Question

Following is a balanced equation for bromination of toluene. \(\begin{array}{ll}\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HBr} \\ \text { Toluene } & \text { Benzyl bromide }\end{array}\) (a) Using the values for bond dissociation enthalpies given in Appendix 3, calculate \(\Delta H^{\circ}\) for this reaction. (b) Propose a pair of chain propagation steps, and show that they add up to the observed reaction. (c) Calculate \(\Delta H^{\circ}\) for each chain propagation step. (d) Which propagation step is rate-determining?

Short answer

Also, which step is the rate-determining step?

Step-by-step solution

Calculate the enthalpy change for the reaction

To calculate the enthalpy change, \(\Delta H^\circ\), we need to use the bond dissociation enthalpies provided in the appendix. The equation we will use is \(\Delta H^\circ = \sum{D_{bonds\:broken}} - \sum{D_{bonds\:formed}}\). First, we need to identify the bonds that are broken and formed in the reaction. Bonds Broken: C-H, Br-Br Bonds Formed: C-Br, H-Br Using the bond dissociation enthalpies from Appendix 3: C-H: 413 kJ/mol Br-Br: 193 kJ/mol C-Br: 276 kJ/mol H-Br: 366 kJ/mol Now we can plug these values into the equation: \(\Delta H^\circ = (413 + 193) - (276 + 366) = 606 - 642 = -36 \: \text{kJ/mol}\) (a) The enthalpy change for the reaction is -36 kJ/mol.

Propose a pair of chain propagation steps

We will propose a pair of chain propagation steps and show that they add up to the observed reaction. Step 1: Formation of the benzyl radical and HBr \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} + \mathrm{Br}\cdot \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\cdot + \mathrm{HBr}\) Step 2: Formation of benzyl bromide and a bromine radical \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \cdot + \mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Br} \cdot\) If we sum these two steps, we obtain the overall reaction: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HBr}\) (b) The proposed pair of chain propagation steps add up to the observed reaction.

Calculate the enthalpy change for each chain propagation step

We will now calculate the enthalpy change for each chain propagation step using the bond dissociation enthalpies from Appendix 3. Step 1: \(\Delta H^\circ_1 = D_{C-H} - D_{H-Br} = 413 - 366 = 47 \: \text{kJ/mol}\) Step 2: \(\Delta H^\circ_2 = D_{Br-Br} - D_{C-Br} = 193 - 276 = -83 \: \text{kJ/mol}\) (c) The enthalpy change for the first chain propagation step is 47 kJ/mol, and for the second step, it is -83 kJ/mol.

Identify the rate-determining step

The rate-determining step is the propagation step with the highest activation energy. Since the first step has a positive enthalpy change and the second step has a negative enthalpy change, the first step is more likely to have a higher activation energy. (d) The rate-determining step is the first chain propagation step, which involves the formation of the benzyl radical and HBr.

Key concepts

Bond Dissociation Enthalpy

In chemical reactions, molecules often change as they form or break bonds. The energy needed to break a bond between two atoms in a molecule is known as bond dissociation enthalpy. It reflects how strongly the atoms are held together. Higher values mean stronger bonds.
In the bromination of toluene, we focus on breaking C-H and Br-Br bonds, while C-Br and H-Br bonds are formed. This information helps in calculating the enthalpy change for the reaction.

• C-H bond energy: 413 kJ/mol
• Br-Br bond energy: 193 kJ/mol
• C-Br bond energy: 276 kJ/mol
• H-Br bond energy: 366 kJ/mol

These numbers are crucial to determining whether a reaction releases or absorbs energy. A negative enthalpy change indicates an exothermic reaction, releasing energy into the surroundings.

Chain Propagation Steps

Chain propagation steps are crucial in understanding chemical reactions, especially in processes like bromination. These steps involve radicals, which are highly reactive atoms or molecules with unpaired electrons. In our exercise, we see two specific chain propagation steps.
The first step involves a bromine radical reacting with toluene, forming a benzyl radical and HBr.
The second step has the benzyl radical reacting with a bromine molecule, forming benzyl bromide and regenerating a bromine radical.
These two steps demonstrate a continuous cycle, which perpetuates the reaction. Summing them gives the overall reaction. This highlights the self-sustaining nature of chain reactions, where initial radicals facilitate subsequent reactions.

Rate-Determining Step

In chemical kinetics, the rate-determining step is the slowest step in the reaction pathway. It acts like a bottleneck, controlling the speed at which a chemical reaction proceeds. For bromination of toluene, determining this step helps us understand where the reaction might be slowed down.
Here, we compare the enthalpy change of the two chain propagation steps. The first step, formation of the benzyl radical and HBr, has a positive enthalpy change ( 47 kJ/mol ), indicating that it requires energy. Positive enthalpy steps are usually slower because they have higher activation energies.
The second step involves a negative enthalpy change ( -83 kJ/mol ), making it faster. Hence, the first step is rate-determining, as it dictates the pace of the overall reaction by being the slowest.

Enthalpy Change Calculation

Calculating the enthalpy change of a reaction requires knowing the energies of bonds breaking and forming. This is fundamental to predicting if the reaction will release or absorb energy.
The formula used is: \(\Delta H^\circ = \sum{D_{bonds\:broken}} - \sum{D_{bonds\:formed}}\).
For bromination of toluene, we broke C-H and Br-Br bonds and formed C-Br and H-Br bonds. Plugging in the bond energies:

• C-H and Br-Br add up to 606 kJ/mol
• C-Br and H-Br add up to 642 kJ/mol

Substituting these values gives us the enthalpy change as -36 kJ/mol. A negative value indicates the reaction is exothermic, meaning it releases heat, making the environment of the reaction warmer.