Step 1: Read the reaction:
The starting compound is 1-methylcyclohexene (a cyclohexene ring with a CH3 on one of the double-bond carbons). The reagent is conc. H2SO4 with H2O, which carries out acid-catalysed hydration of an alkene.
Step 2: Protonation (Markovnikov rule):
H+ adds to the double-bond carbon that already has more hydrogens. This puts the positive charge on the carbon that carries the methyl group, giving a tertiary carbocation, the most stable one.
Step 3: Attack by water:
A water molecule attacks the tertiary carbocation and forms an oxonium ion. The oxonium ion then loses H+ to give the alcohol, and the acid is regenerated.
Step 4: Product:
The OH group and the CH3 group end up on the same ring carbon, so A is 1-methylcyclohexanol. This is the structure in option (B).
Step 5: Why the other options are wrong:
Option (A) has OH on the carbon next to the CH3-carbon (2-methylcyclohexanol), which would need a less stable secondary carbocation. Options (C) and (D) contain a CH2OH group, which needs a carbon skeleton change or oxidation of the methyl group; plain hydration does neither.
Final Answer:
Hydration of 1-methylcyclohexene gives 1-methylcyclohexanol, option (B).
\[ \boxed{\text{1-Methylcyclohexanol (option B)}} \]
