# Can hydrobromic acid perform substitution with propane?

I am a student in the 11th grade, we recently had a chemistry exam, I lost marks because of a question that asks about the reaction between $\ce{HBr}$ and $\ce{C3H8}$. According to our teacher $\ce{HBr + C3H8}$ is going to react and form $\ce{C3H7Br + H2}$. In my opinion I don’t think that a reaction is going to happen, because $\ce{HBr}$ is an acid and acids can't react with alkanes. Can you please state your opinion and justification?

• You are right, they are not going to react, because, er, well, why would they? – Ivan Neretin Nov 30 '17 at 9:34
• I’m not sure , he said that my statement is weak. I am trying to back it up with facts. – Al-Muataz Nov 30 '17 at 9:42
• Well, technically, your statement is weak, because certain acids do react with alkanes. But HBr doesn't, that's for sure. – Ivan Neretin Nov 30 '17 at 9:47
• So C3H7Br won’t be formed , am I right? No chemical reaction is going to happen? – Al-Muataz Nov 30 '17 at 9:57
• I wouldn't think this were an acid/base reaction, but more of a (photochemically initiated) radical process. The issue is that a propyl radical would still prefer to abstract hydrogen instead of bromine from hydrogen bromide... I'd have to think about this more, but that might just be a different way to state the stability arguments other posters have made... – Zhe Nov 30 '17 at 14:38

## 1 Answer

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 (ΔHorxn= ΔHfoproducts - ΔHforeactants) 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.

• This is a good answer if an initator is specified as part of tbe reactions conditions, however the conditions as reported by the OP do not mention one – Waylander Dec 2 '17 at 12:07