Question

Which of the following is obtained as major product when excess of methane is treated with limited chlorine in presence of UV light?

• A. Chloromethane
• B. Dichloromethane
• C. Trichloromethane
• D. Tetrachloromethane

Hint:

Excess Alkane $\rightarrow$ Mono-substituted product; Excess Halogen $\rightarrow$ Poly-substituted product.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the major product when methane ($CH_4$) is in excess and reacts with a limited amount of chlorine ($Cl_2$) in the presence of UV light.
Step 2: Detailed Explanation:
Chlorination of methane is a free-radical substitution reaction. The reaction proceeds through multiple stages:
\[ CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl \]
\[ CH_3Cl + Cl_2 \xrightarrow{hv} CH_2Cl_2 + HCl \]
\[ CH_2Cl_2 + Cl_2 \xrightarrow{hv} CHCl_3 + HCl \]
\[ CHCl_3 + Cl_2 \xrightarrow{hv} CCl_4 + HCl \]
When methane is in excess, the probability of a chlorine radical colliding with a methane molecule is much higher than colliding with a chloromethane molecule.
This suppresses further chlorination, making monosubstitution the dominant process.
Step 4: Final Answer:
Under excess methane conditions, chloromethane is the major product.