Question

Homolysis of the indicated C-H bond in propene forms a resonance-stabilized radical.

1. Draw the two possible resonance structures for this radical.
2. Use half-headed curved arrows to illustrate how one resonance structure can be converted to the other.
3. Draw a structure for the resonance hybrid.

Short answer

Answer

a.Resonance structures of Propene

b.

Half-headed arrows showing the interconversion of resonance structures of Propene

c.Resonance Hybrid structure

Step-by-step solution

Step-by-Step SolutionStep 1: Bond dissociation Energy

The Bond Dissociation energy determines the bond strength. The stronger the bond, the more is the bond dissociation energy and vice-versa.

The bond dissociation energy is positive.

Enthalpy change

The difference in the energy of the products and the reactants is called the enthalpy change of the reaction. The enthalpy change determines the relative bond strength of the reactants and the products.

For endothermic reactions, the enthalpy change is positive. In comparison, for the exothermic reactions, the

enthalpy change is negative.

When the bonds break, the $∆\mathrm{H}°$is positive; when the new bonds are formed, the $∆\mathrm{H}°$ is negative.

Enthalpy change of the given compounds

The resonance stabilized radical formed upon the homolytic bond fission of the indicated C-H bond is shown hereunder.

a. The two resonance structures are:

Resonance structures of Propene

b. The half-headed curved arrows to inter-convert the two resonance structures are shown hereunder.

Half-headed arrows showing the interconversion of resonance structures of Propene

c. For the resonance hybrid, the electron density of the single radical electron is shown above the two bonds under the effect of resonance.

Resonance Hybrid structure