Calculate the \(\Delta H^{\circ}\) for the following reaction step. What can you say regarding the possibility of bromination at a vinylic hydrogen? $$ \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{Br} \cdot \longrightarrow \mathrm{CH}_{2}=\mathrm{CH} \cdot \quad+\quad \mathrm{HBr} $$

In the given reaction, there are three types of bonds involved- C=C, Br•, and C-H. The reaction can be represented as breaking a C=C bond and a Br• bond, followed by the formation of a new C-H bond and a new C-Br bond.

We will now use the bond dissociation energies (BDE) of the involved bonds to calculate the \(\Delta H^{\circ}\) for the reaction. The BDE values are as follows (all values in kJ/mol): C=C: 612 Br•: 193 C-H: 413 C-Br: 276 The equation for \(\Delta H^{\circ}\) calculation using BDE values is: $$ \Delta H^{\circ} = \sum \mathrm{BDE}(\text{bonds broken}) - \sum \mathrm{BDE}(\text{bonds formed}) $$

According to the given reaction: Bonds broken: 1 C=C and 1 Br• Bonds formed: 1 C-H and 1 C-Br Using the BDE values, we can now calculate \(\Delta H^{\circ}\): $$ \Delta H^{\circ} = [(1\times 612) + (1 \times 193)] - [(1 \times 413) + (1 \times 276)] = (612 + 193) - (413 + 276) = 805 - 689 = 116 \, \text{kJ/mol} $$

We found that the standard enthalpy change (\(\Delta H^{\circ}\)) for this reaction is positive, which means the reaction is endothermic. An endothermic reaction requires energy to be added to the system for the reaction to proceed. Thus, bromination at a vinylic hydrogen is less likely to occur spontaneously under standard conditions based on the positive \(\Delta H^{\circ}\). However, that does not mean the reaction is impossible; rather, it requires additional energy, such as external heat or a catalyst, to overcome the energy barrier and proceed. In conclusion, bromination at a vinylic hydrogen is less likely to occur spontaneously under standard conditions, but it can still be achieved under certain conditions by providing the required energy input or using a catalyst.