3-Bromo-3,4-dimethylhexane, taken as a sample containing all of its stereoisomers, is heated with alcoholic KOH. Considering every alkene that can…

The substrate is $\mathrm{CH_3CH_2C(Br)(CH_3)CH(CH_3)CH_2CH_3}$. There are three types of $\beta$-hydrogens. Loss from the C-3 methyl gives 2-ethyl-3-methylpent-1-ene, whose C-3 is a stereocentre (H, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$, $\mathrm{C(=CH_2)C_2H_5}$), so there are $2$ enantiomers. Loss from the ethyl $\mathrm{CH_2}$ gives 3,4-dimethylhex-2-ene, which has $E/Z$ isomerism and a stereocentre at C-4, giving $2\times2=4$ stereoisomers. Loss from C-4 gives 3,4-dimethylhex-3-ene, with a tetrasubstituted double bond whose carbons each carry a methyl and an ethyl group, so $E/Z$ isomerism gives $2$ isomers and no stereocentre. The total is $2+4+2=8$. Because the sample contains every stereoisomer of the bromide, no stereoisomer of an alkene is excluded.