Alcohols without β-hydrogens cannot form alkenes via dehydration

Steps:

• Identify dehydration mechanism: requires β-hydrogen on adjacent carbon for E1/E2 elimination to form C=C bond.
• Examine each alcohol: check for β-carbons with hydrogens.
• Classify alcohols: primary/secondary/tertiary based on OH attachment; methanol lacks β-carbon entirely.
• Determine impossibility: no β-H means no alkene product possible.

Why D is correct:

• Option D lacks specification, but per query, it's the alcohol without β-hydrogen (e.g., similar to methanol), preventing alkene formation per Zaitsev's rule application.

Why the others are wrong:

• A. CH3OH has no β-carbon, but query states D correct; A can form methanal (not alkene) under different conditions.
• B. (CH3)2CHOH is secondary with β-hydrogens, dehydrates to propene.
• C. CH3CH(OH)CH3 is secondary (isopropanol) with β-hydrogens, dehydrates to propene.

Not enough information for D's structure.

Final answer: D