Free Free Radical Substitution MCQs with Answers

1. The chlorination of methane in the presence of ultraviolet light proceeds by

• A. nucleophilic substitution
• B. electrophilic addition
• C. free radical substitution
• D. elimination

Explanation: Ultraviolet light splits the chlorine molecule homolytically into two radicals, which then abstract hydrogen from methane and set up a chain reaction of propagation steps until two radicals combine and terminate it. Because the chain continues, the reaction does not stop at chloromethane but produces a mixture of all four chlorinated products. Substitution is characteristic of saturated compounds, while addition belongs to unsaturated ones.

2. In the free radical substitution of methane, the initiation step is

• A. the reaction of a chlorine radical with methane
• B. the combination of two methyl radicals
• C. the reaction of a methyl radical with chlorine
• D. the homolytic splitting of a chlorine molecule into two chlorine radicals by ultraviolet light

Explanation: Initiation is the step that creates radicals where there were none, and only the chlorine to chlorine bond is weak enough to be broken by ultraviolet light. Propagation steps then consume one radical and produce another, so the chain continues, while termination removes radicals by combining them. Identifying which step is which by counting radicals on each side is the reliable method.

3. What is general requirement for free radical substitution reaction of ethane and halogen?

• A. Low pressure
• B. Low temperature
• C. High pressure
• D. Heat and UV light

Explanation: Initiation requires the halogen molecule to split homolytically into two radicals, and only ultraviolet light or a high temperature supplies enough energy to break that bond. In the dark and in the cold, an alkane and a halogen simply do not react, which is a standard demonstration of how unreactive alkanes are. Once started, the chain propagates on its own until two radicals combine.