CHM 2046C Module 9 Sample Exam Answers

Module 9 Part C: Free Radical Mechanism Example 10 points

The free radical mechanism for the halogenation of alkane has been established as a chain reaction. Demonstrate this mechanism by showing all four steps in this mechanism using Methane and Bromine in Ultraviolet Light

CH₄ + Br₂ + uv light à CH₃Br + HBr

a. Chain initiation step:

Br₂ + uv light à Br· + Br·

(Create Bromine Free Radical from one photon of uvlight)

b. Chain Propogation Steps

Br· + CH₄ à CH₃· + HBr

(Creates Methyl Free Radical and a molecule of Byproduct)

CH₃· + Br₂ à CH₃Br + Br·

(Creates main product plus generates another bromine free radical to repeat the first step in Part b.)

These two steps will go over and over until about 4000 molecules of product are formed for each photon of light used)

c. Chain Termination Steps:

Br· + Br· à Br₂

CH₃· + Br· à CH₃Br

CH₃· + CH₃· à CH₃CH₃

d. What evidences did the chemist record which lead to this suggested mechanism?

1. One photon created about 4000 molecules of product suggesting a Chain reaction with propagating steps.

2. Trace Ethane was found in the prouct which was not there in the beginning