The HYDRATION of propene ($\mathrm{CH_3{-}CH{=}CH_2}$) with $\mathrm{H_2O}/\mathrm{H}^+$ follows Markovnikov's rule. The major product is

1. MARKOVNIKOV's rule: in the acid-catalysed addition of $\mathrm{H{-}X}$ (or $\mathrm{H{-}OH}$) across an unsymmetric alkene, the $\mathrm{H}$ goes to the carbon ALREADY bearing MORE H atoms, leaving the $\mathrm{X}$ (or $\mathrm{OH}$) on the more substituted carbon. 2. Mechanism: protonation of the alkene gives the more stable CARBOCATION (here, secondary at C2 vs primary at C1). Water then attacks the cation. 3. Propene: $\mathrm{CH_3{-}CH{=}CH_2}$. C1 (= the $\mathrm{CH_2}$ end) has 2 H's; C2 (the middle) has 1 H. So $\mathrm{H}^+$ adds to C1, leaving the carbocation on C2. 4. Water attacks C2 → propan-2-ol $\mathrm{CH_3CH(OH)CH_3}$. 5. Option A is the ANTI-Markovnikov product (formed by radical addition / peroxide effect — for HBr, not for water). Option C is propane (no OH). Option D is a diol (would need 2 water additions). _Source: NCERT Class 12 Chemistry Part 2, Ch 7, §7.3 (Preparation — Markovnikov hydration of alkenes), p. 6–7._