An optically active alcohol C5H12O gives a yellow precipitate with I2/NaOH. On acid-catalysed dehydration it gives three different alkenes (counting…

The iodoform test needs $\mathrm{CH_3CH(OH)-}$, which rules out 2-methylbutan-1-ol and pentan-3-ol (also achiral). Both pentan-2-ol and 3-methylbutan-2-ol are chiral with this group. Dehydration of pentan-2-ol gives pent-1-ene, (E)-pent-2-ene and (Z)-pent-2-ene, i.e. three alkenes. 3-Methylbutan-2-ol gives only 3-methylbut-1-ene and 2-methylbut-2-ene (two). Hence the alcohol is pentan-2-ol.