Part 2. The chemistry of ALKENES - unsaturated hydrocarbons

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Part 2.7 The reaction of alkenes with aqueous potassium manganate(VII) - oxidation to form diol alcohols

A redox reaction involving oxidation of an organic compound and the reduction of an inorganic compound.

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2.7 The reaction of alkenes with aqueous potassium manganate(VII)

If alkenes are bubbled into, or the liquid shaken with, a neutral aqueous potassium manganate(VII), KMnO₄(aq), an oxidation reaction occurs and a diol alcohol is formed at room temperature.

The potassium manganate(VII) solution decolourises and a brown/black precipitate of manganese dioxide, MnO₂ forms.

This is an oxidation reaction of an organic compound.

The Mn in KMnO₄ is reduced from a +7 oxidation state to a +4 oxidation state in MnO₂.

The simplified equation for ethane is shown below, which assumes the oxygen atom originates from the oxidising agent.

(1) H₂C=CH₂  +  [O]  +  H₂O  ===>  HOCH₂-CH₂OH

The product is called ethane-1,2-diol,

The equation for propene is:

(2) CH₃CH=CH₂  +  [O]  +  H₂O  ===>  CH₃CH(OH)CH₂OH

The product is called propane-1,2-diol, ,

The equations for but-1-ene and but-2-ene are:

(3) CH₃CH₂CH=CH₂  +  [O]  +  H₂O  ===> CH₃CH₂CH(OH)CH₂OH

(3) CH₃CH=CHCH₃  +  [O]  +  H₂O  ===> CH₃CH(OH)CH(OH)CH₃

The products are butane-1,2-diol  and  butane-2,3-diol respectively.

See also the structure and naming of alcohols

 

The half-reaction equation for the potassium manganate(VII) is ...

MnO₄^-(aq)  +  4H⁺(aq)  + 3e^-  ===>  MnO₂(s)  +  H₂O(l)

The half-cell potential for this half-reaction is E^θ = +1.67 V, so the manganate(VII) ion is a strong oxidising agent.

See also the chemistry of the 3d block transition metal manganese

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