel.

See too

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