AS Chemistry
Unit 4: Organic Chemistry 1
Part 6: Bonding in methane, ethane and ethene (sigma and pi bonds)
We have seen that cracking leads to the formation of alkenes. These are a family
of hydrocarbons that contain only one carbon-carbon double bond. The simplest
alkene is ethene, CH2=CH2. The general formula of the homologous series of
alkenes is CnH2n.
Task 1
Draw structural formulae for the following alkenes:
a) pent-2-ene
b) hex-3-ene
c) 2,3-dimethylpent-2-ene
d) cyclopenta-1,3-diene
e) 3-ethylhept-1-ene
To understand the origin of the double bond in these molecules, and hence the
reactivity of this family, we need an understanding of the bonding in methane and
ethane.
Methane, CH4
The simple view of the bonding in methane
Task 2
a) Can you draw a dot-and-cross diagram for methane?
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
b) Write the electronic configuration for a carbon atom………………………………………….
c) Write this again using the electrons-in-boxes notation. Only the outer
electrons should be shown.
2s
2px
2py
2pz
d) Why does this more sophisticated model not match the dot-and-cross
diagram you drew earlier?
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………
Promotion of an electron and hybridisation
When bonds are formed, energy is released and the system
becomes more stable. If carbon forms 4 bonds rather than
2, twice as much energy is released and so the resulting
molecule becomes even more stable.
There is only a small energy gap between the 2s and 2p
orbitals, and so it pays the carbon to provide a small amount
of energy to promote an electron from the 2s to the empty
2p to give 4 unpaired electrons. The extra energy released
when the bonds form more than compensates for the initial
input.
and three p orbitals).
The electrons now rearrange themselves again in a process
called hybridisation. This reorganises the electrons into four
identical hybrid orbitals called sp3 hybrids
(because they are made from one s orbital
These arrange themselves in space so that they are as far apart
as possible (remember VSEPR theory) to form a tetrahedron.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
In methane four molecular orbitals are formed
with hydrogen. These look rather like the original
sp3 hybrids, but with a hydrogen nucleus
embedded in each lobe.
any covalently-bonded molecule.
The principles involved - promotion of electrons if
necessary, then hybridisation, followed by the
formation of molecular orbitals - can be applied to
Ethane, C2H6
The formation of molecular orbitals in ethane
Ethane is a simple example of how a carbon-carbon single bond is formed.
Each carbon atom in the ethane promotes an electron and then forms sp 3 hybrids
exactly as we've described in methane. So just before bonding, the atoms look like
this:
The hydrogen atoms bond with the two carbons to produce molecular orbitals just
as they did with methane. The two carbon atoms bond by merging their remaining
sp3 hybrid orbitals end-to-end to make a new molecular orbital. The bond formed
by this end-to-end overlap is called a sigma bond. A σ-bond is symmetrical with
respect to rotation about the bond axis. The bonds between the carbons and
hydrogens are also therefore sigma bonds.
In any sigma bond, the most likely place to find the pair of electrons is on a line
between the two nuclei.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Free rotation about the carbon-carbon single bond
The two ends of this molecule can spin quite freely about the sigma bond so that
there are, in a sense, an infinite number of possibilities for the shape of an ethane
molecule. Some possible shapes are:
In each case, the left hand CH3 group has been kept in a constant position so that
you can see the effect of spinning the right hand one.
Other alkanes
All other alkanes will be bonded in the same way:



The carbon atoms will each promote an electron and then hybridise to give
sp3 hybrid orbitals.
The carbon atoms will join to each other by forming sigma bonds by the endto-end overlap of their sp3 hybrid orbitals.
Hydrogen atoms will join on wherever they are needed by overlapping their
1s1 orbitals with sp3 hybrid orbitals on the carbon atoms.
There will therefore be a tetrahedral arrangement of bonds around a 4coordinated sp3 carbon atom.
Ethene, C2H4
The simple view of the bonding in ethene
At a simple level, we have drawn ethene showing two bonds
between the carbon atoms. Each line in this diagram represents
one pair of shared electrons.
Ethene is actually much more interesting than this.
An orbital view of the bonding in ethene
Ethene is built from hydrogen atoms (1s1) and carbon atoms (1s22s22px12py1).
The carbon atom doesn't have enough unpaired electrons to form the required
number of bonds, so it needs to promote one of the 2s2 pair into the empty 2pz
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
orbital. This is exactly the same as happens whenever carbon forms bonds whatever else it ends up joined to.
Now there's a difference, because each carbon is only joining to three other atoms
rather than four - as in methane or ethane. This time, when the carbon atoms
hybridise their outer orbitals before forming bonds, they only hybridise three of
the orbitals rather than all four. They use the 2s electron and two of the 2p
electrons, but leave the other 2p electron unchanged.
Task 3
What do you think these hybrid orbitals might be called?.................................................
These new orbitals look rather like sp3 orbitals except that they are shorter and
fatter (they have more s character and less p character). The orbitals arrange
themselves as far apart as possible - which is at 120° to each other in a plane (i.e. a
trigonal planar arrangement). The remaining p orbital is at right angles to them.
The two carbon atoms and four hydrogen atoms would look like this before they
joined together:
The various atomic orbitals which are pointing towards each other now merge to
give molecular orbitals, each containing a bonding pair of electrons. These are
sigma bonds - just like those formed by end-to-end overlap of atomic orbitals in,
say, ethane.
Notice that the p orbitals are so close that they are overlapping sideways.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
This sideways overlap also creates a molecular orbital, but of a different kind. In
this one the electrons aren't held on the line between the two nuclei, but above
and below the plane of the molecule. A bond formed in this way is called a pi bond.
Be clear about what a pi bond is. It is a region of space in which you can find the
two electrons which make up the bond. Those two electrons can live anywhere
within that space. It would be quite misleading to think of one living in the top and
the other in the bottom.
The pi bond dominates the chemistry of ethene. It is very vulnerable to attack - a
very negative region of space above and below the plane of the molecule. It is also
somewhat distant from the control of the nuclei and so is a weaker bond than the
sigma bond joining the two carbons.
All double bonds (whatever atoms they might be joining) will consist of a sigma
bond and a pi bond.
Task 4
Dehydration of pentan-1-ol produces pent-1-ene. Sketch on the diagram below the
orbital overlap between the two carbon atoms. Label the bonds.
References
A-level Chemistry pages 300-303
Chemistry in Context pages 419
Learning Objectives
Candidates should be able to
 describe covalent bonding in terms of orbital overlap, giving  and  bonds.
 explain the shape of, and bond angles in, ethane and ethane molecules in
terms of  and  bonds.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 7: What is stereoisomerism?
Part 1: Geometric Isomerism
ISOMERISM
STRUCTURAL ISOMERISM
STEREOISOMERISM
GEOMETRIC ISOMERISM
OPTICAL ISOMERISM
In stereoisomerism, the atoms making up the isomers are joined up in the same
order, but still manage to have a different arrangement in space
Geometric (cis / trans) isomerism
Geometric isomerism, also known as cis-trans isomerism, is one type of
stereoisomerism.
Task 1
Look at the two displayed formula below. Do these diagrams represent isomers?
Explain your answer.
………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………..
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
But what happens if you have a carbon-carbon double bond - as in 1,2dichloroethene?
These two molecules aren't the same. The carbon-carbon double bond won't rotate
and so you would have to take the models to pieces in order to convert one
structure into the other one. That is a simple test for isomers. If you have to take
a model to pieces to convert it into another one, then you've got isomers. If you
merely have to twist it a bit, then you haven't!
Drawing structural formulae for the last pair of models gives two possible isomers.
In one, the two chlorine atoms are locked on opposite sides of the double bond.
This is known as the trans isomer. (trans : from latin meaning "across" - as in
transatlantic).
In the other, the two chlorine atoms are locked on the same side of the double
bond. This is know as the cis isomer. (cis : from latin meaning "on this side")
Task 2
The most likely example of geometric isomerism you will meet at A-level is but-2ene. Can you draw and name its two geometric isomers?
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
The importance of drawing geometric isomers properly
It's very easy to miss geometric isomers in exams if you take short-cuts in drawing
the structural formulae. For example, it is very tempting to draw but-2-ene as:
CH3CH=CHCH3
If you write it like this, you will almost certainly miss the fact that there are
geometric isomers. For many situations the formulae given above is fine. However,
if there is even the slightest hint in a question that isomers might be involved,
always draw compounds containing carbon-carbon double bonds showing the correct
bond angles (120 °) around the carbon atoms at the ends of the bond. In other
words, use the format shown for 1,2-dichloroethene earlier.
What needs to be attached to the carbon-carbon double bond?
Think about this case:
Although we've swapped the right-hand groups around, these are still the same
molecule. To get from one to the other, all you would have to do is to turn the
whole model over.
You won't have geometric isomers if there are two identical groups on one end of
the bond. So . . . there must be two different groups on the left-hand carbon and
two different groups on the right-hand one. The cases we've been exploring earlier
are like this:
Here, the two different groups are either on the same side of the bond or the
opposite side.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Summary
To get geometric isomers you must have:


restricted rotation (involving a carbon-carbon double bond for A-level
purposes);
two different groups on the left-hand end of the bond and two different
groups on the right-hand end. It doesn't matter whether the left-hand
groups are the same as the right-hand ones or not.
Geometrical isomers normally have similar chemical properties but often their
physical properties are markedly different. For example the table below shows the
melting point and boiling point of the cis and trans isomers of 1,2-dichloroethene.
isomer
melting point (°C)
boiling point (°C)
cis
-80
60
trans
-50
48
In each case, the higher melting or boiling point is shown in bold.
You will notice that:


the trans isomer has the higher melting point;
the cis isomer has the higher boiling point.
This is common. You can see the same effect with other cis and trans isomers. The
differences are due to two factors:


the relative strengths of the intermolecular forces;
how well the molecules can pack together in the solid.
Task 3
Complete the worksheet ‘Geometrical Isomerism’.
References
A-level Chemistry page 281
Chemistry in Context pages 418-419
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Learning Objectives
Candidates should be able to
 describe cis-trans isomerism in alkenes, and explain its origin in terms of
restricted rotation due to the presence of π bonds.
 deduce the possible isomers for an organic molecule of known molecular
formula.
 identify cis-trans isomerism in a molecule of given structural formula.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 8: What is stereoisomerism?
Part 2: Optical Isomerism
Optical isomerism arises because of the different ways in which you can arrange
four groups around a carbon atom. It is a type of stereoisomerism.
Four single bonds around a carbon atom are arranged tetrahedrally. When four
different atoms or groups are attached to these four bonds, the molecules can
exist in two isomeric forms.
Look at the diagram below. This illustrates the two optical isomers of the amino
acid alanine (2-aminopropanoic acid).
They are different because they are mirror images. All molecules have mirror
images, but they don’t all exist as two isomers. What makes alanine exist in two
forms is that the mirror image and the original molecule are non-superimposable.
If you move the mirror image across to the left, the H atom, C atom and CH 3 group
will coincide, but NH2 and COOH will be in the wrong places. No amount of twisting
and turning will put things right! The only way you can make them superimpose is to
break bonds and swap the two groups around.
Key Terms
Optical isomers are named according to their effect on plane polarised light. There
are a number of different ways of naming the enantiomers, e.g. (+) or (-), L- and D-,
or R- and S-. At A-level you will not be expected to name them.
Molecules such as the two forms of alanine are called optical isomers (or
enantiomers). You don’t have to build models to find them. Whenever a molecule
contains a carbon atom that is surrounded by four different atoms or groups of
atoms there will be optical isomerism.
Molecules that are not superimposable on their mirror images are called chiral
molecules. A carbon atom that is surrounded by four different groups is called a
chiral centre.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
E.g. Butan-2-ol
The asymmetric carbon atom in a compound (the one with four different groups
attached) is often shown by a star.
It's extremely important to draw the isomers correctly. Draw one of them using
standard bond notation to show the 3-dimensional arrangement around the chiral
carbon atom. Then draw the mirror to show the examiner that you know what you
are doing, and then the mirror image.
Task 1
Can you draw the two optical isomers of butan-2-ol
Notice that you don't literally draw the mirror images of all the letters and
numbers! It is, however, quite useful to reverse large groups.
It doesn't matter in the least in what order you draw the four groups around the
central carbon. As long as your mirror image is drawn accurately, you will
automatically have drawn the two isomers.
How do enantiomers differ?
Enantiomers behave identically in ordinary test-tube chemical reactions. Most of
their physical properties, such as melting point, density and solubility, are also the
same. But enantiomers behave differently in the presence of other chiral
molecules.


The proteins in our body are built up from only one enantiomer of each acid.
Enantiomers can interact differently with the chiral ‘taste buds’ on your
tongue. One form of amino acids all taste sweet, while the other is often
tasteless or bitter.
Cambridge A-level Centre
AS Chemistry


Unit 4: Organic Chemistry 1
Enantiomers can smell different. For example the different smells of
oranges and lemons are caused by enantiomers.
Many beneficial medicines have enantiomers which have little or no
pharmacological effect. In the case of one medicine, thalidomide, the
apparently non-active enantiomer was found to damage unborn children when
the drug was taken during pregnancy. A similar situation arises with L-dopa,
used in the treatment of Parkinson’s Disease.
Task 2
Question 15.11 on page 282 of A-level Chemistry.
Task 3
Complete question from worksheet ‘Optical Isomers’.
References
A-level Chemistry pages 282-285
Chemistry in Context pages 392-393
Learning Objectives
Candidates should be able to:
 explain what is meant by a chiral centre and that such a centre gives rise to
optical isomerism.
 deduce the possible isomers for an organic molecule of known molecular
formula.
 identify chiral centres in a molecule of given structural formula.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 9: Electrophilic Addition?
Task 1
Can you complete the gaps in the passage below?
Most alkene reactions involve __________the  bond. This is __________ than
the C-C __________bond and reacts with a variety of reagents. The
characteristic __________ of an alkene involves a simple __________ (such as
hydrogen, water or bromine) joining across the __________ bond to form a
__________ product. Such reactions are called __________ reactions.
The structure of ethene
We are going to start by looking at ethene, because it is the simplest molecule
containing a carbon-carbon double bond. What is true of C=C in ethene will be
equally true of C=C in more complicated alkenes.
Ethene, C2H4, is often modelled as shown on the right. The
double bond between the carbon atoms is, of course, two pairs
of shared electrons.
What the diagram doesn't show is that the two pairs aren't the same as each
other.
One of the pairs of electrons is held on the line between the two carbon nuclei, a
sigma bond, but the other is held in a molecular orbital above
and below the plane of the molecule. This bond is called a pi
bond. The electrons in the pi bond are free to move around
anywhere in this shaded region and can move freely from one
half to the other.
The pi electrons are not as fully under the control of the
carbon nuclei as the electrons in the sigma bond and, because they lie exposed
above and below the rest of the molecule, they are relatively open to attack by
other things.
Electrophiles
An electrophile is something which is attracted to electron-rich regions in other
molecules or ions. Because it is attracted to a negative region, an electrophile must
be something which carries either a full positive charge, or has a slight positive
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
charge on it somewhere. Ethene and the other alkenes are attacked by
electrophiles.
The electrophilic addition of bromine to ethene
The facts
Alkenes react in the cold with both pure liquid bromine, or with a bromine solution.
The double bond breaks, and a bromine atom becomes attached to each carbon
atom. The bromine loses its original red-brown colour to give a colourless liquid. In
the case of the reaction with ethene, 1,2-dibromoethane is formed.
Task 2
Write an equation for the reaction above using structural formulae
This decolourisation of bromine is often used as a test for a carbon-carbon double
bond. The other halogens, apart from fluorine, behave similarly. (Fluorine reacts
explosively with all hydrocarbons - including alkenes - to give carbon and hydrogen
fluoride.)
The mechanism for the reaction between ethene and bromine
The reaction is an example of electrophilic addition.
Bromine as an electrophile
The bromine is a very "polarisable" molecule and the approaching  bond in the
ethene induces a dipole in the bromine molecule. If you draw this mechanism in an
exam, write the words "induced dipole" next to the bromine molecule - to show
that you understand what's going on.
The simplified version of the mechanism
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Summary notes on Electrophilic additions




Alkenes can become saturated by the addition of small molecules across the
double bond.
The most common reactions of the alkenes are described as electrophilic
additions.
You should be able to write a mechanism for the reaction of ethene with
bromine water (a test for alkenes).
You should also know the reactions (no mechanisms required) and conditions for
the reaction of alkenes with both hydrogen, steam and hydrogen halides, and
other halides.
References
A-level Chemistry page 301-303
Chemistry in Context pages 421-424
Learning Objectives
Candidates should be able to:
 describe the mechanism of electrophilic addition in alkenes, using
bromine/ethene as an example.
 describe the chemistry of alkenes as exemplified, where relevant, by the
following reactions of ethene: addition of hydrogen, steam, hydrogen halides
and halogens.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 10: Electrophilic addition to unsymmetrical alkenes
Task 1
So far we have only considered electrophilic additions involving ethene. In the
space below write a mechanism for the reaction of propene with hydrogen bromide.
You should have found that two products are possible. This situation occurs when
an asymmetrical alkene reacts with another asymmetrical molecule. When this
reaction is actually carried out, it is found that one product is more abundant, and
therefore more stable, than the other.

The major product is the one formed via the more stable carbocation.

The order of stability of carbocations is:
tertiary (3o) > secondary (2o) > primary (1o).

A primary carbocation has one alkyl group attached to C+. A secondary
carbocation has two alkyl groups attached to C+. A tertiary carbocation has
three alkyl groups attached to C+.

The increased stability of the tertiary carbocation is due to the electronreleasing (inductive) effect of the attached alkyl groups. Alkyl groups tend
to push electrons slightly towards any carbon atoms to which they are
attached.
Task 2
Draw the four possible carbocations of C4H9+ and label each as 1o, 2o and 3o.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Task 3
Use the information above to deduce the major product of the reaction between
propene and hydrogen bromide.
Task 4
Questions 3-5 on page 307 of A-level Chemistry
References
Chemistry in Context pages 423-424
A-level Chemistry page 303
Learning Objectives
Candidates should be able to:

describe the chemistry of alkenes as exemplified, where relevant, by the
following reactions of ethene: addition of hydrogen, steam, hydrogen halides
and halogens.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 11: The Polymerisation of alkenes
Here we look at the polymerisation of alkenes to produce polymers like
poly(ethene) (usually known as polythene) and PVC.
Poly(ethene) (polythene or polyethylene)
Manufacture
During polymerisation, an alkene molecule undergoes an addition reaction to itself.
As we have seen an addition reaction is one in which two or more molecules join
together to give a single product. A polymer is a long molecule (or macromolecule)
made up from lots of small molecules called monomers.
During the polymerisation of ethene, thousands of ethene molecules join together
to make poly(ethene) - commonly called polythene.
Task 1
Can you write an equation for the polymerisation of ethene?
The number of molecules joining up is very variable, but is in the region of 2000 to
20000.
Conditions
Temperature:
about 200°C
Pressure:
about 2000 atmospheres
Initiator:
often a small amount of oxygen as an impurity
The mechanism
The over-all process is known as a free radical addition.
Chain initiation
The chain is initiated by free radicals, Ra , produced by reaction between some of
the ethene and the oxygen initiator. You are not expected to know any details.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Chain propagation
Each time a free radical hits an ethene molecule a new longer free radical is
formed.
etc.
Chain termination
Eventually two free radicals hit each other producing a final molecule. The process
stops here because no new free radicals are formed.
Because chain termination is a random process, poly(ethene) will be made up of
chains of all sorts of different lengths.
Properties and uses
Poly(ethene) made in this way is called low density poly(ethene). One polymer chain
is held to its neighbours in the solid structure by weak van der Waals dispersion
forces (temporary dipole – induced dipole interactions). Poly(ethene) has quite a lot
of branching along the hydrocarbon chains, and this prevents the chains from lying
very close to each other. As a result the material is soft and flexible. Low density
poly(ethene) is used for familiar things like plastic carrier bags and other similar
low strength and flexible sheet materials.
Another method of making polythene was developed by Ziegler in the 1950s. This
process uses catalysts at lower temperatures and pressure. The molecules
produced have few branches and can pack closely together. This is known as highdensity polythene (HDPE). It is more rigid and has a higher melting point. It is used
for moulding rigid articles such as crates.
Poly(chloroethene) (polyvinyl chloride): PVC
Poly(chloroethene) is commonly known by the initials of its old name, PVC.
Structure
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Poly(chloroethene) is made by polymerising chloroethene, CH2=CHCl. Working out
its structure is no different from working out the structure of poly(ethene) (see
above). As long as you draw the chloroethene molecule in the right way, the
structure is pretty obvious.
The equation is usually written:
It doesn't matter which carbon you attach the chlorine to in the original molecule.
Just be consistent on both sides of the equation.
Properties and uses
Pure poly(chloroethene) tends to be rather hard and rigid. This is because of the
presence of additional dipole-dipole interactions due to the polarity of the carbonchlorine bonds. Chlorine is more electronegative than carbon, and so attracts the
electrons in the bond towards itself. That makes the chlorine atoms slightly
negative and the carbons slightly positive.
Plasticisers are added to the poly(chloroethene) to reduce the effectiveness of
these attractions and make the plastic more flexible. The more plasticiser you add,
the more flexible it becomes.
Poly(chloroethene) is used to make a wide range of things including guttering,
plastic window frames, electrical cable insulation, sheet materials for flooring and
other uses, footwear, clothing, and so on and so on.
Task 2
Can you write balanced equations for the polymerisation of propene and
tetrafluoroethene?
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Disposal of polymers
Polymers are strong and lightweight. As a result they have found many uses in our
modern world. However, there are problems associated with the widespread use of
polymers because they are difficult to dispose of. They are resistant to chemical
attack and to bacteria (non-biodegradable).
Task 3
Using pages 305-307 of your textbook can you summarise some of the advantages
and disadvantages of the various methods used to dispose of polymers.
Method
Comments
Landfill
Incineration
Recycling
Feedstock
recycling
References
Chemistry in Context pages
A-level Chemistry page
Learning Objectives
Candidates should be able to:
 describe the chemistry of alkenes including polymerisation.
 describe the characteristics of addition polymerisation as exemplified by
poly(ethene) and PVC.
 Recognize the difficulty of the disposal of poly(alkene)s, i.e. nonbiodegradability and harmful combustion products.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 12: Oxidation of alkenes
Reactions of alkenes with manganate (VII) ions, MnO4Manganate (VII) ions are oxidizing agents. Alkenes undergo different reactions
with manganate (VII) ions, depending on the conditions.
1. In the presence of dilute (acidified or alkaline) potassium manganate (VII).
Alkenes react readily at room temperature (i.e. in the cold).
The purple colour disappears and a diol is formed.


CH2=CH2
+
H 2O
+
[O]
HOCH2CH2OH
ethane – 1,2-diol

2. In the presence of a hot, concentrated solution of acidified potassium
manganate (VII), any diol formed is split into two fragments which are oxidized
further to carbon dioxide, a ketone or a carboxylic acid. Analysis of the products
can indicate where the carbon-carbon double bond was in the alkene.
Fragment
Product
=CH2
CO2
R-CH=
Aldehyde
R2C=
→
carboxylic acid
Ketone
Task
Some alkenes were heated in the presence of an acidified solution of potassium
manganate (VII). The products obtained in each reaction are listed below. In each
case identify the initial alkene and give its structural formula.
1.
CO2
2.
Cambridge A-level Centre
+
CO2
AS Chemistry
3.
Unit 4: Organic Chemistry 1
+
CO2
4.
References
A-level Chemistry page 304
Learning Objectives
Candidates should be able to describe the oxidation of alkenes by:
 cold, dilute, acidified manganate(VII) ions to form the diol, and
 hot, concentrated, acidified manganate(VII) ions leading to the
rupture of the carbon-to-carbon double bond in order to determine the
position of alkene linkages in larger molecules.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 13: Halogenoalkanes (haloalkanes)
The simple halogenoalkanes are a homologous series of compounds in which one or
more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine,
chlorine, bromine or iodine). They have the general formula CnH2n+1X. They do not
occur much in nature, but they are useful for all sorts of human purposes, so
chemists make and use them a lot. As with all functional groups, the halogen atom
modifies the properties of the relatively unreactive hydrocarbon chain.
They are named after the parent alkanes, with the halogen atom added as the
prefix fluoro-, chloro-, bromo- or iodo-.
Task 1
Use systematic nomenclature to name the following halogenoalkanes. Draw the
displayed formula of each.
a.
b.
c.
CHCl3
CH3CHClCH3
CF3CCl3
Typically, halogenoalkanes are volatile liquids that do not mix with water.
Task 2
a) Explain why 1-chloropropane, C3H7Cl, is a liquid at room temperature whereas
butane is a gas.
b) Why is it that halogen compounds such as 1-chloropropane do not mix with
water?
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Task 3
Can you solve the anagrams (words in bold) to complete the notes below?
Nucleophilic substitution
Halogen atoms are negotiate clever. As a result, the carbon-halogen bond is alp or.
The electrons in the C-X bond are cadet tart towards the halogen atom which
gains a slight eat given charge, leaving the carbon atom electron enticed if.
The + carbon is susceptible to attack by chenille soup, i.e. ions or molecules with a
lone pair of electrons. The C-X bond is broken releasing the had lie ion. A
nucleophilic stubs tuition reaction has occurred.
Task 4
Using the notes above as a guide can you write a general reaction mechanism for
this type of reaction. You should represent the nucleophile as Nu -. Please note,
nucleophiles do not have to carry a negative charge.
This



is known as an SN2 reaction.
S stands for substitution,
N for nucleophilic, and
2 because the initial stage of the reaction involves two species.
All reactions of the haloalkanes involve breaking the C-X bond. The stronger the
bond, the more difficult it is to break, and the lower the rate of reaction.
Task 5
Look at the table of bond strengths and electronegativity values below. Which
halogenoalkane would you expect to be the most reactive and why?
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Halogen
Electronegativity
Bond strength (C-X) kJ mol-1
F
4.0
484
Cl
3.0
338
Br
2.8
276
I
2.5
238
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
Task 6
How could you measure the relative rates of these reactions?
Nucleophilic substitution reactions
Task 7
Using the general equation as your guide, can you write a mechanism for the
following reaction of bromoethane.
1. Hydroxide ions
When haloalkanes are warmed with NaOH (aq) or KOH(aq), alcohols are formed. N.B.
halogenoalkanes do not mix with water, so they are first mixed with a little
ethanol.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
This process is sometimes called hydrolysis.
A similar reaction occurs more slowly with water.
CH3CH2Br
+
H 2O

CH3CH2OH
+
HBr
2. Cyanide ions
Nucleophilic substitutions with cyanide ions add an extra carbon atom to the chain.
This reaction can, therefore, be useful in organic synthesis. The compounds formed
are known as nitriles, RCN. The reaction is carried out in a warm ethanolic solution.
Task 8
Can you write an equation for the reaction of bromoethane with KCN?
3. Ammonia
Primary amines are formed when an ethanolic solution of the haloalkanes are
warmed with an excess of ammonia in a sealed container (i.e. under pressure).
Task 9
Can you write equations for the two reactions below?
For example, bromoethane forms ethylamine:
Since the acid HBr immediately reacts with the base NH3, this equation is more
correctly written as:
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Task 10
Past paper question
References
A-level Chemistry pages 328 - 331
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Learning Objectives
Candidates should be able to recall the chemistry of halogenoalkanes as
exemplified by the following nucleophilic substitution reactions of bromoethane:
hydrolysis; formation of nitriles; and the formation of primary amines by reaction
with ammonia.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 14: Substitution vs. Elimination
The halogenoalkanes can be divided into three groups depending on the position of
the halogeno- group along the carbon chain.
Task 1
Can you draw appropriate structures to illustrate the three classes of
halogenoalkanes in the table below?
Type of halogenoalkane
Position of halogenogroup
Example
primary
at end of chain:
bromoethane
secondary
in middle of chain:
2-bromopropane
tertiary
attached to a carbon 2-bromo-2-methylpropane
atom which carries no H
atoms:
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Nucleophilic substitution
The mechanism of nucleophilic substitution in a tertiary halogenoalkane is
different from that discussed above for a primary halogenoalkane.
Task 2
Read the description below carefully and use the information to draw a reaction
mechanism for this hydrolysis reaction.
The hydrolysis of 2-bromo-2-methylpropane, a tertiary halogenoalkane, involves
the breaking of the C-Br bond followed by attack by OH- ions on the resulting
carbocation.
Explanation: The electron-releasing methyl groups enhance the stability of the
carbocation, thus changing the mechanism of the hydrolysis. As the first step
involves only 1 species this mechanism is described as SN1.
The nucleophilic
mechanisms.
substitution
of
secondary
halogenoalkanes
involves
both
Elimination
We have already seen how the hydroxide ion can act as a nucleophile in the
reaction with a haloalkane to form an alcohol. You need to be aware however, that
this ion can also act as a strong base. An alternative reaction can take place in
which HBr is removed and an alkene is formed. This is known as elimination.
E.g.
CH3CH2Br
+
Cambridge A-level Centre
NaOH

CH2=CH2
+
NaBr
+
H 2O
AS Chemistry
Unit 4: Organic Chemistry 1
Task 3
Can you copy this mechanism into the space below?
Substitution vs. Elimination
Which mechanism is most prevalent depends on a number of factors:

The structure of the haloalkane
1o haloalkanes - predominantly substitution
2o haloalkanes – both reactions occur
3o haloalkanes – predominantly elimination

The reaction conditions
Higher temperatures favour elimination



The choice of solvent
Elimination is favoured by hot ethanolic conditions
Substitution is favoured by warm aqueous conditions.
References
A-level Chemistry pages 328 - 331
Learning Objectives
Candidates should be able to


recall the chemistry of halogenoalkanes as exemplified by the elimination of
hydrogen bromide from 2-bromopropane.
describe the mechanism of nucleophilic substitution (by both SN1 and SN2
mechanisms) in halogenoalkanes.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Part 15: Pros and cons of the halogenoalkanes
Task 1
Read the passage below on the uses of halogenoalkanes (and some related
compounds). See if you can find a replacement for each of the words/phrases in
bold which explains their meaning.
CFCs and their replacements
CFCs are chlorofluorocarbons. Two common CFCs are:
CFC-11
CCl3F
CFC-12
CCl2F2
How to work out the formula from the CFC number: You add 90 to the CFC
number to give a new number. For example, CFC-11 generates the number 101
(11+90)
The first digit of the new number tells you how many carbon atoms there are. The
second number tells you the number of hydrogens, and the third the number of
fluorines. You calculate the number of chlorines from the formula Cl = 2(C+1)-H-F.
Uses of CFCs
CFCs are non-flammable and not very toxic. They therefore had a large number of
uses.
They were used as refrigerants, propellants for aerosols, for generating foamed
plastics like expanded polystyrene or polyurethane foam, and as solvents for dry
cleaning and for general degreasing purposes.
Unfortunately, CFCs are largely responsible for destroying the ozone layer. In the
high atmosphere, the carbon-chlorine bonds break to give chlorine free radicals.
It is these radicals which destroy ozone. CFCs are now being replaced by less
environmentally harmful compounds.
The destruction of atmospheric ozone
Ozone, O3, is constantly being formed and broken up again in the high atmosphere
by the action of ultraviolet light. Ordinary oxygen molecules absorb ultraviolet
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
light and break into individual oxygen atoms. These have unpaired electrons, and
are known as free radicals. They are very reactive.
The oxygen radicals can then combine with ordinary oxygen molecules to make
ozone.
Ozone can also be split up again into ordinary oxygen and an oxygen radical by
absorbing ultraviolet light.
This formation and breaking up of ozone is going on all the time. Taken together,
these reactions stop a lot of harmful ultraviolet radiation penetrating the
atmosphere to reach the surface of the Earth.
The catalytic reaction we are interested in destroys the ozone and so stops it
absorbing UV in this way.
The slow breakdown of chlorofluorocarbons (CFCs) like CF2Cl2 in the atmosphere
produces chlorine atoms - chlorine free radicals. These catalyse the destruction
of the ozone.
This happens in two stages. In the first, the ozone is broken up and a new free
radical is produced.
The chlorine radical catalyst is regenerated by a second reaction. This can happen
in two ways depending on whether the ClO radical hits an ozone molecule or an
oxygen radical.
If it hits an oxygen radical (produced from one of the reactions we've looked at
previously):
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Or if it hits an ozone molecule:
Because the chlorine radical keeps on being regenerated, each one can destroy
thousands of ozone molecules.
CFCs can also cause global warming. One molecule of CFC-11, for example, has a
global warming potential about 5000 times greater than a molecule of carbon
dioxide.
On the other hand, there is far more carbon dioxide in the atmosphere than CFCs,
so global warming isn't the major problem associated with them.
Replacements for CFCs
These are still mainly halogenoalkanes, although simple alkanes such as butane can
be used for some applications (for example, as aerosol propellants).
Hydrochlorofluorocarbons, HCFCs
These are carbon compounds which contain hydrogen as well as halogen atoms. For
example:
HCFC-22
CHClF2
The formula can be worked out from the number in the name in exactly the same
way as for CFCs.
These have a shorter life in the atmosphere than CFCs, and much of them is
destroyed in the low atmosphere and so doesn't reach the ozone layer. HCFC-22
has only about one-twentieth of the effect on the ozone layer as a typical CFC.
Hydrofluorocarbons, HFCs
These are compounds containing only hydrogen and fluorine attached to carbon. For
example:
HFC-134a
Cambridge A-level Centre
CH2F-CF3
AS Chemistry
Unit 4: Organic Chemistry 1
Because these HCFCs don't contain any chlorine, they have zero effect on the
ozone layer. HFC-134a is now widely used in refrigerants, for blowing foamed
plastics and as a propellant in aerosols.
Hydrocarbons
Again, these have no effect on the ozone layer, but they do have a down-side. They
are highly flammable and are involved in environmental problems such as the
formation of photochemical smog.
Other uses of organic halogeno compounds
In making plastics
Strictly speaking, the compounds we are talking about here are halogenoalkenes,
not halogenoalkanes.
Chloroethene, CH2=CHCl, is used to make poly(chloroethene) - usually known as PVC.
Tetrafluoroethene, CF2=CF2, is used to make poly(tetrafluoroethene) - PTFE.
Lab uses of halogenoalkanes
Halogenoalkanes are useful in the lab as intermediates in making other organic
chemicals.
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
Additional uses of halogenoalkanes
Task 2
Use page 452 of ‘Chemistry in Context’ and pages 331-332 of ‘AS Level Chemistry’
to answer the following questions.
1. Anaesthetics
The substance 2-bromo-2-chloro-1,1,1-trifluoroethane (or Halothane) has been
widely used as an anaesthetic in hospitals since its discovery in 1956.
a. Draw the full structural formula of halothane in the space below.
b. Halothane contains fluorine, chlorine and bromine atoms. What properties
do these atoms lend to the molecule:
Fluorine:
Chlorine:
Bromine:
2. Flame retardants
BCF was widely used as a fire-fighting agent.
a. Which compound is BCF?____________________________________
b. Draw the full structural formula of BCF below:
Cambridge A-level Centre
AS Chemistry
Unit 4: Organic Chemistry 1
c. Why is BCF good at extinguishing fires?
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
d. Why is BCF no longer in general use?
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
References
A-level Chemistry pages 331-333
Learning Objectives
Candidates should be able to



interpret the different reactivities of halogenoalkanes e.g. CFCs;
anaesthetics; flame retardants; plastics with particular reference to
hydrolysis and to the relative strengths of the C-Hal bonds;
explain the uses of fluoroalkanes and hydrofluorooalkanes in terms of their
relative chemical inertness;
recognise the concern about the effect of chlorofluoroalkanes on the ozone
layer.
Cambridge A-level Centre