Can hydrobromic acid perform substitution with propane?
I am afraid that you cannot support your argument. Here are two reasons why as I expand on my comment above. If this process were to be a radical chain reaction, the following process should be considered. Hydrogen bromide has a weaker bond then any of the C-H bonds in propane. An initiation step (heat, light?) would be expected to generate atoms of hydrogen and bromine. If you use a bromine atom as a reactant in propagation step 1 (P1), you will produce a hydrogen atom at the end of P2. So we start with a hydrogen and abstract a secondary hydrogen (primary hydrogen is higher in bond strength but it will only make the analysis less favorable) to produce a secondary propyl radical. The bond dissociation energies (BDE's in kcal/mol) are listed over the bonds of interest. The first step is good news. Exothermic by -5 kcal/mol. But P2 is endothermic by +14 kcal/mol. If you add P1 + P2 you get the overall reaction that has a heat (enthalpy) of reaction of +9 kcal/mol. The reaction is endothermic and uphill. The initiation step, the hydrogen atom and the propyl radical do not participate in the computation. The second method is to use heats of formation (in red). Using Hess's Law (&Delta;Horxn= &Delta;Hfoproducts - &Delta;Hforeactants) gives the same result, give or take a little. However, the reaction will be exothermic in the reverse direction given the right conditions. Sorry not to have better news.