Question #880a8

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Question #880a8

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Answer 1 · 2015-01-03T00:14:48

Under alkaline conditions manganate(VII) will oxidise unsaturated hydrocarbons such as alkenes to a diol.

I'll use ethene as an example

The first step is that purple Mn(VII) are reduced to green Mn(VI) ions:

$C H_{2} = C H_{2} + 2 M n O_{4}^{-} + 2 O H^{-} \to H O C H_{2} C H_{2} O H + 2 M n O_{4}^{2 -}$ $CH_2=CH_2 +2MnO_4^(-)+2OH^(-)rarrHOCH_2CH_2OH+2MnO_4^(2-)$

Ethane - 1,2 - diol is formed and green $M n O_{4}^{2 -}$ $MnO_4^(2-)$ .

The reaction then continues to give dark brown solid manganese(IV) oxide:

$3 C H_{2} C H_{2} + 2 M n O_{4}^{2 -} + 4 H_{2} O \to 3 (H O C H_{2} C H_{2} O H) + 2 M n O_{2} + 2 O H^{-}$ $3CH_2CH_2+2MnO_4^(2-)+4H_2Orarr3(HOCH_2CH_2OH)+2MnO_2+2OH^-$

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Question #880a8

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