Oxidation of alkenes with hot concentrated acidified potassium manganate(VII) solution
The carbon-carbon double bonds in alkenes such as ethene react with potassium manganate(VII) solution (potassium permanganate solution).
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We”ll look at the reaction with ethene. Other alkenes react in just the same way. Manganate(VII) ions are a strong oxidizing agent, and in the first instance oxidize ethene to ethane-1,2-diol (old name: ethylene glycol). Looking at the equation purely from the point of view of the organic reaction:
The diols, such as ethane-1,2-diol, which are the products of the reaction with cold dilute potassium manganate(VII), are themselves quite easily oxidized by manganate(VII) ions. That means that the reaction will not stop at this point unless the potassium manganate(VII) solution is very dilute, very cold, and preferably not under acidic conditions. If you are using hot concentrated acidified potassium manganate(VII) solution, what you finally end up with depends on the arrangement of groups around the carbon-carbon double bond. Xem thêm: Công Thức Sữa Hạt Giảm Cân, Bạn Đã Biết Làm Sữa Hạt Giảm Cân Chưa The formula below represents a general alkene. In organic lize.vnistry, the symbol R is used to represent hydrocarbon groups or hydrogen in a formula when you don”t want to talk about specific compounds. If you use the symbol more than once in a formula (as here), the various groups are written as R1, R2, etc. api/deki/files/50350/splitcc.gif?revision=1&size=bestfit&width=427&height=68″ />Oxidation of alkenes with hot concentrated acidified potassium manganate(VII) solution
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The products are known as carbonyl compounds because they contain the carbonyl group, C=O. Carbonyl compounds can also react with potassium manganate(VII), but how they react depends on what is attached to the carbon-oxygen double bond. So we need to work through all the possible combinations.
Carbonyl compounds which have two hydrocarbon groups attached to the carbonyl group are called ketones. Ketones aren”t that easy to oxidize, and so there is no further action. (But see note in red below.) If the groups attached either side of the original carbon-carbon double bond were the same, then you would end up with a single ketone. If they were different, then you would end up with a mixture of two. For example:
Na2Co3 + H3Po4 = Na3Po4 – Naoh + H3Po4 = Na3Po4 + H2O
For example, suppose the first stage of the reaction was:
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