organic chemistry - How would substitution reaction results differ between reacting halides with a tertiary bromoalkane vs reacting it with a primary bromoalkane?

Say, you conducted two separate reactions; in one, you combined $\mathrm{50~mmol}$ saturated $\ce{KCl}$ solution with $\mathrm{10~mmol}$ 1-bromooctane (assume heating), and in the other you combined $\mathrm{10~mmol}$ saturated $\ce{KI}$ solution with $\mathrm{10~mmol}$ 1-bromoctane. In both cases $\mathrm{0.5~mmol}$ hexadecyltributylphosphonium bromide was used as a catalyst.

So, due to relative charge densities and the fact that the electrophillic carbon is primary, you would have a low yield of 1-chloroocatane vs starting material in reaction 1 and you would have a high yield of 1-iodooctane vs starting material in reaction 2, both cases obtained via SN2 reaction.

Now, suppose you would have repeated the same two reactions, only this time you would have reacted the $\ce{KCl}$ and $\ce{KI}$ with 2-bromo, 2-methyloctane, how exactly would the results differ from the results of the previous two reactions, or would they have been the same?

I know that the reaction would have no "choice" but to react via the SN1 (or perhaps E1) route, as the electrophillic carbon is tertiary and thus very sterically hindered. Also, I know that $\ce{I-}$ ions are good nucleophiles, but that doesn't matter in this case, does it?

Perhaps this reaction would go slower than the previous, owing to the fact that it requires the $\ce{Br-}$ to dissociate, and therefore there would be lower yields of both haloalkane products. But then again, wouldn't it go faster owing to the fact that it is in a polar protic solvent, and that the formed carbocation would be tertiary?