Chemguide: Support for CIE A level Chemistry

Learning outcome 14: Hydrocarbons

14.2: Alkenes

Statement 14.2.2

This statement asks you to know about a number of reactions of alkenes such as ethene and propene.

Before you go on, you should find and read the statement in your copy of the syllabus.

Although we mainly talk about ethene and propene, you must be prepared to answer questions about the reactions of carbon-carbon double bonds in a whole lot of other molecules as well - some very complicated.

However, if you understand the reactions of a carbon-carbon double bond in simple molecules like ethene or propene, you can work out the more difficult ones if you need to.

Before you do anything else, read the page introducing alkenes.

Statement 14.2.2(a)(i): Addition of hydrogen

Addition of hydrogen is called hydrogenation. All you will need is in the first couple of paragraphs of the hydrogenation of alkenes page. You can ignore the rest of the page about making margarine.

All you need to remember is that you can hydrogenate a carbon-carbon double bond by reacting it with hydrogen gas in the presence of a nickel catalyst at about 150°C. You can also use platinum as the catalyst.

Statement 14.2.2(a)(ii): Addition of water (steam)

Addition of steam or water is called hydration. Take care not to confuse this with hydrogenation. You will find this on the hydration of alkenes page.

You will notice that only the industrial process is given. This reaction is never done in the lab - the conditions needed are too extreme.

When you add water across the double bond in ethene, a hydrogen joins to one of the carbon atoms, and an OH group to the other. With propene, you could in theory get one of two products, depending on which carbon got the hydrogen and which the OH.

Markovnikov's Rule gives you a way of working out which way round something will add in these circumstances. It is very easy to remember in the form that I have used on the page you have just looked at.

Note:  You will find all sorts of other spellings for Markovnikov. Markovnikov wouldn't actually have recognised any of them, because he was a Russian and used an entirely different alphabet. The English spellings are just an approximation to the sound of the word.

Statement 14.2.2(a)(iii): Addition of hydrogen halides

These reactions are covered on the page alkenes and hydrogen halides.

You can probably safely ignore the section headed "Variation of rates when you change the alkene".

You can also ignore any references to the mechanisms for the reactions for now. These come up in statements 14.2.4 and 14.2.5.

The syllabus isn't clear about whether you need to know about the anti-Markovnikov addition of HBr to propene in the presence of organic peroxides or oxygen. Make sure you know about both possibilities.

If this comes up in an exam, and it isn't made clear which set of conditions are being asked about, assume that the HBr is pure, and give the Markovnikov addition. It would help to add the words "assuming the HBr is pure" to whatever you write (even if it is only an equation or structure).

Statement 14.2.2(a)(iv): Addition of halogens

These reactions are covered on the page halogenation of alkenes.

Ignore the ethene-fluorine reaction. And don't worry about the equation in the section on bromine water which forms 2-bromoethanol. You can just use the simple equation to form 1,2-dibromoethane.

Statements 14.2.2(b) and 14.2.2(c)

These statements are both about the oxidation of alkenes using acidified potassium manganate(VII) solution under two different sets of conditions.

You will find more than you will need on the page about alkenes and potassium manganate(VII).

Statement 14.2.2(b)

This is about the reaction with cold dilute potassium manganate(VII) solution.

You can ignore all the complete equations for these reactions. All you need for CIE purposes are the simple equations involving an oxygen in square brackets - like the first equation on the page. You can also ignore any reference to alkaline conditions. The syllabus talks about "acidified manganate(VII) ions".

Statement 14.2.2(c)

This is about the reaction with hot concentrated potassium manganate(VII) solution.

The reaction with hot concentrated, acidified potassium manganate(VII) is a bit of a problem if you are meeting this early in your organic chemistry course (as you probably are). It includes some really quite complicated chemistry, and discusses compounds that you won't have heard of.

If you try to understand this now, it will totally confuse you unless you are quite confident about your chemistry.

Read the section headed "Oxidation of alkenes with hot concentrated acidified potassium manganate(VII) solution", down as far as the green box which says "Warning!".

Then what I suggest you do for now is to make a note of the summary further down the page (which I will repeat below), but also make a note to come back and look at this again after you have met aldehydes, ketones and carboxylic acids later on in the year.

You must come back to this, though, because CIE do ask questions about it.

To repeat the summary from the page I pointed you at:


Think about both ends of the carbon-carbon double bond separately, and then combine 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 end of the bond, that part of the molecule will give a carboxylic acid.

  • If there are two hydrogens at one end of the bond, that part of the molecule will give carbon dioxide and water.

Two words that you may not be familiar with:

  • A ketone is a compound with a hydrocarbon group either side of a carbon-oxygen double bond. Look back at some structures on that page to see what I mean.

  • A carboxylic acid contains the group -COOH.

For now, just learn this, without trying too hard to understand it. This goes against my principles, but trying to get you to understand this now is likely to put you off chemistry - unless, of course, you find chemistry relatively easy.

Once you have made a note of this, try the examples in the very last section of the page, headed "What is the point of all this?". All you need to do is to follow the rules in the summary to get the right answers. If you get this right, well done! This is fairly typical of what you will need to be able to do in a CIE exam.

Statement 14.2.2(d)

This is about the polymerisation of ethene and propene. You will also meet this again in Section 20, with some other examples of the same thing. We will look at this in more detail then.

You will find far more than you will need on the page about the polymerisation of alkenes.

A fairly common question asks about the bond angles in the polymer as compared with the monomer. In the monomer, the angle around the carbon-carbon double bond is about 120°. In the polymer, where each carbon is joined by single bonds, it will be about 109.5°.

There is no hint in the syllabus that you will be expected to know about the various sorts of poly(propene). Neither has anything been asked so far about the particular properties and uses of the various polymers.

In summary

Make sure that you can write a simple equation for the formation of poly(ethene) and poly(propene), and draw the repeating units in the two polymers accurately.

Be able to quote a set of conditions for the production of both polymers.

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© Jim Clark 2020