Counting isomeric enediols and allylic diols in C4H6O2 Steps:

• Consider carbon skeletons: linear (1-butene, 2-butene) and branched (2-methylpropene) for one C=C.
• Place two -OH groups: on sp3 for alcohols (but-3-ene-1,2-diol, but-2-ene-1,4-diol, 2-methylenepropane-1,3-diol) or one on sp2 for enols (but-2-ene-1,3-diol).
• Apply stereoisomerism: E/Z for but-2-ene-1,4-diol (2) and but-2-ene-1,3-diol (2); enantiomers for but-3-ene-1,2-diol (2); none for 2-methylenepropane-1,3-diol (1).
• Sum isomers: 2 + 2 + 1 + 2 = 7 total. Why C is correct:
• Molecular formula C4H6O likely intends C4H6O2 (standard for one C=C and two -OH); 7 accounts for all structural and stereo variants per isomerism definitions. Why the others are wrong:
• A: Excludes enediol E/Z pair.
• B: Omits branched structural isomer.
• D: Includes non-existent stereoisomers like chiral branched form. Final answer: C