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?

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.

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.

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.

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.