Chemical tests identify Q as an aldehyde

Steps:

• Red precipitate with 2,4-dinitrophenylhydrazine confirms a carbonyl group (C=O) in Q, so Q is propanal or propanone.
• Orange precipitate with Fehling's reagent shows Q reduces Cu²⁺ to Cu₂O, a property of aldehydes but not ketones, identifying Q as propanal.
• Yellow precipitate with alkaline aqueous iodine (iodoform test) is consistent with structures having CH₃C=O or CH₃CH(OH), but the prior tests pinpoint propanal as the match.
• Thus, Q is the aldehyde propanal (CH₃CH₂CHO).

Why C is correct:

• Propanal (CH₃CH₂CHO) is an aldehyde that forms a hydrazone (red ppt) with 2,4-DNP per the carbonyl test and reduces Fehling's via aldehyde oxidation to carboxylic acid.

Why the others are wrong:

• A: Ethanol (CH₃CH₂OH) lacks a carbonyl, so no reaction with 2,4-DNP.
• B: Propan-1-ol (CH₃CH₂CH₂OH) lacks a carbonyl, so no reaction with 2,4-DNP.
• D: Propanone (CH₃COCH₃) is a ketone that does not reduce Fehling's reagent.

Final answer: C