The compound shown is pent-2-ene. How many open-chain alkenes of molecular formula $\mathrm{C}{}_5\mathrm{H}{}_{10}$ exist in total, counting constitutional isomers and geometrical isomers (cis/trans) separately? Give the answer as an integer.

The constitutional isomers of acyclic $\mathrm{C}{}_5\mathrm{H}{}_{10}$ alkenes are pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene (5). Only pent-2-ene has two different groups on each alkene carbon, so only it shows geometrical isomerism (cis and trans). 2-Methylbut-2-ene has two methyls on one carbon and so does not. No stereocentre is present in any of them. Total = 5 + 1 = 6.