Vicinal diol with methyl carbinol group for iodoform and cleavage to aldehyde Steps:

• Yellow precipitate with alkaline I2 shows iodoform test positive for CH3CH(OH)R alcohols.
• Oxidation yields mixture including Z; Z's red-brown Fehling's precipitate indicates aldehyde.
• Eliminate D (ether, no OH for tests); C (butan-2-ol, oxidizes to ketone CH3COCH2CH3, no aldehyde).
• A and B have diol with CH3CH(OH)-; A is vicinal (adjacent OH), cleaves to CH3CHO + HCHO (both aldehydes); B is 1,3-diol (non-adjacent), oxidizes to non-cleaving products like CH3COCH2CH2OH (ketone) + CH3CH(OH)CH2CHO (aldehyde, but no cleavage mixture of simple aldehydes). Why A is correct:
• Propane-1,2-diol's vicinal OH enables oxidative cleavage (e.g., by I2/OH-) to acetaldehyde and formaldehyde, both reducing aldehydes per Fehling's mechanism. Why the others are wrong:
• B: Non-vicinal diol; no cleavage, primary oxidation yields beta-hydroxy aldehyde but lacks defining mixture of simple reducing carbonyls.
• C: Secondary alcohol only; oxidizes to methyl ketone (no Fehling's), single product not mixture.
• D: No alcohol group; fails iodoform and oxidation.

Final answer: A