The methyl ether shown is treated with one equivalent of HI at 0 degrees C. Which products are obtained?

The compound is tert-butyl methyl ether. Protonation of the oxygen is followed by cleavage of the bond that gives the most stable cation: the tertiary C-O bond breaks ($\mathrm{S}{}_{\mathrm{N}}1$) to give the tert-butyl cation, captured by iodide. The products are 2-iodo-2-methylpropane and methanol.