Compound Q is a chiral reducing aldehyde with a methyl carbinol group Steps:

• Fehling's reagent detects aldehydes (reducing carbonyls), so Q must contain -CHO group.
• Alkaline aqueous iodine (iodoform test) detects CH3CH(OH)- or CH3CO- groups via yellow iodoform precipitate.
• Chiral center requires a carbon with four different substituents.
• Match structures: only option with -CHO, CH3CH(OH)-R, and chiral C fits.

Why C is correct:

• 3-Hydroxybutanal (CH3CH(OH)CH2CHO) has -CHO for Fehling's, CH3CH(OH)- for iodoform (oxidizes to CH3COCH2CHO, a methyl ketone), and chiral C at CH(OH) (groups: CH3, OH, H, CH2CHO).

Why the others are wrong:

• A: 1-Hydroxybutanone (likely HOCH2COCH2CH3) lacks chiral center (no asymmetric C); not methyl ketone for iodoform.
• B: 2-Hydroxybutanal (CH3CH2CH(OH)CHO) has -CHO and chiral C, but CH3CH2CH(OH)- oxidizes to CH3CH2COCHO (not methyl ketone), so no iodoform.
• D: 3-Hydroxybutanone (CH3COCH(OH)CH3) is a ketone (no Fehling's), despite chiral C and iodoform.

Final answer: C