Write a pair of chain propagation steps for the radical bromination of propane to give 1-bromopropane, and calculate \(\Delta H^{\circ}\) for each propagation step and for the overall reaction.

The overall reaction for the radical bromination of propane to give 1-bromopropane is: $$\text{C}_3 \text{H}_8 + \text{Br}_2 \to \text{C}_3 \text{H}_7 \text{Br} + \text{HBr}$$ The initiation step involves the homolytic cleavage of the bromine molecule, which leads to the formation of two bromine radicals: $$\text{Br}_2 \to 2 \cdot \text{Br}$$

There are two propagation steps in the radical bromination of propane: 1. The propane molecule reacts with a bromine radical to form a 1-propyl radical and HBr: $$\text{C}_3 \text{H}_8 + \cdot \text{Br} \to \cdot \text{C}_3\text{H}_7 + \text{HBr}$$ 2. The 1-propyl radical reacts with a bromine molecule to form 1-bromopropane and a bromine radical: $$\cdot \text{C}_3 \text{H}_7 + \text{Br}_2 \to \text{C}_3 \text{H}_7\text{Br} + \cdot \text{Br}$$

We can calculate \(\Delta H^{\circ}\) for each propagation step using bond dissociation energies (BDEs). The values for each bond are as follows: - C-H bond in propane: 410 kJ/mol - Br-Br bond: 193 kJ/mol - C-Br bond in 1-bromopropane: 280 kJ/mol - H-Br bond: 366 kJ/mol For the first propagation step: $$\Delta H^{\circ}_1 = \text{BDE}(\text{C-H}) + \text{BDE}(\text{Br-Br}) - \text{BDE}(\text{C-Br}) - \text{BDE}(\text{H-Br})$$ $$\Delta H^{\circ}_1 = 410 + 193 - 280 - 366$$ $$\Delta H^{\circ}_1 = -43 \text{kJ/mol}$$ For the second propagation step: $$\Delta H^{\circ}_2 = \text{BDE}(\text{C-H}) + \text{BDE}(\text{Br-Br}) - \text{BDE}(\text{C-H}) - \text{BDE}(\text{Br-Br})$$ $$\Delta H^{\circ}_2 = 410 + 193 - 410 - 193$$ $$\Delta H^{\circ}_2 = 0 \text{kJ/mol}$$

To calculate the \(\Delta H^{\circ}\) for the overall reaction, we add up the enthalpy changes of the propagation steps: $$\Delta H^{\circ}_\text{overall} = \Delta H^{\circ}_1 + \Delta H^{\circ}_2$$ $$\Delta H^{\circ}_\text{overall} = -43 + 0$$ $$\Delta H^{\circ}_\text{overall} = -43 \text{kJ/mol}$$ In conclusion, the chain propagation steps for the radical bromination of propane to give 1-bromopropane are as follows: 1. \(\text{C}_3 \text{H}_8 + \cdot \text{Br} \to \cdot \text{C}_3\text{H}_7 + \text{HBr}\) with \(\Delta H^{\circ} = -43 \text{kJ/mol}\) 2. \(\cdot \text{C}_3 \text{H}_7 + \text{Br}_2 \to \text{C}_3 \text{H}_7\text{Br} + \cdot \text{Br}\) with \(\Delta H^{\circ} = 0 \text{kJ/mol}\) The overall reaction has a \(\Delta H^{\circ} = -43 \text{kJ/mol}\).