INTRODUCING HALOGENOALKANES (haloalkanes or alkyl halides)

Halogenoalkanes are also known as
haloalkanes or alkyl halides. This page explains
what they are and discusses their physical
properties. It also takes an initial look at their
chemical reactivity. Details of the chemical
reactions of halogenoalkanes are described on
separate pages.
What are halogenoalkanes?
Examples
Halogenoalkanes are compounds in which one
or more hydrogen atoms in an alkane have
been replaced by halogen atoms (fluorine,
chlorine, bromine or iodine). For the purposes
of UK A level, we will only look at compounds
containing one halogen atom.
For example:
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The different kinds of halogenoalkanes
Halogenoalkanes fall into different classes
depending on how the halogen atom is
positioned on the chain of carbon atoms. There
are some chemical differences between the
various types.
Primary halogenoalkanes
In a primary (1°) halogenoalkane, the carbon
which carries the halogen atom is only attached
to one other alkyl group.
Note: An alkyl group is a group such as
methyl, CH 3 , or ethyl, CH 3 CH2 . These are
groups containing chains of carbon atoms
which may be branched. Alkyl groups are given
the general symbol R.
Some examples of primary halogenoalkanes
include:
Notice that it doesn't matter how complicated
the attached alkyl group is. In each case there
is only one linkage to an alkyl group from the
CH 2 group holding the halogen.
There is an exception to this. CH 3 Br and the
other methyl halides are often counted as
primary halogenoalkanes even though there are
no alkyl groups attached to the carbon with the
halogen on it.
Secondary halogenoalkanes
In a secondary (2°) halogenoalkane, the carbon
with the halogen attached is joined directly to
two other alkyl groups, which may be the same
or different.
Examples:
Tertiary halogenoalkanes
In a tertiary (3°) halogenoalkane, the carbon
atom holding the halogen is attached directly to
three alkyl groups, which may be any
combination of same or different.
Examples:
Physical properties of halogenoalkanes
Boiling Points
The chart shows the boiling points of some
simple halogenoalkanes.
Notice that three of these have boiling points
below room temperature (taken as being about
20°C). These will be gases at room
temperature. All the others you are likely to
come across are liquids.
Remember:
the only methyl halide which is a liquid is
iodomethane;
chloroethane is a gas.
The patterns in boiling point reflect the patterns
in intermolecular attractions.
Note: If you aren't happy about intermolecular
forces (particularly van der Waals dispersion
forces and dipole-dipole interactions) then you
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van der Waals dispersion forces
These attractions get stronger as the molecules
get longer and have more electrons. That
increases the sizes of the temporary dipoles
that are set up.
This is why the boiling points increase as the
number of carbon atoms in the chains
increases. Look at the chart for a particular
type of halide (a chloride, for example).
Dispersion forces get stronger as you go from 1
to 2 to 3 carbons in the chain. It takes more
energy to overcome them, and so the boiling
points rise.
The increase in boiling point as you go from a
chloride to a bromide to an iodide (for a given
number of carbon atoms) is also because of
the increase in number of electrons leading to
larger dispersion forces. There are lots more
electrons in, for example, iodomethane than
there are in chloromethane - count them!
van der Waals dipole-dipole attractions
The carbon-halogen bonds (apart from the
carbon-iodine bond) are polar, because the
electron pair is pulled closer to the halogen
atom than the carbon. This is because (apart
from iodine) the halogens are more
electronegative than carbon.
The electronegativity values are:
C 2.5 F 4.0
Cl 3.0
Br 2.8
I 2.5
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This means that in addition to the dispersion
forces there will be forces due to the attractions
between the permanent dipoles (except in the
iodide case).
The size of those dipole-dipole attractions will
fall as the bonds get less polar (as you go from
chloride to bromide to iodide, for example).
Nevertheless, the boiling points rise! This
shows that the effect of the permanent dipole-
dipole attractions is much less important than
that of the temporary dipoles which cause the
dispersion forces.
The large increase in number of electrons by
the time you get to the iodide completely
outweighs the loss of any permanent dipoles in
the molecules.
Boiling points of some isomers
The examples show that the boiling points fall
as the isomers go from a primary to a
secondary to a tertiary halogenoalkane. This is
a simple result of the fall in the effectiveness of
the dispersion forces.
The temporary dipoles are greatest for the
longest molecule. The attractions are also
stronger if the molecules can lie closely
together. The tertiary halogenoalkane is very
short and fat, and won't have much close
contact with its neighbours.
Solubility of halogenoalkanes
Solubility in water
The halogenoalkanes are at best only very
slightly soluble in water.
In order for a halogenoalkane to dissolve in
water you have to break attractions between
the halogenoalkane molecules (van der Waals
dispersion and dipole-dipole interactions) and
break the hydrogen bonds between water
molecules. Both of these cost energy.
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Energy is released when new attractions are set
up between the halogenoalkane and the water
molecules. These will only be dispersion forces
and dipole-dipole interactions. These aren't as
strong as the original hydrogen bonds in the
water, and so not as much energy is released
as was used to separate the water molecules.
The energetics of the change are sufficiently
"unprofitable" that very little dissolves.
Note: To be accurate, you also have to
consider entropy changes when things
dissolve. If you don't yet know about entropy,
don't worry about it!
Solubility in organic solvents
Halogenoalkanes tend to dissolve in organic
solvents because the new intermolecular
attractions have much the same strength as
the ones being broken in the separate
halogenoalkane and solvent.
Chemical reactivity of halogenoalkanes
The importance of bond strengths
The pattern in strengths of the four carbon-
halogen bonds are:
Notice that bond strength falls as you go from
C-F to C-I, and notice how much stronger the
carbon-fluorine bond is than the rest.
Note: You will find almost as many different
values for bond strengths (or bond enthalpies
or bond energies) as there are different
sources! Don't worry about this - the pattern is
always the same. This is why you have got a
chart here rather than actual numbers.
In order for anything to react with the
halogenoalkanes, the carbon-halogen bond
has got to be broken. Because that gets easier
as you go from fluoride to chloride to bromide
to iodide, the compounds get more reactive in
that order.
Iodoalkanes are the most reactive and
fluoroalkanes are the least. In fact,
fluoroalkanes are so unreactive that we shall
pretty well ignore them completely from now on
in this section!
The influence of bond polarity
Of the four halogens, fluorine is the most
electronegative and iodine the least. That
means that the electron pair in the carbon-
fluorine bond will be dragged most towards the
halogen end.
Looking at the methyl halides as simple
examples:
The electronegativities of carbon and iodine are
equal and so there will be no separation of
charge on the bond.
Note: In the diagram, " " (read as "delta")
means "slightly" - so + means "slightly
positive". The larger the symbols, the more
positive or negative the atoms are intended to
be.
If this isn't reasonably familiar to you, then you
ought to have read the page about
electronegativity and polar bonds mentioned
above!
One of the important set of reactions of
halogenoalkanes involves replacing the
halogen by something else - substitution
reactions. These reactions involve either:
the carbon-halogen bond breaking to give
positive and negative ions. The ion with the
positively charged carbon atom then reacts
with something either fully or slightly negatively
charged.
something either fully or negatively charged
attracted to the slightly positive carbon atom
and pushing off the halogen atom.
You might have thought that either of these
would be more effective in the case of the
carbon-fluorine bond with the quite large
amounts of positive and negative charge
already present. But that's not so - quite the
opposite is true!
The thing that governs the reactivity is the
strength of the bonds which have to be broken.
If is difficult to break a carbon-fluorine bond,
but easy to break a carbon-iodine one.