A certain free radical chlorination reaction of 2,3-dimethylbutane leads to the production of two major monochlorinated products 1-chloro-2,3-dimethylbutane and 2-chloro-2,3-dimethylbutane. When run through a Gas chromatograph, it was determined that the product distribution was 51.7% 1-chloro-2,3-dimethylbutane and 48.3% 2-chloro-2,3-dimethylbutane. If we assign the primary H's a reactivity of 1.0, what is the relative reactivity of the tertiary H's?

Question

Quizplus · Accepted Answer

The relative reactivity of the different types of H's in a molecule during free radical chlorination is inversely proportional to their degree of substitution. Since the tertiary H's are the most substituted, they will be the most reactive. In this case, the ratio of the products indicates that the 2-chloro isomer was formed slightly more than the 1-chloro isomer. This is because the tertiary H's are slightly more reactive than the secondary H's. To calculate the exact difference in reactivity, we can use the ratio of the product distribution:

2-chloro / 1-chloro = (relative reactivity of secondary H's / relative reactivity of tertiary H's) ^ 2

Substituting the given values for the ratio of the products:

48.3 / 51.7 = (relative reactivity of secondary H's / relative reactivity of tertiary H's) ^ 2

Solving for the relative reactivity of the tertiary H's:

(relative reactivity of tertiary H's) ^ 2 = 51.7 / 48.3
(relative reactivity of tertiary H's) ^ 2 = 1.071

(relative reactivity of tertiary H's) = sqrt(1.071) = 1.035

Therefore, the relative reactivity of the tertiary H's is 1.035 times that of the secondary H's, or 5.6 times more reactive overall (since the question asks for relative reactivity compared to the primary H's, which have a value of 1).

A Certain Free Radical Chlorination Reaction of 2,3-Dimethylbutane Leads to the Production