Nucleophilic substitution of halogenoalkanes
Halogenoalkanes are basically alkanes that have one or more hydrogen atoms substituted by halogen atoms (Group IV elements such as fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). However, compared to the reaction of the alkanes, the reactivity of halogenoalkanes is greatly affected by the presence of the halogen atom.
The general formula of simple halogenoalkanes, whose molecules contain just one halogen atom, is CnH2n + 1X, where X is an atom of F, Cl, Br or I.
Nucleophiles are reagents that attack and form bonds with positively or partially positively charged carbon atoms.
- A nucleophile is either a negatively charged ion or has an atom with a partial negative charge (δ-).
- It has a lone pair of electrons with which it can use to form a covalent bond.
Examples of common nucleophiles are:
- hydroxide ion, OH–
- ammonia, NH3
- cyanide ion, CN.
Mechanism of nucleophilic substitution in halogenoalkanes
Most of the reactions of halogenoalkanes are nucleophilic substitutions in which the nucleophile attacks the carbon atom bonded to the halogen.
The carbon-halogen (C-X) bond is polarised due to the fact that the halogen is more electronegative than carbon. This results in the carbon atom carrying a partial positive charge as the electron pair in the carbon-halogen bond is drawn nearer to the halogen atom.
The halogen atom is then replaced by the nucleophile in the substitution reaction.
Nucleophilic substitution with aqueous alkali, OH–(aq)
When an aqueous solution of sodium hydroxide reacts with a halogenoalkane, a nucleophilic substitution reaction takes place and the halogen atom in the halogenoalkane is replaced by an OH, hydroxyl group, to form an alcohol, as follows:
bromoethane + sodium hydroxide → ethanol + sodium bromide
CH3CH2Br + NaOH → CH3CH2OH + NaBr
The ionic form of this reaction is:
CH3CH2Br + OH– → CH3CH2OH + Br–
The aqueous hydroxide ion behaves is the nucleophile here, because it donates a pair of electrons to the carbon atom bonded to the halogen in the halogenoalkane. This is the reason why this reaction is called a nucleophilic substitution.
Nucleophilic substitution reaction of halogenoalkanes involve the breaking of the carbon-halogen bond. The more reactive the halogen in the bond, the stronger the carbon-halogen bond.
The table below shows the bond energies of carbon-halogen bonds
|Bond||Bond energy in kJ/mol|
|C-F||467 (strongest bond)|
|C-I||228 (weakest bond)|
The above shown trend in carbon-halogen bond energies helps us to explain the rates of reaction of the halogenoalkanes as follows:
- fastest nucleophilic substitution reactions take place with the iodoalkanes
- slowest nucleophilic substitution reactions take place with the fluoroalkanes
The weaker the bond the more easily it is broken and the faster the reaction.
Nucleophilic substitution with cyanide ions, CN–
This time the nucleophile is the cyanide, CN–, ion. The reaction is carried out using a solution of potassium cyanide, KCN, in ethanol (known as an ethanolic or alcoholic solution of potassium cyanide).
The mixture of the halogenoalkane and the alcoholic solution of potassium cyanide is heated under reflux. An ionic equation for this reaction is:
CH3CH2Br + CN– → CH3CH2CN + Br–
bromoethane + cyanide ions → propanenitrile + bromide ions
Take note that :
- the reaction started with bromoethane, which has 2 carbon atoms in its chain.
- during the reaction, the cyanide ion added an extra carbon atom to the original halogenoalkane chain.
- the product is propanenitrile, which has 3 carbon atoms in its chain.
The importance of this reaction is on situations like when chemists need to make a new compound with one more carbon atom than the best available organic raw material. The stages are as follows:
- first the chemists convert the starting compound to a halogenoalkane
- then they reflux the halogenoalkane with ethanolic KCN to make a nitrile.
- the nitrile is an intermediate product with the correct number of carbon atoms (one more than the raw material).
- finally the nitrile is converted to the required product , such as a carboxylic acids or amine.
Nucleophilic substitution with ammonia, NH3
If a halogenoalkane is heated with an excess of ammonia dissolved in ethanol under pressure, an amine is formed.
bromoethane + ammonia → ethylamine + hydrogen bromide
CH3CH2Br + NH3 → CH3CH2NH2 + HBr
In this reaction the ammonia molecule is the nucleophile. Ethylamine is a primary amine because the nitrogen atom is attached to only one alkyl group.
However if the reaction above where to be performed with the ammonia not in excess, a mixture of amine products will be formed. The reason being that the primary amine product will act as a nucleophile itself and attack the extra halogenoalkane molecules, forming secondary amines. The secondary amine formed above would be (CH3CH2)2NH, called diethylamine.