Carbocation stability in secondary alcohols maximizes alkene diversity in dehydration.

Steps:

• Classify isomers: A (2-methylpropan-1-ol, primary), B (butan-2-ol, secondary), C (butan-1-ol, primary), D (2-methylpropan-2-ol, tertiary).
• Primary alcohols (A, C) dehydrate mainly via E2, producing one alkene each without stereoisomers.
• Secondary (B) and tertiary (D) dehydrate via E1, forming carbocations; B's is unsymmetric, D's symmetric.
• For B, CH3CH2CH+CH3 loses H+ from methyl (1-butene) or methylene (cis-/trans-2-butene), yielding three alkenes.

Why B is correct:

• Unsymmetric secondary carbocation enables elimination to one constitutional isomer without stereo (1-butene) and one with E/Z stereoisomers (2-butene), per E1 mechanism rules.

Why the others are wrong:

• A: E2 or rearranged E1 yields only 2-methylpropene (no stereo).
• C: E2 yields only 1-butene (no stereo; minor E1 ignored for primary).
• D: Symmetric tertiary carbocation yields only 2-methylpropene (equivalent betas, no stereo).

Final answer: B