Question

(a) Draw the structure of the transition state for the second propagation step in the chlorination of methane.

Show whether the transition state is product-like or reactant-like and which of the two partial bonds is stronger.

(b) Repeat for the second propagation step in the bromination of methane.

Short answer

a)

second propagation step in the chlorination of methane

transition state

The transition state is more reactant-like.

(b)

Step-by-step solution

Free radicals

An atom or group of atoms containing odd or unpaired electron is known as the free radical. The unpaired electron is represented by a single unpaired dot in the formula. Free radicals are electrically neutral. They are highly reactive species formed by homolytic fission of a covalent bond.

Propagation

The formation of products with regeneration of reactive species takes place in the propagation step.

Transition state or activated complex

The highest energy state in a molecular collision that leads to reaction is the transition state. Transition state is most unstable. The transition state or activated complex gets converted into product molecules. During the formation of this complex, old bonds start breaking and the new bonds start making.

Hammond’s postulate

According to Hammond’s postulate, related species which are closer in energy are also closer in structure and the structure of the transition state resembles the structure of the closest stable species.

Explanation

(a)

second propagation step in the chlorination of methane

transition state

From Hammond’s postulate, it is known that the transition state in the exothermic chlorination must resemble the reactants more than the products. Hence, the transition state is more reactant-like.

(b)

second propagation step in the bromination of methane

transition state