The alcohol shown is dehydrated with acid, and the secondary cation can undergo a 1,2-hydride shift to a tertiary cation. Counting E and Z isomers as different compounds and ignoring optical isomerism, how many distinct alkenes can form in total?

The alcohol is 3-methylpentan-2-ol. The unrearranged cation gives 3-methylpent-1-ene and (E)- and (Z)-3-methylpent-2-ene. The hydride shift gives the tertiary 3-methylpent-3-yl cation, which gives the same 3-methylpent-2-ene isomers plus 2-ethylbut-1-ene. The total is 3 + 1 = 4.