4.1.2 (d,e,f,g) Reactions of Alkanes

Syllabus

(d) the low reactivity of Alkanes with many reagents in terms of the high bond enthalpy and very low polarity of the σ-bonds present (see also 2.2.2 j)

{Use of ideas about enthalpy and polarity to explain macroscopic properties of Alkanes.}

(e) complete combustion of Alkanes, as used in fuels, and the incomplete combustion of Alkane fuels in a limited supply of Oxygen with the resulting potential dangers from CO

(f) the reaction of Alkanes with chlorine and bromine by radical substitution using ultraviolet radiation, including a mechanism involving homolytic fission and radical reactions in terms of initiation, propagation and termination (see also 4.1.1 f–g)

{Learners are not required to use ‘half curly arrows’ in this mechanism.}

{Equations should show which species are radicals using a single ‘dot’, •, to represent the unpaired electron.}

(g) the limitations of radical substitution in synthesis by the formation of a mixture of organic products, in terms of further substitution and reactions at different positions in a Carbon chain.

What does this mean?

Why aren't Alkanes Reactive?

Alkanes are the most basic homologous series - they have no functional group.

This means that they have no centre where reactions tend to happen.

Why not?

Because C-C and C-H bonds are both strong and non-polar.

They're not easily broken and also not very attractive to other reactants.

A polar bond would attract other charged or partially charged molecules.

C-C bonds are obviously non-polar because both atoms have the same electronegativity; C-H bonds are considered non-polar because the difference in electronegativity is very small

Weaker bonds would mean that energy would be easily released when a C-H or C-C bond was broken and replaced by a C-X bond - (where X is another atom).

Combustion of Alkanes

One of the few substances that Alkanes react with well is Oxygen.

We call this reaction Combustion and it is the basis of the whole Oil and Gas industry that powers the world.

Even partial combustion of alkanes is exothermic.

This is a problem because it means that combustion may continue in low Oxygen conditions producing Carbon Monoxide which is toxic due to the strength with which it binds to red blood cells, stopping them transporting Oxygen. (And, yes. that is Biology but it was on the exam one year).

The reason for these exothemic reactions are that the C=O bonds (and O-H bonds) produced are sufficiently strong that although it takes a great deal of energy to break the C-H and C-C bonds in the Alkane (and O=O bonds in the Oxygen) more energy is released than taken-in overall.

As we can see from the graph, the longer the Alkane the more energy is released when 1 mole is burned.

But the longer the Alkane the more electrons it will have and the stronger its van der Waals forces.

Consequently, long Alkanes have high boiling points.

And it is only Alkanes as vapours that burn.

Methane, Ethane, Propane and Butane are gases at room temperature so therefore burn particularly easily even if their energy content is moderate.

So the Natural Gas fraction is highly valued for cooking and heating.

You can run car on gas but its more convenient to have a liquid fuel that easily vaporises - Petrol.

This means alkanes that are long enough to be liquid but not so long that they require very much heat to turn into a gas.

The reason that Diesel is used is that it has a higher energy content - making it good for fuel economy.

On the downside, it has to be heated to vaporise, so the engines are a little more complicated and therefore generally a bit more expensive to manufacture and to buy.

Jet engines need even more energy and so use even longer alkanes with even higher boiling points.

And really vast engines, in Supertankers etc, use thick oils to power them that require a great deal of heating to vaporise.

Radical Substitution.

Initiation - creating radicals

To get Alkanes to react with anything other than Oxygen requires us to "force them into it".

Physicists will tell you that E = hxf - in other words the Energy of a wave is proportional to its frequency.

As it happens, Ultraviolet light (UV) has a frequency that matches the strength of Cl-Cl bonds.

The bonds break by homolytic fission - this means that each Chlorine atom gets one of the bonding pair of electrons.

They become radicals - species that react with just about anything they collide with.

Notice that each radical is shown with a dot to represent their unpaired electron.

This is called the initiation phase because nothing further can happen unless this reaction happens first.

Older textbooks may show homolytic fission with two half-curly arrows.

Since a curly arrow represents a movement of a pair of electrons, a half-curly arrow represented one electron moving.

You don't have to write this any more.

But it isn't wrong.

Propagation - swapping radicals

When a radical collides with a molecule it will probably react.

All the electrons in the molecule were in pairs.

So when the radical reacts and forms a bond it will "pass on" its unpaired electron - forming a new radical.

And if this new radical collides with another molecule it will pass on its unpaired electron forming another radical.

Etc.

Each reaction above is a propagation because each starts with a radical and makes a new one,

Again, you don't have to include the half-arrows but you would have to include the dots.

Since the collisions are uncontrollable you can't use this method to make one particular product.

Exam questions often ask you to draw the first two of these propagation reactions to form CH₃Cl.

But once that has been successfully done more collisions are inevitable because the remaining Chlorine radicals can't be stopped from colliding with this desired product.

So further substitutions will happen and dichloro-, trichloro and tetrachloromethane will form, creating a mixture of products.

This would mean that messy and expensive separation would be required.

Termination - destroying radicals

There's nothing to stop two radicals colliding.

Their unpaired electrons will pair forming a molecule that is not a radical.

This is how the reaction eventually comes to an end - which is why it is called termination.

These are the radicals that form in the propagation phases so these are the three obvious products.

Of course, if there are still chlorine radicals around (and there may well be) even these products could end up with further substitutions.

So any chlorinated ethane might form, or longer alkanes may form too.

Examiners don't ask for these.

They do sometimes ask for an explanation of the word "substitution" however.

Just as in football it means replacing one with another.

In this case exchanging one of the Hydrogen atoms in CH₄ with a chlorine atom to make CH₃Cl, for example.

Methane is the most commonly used Alkane in exam questions but it could be any Alkane.

Chlorine is the most commonly used halogen but it could equally be Bromine.

It won't be Fluorine because F-F bonds are too strong to be broken by UV light.

Video

Exam-style questions

1. a) Butane, C₄H₁₀, is a hydrocarbon which is used as a fuel.

(i) Explain what is meant by the term hydrocarbon.

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(ii) Explain what is meant by the term fuel.

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(iii) Write an equation for the complete combustion of Butane.

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(iv) Write an equation for the incomplete combustion of Butane to produce Carbon Monoxide and water.

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(v) Under what conditions would you expect incomplete combustion to occur?

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b) The burning of fossil fuels can produce atmospheric pollutants.

The combustion of petrol in internal combustion engines can lead to the formation of Carbon Monoxide, CO, & Nitrogen Monoxide, NO.

(i) Write an equation for the incomplete combustion of Octane, C₈H₁₈, to produce CO and water only.

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2. Chloromethane can be prepared by a reaction between Methane and Chlorine in the presence of ultraviolet radiation.

(i) Outline the mechanism for this reaction.

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(ii) Name the first step in this mechanism.

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(iii) Suggest, giving a reason, how the reaction would be carried out in order to ensure that Chloromethane would be the major organic product.

Suggestion...........................................................................................................................................................

Reason.................................................................................................................................................................

(Total 7 marks)

Answers

1. (a) (i) A molecule/compound/it consists/it is composed/it is made up of Hydrogen/H and Carbon/C only (1)

(ii) release (heat) energy (when burned) (1)

OR provides a (useable form of) energy

OR is a source of energy

Accept heat º energy

NOT is energy / is heat

NOT burns exothermically

(iii) C₄H₁₀ + 6 ½ O₂ ® 4CO₂ + 5H₂O (1)

OR 2C₄H₁₀ + 13 O₂ ® 8CO₂ + 10H₂O

ignore state symbols

(iv) C₄H₁₀ + 4½ O₂ ® 4CO + 5H₂O (1)

OR 2C₄H₁₀ + 9 O₂ ® 8CO + 10 H₂O

ignore state symbols

(iii) and (iv) if not C₄H₁₀ = CE

(v) Limited or reduced supply of air / Oxygen (1)

OR low temperature OR poor mixing

OR insufficient oxygen / air OR shortage of O₂

NOT no oxygen / lack of oxygen / not in excess

(b) C₈H₁₈ + 8½ O₂ ® 8CO + 9H₂O (1)

OR double this equation

2. (i) Cl2 ® 2 Cl· (1)

CH₄ + Cl· ® CH₃· + HCl (1)

CH₃· + Cl2 ® CH₃ Cl + Cl· (1)

Cl· + Cl· ® Cl₂

or CH₃· + CH₃· ® C₂H₆

or CH₃· + Cl· ® CH₃Cl

etc.(1)

(ii) Initiation (1)

(iii) Suggestion Use excess methane or limit the amount of Cl2 (1)

Reason Idea of - to minimise multiple substitutions (1)

or to minimise or prevent further reaction or substitution (of CH3Cl)

or more chance of Cl· colliding with CH4 or methane

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