See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

To determine whether a carbon is oxidized or reduced, we assign oxidation states based on the rule: each C-H bond contributes -1 to the carbon's oxidation state, each C-C bond contributes 0, and each C-O, C-Br, or C-heteroatom bond contributes +1. An increase in oxidation state = oxidized; a decrease = reduced. Reaction A: CH3CH=CH2 + Br2 -> CH3CHBrCH2Br - C12 (middle carbon, CH=): In propene, C12 is =CH-, bonded to 1H, 1C (double bond counts as 2 bonds to same C but each bond to heteroatom/H/C is counted individually). Oxidation state of C12 in propene: bonds are 1H (-1), 1C (0 from CH3), 1C (0 from double bond partner). OS = -1. In product CH3CHBrCH2Br: C12 is CHBr, bonded to 1H (-1), 1Br (+1), 2C (0). OS = 0. Change: -1 -> 0, oxidized. - C14 (terminal carbon, =CH2): In propene, bonded to 2H (-2), 1C (double bond, 0). OS = -2. In product CH2Br, bonded to 2H (-2), 1Br (+1), 1C (0). OS = -1. Change: -2 -> -1, oxidized. Both C12 and C14 are oxidized in A. Reaction B: CH3CH=CH2 --(i)B2H6, (ii)H2O2,NaOH--> CH3CH2CH2OH (anti-Markovnikov hydroboration-oxidation) - C12 (middle carbon, CH=): In propene OS = -1 (as above). In product CH3CH2-, C12 is now -CH2- bonded to 2H (-2), 2C (0). OS = -2. Change: -1 -> -2, reduced. - C14 (terminal carbon, =CH2): In propene OS = -2 (as above). In product CH2OH, bonded to 2H (-2), 1O (+1), 1C (0). OS = -1. Change: -2 -> -1, oxidized. C12 is reduced, C14 is oxidized in B. Reaction C: CH3CH2*CH=O --NaBH4--> CH3CH2CH2OH - Starred carbon (*CH=O, aldehyde carbon): bonded to 1H (-1), 1O (+1), 1C (0). OS = 0. In product CH2OH: bonded to 2H (-2), 1O (+1), 1C (0). OS = -1. Change: 0 -> -1, reduced. The starred carbon is reduced in C. Reaction D: CH3CH2*CH=O --Ag+, H2O, pH>8--> CH3CH2CO2H - Starred carbon (*CH=O, aldehyde carbon): OS = 0 (as computed above). In product CO2H (carboxylic acid carbon): bonded to 0H, 2O (+2), 1C (0). OS = +2. Wait, let me recount. Carboxylic acid carbon: bonds to 1C (0), 1=O (+1), 1-O-H (+1). OS = +2. Change: 0 -> +2, oxidized. The starred carbon is oxidized in D. Reaction E: CH3CO-CH2-CO2H --Heat--> CH3COCH3 + CO2 12 14 - C12 (CH2 between the two carbonyls): bonded to 2H (-2), 2C (0). OS = -2. In product CH3COCH3, C12 becomes a CH3 group: bonded to 3H (-3), 1C (0). OS = -3. Change: -2 -> -3, reduced. - C14 (CO2H, carboxylic acid carbon): OS = +2 (as computed above for carboxylic acid carbon: 2O bonds +2, 1C bond 0). In product CO2: carbon bonded to 2=O (+2 each) = +4. OS = +4. Change: +2 -> +4, oxidized. C12 is reduced, C14 is oxidized in E. Reaction F: H2C=C(OH)C2H5 --tautomerization--> H3C-CO-C2H5 12 14 - C12 (=CH2 terminal): bonded to 2H (-2), 1C (double bond, 0). OS = -2. In product H3C- (methyl): bonded to 3H (-3), 1C (0). OS = -3. Change: -2 -> -3, reduced. - C14 (=C(OH)-, enol carbon bearing OH): bonded to 0H, 1O (+1), 2C (0). OS = +1. In product C=O (ketone carbonyl): bonded to 0H, 1=O (+1), 2C (0). OS = +1... Let me reconsider. Enol C14: double bond to C12 and single bond to OH and single bond to C2H5. Bonds: 1O (+1), 1C(=C12, 0), 1C(C2H5, 0), no H. OS = +1. Ketone carbonyl C14: double bond to O and single bonds to CH3 and C2H5. Bonds: 1=O (+1 for each bond? No - use oxidation state method): 2 bonds to O = +2, 2 bonds to C = 0. OS = +2. Hmm, that gives oxidized. But the answer says C14 is oxidized. Wait - the answer says C14 oxidized. Let me recheck C12: enol =CH2 has OS -2, product CH3 has OS -3, so reduced. C14: enol C(OH)= - using the standard method: each bond to O gives +1, each bond to H gives -1, each bond to C gives 0. C14 in enol: 1 bond to O (+1), 1 bond to C12 via double bond (treated as 2 bonds, 0 each), 1 bond to C2H5 (0). Total: +1... but double bond means 2 bonds. Actually in oxidation state counting for organic: count number of bonds to more electronegative atoms minus bonds to less electronegative atoms (H). C14 in enol H2C=C(OH)C2H5: bonded to O (x1, +1), C=C (two bonds to C12, 0 each), C-C2H5 (0). OS = +1. C14 in ketone H3C-CO-C2H5: bonded to O via double bond (two bonds to O, +2), C-CH3 (0), C-C2H5 (0). OS = +2. Change: +1 -> +2, oxidized. C12 is reduced, C14 is oxidized in F. Therefore, the correct answer is {"A": {"C12": "oxidized", "C14": "oxidized"}, "B": {"C12": "reduced", "C14": "oxidized"}, "C": "reduced", "D": "oxidized", "E": {"C12": "reduced", "C14": "oxidized"}, "F": {"C12": "reduced", "C14": "oxidized"}}.