Simple Halogenation of alkane

In this post, we will discuss about the halogenation of alkane with a previous year question based on it.

What is the halogenation of alkane?

A halogenation reaction is a chemical reaction between a substance and a halogen in which one or more halogen atoms are incorporated into molecules of the substance. Halogenation of an alkane produces a hydrocarbon derivative in which one or more halogen atoms have been substituted for hydrogen atoms.

Alkanes are famous for their unreactive nature because they are non-polar and lack functional groups at which reactions can take place. Therefore, free radical halogenation provides a method by which alkanes can be functionalized.

However, a severe limitation of radical halogenation is the number of similar C-H bonds that are present in all but the simplest alkanes, so selective reactions are difficult to achieve.

General Reaction of alkane

Halogenation of alkane is an example of a substitution reaction, i.e., a type of reaction that often occurs in organic chemistry. A substitution reaction is a chemical reaction in which part of a small reacting molecule replaces an atom or a group of atoms on a hydrocarbon or hydrocarbon derivative.

Features of Halogenation of alkane

• The notation R-H is a general formula for an alkane. R in this case represents an alkyl group. Addition of a hydrogen atom to an alkyl group produces the parent hydrocarbon of the alkyl group.
• The notation R-X on the product side is the general formula for a halogenated alkane. X is the general symbol for a halogen atom.
• Reaction conditions are noted by placing these conditions on the equation arrow that separates reactants from products. Halogenation of an alkane requires the presence of heat or light.

In halogenation of an alkane, the alkane is said to undergo fluorination, chlorination, bromination or iodination depending on the identity of the halogen reactant. Chlorination and bromination are the two widely used alkane halogenation reactions. Fluorination reactions generally proceed too quickly to be useful and iodination reactions go too slowly.

Halogenation usually result in the formation of a mixture of products rather than a single product. More than one product results because more than one hydrogen atom on an alkane can be replaced with halogen atoms.

The mechanism followed in halogenation of alkane is free radical mechanism.

Mechanism for chlorination of methane

1. Initiation Step- The Cl-Cl bond of the elemental chlorine undergoes homolysis when irradiated with UV light. This process yields two chlorine atoms, which are also called chlorine radicals.

2. Propagation Step- One of the chlorine radicals abstracts a hydrogen atom from methane to form the methyl radical. In turn, the methyl radical abstracts a chlorine atom from one of the chlorine molecules, and then, the formation of chloromethane takes place. Also, the second step of the propagation regenerates a chlorine atom, and these steps repeat several times until the termination happens.

3. Termination Step- The termination happens when a chlorine atom either reacts with the other chlorine atom to generate Cl₂, or a chlorine atom reacts with a methyl radical to produce chloromethane, which constitutes a minor pathway, where the product is made. Also, two methyl radicals can combine to form ethane, which is a minor by-product of this reaction.  At this step, this reaction does not stop. However, the chlorinated methane product can be allowed to react with additional chlorine to form polychlorinated products.

By controlling the reaction conditions, including the ratio of chlorine to methane, it can be possible to favour the formation of either one or another possible chlorinated methane product.

Question: Hydrocarbon (CH₃)₃CH undergoes reaction with Br₂ and Cl₂ in the presence of sunlight, if the reaction with Cl is highly reactive and that with Br is highly selective, so no of possible products respectively is: (i) 2,2 (ii) 2,1 (iii) 1,2 (iv) 1,1

Answer: Chlorine atom is highly reactive so it will react with all type of hydrogen available while the Br atom is highly selective so it will react with that hydrogen which give the highly stabilize tertiary alkyl radical.

So, Cl₂ on reaction form 2 products while Br₂ on reaction form 1 product only. Hence, option (ii) is correct.

Read More: Conversion in Organic Chemistry Class 12

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