Functional group identification via specific chemical tests Steps:

• PCl5 treatment evolves HCl fumes from compounds with -OH groups, like alcohols (ROH + PCl5 → RCl + POCl3 + HCl) or carboxylic acids.
• Acidified K2Cr2O7 oxidation turns green (Cr^{6+} to Cr^{3+}) for oxidizable compounds like primary/secondary alcohols or aldehydes.
• Q must contain -OH (for HCl) and be oxidizable (for green color), pointing to an alcohol.
• CH3CH2OH (primary alcohol) fits both tests perfectly.

Why B is correct:

• CH3CH2OH reacts with PCl5 to give HCl (from -OH) and oxidizes to CH3CHO/CH3COOH, reducing dichromate to green Cr^{3+}.

Why the others are wrong:

• A: CH3CHO (aldehyde) forms gem-dichloride with PCl5 (no HCl); oxidizes to green but fails first test.
• C: CH3COCH3 (ketone) forms gem-dichloride with PCl5 (no HCl) and resists oxidation (no green).
• D: CH3COOH (carboxylic acid) gives HCl with PCl5 but does not oxidize (no green).

Final answer: B