Question

2. Methane reacts with chlorine in the presence of ultraviolet lighta. Outline the mechanism of the reaction chlorine and methane to form chloromethane, clearly describing each step of the mechanism.b. Dichloromethane and ethane are just two of many possible alternative products that could be formed by the reaction of methane with chlorine under these conditions.i. Give and equation for the reaction of methane to produce dichloromethane.ii. Describe how ethane could be formed in this reaction and how this could lead to even more bi-products.iii. Explain why tetrachloromethane may be formed

Answer 1

Step-by-step explanation:

a) Here, we want to get the mechanism of the reaction between chlorine and methane in the presence of ultraviolet light

The first thing that happens here is called radicalization. It simply involves the attack of a chlorine gas molecule by the ultraviolet rays of the sun to produce two chlorine radicals that are ready to attack the methane molecule. This is known as the initiation stage

We have this as:

`Cl_2\text{ }\rightarrow Cl^(\cdot)-Cl^(\cdot)`

This radical then attack the methane molecule to give a CH3 radical alongside Hydrogen chloride

We have that as:

`CH_4+Cl^{\cdot\text{ }}\rightarrow CH^(\cdot)_3\text{ + HCl}`

The CH3 radical then attacks a chlorine gas molecule to give the chloromethane alongside another chlorine radical:

`\text{ CH}_3Cl^(\cdot)+Cl_2\text{ }\rightarrow CH_3Cl+Cl^(\cdot)`

This is known as the propagation stage

For the termination stage, a CH3 radical can react with free chlorine radical as shown below:

`CH^(\cdot)_3+Cl^(\cdot)\text{ }\rightarrow CH_3Cl`

b) i) To produce dichloromethane, a chlorine molecule will be attacked by chloromethane radical:

`CH_3Cl^(\cdot)+Cl_2\text{ }\rightarrow\text{ }CH_2Cl_2\text{ + HCl}`

II) For ethane to be formed, two methyl radicals will combine to give an ethane molecule as shown below:

`CH^(\cdot)_3+CH_{3^{}}\cdot\text{ }\rightarrow CH_3-CH_3`

iii) Tetrachloromethane can be formed when all the hydrogen atoms in methane are replaced by chlorine atoms by a looping activity of propagation and termination

Basically, it happens when a trichloromethane radical attacks a chlorine gas molecule

`\begin{gathered} \text{CHCl}^(\cdot)_3+Cl_2\text{ }\rightarrow_{} \\ \C Cl_4\text{ + HCl} \end{gathered}`