Investigate The Relationship Between Structure And Properties Of Hydrocarbons - Session 1

Halogenation of alkanes is a chemical reaction in which a halogen atom replaces a Hydrogen atom in an alkane molecule. In this process, Carbon Hydrogen bond is broken. After breaking of bond, new Carbon halogen bond is formed. It is a Substitution reaction. Substitution means replacement. Halogenation of alkanes is specifically referred as Free Radical Substitution. This is because it involves the generation and utilization of free radicals. Free radicals are highly reactive chemical species that contain one or more unpaired electrons.

Let us focus on the Chlorination of Methane as an example. The Chlorination of Methane involves four steps. The reaction begins with the initiation step. In this step, a Chlorine molecule is broken-down into two highly reactive Chlorine radicals. This step typically requires energy in the form of heat or ultraviolet light.

Second step is called propagation. In the propagation step, a Chlorine radical reacts with a methane molecule to form a Methyl radical and Hydrogen Chloride. Chlorine radical replaces the Hydrogen atom in Methane. As a result Methyl radical and Hydrogen Chloride is formed.

The next step is chain propagation. In this step, Methyl radical reacts with another Chlorine molecule to form Chloromethane and another Chlorine radical. This chain propagation step continues the reaction. This is because newly formed Chlorine radical can react with additional methane molecules.

The final step is Termination. In Termination, reactive radicals combine to form stable molecules. Here are some examples of termination reactions that can take place during Chlorination of Methane. Two Methyl radicals can combine to form Ethane. A Methyl radical can combine with a Chlorine radical to form Chloromethane. A Methyl radical can combine with a Hydrogen atom to form a stable Methane molecule. These termination reactions help reduce the concentration of reactive radicals and bring the Chlorination reaction to an end.

Hydrohalogenation of alkenes involves the addition of a Hydrogen halide to an alkene. The Hydrogen halide adds across the Carbon Carbon double bond of alkene. The double bond of the alkenes breaks in this process. The product of this reaction is alkyl halide.

The hydrohalogenation of alkenes happen through a stepwise mechanism that involves the formation of a carbocation intermediate. Let us take the example of hydroChlorination of Ethene to illustrate the process. The reaction is initiated by the dissociation of Hydrogen Chloride into its constituent ions. This generates a positively charged Hydrogen Ion and a negatively charged Chloride Ion.

Ethene then reacts with the Hydrogen ion. The Carbon Carbon double bond in the ethene acts as a nucleophile. As we know, nucleophile means electron rich specie. It attacks the Hydrogen ion. The pi bond between the two Carbon atoms breaks. Then the Hydrogen atom bonds to one of the Carbon atoms. The addition of the Hydrogen ion forms a carbocation intermediate. The carbocation is a Carbon atom with a positive charge and three bonds. It lacks one electron.

After this, The negatively charged halide ion attacks the positively charged Carbon in the carbocation. The negative charged halide ion acts as a nucleophile. This nucleophilic attack leads to the formation of a new Carbon Chloride bond. The reaction is terminated by the formation of the alkyl halide product. In this case the alkyl halide is Chloroethane. The alkyl halide contains a new Carbon-Halogen bond and retains the original Carbon skeleton of the alkene.

During the hydrohalogenation with alkene with greater number of Carbon atoms, there is possibility of formation of two types of products. For example, during addition of Hydrogen bromide to propene, two different products can be formed. These are one bromopropane and two bromopropane. Formation of these products is explained by Markovnikov's Rule and Anti Markovnikov's Rule.

According to Markovnikov's Rule, the electrophilic component of the reagent attaches itself to the Carbon atom with greater number of Hydrogen atoms attached to it. The nucleophilic component of reagent adds to the Carbon atom with less number of Hydrogen atoms attached to it. For example during addition of Hydrogen bromide to propene, the Hydrogen ion is added to Carbon number one. This is because Carbon number one has greater number of Hydrogen atoms attached to it as compared to Carbon number two. The bromide ion is added to Carbon number two.

According to Anti Markovnikov's Rule, the electrophilic component of the reagent attaches itself to the Carbon atom with less number of Hydrogen atoms attached to it. The nucleophilic component of reagent adds to the Carbon atom with greater number of Hydrogen atoms attached to it. For example during addition of Hydrogen bromide to propene, the Hydrogen ion is added to Carbon number two. This is because Carbon number two has less number of Hydrogen attached to it as compared to Carbon number one. The bromide ion is added to Carbon number one. This addition is according to Anti Markovnikov's Rule.

As we know, two types of products are formed during hydrohalogenation of alkenes. One is formed according to Markovnikov’s rule. Other product is formed according to Anti Markovnikov’s rule. Can you tell which is the more likely product to be formed? Well, the answer lies in the stability of carbocation intermediate formed in hydrohalogenation of alkenes. Stable carbocation forms stable product. First we shall discuss the types of carbocations that can be formed during hydrohalogenation of alkenes.

There are three types of carbocations. Primary carbocation, secondary carbocation and tertiary carbocation. A primary carbocation has only one alkyl group bonded to the positively charged Carbon atom. A secondary carbocation has two alkyl groups bonded to the positively charged Carbon atom. A tertiary carbocation has three alkyl groups bonded to the positively charged Carbon atom.

Now we shall discuss which carbocation is most stable. The alkyl groups attached to the positively charged Carbon donate electrons to that positively charged Carbons. They contribute in increasing the electron density at positively charged Carbon. Positive Character of positively charged Carbon is reduced as a result. It becomes stable. This means that most stable carbocation is the one that has greater number of alkyl groups attached to positive charged Carbon. So, the tertiary carbocation is most stable. The increasing order of stability of carbocation shown here.

Now, lets talk about the products that were formed during hydrobromination of propene. First product is two bromopropane that followed Markovnikov’s rule. Second Product is one bromopropane that followed Anti Markovnikov’s rule. The carbocation intermediate of two bromopropane was secondary carbocation. The carbocation intermediate of one bromopropane was primary carbocation. We can see secondary carbocation is more stable. So the major product will be two bromopropane.