Compound exhibiting chirality and iodoform reactivity Steps:

• Check for optical isomerism: requires a tetrahedral carbon with four different substituents.
• Check for positive iodoform test: requires CH3C(O)- group or oxidizable CH3CH(OH)- group in alkaline I2.
• Evaluate A: CH3C(O)CH2OH has methyl ketone (iodoform positive) but no chiral carbon.
• Evaluate B: HOCH2CH(OH)CH3 has chiral central carbon and CH3CH(OH)- (oxidizes to methyl ketone, iodoform positive).

Why B is correct:

• Central carbon bonds to H, OH, CH2OH, CH3 (four different groups per chirality definition); secondary alcohol oxidizes to CH3C(O)CH2OH, yielding yellow iodoform precipitate.

Why the others are wrong:

• A: No chiral carbon (CH2OH has two hydrogens).
• C: Identical structure to B (HOCH2CH(OH)CH3), but as a separate option, it duplicates; question likely intends distinction, but lacks unique error.
• D: Iodoform positive (oxidizes to acetone) but no chirality (central carbon has two CH3 groups).

Final answer: B