Dehydration of pentan-3-ol yields geometric isomers of the alkene Steps:

• Pentan-3-ol is a secondary alcohol that forms a symmetric carbocation upon protonation and loss of water under acidic conditions.
• The carbocation eliminates a proton from an adjacent carbon to form pent-2-ene.
• Pent-2-ene has a disubstituted double bond with different groups on each carbon (CH3 vs H on C2; CH2CH3 vs H on C3), allowing E and Z isomers.
• The elimination produces both isomers indiscriminately due to the planar carbocation intermediate.

Why B is correct:

• Ethanoic acid with little concentrated H2SO4 creates highly acidic conditions where acetic acid is a poor nucleophile (protonated in acid), favoring elimination over esterification and yielding a mixture of (E)- and (Z)-pent-2-ene (geometric stereoisomers).

Why the others are wrong:

• A: Oxidizes to achiral pentan-3-one; no stereoisomers.
• C: Ethanol acts as a good nucleophile, trapping the carbocation to form achiral ether (1-ethoxybutane); minimal elimination.
• D: Cl⁻ is a strong nucleophile, favoring SN1 substitution to achiral 3-chloropentane; little elimination.

Final answer: B