As we learned in Chapter 4, propane $\left({\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3\right)$has both $\mathbf{1}\mathbf{°}$and $\mathbf{2}\mathbf{°}$hydrogens.

1. Draw the carbon radical formed by homolysis of each type of C-H bond.
2. Use the values in Table 6.2 to determine which C-H bond is stronger.
3. Explain how this information can be used to determine the relative stability of the two radicals formed. Which radical formed from propane is more stable?

1° and 2° hydrogens are the hydrogen atoms attached to primary and secondary carbon atoms, respectively.

The homolytic bond fission is the initiation step for radical formation. The bond breaks off in such a manner that the shared electrons are distributed equally between the two bonded atoms.

This type of bond breakage is represented by two half-headed curved arrows pointing towards each atom bonded.

The bond dissociation energy determines the strength of a bond as the bond dissociation energy is the energy given to the molecule to break off that bond.

The more energy given to the bond to break it off, the more is the energy given to form that bond.

a. The type of hydrogen atoms in propane is represented hereunder:Types of hydrogen atoms in propane

The carbon radicals formed by the homolysis of each of hydrogen atoms in propane are shown hereunder.

Homolysis of $2°$hydrogen

Homolysis of $1°$Hydrogen

b.The bond dissociation energy to form radical A is higher. Hence the bond is stronger.

Bond dissociation energy for the homolysis of $1°$Hydrogen

Bond dissociation energy for the homolysis of $2°$Hydrogen

Radical A is less stable.