nucleophilic substitution

Nucleophilic substitution involving halogenalkanes

We met the halogenalkanes earlier. The halogenalkanes 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 below:

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 of electrons 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

C-X bond Bond enthalpy kJmol-1
C-F 484
C-Cl 338
C-Br 276
C-I 238
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.

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 to imagine 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, therfore 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, therfore 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 quicker 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 simply because 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 the solvent. This hydrolysis reaction can be thought of as occurring in two steps.

step 1:

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.

step 2:

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

colours of silver halide precipitates after addition of silver nitrate

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 silver halide. 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.

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.

experimental set-up for the hydrolysis of a halogen alkane


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 which reacts faster than chloro halogenalkanes.

Key Points

Practice questions

Check your understanding - Questions on Nucleophilic substitution and hydrolysis.