Chapter 13.5 - Mechanism of Halogenation

In the previous section, we completed a discussion on the physical properties of alkanes. We also saw the basics of a halogenation, which is a chemical property. In this section, we will see the mechanism of halogenation.

The mechanism involves three steps: Initiation, Propagation and Termination.
A. Initiation
This can be written in 3 steps:
1. Suppose that in the reaction mixture there is CH₄ and Cl₂.
• Then in the initial stage, there will be three types of bonds: C-C bonds, C-H bonds and Cl-Cl bonds.
2. When light or heat is applied, it is the Cl-Cl bond that breaks first. This is because, Cl-Cl bond is the weakest among the three.
3. The fission of Cl-Cl bond will be a homolysis. This is shown using Lewis structures in fig.13.32 below:

Fig.13.32

• So the result will be two Cl atoms with one unpaired electron. That means, we get two chlorine free radicals.
• This process can be written in a condensed form as shown below:
$\rm{Cl_2~ \color {green}{\xrightarrow[{homolysis}]{h \nu}} ~ \overset {•}{Cl}~+~\overset {•}{Cl}}$

B. Propagation
This can be written in 4 steps:
1. One of the chlorine free radicals obtained in the initiation step, attacks a CH₄ molecule.
• A C-H bond in the CH₄ molecule, undergoes homolysis. Thus we get a methyl free radical and a hydrogen free radical. This is shown using Lewis structures in fig.13.33(a) below:

Fig.13.33

• The hydrogen free radical combines with the chlorine free radical to give a molecule of HCl. This is shown in fig.b
• This process can be written in a condensed form as shown below:
$\rm{CH_4~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{h \nu}} ~ \overset {•}{C}H_3~+~H-Cl}$
2. The methyl free radical thus formed, attacks a second chlorine molecule.
• The Cl-Cl bond in that molecule undergoes homolysis and thus we get two chlorine free radicals. This is shown using Lewis structures in fig.13.34(a) below:

Fig.13.34

• One of those chlorine free radicals combine with the methyl free radical to form a chloromethane molecule. This is shown in fig.b
Thus we get one chloromethane molecule and one chlorine free radical.
• This process can be written in a condensed form as shown below:
$\rm{\overset {•}{C}H_3~+~Cl-Cl~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 - Cl~+~\overset {•}{Cl}}$
• Chloromethane is our required end product.
3. Remember that, the initiation step was possible due to the formation of chlorine free radical. But that radical was obtained with the help of sunlight.
• The chlorine free radical that we obtained just now was with the help of a methyl free radical.
• This new chlorine free radical will produce a new methyl free radical.
• The new methyl free radical will in turn produce a new chlorine free radical. So now a chain reaction will begin. Thus the reaction will propagate.
4. The condensed equations that we wrote in (1) and (2) represent the propagation steps. These steps give the principal products CH₃Cl and HCl.
• However, many other propagation steps are also possible. Two of them are written below:
(i) $\rm{CH_3 Cl~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{h \nu}} ~ \overset {•}{C}H_2 Cl~+~H-Cl}$
(ii) $\rm{\overset {•}{C}H_2 Cl~+~Cl-Cl~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_2 Cl_2~+~\overset {•}{Cl}}$
• The above propagation step (i) indicates that, the chlorine free radical, attacks a newly formed chloromethane instead of attacking a methane.
• Fig.13.35 below, shows the propagation step (i) using Lewis structures.

Fig.13.35

• Fig.13.36 below, shows the propagation step (ii) using Lewis structures.

Fig.13.36

• In these propagation steps, the principal products are: CH₂Cl₂ and HCl
• So we see that, highly halogenated products are also possible.

C. Termination
This can be written in 5 steps:
1. The reaction stops after some time due to the consumption of reactants.
2. The reaction may stop due to side reactions also.
3. One of the possible side reaction is shown below:
$\rm{\overset {•}{Cl}~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{{}}} ~ Cl-Cl}$
• If this reaction, occurs, there will be no more chlorine free radicals available for the propagation step.
• In such a situation, the reaction will stop even if the reactants are available.   
4. Another possible side reaction is shown below:
$\rm{H_3 \overset {•}{C}~+~\overset {•}{C}H_3~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CH_3}$
• If this reaction, occurs, there will be no more methyl free radicals available for the propagation step.
• In such a situation also, the reaction will stop even if the reactants are available.   
• This reaction helps us to explain the formation of ethane during the chlorination of methane.
5. Yet another possible side reaction is shown below:
$\rm{H_3 \overset {•}{C}~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - Cl}$
• If this reaction, occurs, there will be neither of chlorine free radicals or methyl free radicals available for the propagation step.
• In such a situation also, the reaction will stop even if the reactants are available.
• In this reaction, we get an useful product chloromethane. But not much chloromethane will be produced because, the reaction does not propagate.

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II. Combustion
• Combustion is the second chemical property of alkanes. It can be written 5 in steps:
1. Alkanes can be heated in the presence of air or dioxygen.
• Then a reaction will take place between the dioxygen and the alkane.
• The products are: CO₂, H₂O and large amounts of heat.
   ♦ Due to the intense heat, the H₂O will be in the gaseous state.
2. Let us see some examples:
(i) $\rm{CH_4 (g)~+~2O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ CO_2 (g)~+~2H_2 O (g);~\Delta_c H^\circleddash = -890~kJ~{mol}^{-1}}$
(ii) $\rm{C_4 H_{10} (g)~+~13/2 \; O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ 4\;CO_2 (g)~+~5\;H_2 O (g);~\Delta_c H^\circleddash = -2875.84~kJ~{mol}^{-1}}$
• These are thermochemical equations. We saw the details about such equations in section 6.6
3. The general combustion equation for any alkane is:
$\rm{C_n H_{2n+2} (g)~+~\left(\frac{3n+1}{2} \right) \; O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ n\;CO_2 (g)~+~(n+1)\;H_2 O (g)}$
• The validity of this general equation can be checked by putting n = 4. We will get the same equation in 2(ii).
4. Due to the evolution of large quantities of heat during combustion, alkanes are can be used as fuels.
• Recall that petrol is a mixture of alkanes. It is used as fuel in the internal combustion engines in automobiles. The large heat produced, causes expansion of the gaseous mixture inside the cylinder. The expanding gas pushes the piston of the cylinder, thereby rotating the wheel of the automobile.
• Kerosene is also a mixture of alkanes. The large heat produced is utilized to cook food.
5. We know that dioxygen is essential for the combustion of alkanes.
• If enough air or dioxygen is not available, all the carbon atoms in the alkane will not be oxidized to CO₂. Some C atoms will remain as such. This can be explained in 5 steps:
(i) During combustion of an alkane in the presence of insufficient dioxygen, all the H atoms will be oxidized to H₂O
• So the C atoms are now free.
(ii) There is not enough dioxygen for those C atoms.
• Some of them will be oxidized to CO₂
• The remaining C atoms form soot. Soot is also known as carbon black. Some images can be seen here.
(iii) This process is known as incomplete combustion.
• The equation for the incomplete combustion of methane is:
$\rm{CH_4 (g)~+~O_2 (g)~ \color {green}{\xrightarrow[{combustion}]{Incomplete}} ~ C (s)~+~2H_2 O (g)}$
(iv) Carbon black is used in the manufacture of ink, printer ink, pigments etc., It is also used in filters.
(v) We saw that, all H atoms will be oxidized even if there is insufficient dioxygen.
• This is because, oxidation of H atoms releases greater energy than the oxidation of C atoms. So the system can attain greater stability when H atoms are oxidized. That is the reason why the H atoms get preference.

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In the next section we will see the controlled oxidation.

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