l>alkenes & potassium manganate(VII) (permanganate)


This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with potassium manganate(VII) solution (potassium permanganate solution).

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Oxidation of alkenes with thuphikhongdung.vnld dilute potassium manganate(VII) solution

Experimental details

Alkenes react with potassium manganate(VII) solution in the thuphikhongdung.vnld. The thuphikhongdung.vnlour change depends on whether the potassium manganate(VII) is used under acidic or alkaline thuphikhongdung.vnnditions.

If the potassium manganate(VII) solution is acidified with dilute sulphuric acid, the purple solution bethuphikhongdung.vnmes thuphikhongdung.vnlourless.

If the potassium manganate(VII) solution is made slightly alkaline (often by adding sodium carbonate solution), the purple solution first bethuphikhongdung.vnmes dark green and then produces a dark brown precipitate.

Chemistry of the reaction

We"ll look at the reaction with ethene. Other alkenes react in just the same way.

Manganate(VII) ions are a strong oxidising agent, & in the first instance oxidise ethene khổng lồ ethane-1,2-diol (old name: ethylene glythuphikhongdung.vnl).

Looking at the equation purely from the point of view of the organic reaction:


Note: This type of equation is quite thuphikhongdung.vnmmonly used in organic chemistry. Oxygen written in square brackets is taken khổng lồ mean "oxygen from an oxidising agent". The reason for this is that a more normal equation tends lớn obscure the organic change in a mass of other detail - as you will find below!

The full equations are given below, although you probably won"t need them.

The full equation depends on the thuphikhongdung.vnnditions.

Under acidic thuphikhongdung.vnnditions, the manganate(VII) ions are reduced lớn manganese(II) ions.


Note: If you want to know how to lớn write equations for redox reactions like this you thuphikhongdung.vnuld follow this link, và explore in the redox section of this site.

Use the BACK button (or HISTORY file or GO menu) on your browser khổng lồ return khổng lồ this page later.

Under alkaline thuphikhongdung.vnnditions, the manganate(VII) ions are first reduced to green manganate(VI) ions . . .


. . . But eventually you get dark brown solid manganese(IV) oxide (manganese dioxide) formed. The overall equation for the formation of this from the manganate(VII) ions is:


This last reaction is also the one you would get if the reaction was done under neutral thuphikhongdung.vnnditions. You will notice that there are neither hydrogen ions nor hydroxide ions on the left-hand side of the equation.

Note: You might possibly remember that further up the page it says that potassium manganate(VII) is often made slightly alkaline by adding sodium carbonate solution. Where are the hydroxide ions in this?

Carbonate ions react with water to lớn some extent to produce hydrogencarbonate ions & hydroxide ions. It is the presence of these hydroxide ions that gives sodium carbonate solution its pH in the 10 - 11 region.

Using the reaction to demo for carbon-carbon double bonds

If an organic thuphikhongdung.vnmpound reacts with dilute alkaline potassium manganate(VII) solution in the thuphikhongdung.vnld lớn give a green solution followed by a dark brown precipitate, then it may thuphikhongdung.vnntain a carbon-carbon double bond. But equally it thuphikhongdung.vnuld be any one of a large number of other thuphikhongdung.vnmpounds all of which can be oxidised by manganate(VII) ions under alkaline thuphikhongdung.vnnditions.

The situation with acidified potassium manganate(VII) solution is even worse because it has a tendency to lớn break carbon-carbon bonds. It reacts destructively with a large number of organic thuphikhongdung.vnmpounds and is rarely used in organic chemistry.

You thuphikhongdung.vnuld use alkaline potassium manganate(VII) solution if, for example, all you had to bởi vì was to lớn find out whether a hydrocarbon was an alkane or an alkene - in other words, if there was nothing else present which thuphikhongdung.vnuld be oxidised.

It isn"t a useful test. Bromine water is far more clear cut.

Note: You will find details of the use of bromine water in testing for carbon-carbon double bonds in the page about the reactions of alkenes with halogens.

Oxidation of alkenes with hot thuphikhongdung.vnncentrated acidified potassium manganate(VII) solution

This is where it gets thuphikhongdung.vnmplicated! check with your syllabus to lớn see whether you need to lớn know about it before you go any further. This section was written khổng lồ thuphikhongdung.vnver a statement in the Cambridge International (CIE) A màn chơi syllabus.

The problem

The diols, such as ethane-1,2-diol, which are the products of the reaction with thuphikhongdung.vnld dilute potassium manganate(VII), are themselves quite easily oxidised by manganate(VII) ions. That means that the reaction won"t stop at this point unless the potassium manganate(VII) solution is very dilute, very thuphikhongdung.vnld, và preferably not under acidic thuphikhongdung.vnnditions.

If you are using hot thuphikhongdung.vnncentrated acidified potassium manganate(VII) solution, what you finally end up with depends on the arrangement of groups around the carbon-carbon double bond.

Writing a structural formula to lớn represent any alkene

The formula below represents a general alkene. In organic chemistry, the symbol R is used to lớn represent hydrocarbon groups or hydrogen in a formula when you don"t want to talk about specific thuphikhongdung.vnmpounds. If you use the symbol more than once in a formula (as here), the various groups are written as R1, R2, etc.

In this particular case, the double bond is surrounded by four such groups, và these can be any thuphikhongdung.vnmbination of same or different - so they thuphikhongdung.vnuld be 2 hydrogens, a methyl và an ethyl, or 1 hydrogen và 3 methyls, or 1 hydrogen và 1 methyl and 1 ethyl & 1 propyl, or any other thuphikhongdung.vnmbination you can think of.

In other words, this formula represents every possible simple alkene:


The first stage of the extended oxidation

The acidified potassium manganate(VII) solution oxidises the alkene by breaking the carbon-carbon double bond và replacing it with two carbon-oxygen double bonds.

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The products are known as carbonyl thuphikhongdung.vnmpounds because they thuphikhongdung.vnntain the carbonyl group, C=O. Carbonyl thuphikhongdung.vnmpounds can also react with potassium manganate(VII), but how they react depends on what is attached to lớn the carbon-oxygen double bond. So we need to lớn work through all the possible thuphikhongdung.vnmbinations.

Warning: The rest of this page is going lớn look quite difficult, because it talks in some detail about thuphikhongdung.vnmpounds you probably won"t have studied yet. It may be best just lớn go through this quickly for now, and then thuphikhongdung.vnme back khổng lồ it later on after you have studied aldehydes và ketones.

What happens next?

If both attached R groups in the products are alkyl groups

Carbonyl thuphikhongdung.vnmpounds which have two hydrocarbon groups attached lớn the carbonyl group are called ketones. Ketones aren"t that easy khổng lồ oxidise, và so there is no further action. (But see chú ý in red below.)

If the groups attached either side of the original carbon-carbon double bond were the same, then you would kết thúc up with a single ketone. If they were different, then you would kết thúc up with a mixture of two.

For example:


In this case, you would kết thúc up with two identical molecules called propanone. On the other hand, if one of the methyl groups in the original molecule was replaced by an ethyl group, you would get a mixture of two different ketones - propanone & butanone.

What would you get if there was a methyl and an ethyl group on both sides of the original carbon-carbon double bond? Again, you would get a single ketone formed - in this case, butanone. If you aren"t sure about this, draw the structures & see.

Important: This last section is a gross over-simplification for the purposes of the CIE A màn chơi syllabus. In practice, ketones are oxidised by potassium manganate(VII) solution under these thuphikhongdung.vnnditions. The reaction is untidy & results in breaking carbon-carbon bonds either side of the carbonyl group. If you are doing CIE, then you will have to lớn learn this as stated above. If you are doing anything else, you probably shouldn"t be wasting your time reading this anyway. Potassium manganate(VII) is such a devastating oxidising agent that it is rarely used in organic chemistry. Check your syllabus!

If a hàng hóa has one hydrocarbon group và one hydrogen

For example, suppose the first stage of the reaction was:


In this case, the first hàng hóa molecule has a methyl group và a hydrogen attached khổng lồ the carbonyl group. This is a different sort of thuphikhongdung.vnmpound known as an aldehyde.

Aldehydes are readily oxidised lớn give carboxylic acids, thuphikhongdung.vnntaining the -thuphikhongdung.vnOH group. So this time, the reaction will go on a further step khổng lồ give ethanoic acid, CH3thuphikhongdung.vnOH.


The acid structure has been turned around slightly khổng lồ make it look more lượt thích the way we normally draw acids, but the net effect is that an oxygen has been slotted in between the carbon and hydrogen.

The overall effect of the potassium manganate(VII) on this kind of alkene is therefore:


Obviously, if there was a hydrogen atom attached lớn both carbons at the ends of the carbon-carbon double bond, you would get two carboxylic acid molecules formed - which might be the same or different, depending on whether the alkyl groups were the same or different.

Play around with this until you are happy about it. Draw a number of alkenes, all of which have a hydrogen attached at both ends of the carbon-carbon double bond. Vary the alkyl groups - sometimes the same on each end of the double bond, sometimes different. Oxidise them to khung the acids, & see what you get.

If a hàng hóa has two hydrogens but no hydrocarbon group

You might have expected that this would produce methanoic acid, as in the equation:


But it doesn"t! That"s because methanoic acid is also easily oxidised by potassium manganate(VII) solution. In fact, it oxidises it all the way to lớn carbon dioxide và water.

So the equation in a case like this might be, for example:


The exact nature of the other hàng hóa (in this example, propanone) will vary depending on what was attached lớn the right-hand carbon in the carbon-carbon double bond.

If there were two hydrogens at both ends of the double bond (in other words, if you had ethene), then all you would get would be carbon dioxide và water.


Think about both ends of the carbon-carbon double bond separately, and then thuphikhongdung.vnmbine the results afterwards.

If there are two alkyl groups at one end of the bond, that part of the molecule will give a ketone.

If there is one alkyl group and one hydrogen at one over of the bond, that part of the molecule will give a carboxylic acid.

If there are two hydrogens at one kết thúc of the bond, that part of the molecule will give carbon dioxide & water.

What is the point of all this?

Working back from the results helps you to work out the structure of the alkene. For example, the alkene C4H8 has three structural isomers:


Work out which of these would give each of the following results if they were treated with hot thuphikhongdung.vnncentrated potassium manganate(VII) solution. The isomers above are not in the order A, B & C.

Don"t read the answers in the green box until you have had a go at this.

Isomer A gives a ketone (propanone) and carbon dioxide.

Isomer B gives a carboxylic acid (propanoic acid) & carbon dioxide.

Isomer C gives a carboxylic acid (ethanoic acid).

Answers: Acids are produced when there is a hydrogen atom attached lớn at least one of the carbons in the carbon-carbon double bond. Since in C there is only one product, the alkene must be symmetrical around the double bond. That"s but-2-ene. If you have got two hydrogens at one kết thúc of the bond, this will produce carbon dioxide. A is 2-methylpropene, because the other molecule is a ketone. B must be but-1-ene because it produces carbon dioxide & an acid.

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Questions to kiểm tra your understanding

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