Dehydration of 3-methylpentan-2-ol with concentrated H2SO4 proceeds through carbocations that can rearrange by 1,2-hydride shift. The total number of…

Protonation and loss of water gives the secondary cation $\mathrm{CH_3\overset{+}{C}H{-}CH(CH_3)CH_2CH_3}$. Elimination from it gives: 3-methylpent-1-ene, $\mathrm{CH_2{=}CH{-}CH(CH_3)CH_2CH_3}$, which has a stereocentre at C-3 (2 enantiomers); and 3-methylpent-2-ene, $\mathrm{CH_3CH{=}C(CH_3)CH_2CH_3}$, with $E$ and $Z$ forms (2 isomers). A 1,2-hydride shift from C-3 to C-2 converts it to the tertiary cation $\mathrm{(CH_3CH_2)_2\overset{+}{C}CH_3}$ (3-methylpentan-3-yl), which loses a proton from either ethyl group to give the same 3-methylpent-2-ene ($E$ and $Z$, already counted), or from the methyl group to give 2-ethylbut-1-ene, $\mathrm{CH_2{=}C(CH_2CH_3)_2}$, which has no stereoisomer (1 new alkene). Total $=2+2+1=5$.