See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the key observations: • Compound A has formula C6H12O3. • Positive iodoform test: indicates a methyl ketone (CH3CO-) group or a secondary alcohol that can be oxidized to a methyl ketone (CH3CHOH-). • Negative Tollens test: no free aldehyde group present initially. • After acid hydrolysis (H2O / H2SO4), gives positive Tollens test: a masked aldehyde (acetal) is revealed upon hydrolysis. Step 2 - Interpret the iodoform test: A positive iodoform test requires a CH3C(=O)- group. Options (a), (b), and (c) all contain CH3C(=O)-, while option (d) contains an aldehyde (H-C=O) directly - but option (d) also has a free aldehyde which would give a positive Tollens test directly, contradicting the negative Tollens test. So (d) is eliminated. Step 3 - Interpret the negative Tollens test: The compound has no free aldehyde. Options (a), (b), and (c) all have a ketone (no free aldehyde), so they would all give negative Tollens test. This is consistent with (a), (b), and (c). Step 4 - Interpret the positive Tollens test after hydrolysis: Upon acid hydrolysis, an acetal group (-CH(OCH3)2 or -C(OCH3)2-) is hydrolyzed to an aldehyde (-CHO) or a ketone. Only if the acetal produces an ALDEHYDE will the Tollens test become positive. • Option (b): CH3-C(=O)-C(OCH3)2-CH3. Hydrolysis of the ketal -C(OCH3)2- gives CH3-CO-CO-CH3 (a diketone / pyruvaldehyde dimethyl ketal hydrolysis gives a ketone), no aldehyde formed. Tollens would remain negative. • Option (a): CH3-CO-CH(OCH3)-CH2-OCH3. This is not a full acetal; it has two separate ether groups and would not straightforwardly yield an aldehyde on simple hydrolysis. • Option (c): CH3-C(=O)-CH2-CH(OCH3)2. This contains a dimethyl acetal group -CH(OCH3)2. Acid hydrolysis of this acetal gives CH3-CO-CH2-CHO (3-oxobutanal / acetoacetaldehyde). The product has an aldehyde, giving a positive Tollens test. The original compound has a methyl ketone (CH3CO-) giving positive iodoform, and no free aldehyde giving negative Tollens. This is fully consistent. Step 5 - Verify molecular formula for option (c): CH3-CO-CH2-CH(OCH3)2: C atoms: 1(CH3)+1(CO)+1(CH2)+1(CH)+2(OCH3) = 1+1+1+1+2 = 6 C. H atoms: 3+0+2+1+6 = 12 H. O atoms: 1(C=O)+2(OCH3) = 3 O. Formula = C6H12O3. Correct. Step 6 - Why other options fail: (a) Not a proper acetal; does not yield an aldehyde cleanly on hydrolysis. (b) Ketal hydrolysis gives a ketone, not an aldehyde; Tollens remains negative. (d) Has a free aldehyde, giving positive Tollens before hydrolysis - contradicts the negative Tollens condition. Therefore, the correct answer is C.