See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify compound A via retro-aldol. The starting material is 3-hydroxybutanal: CH3-CH(OH)-CH2-CHO. Under basic conditions (HO^-), a retro-aldol reaction cleaves the beta-hydroxy aldehyde. The retro-aldol of CH3-CH(OH)-CH2-CHO breaks the C2-C3 bond (between the alpha and beta carbons relative to the carbonyl), giving acetaldehyde (CH3CHO) as compound A and formaldehyde... Actually, let us re-examine. The structure given is CH3-CH(OH)-CH2-CHO. In a retro-aldol, the bond between the alpha-carbon and beta-carbon (beta-hydroxy carbonyl) is cleaved. Here the beta-hydroxy group is on C3 (numbering from CHO as C1): C1=CHO, C2=CH2, C3=CH(OH), C4=CH3. The beta-hydroxy aldehyde undergoes retro-aldol at the C2-C3 bond, giving: - Fragment 1: CH2=O (formaldehyde) from C1-C2? No. Let us reconsider: retro-aldol of a beta-hydroxy carbonyl compound breaks the Calpha-Cbeta bond. C1(CHO)-C2(H2)-C3(OH)(H)-C4(H3). The Cbeta is C3 (bears OH), Calpha is C2. Cleavage of C2-C3 gives: CHO-CH2^- (enolate of acetaldehyde, i.e., CH3CHO after protonation) + CH3CHO (from C3-C4 as acetaldehyde). Wait: C3-C4 fragment is CH3-CH=O = acetaldehyde, and C1-C2 fragment is H-CHO + CH2 ... More carefully: cleavage of C2-C3 bond gives [CH2=CH-OH]^- (enolate, equivalent to CH3CHO) from C1-C2 side, and CH3CHO from C3-C4 side. So A = CH3CHO (acetaldehyde). Step 2 - Reaction of A (acetaldehyde, CH3CHO) with 3 equivalents of HCHO under Na2CO3 (Tollens-type crossed aldol / Cannizzaro conditions). This is the classic reaction of an aldehyde bearing alpha-hydrogens with excess formaldehyde under basic conditions: three successive aldol additions of HCHO to the alpha-carbons of CH3CHO, followed by an intramolecular Cannizzaro (crossed Cannizzaro) of the remaining aldehyde with HCHO. CH3CHO + 3 HCHO (aldol additions at alpha position): - First aldol: HCHO adds to CH3CHO -> HOCH2-CH2-CHO - Second aldol: HCHO adds to alpha of above -> HOCH2-C(CH2OH)(H)-CHO - Third aldol: HCHO adds to alpha -> HOCH2-C(CH2OH)2-CHO (tris-hydroxymethyl acetaldehyde) Now the product HOCH2-C(CH2OH)2-CHO has no more alpha-hydrogens (the alpha carbon is fully substituted: it bears CHO, and three CH2OH groups... wait, let me recount. CH3CHO: alpha carbon = CH3, so 3 H's available. After 1st HCHO aldol addition: HOCH2-CH2-CHO (alpha carbon now has 2 H's) After 2nd HCHO aldol addition: (HOCH2)2CH-CHO (alpha carbon has 1 H) After 3rd HCHO aldol addition: (HOCH2)3C-CHO (alpha carbon has 0 H's) The product (HOCH2)3C-CHO = 3-(hydroxymethyl)-3-(bis-hydroxymethyl)propanal, i.e., the central carbon bears CHO and three CH2OH groups. This matches option (c): HO-CH2-C(CH2OH)(CH2OH)-CHO, which shows a central carbon with two CH2OH substituents shown explicitly on the sides plus one CH2OH on top and one CHO at bottom (three CH2OH and one CHO total on the central carbon). Note: In a full Tollens condensation with 4 equivalents of HCHO, the final step would be a crossed Cannizzaro reducing the CHO to CH2OH giving pentaerythritol (option a). But here only 3 HCHO are used, so the Cannizzaro step does not occur and the aldehyde remains, giving (HOCH2)3C-CHO. Why other options fail: - (a) Pentaerythritol would require 4 HCHO (3 aldol + 1 Cannizzaro); only 3 HCHO used here. - (b) Shows two CHO groups on the central carbon, which is not consistent with the reaction. - (d) Shows a linear chain structure inconsistent with the aldol addition pattern. Therefore, the correct answer is C.