CHM 2210C Chapter 5 Sample Exam

 

Chapter 5 Part M_1: 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 CH3Br + HBr

 

a.   Chain initiation step:

 

Br2 + uv light Br + Br

(Create Bromine Free Radical from one photon of uvlight)

 

 

b.   Chain Propogation Steps

 

Br + CH4 CH3 + HBr

(Creates Methyl Free Radical and a molecule of Byproduct)

 

CH3 + Br2 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 Br2

CH3 + Br CH3Br

CH3 + CH3 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