Preparation of Haloalkanes
from Hydrocarbons
Preparation of Haloalkanes by Free Radical Halogenation of Alkanes
When the mixture of hydrocarbon and halogen is heated at 520-670 K in dark or is subjected to ultraviolet light at room temperature, the free radical substitution reaction takes place.
The reactivity of hydrogen towards free radical substitution is 3° > 2° > 1°. For example, when butane is used in free radical substitution reaction, the products we get are 1-Chlorobutane and 2-Chlorobutane. 2-Chlorobutane being 2° is the major product.
Mechanism of free radical substitution reaction
Free radical substitution reaction is a chain reaction. Let's take the example of CH4 and Cl2 to understand the mechanism. The free radical substitution reaction consists of the following steps :
Initiation step
When a mixture of CH4 and Cl2 is heated at 520-670 K or is subjected to ultraviolet light at room temperature, Cl2 absorbs energy and undergoes homolytic fission.
Propagation step
Propagation step consists of two sub steps
- The free radical of chlorine (formed in initiation step) attacks the CH4 molecule and removes a hydrogen atom
from CH4 forming ∙CH3 and HCl.
- ∙CH3 thus produced, reacts with a molecule of Cl2 which results in the formation of methyl chloride.
- This reaction continues until all the hydrogen atoms of methane are replaced by halogen atoms.
Termination step
The chain reaction may terminate if two of the same or different free radicals combine among themselves without producing new free radicals.
Preparation of Haloalkanes from Alkenes
Alkenes react with halogen acids to form haloalkanes. Markovnikov's rule is applied in case of unsymmetrical alkenes.
Markovnikov's rule
Markovnikov's rule states that in the addition reactions of unsymmetrical alkenes, the negative part (e.g. halogen) of reagent attaches to the carbon having less number of hydrogen atoms. For example, prop-1-ene reacts with HBr to give 2-bromopropane as a major product.
Peroxide effect (Kharash effect)
If HBr is added to alkene in the presence of peroxide, the negative part (i.e. Br-) attaches to the carbon having more number of
hydrogen atoms. Since the peroxide effect is contrary to Markovnikov's rule, it is also known as antimarkovnikov's rule. For example,
Prop-1-ene reacts with HBr in the presence of peroxide to give 1-bromopropane as a major product.
A hydrocarbon C5H12 gives only one monochlorination product. Identify the hydrocarbon.