Question

When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as unreacted methane.

(a) Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the formation of these compounds from chloromethane.

(b) How would you run this reaction to get a good conversion of methane toCH₃Cl? Of methane to CCl₄?

Short answer

a)

(b) The amount of Cl₂ (chlorine) reacted with CH₄(methane) should be less for favoring the production of CH₃Cl. This is to make certain that the reaction progress all the way such that all the chlorine atoms would more likely react with methane to form CH₃Cl and not with CH₃Cl to form CH₂Cl₂ .

The amount of Cl₂ (chlorine) reacted with CH₄ (methane) should be in excess for favoring the production of CCl₄. This is to make certain that the reaction progress all the way such that all the C-H bonds are replaced with C-Cl bonds.

Step-by-step solution

Free radicals

An atom or group of atoms containing odd or unpaired electron is known as the free radical. The unpaired electron is represented by a single unpaired dot in the formula. Free radicals are electrically neutral. They are highly reactive species formed by homolytic fission of a covalent bond.

Steps involved in free radical chain reaction

In a free-radical chain reaction, free radicals are generally created in the initiation steps. A free radical and a reactant is combined to yield a product and another free radical in the propagation steps. Lastly, the number of free radicals generally decrease in the termination steps.

Mechanism and explanation

(a) The mechanism for the formation of di-, tri-, and tetrachloromethane consists of three parts which are initiation step, propagation step I and propagation step II.

Mechanism for the formation of di-, tri-, and tetrachloromethane

(b)The amount of Cl₂ (chlorine) reacted with CH₄(methane) should be less for favoring the production of CH₃Cl. This is to make certain that the reaction progress all the way such that all the chlorine atoms would more likely react with methane to form CH₃Cl and not with CH₃Cl to form CH₂Cl₂ .

The amount of Cl₂ (chlorine) reacted with CH₄ (methane) should be in excess for favoring the production of CCl₄. This is to make certain that the reaction progress all the way such that all the C-H bonds are replaced with C-Cl bonds.