See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Concept: We need to differentiate four compounds - Salicylic acid, Ethyl salicylate, Ethyl acetyl salicylate, and Acetyl salicylic acid - using chemical tests in decreasing order of reactivity/selectivity. Step 1: Identify the functional groups present in each compound. - Salicylic acid: free -OH (phenolic) and free -CO2H (carboxylic acid) - Ethyl salicylate: free -OH (phenolic) and ester (-CO2Et) - Ethyl acetyl (aspirin ethyl ester): acetylated -OH (no free phenol) and ester (-CO2Et) - Acetyl salicylic acid (aspirin): acetylated -OH (no free phenol) and free -CO2H Step 2: Determine which reagent differentiates which compound first. Test 1 - aq. NaHCO3: Only compounds with free carboxylic acid (-CO2H) will effervesce (give CO2). Salicylic acid and Acetyl salicylic acid both have -CO2H, but salicylic acid with the adjacent free -OH does not react differently here - both would give CO2. However, NaHCO3 specifically reacts with -CO2H groups (stronger acids) but NOT with phenolic -OH. So NaHCO3 identifies the carboxylic acid-containing compounds (Salicylic acid and Acetyl salicylic acid) from the esters. Actually, rethinking the decreasing order: We want to differentiate all four compounds stepwise. Test 1 - aq. NaHCO3: Reacts with free -COOH (Salicylic acid and Acetyl salicylic acid give CO2 effervescence). Ethyl salicylate and Ethyl acetyl do NOT react. This separates the group into two pairs. Test 2 - FeCl3: Reacts with free phenolic -OH to give characteristic violet/purple color. Among all four, Salicylic acid and Ethyl salicylate have free -OH. This distinguishes Salicylic acid from Acetyl salicylic acid (in the COOH group), and Ethyl salicylate from Ethyl acetyl (in the ester group). Test 3 - NaOH: NaOH is a stronger base that can hydrolyze esters (saponification) or react with phenols. This can further differentiate remaining compounds. Step 3: Evaluate option (b) aq. NaHCO3, FeCl3, NaOH: - First use NaHCO3: identifies free COOH groups - separates Salicylic acid & Acetyl salicylic acid (positive) from Ethyl salicylate & Ethyl acetyl (negative) - Then FeCl3: identifies free phenolic OH - among those that tested positive with NaHCO3, Salicylic acid gives violet color (has free OH) while Acetyl salicylic acid does not; among those negative with NaHCO3, Ethyl salicylate gives violet color while Ethyl acetyl does not - Finally NaOH: can hydrolyze esters and acetyl groups to further confirm This sequence correctly differentiates all four compounds in a logical decreasing order of specificity. Step 4: Why other options fail: - Option (a) NaOH, FeCl3, NaHCO3: NaOH first would hydrolyze esters and acetyl groups, destroying the distinguishing features before FeCl3 and NaHCO3 tests can be applied selectively. - Option (c) NaOI, NaOH, NaHCO3: NaOI (iodine in NaOH) is used for iodoform test; not relevant here as primary differentiating reagent. - Option (d) NaOH, Na, NaHCO3: Na metal reacts with both -OH and -COOH; not selective enough as first test, and NaOH first causes hydrolysis issues. Therefore, the correct answer is B.