Vicinal diol identification via iodoform and oxidative cleavage Steps:

• Yellow precipitate with alkaline iodine indicates positive iodoform test, requiring CH3CH(OH)- group in the alcohol.
• Substance Z gives red-brown precipitate with Fehling's solution, confirming Z is an aldehyde.
• Oxidation yields a mixture including Z, indicating oxidative cleavage of a 1,2-diol to carbonyl compounds, one being an aldehyde.
• Only option A, CH3CH(OH)CH2OH (propane-1,2-diol), is a vicinal diol with the required CH3CH(OH)- structure.

Why A is correct:

• CH3CH(OH)CH2OH undergoes iodoform reaction due to CH3CH(OH)- and cleaves oxidatively to CH3CHO and HCHO, both aldehydes (Z reduces Fehling's).

Why the others are wrong:

• B: CH3CH(OH)CH2CH2OH is a 1,3-diol (non-vicinal); no cleavage, oxidation yields ketone primarily, not aldehyde mixture.
• C: CH3CH(OH)CH2CH3 is secondary alcohol only; oxidizes to butanone (ketone), no aldehyde.
• D: CH3OCH2CH2CH3 is an ether; lacks OH group, fails both tests.

Final answer: A