Question

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.

Short answer

The chain propagation steps are: 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 \(\Delta H^{\circ} = -43 \text{kJ/mol}\).

Step-by-step solution

Write the overall reaction and the initiation step.

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: $$\text{Br}_2 \to 2 \cdot \text{Br}$$

Write the chain propagation steps.

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: $$\cdot \text{C}_3 \text{H}_7 + \text{Br}_2 \to \text{C}_3 \text{H}_7\text{Br} + \cdot \text{Br}$$

Calculate \(\Delta H^{\circ}\) for each propagation step.

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: $$\Delta H^{\circ}_2 = \text{BDE}(\text{C-H}) + \text{BDE}(\text{Br-Br}) - \text{BDE}(\text{C-H}) - \text{BDE}(\text{Br-Br})$$ $$\Delta H^{\circ}_2 = 410 + 193 - 410 - 193$$ $$\Delta H^{\circ}_2 = 0 \text{kJ/mol}$$

Calculate \(\Delta H^{\circ}\) for the overall reaction.

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 \(\Delta H^{\circ} = -43 \text{kJ/mol}\).

Key concepts

Chain Propagation Steps in Radical Bromination

Radical bromination is an essential reaction for introducing bromine atoms into hydrocarbons. The process is broken down into distinct steps: initiation, propagation, and termination. In the context of propane being transformed to 1-bromopropane, the focus is on chain propagation steps. These are crucial because they sustain the reaction, turning reactants into products continually.

During the chain propagation phase, two main reactions occur in sequence. First, a bromine radical (\( \cdot \text{Br} \)) reacts with propane (\( \text{C}_3 \text{H}_8 \)), abstracting a hydrogen atom and forming hydrogen bromide (\( \text{HBr} \)) and creating a 1-propyl radical (\( \cdot \text{C}_3 \text{H}_7 \)). This can be represented by:
\[\text{C}_3 \text{H}_8 + \cdot \text{Br} \to \cdot \text{C}_3\text{H}_7 + \text{HBr}\]

Next, in the second propagation step, the reactive 1-propyl radical interacts with a bromine molecule (\( \text{Br}_2 \)), resulting in the formation of 1-bromopropane (\( \text{C}_3 \text{H}_7\text{Br} \)) and another bromine radical to perpetuate the cycle:
\[\cdot \text{C}_3 \text{H}_7 + \text{Br}_2 \to \text{C}_3 \text{H}_7\text{Br} + \cdot \text{Br}\]

These steps ensure that the radical bromination process continues efficiently, converting the starting materials into the desired brominated product.

Understanding Bond Dissociation Energy (BDE)

Bond dissociation energy (BDE) is a crucial concept in understanding the energy changes during chemical reactions, including radical bromination.

BDE refers to the amount of energy needed to break a specific bond in a molecule into two atoms or radicals. In our exercise, different bonds contribute to the energy changes during the propagation steps. For example:

• C-H bond in propane: This bond has a BDE of 410 kJ/mol, indicating the energy needed to break a carbon-hydrogen bond in propane and form a radical.
• Br-Br bond: With a BDE of 193 kJ/mol, this represents the energy required to dissociate a bromine molecule into bromine radicals.
• C-Br bond in 1-bromopropane: The BDE here is 280 kJ/mol, needed to break the carbon-bromine bond in the product.
• H-Br bond: This bond's BDE is 366 kJ/mol, related to forming hydrogen bromide.

The differences in BDE values help determine whether a chemical reaction releases or absorbs energy, playing a crucial role in predicting the reaction's behavior.

Calculating Enthalpy Changes in Radical Reactions

Enthalpy calculation is vital for evaluating the energetic feasibility of chemical reactions, like radical bromination. It provides insight into whether a reaction is exothermic (releasing heat) or endothermic (absorbing heat).

To find the enthalpy change (\( \Delta H^{\circ} \)) for each propagation step, we use bond dissociation energies from earlier. For the first step, where a bromine radical reacts with propane, the formula is:
\[\Delta H^{\circ}_1 = \text{BDE of C-H (410 kJ/mol)} + \text{BDE of Br-Br (193 kJ/mol)} - \text{BDE of C-Br (280 kJ/mol)} - \text{BDE of H-Br (366 kJ/mol)}\]
\[\Delta H^{\circ}_1 = 410 + 193 - 280 - 366 = -43 \text{kJ/mol}\]

The second step, involving a 1-propyl radical and bromine molecule, completes the cycle:
\[\Delta H^{\circ}_2 = 410 + 193 - 410 - 193 = 0 \text{kJ/mol}\]

To find the overall reaction enthalpy, sum these values:
\[\Delta H^{\circ}_\text{overall} = \Delta H^{\circ}_1 + \Delta H^{\circ}_2 = -43 + 0 = -43 \text{kJ/mol}\]

This negative enthalpy indicates that the overall process of radical bromination of propane to form 1-bromopropane is exothermic, making it energetically favorable.