Nucleophilic substitution involving halogenalkanes
We met the halogenalkanes earlier.
form a homologous series of compounds with the general formula CnH2n+1X, where X is one
of the halogens
(F, Cl, Br or I). Some simple examples of halogenalkane molecules are shown
The halogens are generally electronegative
elements which means that the C-X bond in the halogenalkane molecules
will be a polar one, with
the carbon atom in the C-X bond having a partial positive charge (δ+)
and the halogen atom having a
partial negative charge (δ-).
A nucleophile is an electron rich species.
Nucleophiles can be
neutral molecules or ions with lone pairs
which they can donate to an electron deficient molecule/atom,
that is an electrophile. Examples of nucleophiles
include hydroxide ions (OH-), cyanide ions (CN-),
ammonia (NH3) and water (H2O). All these molecules,
whether charged or neutral are able to act as nucleophiles
simply because they have lone pairs of electrons.
The electron deficient carbon atom attached the halogen in a
halogenalkane molecule is
susceptible to attack by
electron rich spieces, that is nucleophiles with lone
pairs of electrons. In the diagram shown below a nucleophile uses a
lone pair of electrons to form a covalent bond to the
carbon atom and at the same time the carbon-iodine bond breaks and an
iodide ion leaves. The end result is that the iodine is replaced or
substituted by the nucleophile, this is nucleophilic
substitution. As an example consider the diagram below which shows a nucleophile
using its lone pair of electrons to attack a polar C-I bond.
Factors affecting the rate of nucleophilic substitution reactions
One of the main factors that will influence how fast these
nucleophilic substitution reaction will occur is the
bond strength of
the carbon-halogen bond (C-X). Since the halogens
obviously increase in size as we go down group 7 this means that
the C-X bond length will be getting longer, this means a
decrease in bond strength for the carbon-halogen bond and a decrease in the
bond enthalpies from the strong C-F bond to the relatively weak C-I bond.
| C-X bond
||Bond enthalpy kJmol-1
One other factor that could influence the rate of these substitution reactions is
the polarity of the C-X bond. The C-F bond is obviously much more polar than the C-I bond, so it would be easy
that in a C-F
bond the size of the partial positive charge (δ+) on the carbon
atom is larger than that
in say a C-I bond, therefore the larger partial positive charge on the
carbon atom should encourage the nucleophile
to attack more quickly. However before the nucleophile can bond to
the carbon atom in the C-X bond, the halogen
atom must leave, therefore the strength of the C-X bond will be the
main factor in determing how quickly
these nucleophilic substitution reactions occur. The
weaker the C-X bond the easier
and the more quickly it will be to break and the faster the reaction will be.
Reaction of primary halogenalkanes with water - Preparation of alcohols by nucleophilic substitution
It is possible to prepare alcohols by hydrolysing
halogenalkanes with water, however the reaction is slow
water is not a particularly good nucleophile and the
halogenalkanes are barely soluble
or not soluble at all in water, so ethanol is often added since the halogenalknes
will dissolve in it, this
increases the rate of the hydrolysis reaction.
halogenalkane(aq) + water(l) → alcohol(aq) + acid(aq)
RX(aq) + H2O(l) → ROH(aq) + HX (aq)
The acid produced, HX(aq) will be either hydrochloric acid (HCl), hydrobromic acid (HBr) or hydroiodic acid (HI) depending on
which halogen is present in the halogenalkane used. It is also worth mentioning that the (aq) state symbol used
after the halogenalkane
does not mean it is dissolved in water, as halogenalkanes are largely
insoluble in water, it simply means it is in solution, with ethanol as
This hydrolysis reaction can be thought of as occurring in two steps.
Here the water molecule uses its lone pair of electrons
to act as a nucleophile and attack the partially charged carbon atom (Cδ+). In
the example below bromoethane is being hydrolysed with water. An intermediate is formed which contains an oxygen
atom with a positive charge- an oxonium ion.
To complete the reaction the intermediate has a proton (H+)
removed by a water molecule. Here the water
molecule is acting as a base to
accept a proton (H+) and form the alcohol ethanol and also the oxonium ion or hydroxonium ion (H3O+). This is outlined below.
Measuring the rate of hydroysis of a halogenalkane
In the hydrolysis reaction of a halogenalkane as shown above we can easily measure
the rate of hydrolysis by making use of a
reaction you will have seen before, that is the
reaction of silver nitrate solution with halide ions (Cl-, Br-, I-) to form white,
cream and yellow precipitates of the
This can be shown as:
X-(aq) + Ag+(aq) → AgX(s)
For example you may recall that chloride ions (Cl-) will react with a solution of silver
nitrate to form a white precipitate of silver chloride.
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
Similarly bromide ions (Br-) will give a cream coloured
precipiate of silver bromide and iodide ions (I-) will form a yellow precipiate of silver iodide.
The experimental set-up is shown below, it is fairly straightfoward. You will need 3 labelled test-tubes, each
test-tube will contain:
You will also need three other test-tubes, each of these should contain 5ml of silver nitrate solution. The basic set-up
is shown below, although I have only shown two test-tubes in the water bath, in reality there should be six test-tubes, 3 test-tubes
of the silver nitrate and 3 test-tubes containing each of the halogenalkanes dissolved in hot ethanol.
- Test-tube 1; 5ml of hot aqueous ethanol and 0.5ml of 1-chlorobutane.
- Test-tube 2; 5ml of hot aqueous ethanol and 0.5ml of 1-bromobutane.
- Test-tube 3; 5ml of hot aqueous ethanol and 0.5ml of 1-iodobutane.
Pour the silver nitrate solution into each of the test-tubes and start the stop-clock. Time how long it takes for each of the
coloured white, cream and yellow precipitates of the silver halide to form.
As mentioned above the hydrolysis reaction of halogenalkanes with water is a slow one. However due to the decrease in
bond enthalpy of the C-X bond as we descend group 7, we can order the rate of reaction as:
iodo > bromo > chloro > fluoro
We can use primary, secondary or tertiary haloagenalkanes in this hydrolysis reaction. Here the order of the rate of reaction is
tertiary > secondary > primary
The reasons for this are due to differences in the
mechanisms of these reactions. Though it does not matter if you use
primary, secondary or tertiary halogenalkanes similar trends are seen
in the order that iodo react faster than bromo
faster than chloro halogenalkanes.
- Halogenalkanes can be hydrolysed by water to form alcohol.
The reaction is slow because water is a poor
the halogenalkanes are almost insoluble in water.
- The main factor in the rate of this hydrolysis reaction is the
strength of the C-X bond.
- The rate of the reaction can be measured by timing how long it takes a coloured precipitate of silver halide to form.
An outline of the experimental procedure is:
- warm a solution containing a few cm3 of ethanol, water and silver nitrate solution.
- record the time it takes for the coloured precipiate of silver halide to appear
- The quicker the coloured precipiate appears the faster is the hydrolysis reaction and the weaker the C-X bond.