Aldehydes Ketones and Carboxylic AcidshardNUMERICAL

See imageAldehydes Ketones and Carboxylic Acids Chemistry Question

Question

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Chemistry diagram for: See image
Answer: 17

💡 Solution & Explanation

Step 1: Identify Reaction 1 and Product A. Benzaldehyde undergoes Perkin condensation with acetic anhydride (Ac2O) and sodium acetate (AcONa) with heat, followed by acidic hydrolysis (H3O+, Delta). The Perkin reaction on benzaldehyde gives cinnamic acid (trans-PhCH=CHCOOH) as product A. Molecular formula of cinnamic acid: C9H8O2. Degree of unsaturation (DoU) = (2×9 + 2 - 8)/2 = (18 + 2 - 8)/2 = 12/2 = 6. DoU of A = 6 (4 from benzene ring including 3 C=C and 1 aromatic, actually: benzene ring = 4 DoU, C=C alkene = 1 DoU, C=O of COOH = 1 DoU, total = 6). Step 2: Identify Reaction 2 and Product B. Ph-CH=CH-C(=O)-CH3 (benzalacetone / 4-phenyl-3-buten-2-one) reacts with Al(OCHMe2)3 (aluminum isopropoxide) and isopropanol. This is the Meerwein-Ponndorf-Verley (MPV) reduction, which selectively reduces the carbonyl (ketone) to an alcohol without affecting the C=C double bond. Starting material: Ph-CH=CH-C(=O)-CH3, molecular formula C10H10O. Product B: Ph-CH=CH-CH(OH)-CH3, molecular formula C10H12O. DoU of B = (2×10 + 2 - 12)/2 = (20 + 2 - 12)/2 = 10/2 = 5. DoU of B = 5 (benzene ring = 4 DoU, C=C alkene = 1 DoU, total = 5). Step 3: Identify Reaction 3 and Product C. Ph-CH=CH-CH(OH)-CH3 reacts with (1) NaOI (iodoform reaction conditions) and (2) H+. NaOI is sodium hypoiodite, used in the iodoform reaction. The substrate Ph-CH=CH-CH(OH)-CH3 has a secondary alcohol with a methyl group adjacent (CH(OH)-CH3 pattern = methyl carbinol). The iodoform reaction converts CH3CH(OH)- group: the alpha carbon gets triiodinated and then cleaved. The iodoform reaction on Ph-CH=CH-CH(OH)-CH3 would oxidize the -CH(OH)-CH3 to a ketone Ph-CH=CH-CO-CH3 first, then the iodoform reaction cleaves it to give Ph-CH=CH-COO- (cinnamate) + CHI3. After acidification (H+), product C = Ph-CH=CH-COOH (cinnamic acid). Molecular formula of C (cinnamic acid): C9H8O2. DoU of C = 6 (same as A). Step 4: Sum the degrees of unsaturation. DoU(A) + DoU(B) + DoU(C) = 6 + 5 + 6 = 17. Therefore, the correct answer is 17.

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