See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify Product (A). CH3-C≡CH reacted with HgSO4/dil. H2SO4 undergoes acid-catalyzed hydration (Markovnikov addition of water). For a terminal alkyne, Markovnikov addition places the OH on the internal carbon, giving an enol that tautomerizes to a ketone. Thus (A) = CH3-CO-CH3, acetone (a ketone). Step 2: Identify Product (B). CH3-C≡CH reacted with (1) BH3·THF then (2) H2O2/HO- undergoes hydroboration-oxidation (anti-Markovnikov addition of water). The boron adds to the terminal carbon, so after oxidation the OH ends up on the terminal carbon, giving an enol that tautomerizes to an aldehyde. Thus (B) = CH3-CH2-CHO, propanal (an aldehyde). Step 3: Identify which reagent differentiates (A) a ketone from (B) an aldehyde. (a) 2,4-DNP (2,4-dinitrophenylhydrazine): reacts with both aldehydes AND ketones to form orange/yellow precipitates. Cannot differentiate. (b) NaOI (sodium hypoiodite / iodoform reagent): reacts with methyl ketones (CH3CO-) to give iodoform (CHI3, yellow precipitate). Acetone (CH3COCH3) is a methyl ketone and will give a positive iodoform test. Propanal (CH3CH2CHO) is not a methyl ketone and does not give a positive iodoform test. Therefore NaOI differentiates (A) from (B). (c) Na-metal: reacts with compounds having active hydrogen (like alcohols, not aldehydes or ketones in this context). Neither pure ketone nor pure aldehyde reacts readily with Na-metal to produce H2 gas in the typical sense used for differentiation here. Does not differentiate. (d) NaHSO3: both aldehydes and most ketones (if unhindered) form bisulfite addition products. Acetone also reacts with NaHSO3. Does not cleanly differentiate in this case. Step 4: Conclusion. NaOI (iodoform test) distinguishes acetone (positive) from propanal (negative), thereby differentiating product (A) from product (B). Therefore, the correct answer is B.