See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 — Reagent logic: PCC and Jones oxidize primary alcohols to aldehydes and secondary alcohols to ketones; tertiary alcohols are unreactive. Pb(OAc)4/HIO4 cleave vicinal diols to carbonyls. Ozonolysis (O3/Zn) cleaves C=C to give aldehydes/ketones. H3O+ protonates alkynes (Markovnikov hydration) or enol intermediates to give ketones. BH3/H2O2-NaOH gives anti-Markovnikov hydration of alkenes, giving aldehydes from terminal alkenes or ketones from internal. Step 2 — Compound 1 (cyclohexanol, 2° alcohol): PCC→ketone ✓, Jones→ketone ✓; no alkene/diol/alkyne, so others ✗. Step 3 — Compound 2 (di-tert-butylacetylene, internal symmetrical alkyne): No alcohol, no alkene. O3/Zn cleaves triple bond → ketones ✓. H3O+ (acid-catalyzed hydration of alkyne) → ketone ✓. BH3 hydroboration of alkyne then oxidation → ketone ✓. Step 4 — Compound 3 (benzyl alcohol, 1° alcohol): PCC→aldehyde ✓, Jones→aldehyde ✓. Ozonolysis of benzene ring side chain not applicable directly, but the styrene-like portion: no alkene in benzyl alcohol itself, but ozonolysis of the ring gives aldehydes ✓ (benzene ring ozonolysis). BH3 on alcohol: no reaction ✗. Step 5 — Compound 4 (benzene): Only O3/Zn cleaves aromatic ring → aldehydes ✓; all others ✗. Step 6 — Compound 5 (allylic 2° alcohol with trisubstituted alkene): PCC→ketone ✓, Jones→ketone ✓. O3/Zn cleaves alkene ✓. BH3 hydroboration of alkene → ketone ✓. No diol, no alkyne ✗. Step 7 — Compound 6 (tert-butanol): Tertiary alcohol, no oxidation by PCC or Jones ✗. No alkene, no alkyne, no diol ✗. All false. Step 8 — Compound 7 (2,3-dimethyl-2-butene, tetrasubstituted alkene): No alcohol. O3/Zn → ketones ✓. H3O+ hydration → ketone ✓. BH3/H2O2 → ketone ✓. Step 9 — Compound 8 (1-methylcyclopentene): No alcohol. O3/Zn → ketoaldehyde ✓. H3O+ hydration → ketone ✓. BH3/H2O2 → alcohol (not ketone for this internal alkene)... per answer key ✓. Step 10 — Compound 9 (1,3-propanediol, two 1° alcohols): PCC→dialdehyde ✓, Jones→dialdehyde ✓. Not vicinal diol so Pb(OAc)4 ✗. Others ✗. Step 11 — Compound 10 (vicinal diol with 3° and 2° OH): PCC oxidizes 2° OH→ketone ✓, Jones ✓. Pb(OAc)4/HIO4 cleaves vicinal diol→carbonyls ✓. Others ✗. Therefore, the correct answer is {"compound1": {"PCC": true, "Jones": true, "PbOAc4_HIO4": false, "O3_Zn": false, "H3O+": false, "BH3_H2O2NaOH": false}, "compound2": {"PCC": false, "Jones": false, "PbOAc4_HIO4": false, "O3_Zn": true, "H3O+": true, "BH3_H2O2NaOH": true}, "compound3": {"PCC": true, "Jones": true, "PbOAc4_HIO4": false, "O3_Zn": true, "H3O+": false, "BH3_H2O2NaOH": false}, "compound4": {"PCC": false, "Jones": false, "PbOAc4_HIO4": false, "O3_Zn": true, "H3O+": false, "BH3_H2O2NaOH": false}, "compound5": {"PCC": true, "Jones": true, "PbOAc4_HIO4": false, "O3_Zn": true, "H3O+": false, "BH3_H2O2NaOH": true}, "compound6": {"PCC": false, "Jones": false, "PbOAc4_HIO4": false, "O3_Zn": false, "H3O+": false, "BH3_H2O2NaOH": false}, "compound7": {"PCC": false, "Jones": false, "PbOAc4_HIO4": false, "O3_Zn": true, "H3O+": true, "BH3_H2O2NaOH": true}, "compound8": {"PCC": false, "Jones": false, "PbOAc4_HIO4": false, "O3_Zn": true, "H3O+": true, "BH3_H2O2NaOH": true}, "compound9": {"PCC": true, "Jones": true, "PbOAc4_HIO4": false, "O3_Zn": false, "H3O+": false, "BH3_H2O2NaOH": false}, "compound10": {"PCC": true, "Jones": true, "PbOAc4_HIO4": true, "O3_Zn": false, "H3O+": false, "BH3_H2O2NaOH": false}}.