Iodoform test identifies methyl ketones without chiral centers

Steps:

• Alkaline aqueous iodine performs the iodoform test, positive for methyl ketones (CH3COR) forming yellow iodoform precipitate.
• Identify methyl ketone group in options: only A has CH3C(O)CH2OH.
• Check for optical isomerism: requires tetrahedral carbon with four different substituents.
• A lacks such a carbon, while others either fail the test or have chirality.

Why A is correct:

• Contains CH3C(O)- group, satisfying iodoform reaction where methyl ketones undergo haloform cleavage (CH3COR + 3I2 + 4NaOH → CHI3 + RCOONa + 3NaI + 3H2O), and no chiral carbon.

Why the others are wrong:

• B: Has chiral carbon at CH(OH) (groups: CH3, H, OH, CHO), shows optical isomerism.
• C: Has chiral carbon at CH(OH) (groups: CH3, H, OH, CH2OH), shows optical isomerism.
• D: Lacks CH3C(O)- group, does not give positive iodoform test.

Final answer: A