NCERT Solutions · Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids

Nucleophilic addition to a carbonyl gets harder as the carbonyl carbon gets more crowded and less electron-poor — so more/bulkier alkyl groups on the carbonyl carbon means lower reactivity toward HCN.Addition of \(\displaystyle HCN \) to a carbonyl compound is nucleophilic addition: the cyanide ion, \(\displaystyle CN^- \), attacks the carbonyl carbon, the \(\displaystyle \pi \) bond of \(\displaystyle C=O \) breaks, and the oxygen picks up the negative charge to give a cyanohydrin, \(\displaystyle R_2C(OH)(CN) \). Two things make this attack easier: (a) the carbonyl carbon must be open enough for \(\displaystyle CN^- \) to reach it (steric factor), and (b) the carbonyl carbon must carry enough positive character to attract the nucleophile (electronic factor). Alkyl groups attached to the carbonyl carbon work against both — they donate electron density by the \(\displaystyle +I\) effect (making the carbon less electrophilic) and they physically block the approach of \(\displaystyle CN^- \) the bulkier they are.Compare the four carbonyl compounds by what sits on the carbonyl carbon:Acetaldehyde , \(\displaystyle CH_3CHO \) — one methyl and one hydrogen; smallest, most electrophilic.Acetone , \(\displaystyle CH_3COCH_3 \) — two methyl groups; more \(\displaystyle +I\) donation and more crowding than an aldehyde.Methyl tert-butyl ketone , \(\displaystyle CH_3COC(CH_3)_3 \) — one methyl and one bulky tert-butyl group.Di-tert-butyl ketone, \(\displaystyle (CH_3)_3C\text{-}CO\text{-}C(CH_3)_3 \) — two tert-butyl groups on both sides; maximum electron donation and maximum steric blocking, so \(\displaystyle CN^- \) can barely reach the carbon.So reactivity toward \(\displaystyle HCN \) falls as the groups get bigger and more numerous, giving, in increasing order of reactivity:Di-tert-butyl ketone \(\displaystyle <\) Methyl tert-butyl ketone \(\displaystyle <\) Acetone \(\displaystyle <\) AcetaldehydeA carboxylic acid gets stronger when an electron-withdrawing group sits closer to the \(\displaystyle -COOH \), because that pulls electron density away and stabilizes the carboxylate anion left behind after the acid gives up its proton.Name the four acids and mark where the substituent sits relative to \(\displaystyle -COOH \) (carbon $\displaystyle 1$):\(\displaystyle CH_3CH_2CH(Br)COOH \) — $\displaystyle 2$-bromobutanoic acid , bromine on the carbon next to \(\displaystyle -COOH \) (the \(\displaystyle \alpha \)-carbon).\(\displaystyle CH_3CH(Br)CH_2COOH \) — $\displaystyle 3$-bromobutanoic acid , bromine one carbon further away (the \(\displaystyle \beta \)-carbon).\(\displaystyle (CH_3)_2CHCOOH \) — $\displaystyle 2$-methylpropanoic acid, no halogen, but the \(\displaystyle \alpha \)-carbon carries two methyl groups.\(\displaystyle CH_3CH_2CH_2COOH \) — butanoic acid, no halogen, straight chain.Bromine is electronegative and withdraws electron density inductively (\(\displaystyle -I\) effect), which pulls electron density toward itself and away from the \(\displaystyle -COO^- \) group, spreading out (stabilizing) the negative charge of the conjugate base — this makes the acid stronger. The inductive effect falls off sharply with distance, so bromine on the \(\displaystyle \alpha \)-carbon (right next to \(\displaystyle COOH \)) stabilizes the anion far more than bromine on the \(\displaystyle \beta \)-carbon. Hence $\displaystyle 2$-bromobutanoic acid is a stronger acid than $\displaystyle 3$-bromobutanoic acid.Between the two acids with no halogen, alkyl groups are electron-donating (\(\displaystyle +I\) effect); more branching at the \(\displaystyle \alpha \)-carbon pushes more electron density onto \(\displaystyle -COO^- \), destabilizing the anion and making the acid weaker. \(\displaystyle (CH_3)_2CHCOOH \) has two methyl groups on its \(\displaystyle \alpha \)-carbon (more \(\displaystyle +I\) donation) while \(\displaystyle CH_3CH_2CH_2COOH \) has only a \(\displaystyle -CH_2- \) there, so $\displaystyle 2$-methylpropanoic acid is the weaker of the two.Putting the four together, from weakest acid to strongest acid:\(\displaystyle (CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH \)On the benzene ring, a substituent that pulls electron density out of the ring by resonance strengthens the acid, and one that pushes electron density in by resonance weakens it — because that is exactly what stabilizes or destabilizes the carboxylate anion formed after the proton leaves.The four acids are benzoic acid ( \(\displaystyle C_6H_5COOH \) ), $\displaystyle 4$-nitrobenzoic acid (one \(\displaystyle -NO_2 \) group at the para position), $\displaystyle 3,4$-dinitrobenzoic acid (two \(\displaystyle -NO_2 \) groups), and $\displaystyle 4$-methoxybenzoic acid (one \(\displaystyle -OCH_3 \) group at the para position).The nitro group, \(\displaystyle -NO_2 \), is strongly electron-withdrawing by both induction and resonance: the ring can delocalize electron density into the nitro group's own \(\displaystyle \pi \) system, and this same withdrawal extends to stabilizing the negative charge on the carboxylate oxygen once the acid ionizes. More nitro groups withdraw more electron density, so $\displaystyle 3,4$-dinitrobenzoic acid (two \(\displaystyle -NO_2 \) groups) is a stronger acid than $\displaystyle 4$-nitrobenzoic acid (one \(\displaystyle -NO_2 \) group), and both are stronger than plain benzoic acid.The methoxy group, \(\displaystyle -OCH_3 \), has an oxygen lone pair that can be donated into the ring by resonance (the \(\displaystyle +R\) or \(\displaystyle +M\) effect); at the para position this resonance donation dominates over the oxygen's weak inductive electron-withdrawal. Net electron density is pushed into the ring and toward the carboxylate, which destabilizes the negative charge rather than spreading it out, so $\displaystyle 4$-methoxybenzoic acid ionizes less readily than benzoic acid itself — it is the weakest acid of the four.So, in increasing order of acid strength:$\displaystyle 4$-Methoxybenzoic acid \(\displaystyle <\) Benzoic acid \(\displaystyle <\) $\displaystyle 4$-Nitrobenzoic acid \(\displaystyle <\) $\displaystyle 3,4$-Dinitrobenzoic acidAnswer: (i) Di-tert-butyl ketone \(\displaystyle <\) Methyl tert-butyl ketone \(\displaystyle <\) Acetone \(\displaystyle <\) Acetaldehyde (increasing reactivity to HCN). (ii) \(\displaystyle (CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH \) (increasing acid strength). (iii) $\displaystyle 4$-Methoxybenzoic acid \(\displaystyle <\) Benzoic acid \(\displaystyle <\) $\displaystyle 4$-Nitrobenzoic acid \(\displaystyle <\) $\displaystyle 3,4$-Dinitrobenzoic acid (increasing acid strength).

A carbonyl or acid pair is told apart by finding the one functional-group test that only one partner in the pair can pass — an aldehyde's oxidisable \(\displaystyle -CHO\) hydrogen (Tollens'/Fehling's test), a \(\displaystyle \mathrm{CH_3-CO-}\) fragment's iodoform reaction, or a carboxylic acid's ability to liberate \(\displaystyle \mathrm{CO_2}\) from \(\displaystyle \mathrm{NaHCO_3}\).(i) Propanal , \(\displaystyle \mathrm{CH_3-CH_2-CHO}\), vs propanone , \(\displaystyle \mathrm{CH_3-CO-CH_3}\) Propanal is an aldehyde — its carbonyl carbon still carries a hydrogen, so it is easily oxidised. Warm each with Tollens' reagent, the diammine silver(I) complex \(\displaystyle \mathrm{[Ag(NH_3)_2]^+}\). Propanal reduces \(\displaystyle \mathrm{Ag^+}\) to metallic silver, which deposits as a bright silver mirror on the tube wall, while the aldehyde is itself oxidised to propanoate. Propanone's carbonyl carbon bears only carbon substituents, so it cannot be oxidised this way and gives no mirror. (Fehling's solution gives the same distinction: propanal gives a brick-red precipitate of \(\displaystyle \mathrm{Cu_2O}\); propanone gives none.)(ii) Acetophenone , \(\displaystyle \mathrm{C_6H_5-CO-CH_3}\), vs benzophenone , \(\displaystyle \mathrm{C_6H_5-CO-C_6H_5}\) Acetophenone carries a methyl ketone group, \(\displaystyle \mathrm{CH_3-CO-}\), on the ring; benzophenone has phenyl on both sides of the carbonyl and no such methyl. Treat each with iodine and aqueous sodium hydroxide (the iodoform test). In acetophenone the three hydrogens of that methyl are progressively replaced by iodine, and hydroxide then cleaves the \(\displaystyle \mathrm{C-C}\) bond of the resulting \(\displaystyle \mathrm{C_6H_5-CO-CI_3}\), giving sodium benzoate and triiodomethane, \(\displaystyle \mathrm{CHI_3}\) — a pale-yellow precipitate with a characteristic medicinal smell. Benzophenone has no methyl on its carbonyl carbon, so it gives no precipitate.(iii) Phenol , \(\displaystyle \mathrm{C_6H_5-OH}\), vs benzoic acid , \(\displaystyle \mathrm{C_6H_5-COOH}\) The difference is acid strength: benzoic acid (\(\displaystyle \mathrm{p}K_a \approx 4.2\)) is acidic enough to protonate bicarbonate ion, while phenol (\(\displaystyle \mathrm{p}K_a \approx 10\)) is not. Add aqueous \(\displaystyle \mathrm{NaHCO_3}\) to each. Benzoic acid effervesces briskly — \[\mathrm{C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2\uparrow} \] and the gas turns lime water milky. Phenol produces no bubbles at all; it cannot displace carbonic acid from bicarbonate.(iv) Benzoic acid , \(\displaystyle \mathrm{C_6H_5-COOH}\), vs ethyl benzoate, \(\displaystyle \mathrm{C_6H_5-COO-C_2H_5}\) Same reagent and reasoning as (iii). Benzoic acid has a free, acidic \(\displaystyle -\mathrm{COOH}\) and reacts with \(\displaystyle \mathrm{NaHCO_3}\) with brisk effervescence of \(\displaystyle \mathrm{CO_2}\). Ethyl benzoate has no free \(\displaystyle -\mathrm{OH}\) to donate a proton to bicarbonate — its acidic hydrogen has been replaced by an ethyl group — so it shows no gas evolution at all.(v) Pentan-$\displaystyle 2$-one , \(\displaystyle \mathrm{CH_3-CO-CH_2-CH_2-CH_3}\), vs pentan-$\displaystyle 3$-one, \(\displaystyle \mathrm{CH_3-CH_2-CO-CH_2-CH_3}\) Pentan-$\displaystyle 2$-one is a methyl ketone, so it answers the same iodoform test as in (ii): with \(\displaystyle \mathrm{I_2/NaOH}\) it is converted to sodium propanoate plus a pale-yellow precipitate of \(\displaystyle \mathrm{CHI_3}\). Pentan-$\displaystyle 3$-one is symmetrical, with an ethyl group on each side of the carbonyl carbon and no methyl group directly attached to it, so it cannot undergo the haloform cleavage — no precipitate forms.(vi) Benzaldehyde , \(\displaystyle \mathrm{C_6H_5-CHO}\), vs acetophenone , \(\displaystyle \mathrm{C_6H_5-CO-CH_3}\) Benzaldehyde still has the oxidisable aldehydic hydrogen on its carbonyl carbon, so Tollens' reagent oxidises it to benzoate while \(\displaystyle \mathrm{Ag^+}\) is reduced to a silver mirror. Acetophenone's carbonyl carbon carries only carbon substituents (phenyl and methyl), so it cannot reduce Tollens' reagent and gives no mirror. (Fehling's solution should not be used here — aromatic aldehydes such as benzaldehyde do not reduce it — so Tollens' reagent is the reliable choice.)(vii) Ethanal , \(\displaystyle \mathrm{CH_3-CHO}\), vs propanal , \(\displaystyle \mathrm{CH_3-CH_2-CHO}\) Both are aliphatic aldehydes, so both reduce Tollens' reagent to a silver mirror and Fehling's solution to red \(\displaystyle \mathrm{Cu_2O}\) — those two tests cannot separate them. What differs is the group sitting next to the carbonyl carbon: ethanal has a methyl group directly attached, \(\displaystyle \mathrm{CH_3-CHO}\), while propanal has an ethyl group, \(\displaystyle \mathrm{CH_3-CH_2-CHO}\). Add \(\displaystyle \mathrm{I_2/NaOH}\) (the iodoform test): ethanal's methyl group is halogenated and cleaved by base to sodium formate plus a pale-yellow precipitate of \(\displaystyle \mathrm{CHI_3}\); propanal has no methyl group on its carbonyl carbon, so no precipitate forms.Answer: (i) Tollens'/Fehling's test — propanal gives a silver mirror/red \(\displaystyle \mathrm{Cu_2O}\), propanone gives neither. (ii) Iodoform test — acetophenone gives a yellow \(\displaystyle \mathrm{CHI_3}\) precipitate, benzophenone does not. (iii) \(\displaystyle \mathrm{NaHCO_3}\) test — benzoic acid effervesces \(\displaystyle \mathrm{CO_2}\), phenol does not. (iv) \(\displaystyle \mathrm{NaHCO_3}\) test — benzoic acid effervesces \(\displaystyle \mathrm{CO_2}\), ethyl benzoate does not. (v) Iodoform test — pentan-$\displaystyle 2$-one gives a yellow \(\displaystyle \mathrm{CHI_3}\) precipitate, pentan-$\displaystyle 3$-one does not. (vi) Tollens' test — benzaldehyde gives a silver mirror, acetophenone does not. (vii) Iodoform test — ethanal gives a yellow \(\displaystyle \mathrm{CHI_3}\) precipitate, propanal does not.

A synthesis from benzene is really a directing-group puzzle: decide which one-carbon group you must build first so that the electrophile that goes on next lands at the ring position you actually want, then only afterwards convert that first group into the target functional group. Oxidation of a side chain never moves; only the choice of when you nitrate can change where the nitro group sits.(i) Methyl benzoate, \(\displaystyle \mathrm{C_6H_5COOCH_3}\)Step $\displaystyle 1$ — Friedel–Crafts alkylation with tetrachloromethane, \(\displaystyle \mathrm{CCl_4}\) (one carbon atom), anhydrous \(\displaystyle \mathrm{AlCl_3}\) as the Lewis-acid catalyst that generates the electrophile \(\displaystyle \mathrm{{}^{+}CCl_3}\): \(\displaystyle \mathrm{C_6H_6 + CCl_4 \xrightarrow{anhyd.\ AlCl_3} C_6H_5CCl_3 + HCl}\) Product: (trichloromethyl)benzene, common name benzotrichloride, \(\displaystyle \mathrm{C_6H_5CCl_3}\).Step $\displaystyle 2$ — Hydrolysis of the geminal trihalide. Water molecules successively displace the three chlorines on that one carbon; a carbon cannot hold three \(\displaystyle \mathrm{-OH}\) groups, so as soon as two chlorines are replaced the intermediate collapses, losing water, to give a carbonyl: \(\displaystyle \mathrm{C_6H_5CCl_3 + 2H_2O \xrightarrow{\Delta} C_6H_5COOH + 3HCl}\) Product: benzoic acid , \(\displaystyle \mathrm{C_6H_5COOH}\).Step $\displaystyle 3$ — Fischer esterification with methanol, \(\displaystyle \mathrm{CH_3OH}\) (one carbon atom). Conc. \(\displaystyle \mathrm{H_2SO_4}\) protonates the carbonyl oxygen, making the carbonyl carbon strongly electrophilic; the oxygen lone pair of methanol attacks that carbon, and the tetrahedral intermediate loses water: \(\displaystyle \mathrm{C_6H_5COOH + CH_3OH \xrightarrow{conc.\ H_2SO_4,\ \Delta} C_6H_5COOCH_3 + H_2O}\) Product: methyl benzoate, \(\displaystyle \mathrm{C_6H_5COOCH_3}\).(ii) m-Nitrobenzoic acid, \(\displaystyle \mathrm{3\text{-}O_2N\text{-}C_6H_4\text{-}COOH}\)\(\displaystyle \mathrm{-COOH}\) is a deactivating, meta-directing group, so nitrating benzoic acid itself puts the nitro group exactly meta — no separation of isomers needed.Steps $\displaystyle 1$–$\displaystyle 2$ as above: benzene \(\displaystyle \rightarrow\) \(\displaystyle \mathrm{C_6H_5CCl_3}\) \(\displaystyle \rightarrow\) benzoic acid, \(\displaystyle \mathrm{C_6H_5COOH}\).Step $\displaystyle 3$ — Nitration with the nitrating mixture conc. \(\displaystyle \mathrm{HNO_3}\)/conc. \(\displaystyle \mathrm{H_2SO_4}\) (this pair generates the electrophile \(\displaystyle \mathrm{NO_2^{+}}\)), around $\displaystyle 330$ K. The electron-withdrawing \(\displaystyle \mathrm{-COOH}\) destabilises the arenium-ion intermediate least when the new bond forms meta to it, so that is where substitution occurs: \(\displaystyle \mathrm{C_6H_5COOH + HNO_3 \xrightarrow{conc.\ H_2SO_4} 3\text{-}O_2N\text{-}C_6H_4\text{-}COOH + H_2O}\) Product: m-nitrobenzoic acid, $\displaystyle 3$-nitrobenzoic acid.(iii) p-Nitrobenzoic acid, \(\displaystyle \mathrm{4\text{-}O_2N\text{-}C_6H_4\text{-}COOH}\)Because \(\displaystyle \mathrm{-COOH}\) only ever directs meta, the para isomer can never be reached by nitrating benzoic acid — the ring must still be carrying an ortho,para-director (methyl) at the moment of nitration, and only afterwards should that methyl be oxidised up to \(\displaystyle \mathrm{-COOH}\), since oxidation of a side chain does not touch the substitution pattern already fixed on the ring.Step $\displaystyle 1$ — Friedel–Crafts alkylation with chloromethane, \(\displaystyle \mathrm{CH_3Cl}\) (one carbon atom), anhydrous \(\displaystyle \mathrm{AlCl_3}\): \(\displaystyle \mathrm{C_6H_6 + CH_3Cl \xrightarrow{anhyd.\ AlCl_3} C_6H_5CH_3 + HCl}\) Product: toluene (methylbenzene) , \(\displaystyle \mathrm{C_6H_5CH_3}\).Step $\displaystyle 2$ — Nitration of toluene (conc. \(\displaystyle \mathrm{HNO_3}\)/conc. \(\displaystyle \mathrm{H_2SO_4}\)). \(\displaystyle \mathrm{-CH_3}\) is an ortho,para-director, so the product is a mixture of ortho- and para-nitrotoluene: \(\displaystyle \mathrm{C_6H_5CH_3 + HNO_3 \xrightarrow{conc.\ H_2SO_4} o\text{-}O_2N\text{-}C_6H_4\text{-}CH_3 + p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3}\) The para isomer, being more symmetric, packs more tightly and has the higher melting point, so it is separated from the liquid ortho isomer by fractional distillation/crystallisation. Isolate p-nitrotoluene ($\displaystyle 1$-methyl-$\displaystyle 4$-nitrobenzene).Step $\displaystyle 3$ — Oxidise the benzylic \(\displaystyle \mathrm{-CH_3}\) with hot alkaline \(\displaystyle \mathrm{KMnO_4}\) (or acidified \(\displaystyle \mathrm{K_2Cr_2O_7}\)), then acidify. This oxidation happens at the side-chain carbon only, so the \(\displaystyle \mathrm{NO_2}\) already fixed on the ring stays exactly para: \(\displaystyle \mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3 \xrightarrow{KMnO_4/KOH,\ \Delta;\ then\ H_3O^{+}} p\text{-}O_2N\text{-}C_6H_4\text{-}COOH}\) Product: p-nitrobenzoic acid , $\displaystyle 4$-nitrobenzoic acid.(iv) Phenylacetic acid, \(\displaystyle \mathrm{C_6H_5CH_2COOH}\)This acid has an extra \(\displaystyle \mathrm{-CH_2-}\) between the ring and \(\displaystyle \mathrm{-COOH}\), so it cannot come from oxidising a ring methyl directly (that gives \(\displaystyle \mathrm{-COOH}\) bonded straight to the ring, as in part iii). The chain has to be extended by one carbon using cyanide, then the nitrile hydrolysed.Step $\displaystyle 1$ — as in (iii): benzene + \(\displaystyle \mathrm{CH_3Cl}\)/anhydrous \(\displaystyle \mathrm{AlCl_3}\) \(\displaystyle \rightarrow\) toluene, \(\displaystyle \mathrm{C_6H_5CH_3}\).Step $\displaystyle 2$ — Free-radical halogenation at the benzylic position: toluene with \(\displaystyle \mathrm{Cl_2}\) under ultraviolet light or heat, without any Lewis-acid catalyst, so substitution happens on the side chain (radical mechanism) instead of on the ring (which would need \(\displaystyle \mathrm{FeCl_3}\)/electrophilic substitution): \(\displaystyle \mathrm{C_6H_5CH_3 + Cl_2 \xrightarrow{h\nu\ or\ \Delta} C_6H_5CH_2Cl + HCl}\) Product: benzyl chloride , (chloromethyl)benzene, \(\displaystyle \mathrm{C_6H_5CH_2Cl}\).Step $\displaystyle 3$ — Nucleophilic substitution with cyanide: reflux with \(\displaystyle \mathrm{KCN}\) (an inorganic reagent, so it is not limited by the one-carbon rule). The cyanide ion attacks the benzylic carbon from the side opposite the leaving group, displacing \(\displaystyle \mathrm{Cl^{-}}\) in one step (\(\displaystyle \mathrm{S_N2}\)): \(\displaystyle \mathrm{C_6H_5CH_2Cl + KCN \xrightarrow{alc.,\ \Delta} C_6H_5CH_2CN + KCl}\) Product: phenylacetonitrile (benzyl cyanide), \(\displaystyle \mathrm{C_6H_5CH_2CN}\) — this step is what adds the missing carbon.Step $\displaystyle 4$ — Acidic hydrolysis of the nitrile: reflux with dilute \(\displaystyle \mathrm{HCl}\)/water. Water adds across the \(\displaystyle \mathrm{C \equiv N}\) (via an amide intermediate) and the amide is hydrolysed further to the acid, releasing ammonium ion: \(\displaystyle \mathrm{C_6H_5CH_2CN + 2H_2O \xrightarrow{H_3O^{+},\ \Delta} C_6H_5CH_2COOH + NH_4^{+}}\) Product: phenylacetic acid ($\displaystyle 2$-phenylacetic acid), \(\displaystyle \mathrm{C_6H_5CH_2COOH}\).(v) p-Nitrobenzaldehyde, \(\displaystyle \mathrm{4\text{-}O_2N\text{-}C_6H_4\text{-}CHO}\)The ring positioning is identical to part (iii) — nitrate while an ortho,para-director is present and isolate the para isomer — but the final oxidation must be stopped at the aldehyde stage. \(\displaystyle \mathrm{KMnO_4}\)/\(\displaystyle \mathrm{K_2Cr_2O_7}\) are too strong and run straight through to the acid, so the milder Étard reaction is used instead: it traps the benzylic carbon as a chromium complex before it can be over-oxidised, and only releases the aldehyde on hydrolysis.Steps $\displaystyle 1$–$\displaystyle 2$ as in (iii): benzene \(\displaystyle \rightarrow\) toluene (via \(\displaystyle \mathrm{CH_3Cl}\)/anhyd. \(\displaystyle \mathrm{AlCl_3}\)) \(\displaystyle \rightarrow\) nitrate \(\displaystyle \rightarrow\) isolate p-nitrotoluene, \(\displaystyle \mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3}\).Step $\displaystyle 3$ — Étard reaction: treat p-nitrotoluene with chromyl chloride, \(\displaystyle \mathrm{CrO_2Cl_2}\), in \(\displaystyle \mathrm{CS_2}\) (or \(\displaystyle \mathrm{CCl_4}\)) solution. The benzylic \(\displaystyle \mathrm{C-H}\) bonds are attacked and the side-chain carbon is captured as a chromium(VI)–oxygen complex rather than being oxidised straight through to the acid: \(\displaystyle \mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3 \xrightarrow{CrO_2Cl_2,\ CS_2} [p\text{-}O_2N\text{-}C_6H_4\text{-}CH(OCrOHCl_2)_2]}\) (Étard complex) Step $\displaystyle 4$ — Hydrolyse the complex with water/dilute acid, which cleaves the chromium–oxygen bonds and releases the aldehyde without further oxidation: \(\displaystyle \mathrm{[Étard\ complex] \xrightarrow{H_2O/H_3O^{+}} p\text{-}O_2N\text{-}C_6H_4\text{-}CHO}\) Product: p-nitrobenzaldehyde, $\displaystyle 4$-nitrobenzaldehyde, \(\displaystyle \mathrm{O_2N\text{-}C_6H_4\text{-}CHO}\).Answer: (i) Methyl benzoate: benzene \(\displaystyle \to\) \(\displaystyle \mathrm{C_6H_5CCl_3}\) (\(\displaystyle \mathrm{CCl_4}\)/anhyd. \(\displaystyle \mathrm{AlCl_3}\)) \(\displaystyle \to\) benzoic acid \(\displaystyle \mathrm{C_6H_5COOH}\) (\(\displaystyle \mathrm{H_2O}\), hydrolysis) \(\displaystyle \to\) methyl benzoate \(\displaystyle \mathrm{C_6H_5COOCH_3}\) (\(\displaystyle \mathrm{CH_3OH}\), conc. \(\displaystyle \mathrm{H_2SO_4}\)). (ii) m-Nitrobenzoic acid: benzoic acid nitrated directly (conc. \(\displaystyle \mathrm{HNO_3/H_2SO_4}\)) gives \(\displaystyle \mathrm{3\text{-}O_2N\text{-}C_6H_4COOH}\) since \(\displaystyle \mathrm{-COOH}\) is meta-directing. (iii) p-Nitrobenzoic acid: benzene \(\displaystyle \to\) toluene (\(\displaystyle \mathrm{CH_3Cl}\)/anhyd. \(\displaystyle \mathrm{AlCl_3}\)) \(\displaystyle \to\) nitrate, isolate the para isomer \(\displaystyle \to\) oxidise \(\displaystyle \mathrm{-CH_3}\) with hot \(\displaystyle \mathrm{KMnO_4}\) to give \(\displaystyle \mathrm{4\text{-}O_2N\text{-}C_6H_4COOH}\). (iv) Phenylacetic acid: toluene \(\displaystyle \to\) \(\displaystyle \mathrm{C_6H_5CH_2Cl}\) (\(\displaystyle \mathrm{Cl_2}\), \(\displaystyle h\nu\)) \(\displaystyle \to\) \(\displaystyle \mathrm{C_6H_5CH_2CN}\) (\(\displaystyle \mathrm{KCN}\)) \(\displaystyle \to\) \(\displaystyle \mathrm{C_6H_5CH_2COOH}\) (\(\displaystyle \mathrm{H_3O^{+}}\) hydrolysis). (v) p-Nitrobenzaldehyde: p-nitrotoluene (from iii) \(\displaystyle \xrightarrow{\mathrm{CrO_2Cl_2}}\) Étard complex \(\displaystyle \xrightarrow{\mathrm{H_2O}}\) \(\displaystyle \mathrm{4\text{-}O_2N\text{-}C_6H_4CHO}\).

A carbon skeleton tells you how many steps you need — count the carbons on each side first, and only then pick the reagents. Below, each of the nine conversions is done in exactly two steps, with the electron movement and the intermediate named at every stage.(i) Propanone to Propene — Reduce the carbonyl to an alcohol, then eliminate water; the carbon count already matches, so nothing else needs to change. Propanone, \(\displaystyle \text{CH}_3\text{-CO-CH}_3\) (a ketone with the carbonyl on C-$\displaystyle 2$), and propene, \(\displaystyle \text{CH}_3\text{-CH=CH}_2\), both have three carbons — only the oxygen has to go. \[\text{CH}_3\text{-CO-CH}_3 \xrightarrow{\text{NaBH}_4 \text{ (or LiAlH}_4\text{)}} \text{CH}_3\text{-CH(OH)-CH}_3 \xrightarrow[\Delta]{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{-CH=CH}_2 \] Step $\displaystyle 1$: hydride from \(\displaystyle \text{NaBH}_4\) adds to the electrophilic carbonyl carbon; after work-up the oxygen becomes -OH, giving propan-$\displaystyle 2$-ol . Step $\displaystyle 2$: protonation of the -OH by \(\displaystyle \text{H}_2\text{SO}_4\) turns it into a leaving group (water); loss of water together with a proton from the adjacent \(\displaystyle \text{CH}_3\) forms the new \(\displaystyle \text{C=C}\) bond, giving propene.(ii) Benzoic acid to Benzaldehyde — A carboxylic acid cannot be reduced straight to an aldehyde with \(\displaystyle \mathrm{LiAlH_{4}}\) — it would not stop there — so the acid is first turned into the acid chloride, and that is reduced with a deliberately poisoned catalyst that does stop at the aldehyde. \[\text{C}_6\text{H}_5\text{-COOH} \xrightarrow{\text{SOCl}_2} \text{C}_6\text{H}_5\text{-COCl} \xrightarrow{\text{H}_2 / \text{Pd-BaSO}_4 \text{ (S, quinoline)}} \text{C}_6\text{H}_5\text{-CHO} \] Step $\displaystyle 1$: \(\displaystyle \text{SOCl}_2\) replaces the -OH of benzoic acid with -Cl (byproducts \(\displaystyle \text{SO}_2\) and \(\displaystyle \text{HCl}\) leave as gas, so the product needs no separation), giving benzoyl chloride . Step $\displaystyle 2$ (Rosenmund reduction): hydrogen replaces the chlorine at the same carbon. The catalyst is palladium on barium sulphate poisoned with sulphur/quinoline, which is deliberately too weak to reduce the resulting aldehyde further, giving benzaldehyde, \(\displaystyle \text{C}_6\text{H}_5\text{-CHO}\).(iii) Ethanol to $\displaystyle 3$-Hydroxybutanal — First oxidise to the aldehyde that still has an alpha-hydrogen, then let two molecules of it add to each other (aldol addition, stopped short of dehydration) — that is what plants a second carbonyl-bearing chain onto the first. \[\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH, cold}} \text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO} \] Step $\displaystyle 1$: PCC is a mild oxidant that stops at the aldehyde instead of going on to the acid, giving ethanal . Step $\displaystyle 2$: hydroxide removes an alpha-hydrogen from one ethanal molecule, giving the stabilised carbanion \(\displaystyle \text{CH}_2^-\text{-CHO}\) (an enolate). This carbanion attacks the carbonyl carbon of a second, intact ethanal molecule — the \(\displaystyle \text{C=O}\) pi bond breaks and the oxygen picks up a proton from water. Because the mixture is kept cold and not heated to eliminate water, the product stops at the addition stage: $\displaystyle 3$-hydroxybutanal, \(\displaystyle \text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO}\) (numbering from the aldehyde carbon as C-$\displaystyle 1$, the new -OH sits on C-$\displaystyle 3$).(iv) Benzene to m-Nitroacetophenone — Put the acetyl group on first: Friedel–Crafts acylation needs an unsubstituted ring, and once the group is on, it deactivates the ring and directs the next substituent to the meta position for you. \[\text{C}_6\text{H}_6 \xrightarrow{\text{CH}_3\text{COCl}, \text{ anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{-CO-CH}_3 \xrightarrow{\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4} m\text{-O}_2\text{N-C}_6\text{H}_4\text{-CO-CH}_3 \] Step $\displaystyle 1$: \(\displaystyle \text{AlCl}_3\) pulls chloride off \(\displaystyle \text{CH}_3\text{COCl}\) to generate the electrophilic acylium ion \(\displaystyle \text{CH}_3\text{CO}^+\); this attacks the ring, and loss of a ring proton restores aromaticity, giving acetophenone , \(\displaystyle \text{C}_6\text{H}_5\text{-CO-CH}_3\). Step $\displaystyle 2$: the nitrating mixture generates the electrophile \(\displaystyle \text{NO}_2^+\). The carbonyl of the acetyl group withdraws electron density from the ring by resonance, so -COCH3 is deactivating and meta-directing; \(\displaystyle \text{NO}_2^+\) attacks meta to it, giving m-nitroacetophenone.(v) Benzaldehyde to Benzophenone — Benzaldehyde already carries one phenyl ring on its carbonyl carbon; add a second phenyl with a Grignard reagent, then oxidise the resulting secondary alcohol back up to a ketone. \[\text{C}_6\text{H}_5\text{-CHO} \xrightarrow[\text{ii) H}_3\text{O}^+]{\text{i) C}_6\text{H}_5\text{MgBr}} \text{C}_6\text{H}_5\text{-CH(OH)-C}_6\text{H}_5 \xrightarrow{\text{PCC}} \text{C}_6\text{H}_5\text{-CO-C}_6\text{H}_5 \] Step $\displaystyle 1$: the phenyl carbon of phenylmagnesium bromide (itself nucleophilic) attacks the electrophilic carbonyl carbon of benzaldehyde; the \(\displaystyle \text{C=O}\) pi bond breaks and the oxygen becomes a magnesium alkoxide. Aqueous acid work-up protonates it, giving diphenylmethanol (benzhydrol), \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OH)-C}_6\text{H}_5\). Step $\displaystyle 2$: PCC oxidises this secondary alcohol — it removes the O–H hydrogen and the C–H hydrogen on the same carbon, reforming a \(\displaystyle \text{C=O}\) bond — giving benzophenone, \(\displaystyle \text{C}_6\text{H}_5\text{-CO-C}_6\text{H}_5\).(vi) Bromobenzene to $\displaystyle 1$-Phenylethanol — Turn the aryl halide into a Grignard reagent, then let its nucleophilic carbon attack the carbonyl carbon of acetaldehyde — that one new C–C bond supplies both the ring and the extra carbon $\displaystyle 1$-phenylethanol needs. \[\text{C}_6\text{H}_5\text{Br} \xrightarrow{\text{Mg, dry ether}} \text{C}_6\text{H}_5\text{MgBr} \xrightarrow[\text{ii) H}_3\text{O}^+]{\text{i) CH}_3\text{CHO}} \text{C}_6\text{H}_5\text{-CH(OH)-CH}_3 \] Step $\displaystyle 1$: magnesium inserts into the \(\displaystyle \text{C-Br}\) bond, giving phenylmagnesium bromide — the aryl carbon here behaves like a carbanion. Step $\displaystyle 2$: this carbon attacks the electrophilic carbonyl carbon of acetaldehyde; the \(\displaystyle \text{C=O}\) pi bond breaks and the oxygen becomes a magnesium alkoxide, \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OMgBr)-CH}_3\). Dilute acid work-up protonates it, giving $\displaystyle 1$-phenylethanol, \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OH)-CH}_3\).(vii) Benzaldehyde to $\displaystyle 3$-Phenylpropan-$\displaystyle 1$-ol — Benzaldehyde has no alpha-hydrogen of its own and cannot self-condense, so let acetaldehyde's alpha-carbanion attack it instead in a crossed aldol condensation, then reduce the new C=C and the -CHO together in one step. \[\text{C}_6\text{H}_5\text{-CHO} + \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{-CH=CH-CHO} \xrightarrow{\text{H}_2 / \text{Ni}} \text{C}_6\text{H}_5\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH} \] Step $\displaystyle 1$: since benzaldehyde has no alpha-hydrogen, hydroxide instead removes one from acetaldehyde, giving the enolate \(\displaystyle \text{CH}_2^-\text{-CHO}\). This attacks the carbonyl carbon of benzaldehyde; the resulting beta-hydroxy aldehyde loses water immediately (elimination is favoured because it extends conjugation into the ring), giving cinnamaldehyde, \(\displaystyle \text{C}_6\text{H}_5\text{-CH=CH-CHO}\). Step $\displaystyle 2$: catalytic hydrogenation (\(\displaystyle \text{H}_2/\text{Ni}\)) reduces both the \(\displaystyle \text{C=C}\) double bond and the \(\displaystyle \text{-CHO}\) group in the same operation, giving $\displaystyle 3$-phenylpropan-$\displaystyle 1$-ol, \(\displaystyle \text{C}_6\text{H}_5\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}\).(viii) Benzaldehyde to α-Hydroxyphenylacetic acid — Add cyanide across the carbonyl to install one extra carbon and a masked -COOH at the same time, then hydrolyse that nitrile to the free acid. \[\text{C}_6\text{H}_5\text{-CHO} \xrightarrow{\text{HCN}} \text{C}_6\text{H}_5\text{-CH(OH)-CN} \xrightarrow{\text{dil. HCl}, \Delta} \text{C}_6\text{H}_5\text{-CH(OH)-COOH} \] Step $\displaystyle 1$: cyanide ion attacks the electrophilic carbonyl carbon of benzaldehyde; the \(\displaystyle \text{C=O}\) pi bond breaks and the oxygen picks up a proton, giving the cyanohydrin mandelonitrile, \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OH)-CN}\). Step $\displaystyle 2$: boiling with dilute acid hydrates the nitrile carbon stepwise (through an amide) to a carboxylic acid, with loss of ammonium ion, giving α-hydroxyphenylacetic acid — mandelic acid — \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OH)-COOH}\).(ix) Benzoic acid to m-Nitrobenzyl alcohol — Put the nitro group on first, while -COOH is still there to direct it to the meta position, then reduce the carboxylic acid down to a primary alcohol; \(\displaystyle \mathrm{LiAlH_{4}}\) leaves the ring nitro group alone. \[\text{C}_6\text{H}_5\text{-COOH} \xrightarrow{\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4} m\text{-O}_2\text{N-C}_6\text{H}_4\text{-COOH} \xrightarrow{\text{LiAlH}_4} m\text{-O}_2\text{N-C}_6\text{H}_4\text{-CH}_2\text{OH} \] Step $\displaystyle 1$: the nitronium ion \(\displaystyle \text{NO}_2^+\) is generated by the nitrating mixture. The -COOH group withdraws electron density from the ring by both resonance and induction, so it is deactivating and meta-directing; \(\displaystyle \text{NO}_2^+\) attacks meta to it, giving m-nitrobenzoic acid. Step $\displaystyle 2$: hydride from \(\displaystyle \text{LiAlH}_4\) attacks the carbonyl carbon of -COOH in two successive additions (through the aldehyde oxidation level), reducing it all the way to \(\displaystyle \text{-CH}_2\text{OH}\); the ring nitro group is unaffected under these conditions. Aqueous work-up gives m-nitrobenzyl alcohol, \(\displaystyle m\text{-O}_2\text{N-C}_6\text{H}_4\text{-CH}_2\text{OH}\).Answer: (i) propene, \(\displaystyle \text{CH}_3\text{-CH=CH}_2\) — via propan-$\displaystyle 2$-ol; (ii) benzaldehyde, \(\displaystyle \text{C}_6\text{H}_5\text{-CHO}\) — via benzoyl chloride, Rosenmund reduction; (iii) $\displaystyle 3$-hydroxybutanal, \(\displaystyle \text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO}\) — via ethanal, aldol addition; (iv) m-nitroacetophenone, \(\displaystyle m\text{-O}_2\text{N-C}_6\text{H}_4\text{-CO-CH}_3\) — via acetophenone; (v) benzophenone, \(\displaystyle \text{C}_6\text{H}_5\text{-CO-C}_6\text{H}_5\) — via diphenylmethanol; (vi) $\displaystyle 1$-phenylethanol, \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OH)-CH}_3\) — via phenylmagnesium bromide; (vii) $\displaystyle 3$-phenylpropan-$\displaystyle 1$-ol, \(\displaystyle \text{C}_6\text{H}_5\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}\) — via cinnamaldehyde; (viii) mandelic acid (α-hydroxyphenylacetic acid), \(\displaystyle \text{C}_6\text{H}_5\text{-CH(OH)-COOH}\) — via mandelonitrile; (ix) m-nitrobenzyl alcohol, \(\displaystyle m\text{-O}_2\text{N-C}_6\text{H}_4\text{-CH}_2\text{OH}\) — via m-nitrobenzoic acid.

(i) Acetylation replaces the active hydrogen of an -OH or -NH2 group by the acetyl group, CH3-CO-, turning a reactive alcohol, phenol or amine into a far less reactive ester or amide.The reagent is acetic anhydride, \(\displaystyle \mathrm{(CH_{3}CO)_{2}O}\), or acetyl chloride, \(\displaystyle \mathrm{CH_{3}COCl}\), often with a trace of pyridine or concentrated \(\displaystyle \mathrm{H_{2}SO_{4}}\) as catalyst. The nucleophilic oxygen of an -OH (or nitrogen of an -NH2) attacks the electrophilic carbonyl carbon of the anhydride/acid chloride; the C-O(anhydride) or C-Cl bond then breaks, expelling acetate ion or chloride ion, and loss of the acidic proton gives the neutral acetylated product.Examples: ethanol, CH3-CH2-OH, plus acetic anhydride gives ethyl acetate, CH3-CH2-O-CO-CH3, plus acetic acid. Aniline, C6H5-NH2 , plus acetic anhydride gives acetanilide, C6H5-NH-CO-CH3 , plus acetic acid.Because the free -OH or -NH2 is now tied up as an ester or amide, acetylation is used to protect these groups during a synthesis — for example the -NH2 of aniline is acetylated before nitration so the ring is not over-oxidised — and to moderate the reactivity of phenolic -OH groups, as when salicylic acid, \(\displaystyle \mathrm{C_{6}H_{4}(OH)(COOH)}\), is acetylated to give aspirin (acetylsalicylic acid).(ii) In the Cannizzaro reaction an aldehyde with no alpha-hydrogen disproportionates in concentrated alkali: one molecule is reduced to the alcohol while a second is oxidised to the carboxylate salt, because with no alpha-hydrogen to remove, neither molecule can form the enolate that ordinary aldol chemistry needs.Step $\displaystyle 1$ — hydroxide ion, OH^-, attacks the carbonyl carbon of one aldehyde molecule, R-CHO, adding across the C=O bond to give a tetrahedral gem-diolate intermediate, R-CH(O^-)(OH).Step $\displaystyle 2$ — this intermediate acts as a hydride donor: it transfers a hydride ion, H^-, from its carbon to the carbonyl carbon of a second, unreacted aldehyde molecule. This oxidises the first carbon to a carboxylic acid (instantly deprotonated by the excess base to the carboxylate, R-COO^-) and reduces the carbonyl carbon of the second molecule to an alkoxide, R-CH2-O^-.Step $\displaystyle 3$ — the alkoxide is protonated by the solvent to give the alcohol, R-CH2-OH.Net change: $\displaystyle 2$ R-CHO + NaOH gives R-CH2-OH + R-COONa.Example: formaldehyde, HCHO, has no alpha-carbon at all, so $\displaystyle 2$ HCHO + NaOH gives methanol, \(\displaystyle \mathrm{CH_{3}OH}\), plus sodium formate, HCOONa. Benzaldehyde reacts the same way, since its "alpha-position" is the aromatic ring and carries no removable proton on the required carbon: $\displaystyle 2$ \(\displaystyle \mathrm{C_{6}H_{5}CHO}\) + NaOH gives benzyl alcohol, \(\displaystyle \mathrm{C_{6}H_{5}CH_{2}OH}\) , plus sodium benzoate, \(\displaystyle \mathrm{C_{6}H_{5}COONa}\) .(iii) Cross (mixed) aldol condensation is an aldol condensation run between two different carbonyl compounds instead of two molecules of the same one; when both partners carry alpha-hydrogens, base can form the enolate from either compound and that enolate can attack the carbonyl of either compound, so up to four different condensation products form together.Take acetaldehyde, CH3-CHO (A) , and acetone, CH3-CO-CH3 (B) , with dilute NaOH: Because all four compete at once, crossing two enolisable partners gives a mixture rather than a clean synthesis. Cross aldol condensation only becomes preparatively useful when one partner has no alpha-hydrogen and so can act solely as the electrophile — benzaldehyde, C6H5-CHO , is the standard case. With acetaldehyde and dilute NaOH, base removes an alpha-hydrogen from acetaldehyde's \(\displaystyle \mathrm{CH_{3}}\) group to give the enolate \(\displaystyle \mathrm{CH_{2}}\)=CH-O^-; this attacks the carbonyl carbon of benzaldehyde (which has no alpha-hydrogen of its own to form an enolate from), giving the alkoxide C6H5-CH(O^-)-CH2-CHO, which is protonated to the aldol C6H5-CH(OH)-CH2-CHO. A further base-mediated removal of an alpha-hydrogen and loss of hydroxide (dehydration) then gives a single conjugated product, cinnamaldehyde, C6H5-CH=CH-CHO ($\displaystyle 3$-phenylprop-$\displaystyle 2$-enal).(iv) Decarboxylation is the loss of carbon dioxide from a carboxylic acid, carried out in practice on its sodium salt, giving a hydrocarbon with exactly one carbon fewer than the acid.The sodium salt of the acid, R-COONa, is heated with soda lime (a mixture of NaOH with CaO, the CaO being present only to keep the NaOH from caking). The bond joining the carboxylate carbon to R breaks, the carbon leaving as carbonate (CO2 is trapped by the NaOH present, as Na2CO3), and the resulting carbanion, R^-, is immediately protonated by the NaOH, giving the alkane R-H:R-COONa + NaOH, heated with CaO, gives R-H + Na2CO3.Example: sodium acetate, \(\displaystyle \mathrm{CH_{3}COONa}\), heated with soda lime gives methane, \(\displaystyle \mathrm{CH_{3}COONa}\) + NaOH (CaO, heat) giving \(\displaystyle \mathrm{CH_{4}}\) + \(\displaystyle \mathrm{Na_{2}CO_{3}}\) — the standard laboratory route from a carboxylic acid salt to the alkane with one less carbon than the acid.Answer: (i) Acetylation — introducing the acetyl group, CH3CO-, onto an -OH/-NH2 using acetic anhydride or acetyl chloride, e.g. \(\displaystyle \mathrm{C_{6}H_{5}NH_{2}}\) to C6H5NHCOCH3. (ii) Cannizzaro reaction — an aldehyde with no alpha-hydrogen disproportionates in concentrated alkali into one molecule of alcohol and one of carboxylate salt, e.g. 2HCHO + NaOH to \(\displaystyle \mathrm{CH_{3}OH}\) + HCOONa. (iii) Cross aldol condensation — aldol condensation between two different carbonyl compounds; with alpha-hydrogens on both sides it gives a mixture of up to four products (acetaldehyde + acetone gives crotonaldehyde, mesityl oxide, and two cross enones), while pairing an enolisable aldehyde with one lacking alpha-hydrogen (acetaldehyde + benzaldehyde) gives one clean product, cinnamaldehyde. (iv) Decarboxylation — loss of \(\displaystyle \mathrm{CO_{2}}\) from a carboxylic acid salt on heating with soda lime, giving an alkane with one carbon fewer, e.g. \(\displaystyle \mathrm{CH_{3}COONa}\) + NaOH (CaO, heat) to \(\displaystyle \mathrm{CH_{4}}\) + Na2CO3.

Read every arrow the same way: find the functional group the reagent attacks, name the standard reaction of that group, then draw what that reaction must give. Nothing here needs a new idea — each part is one named reaction from the aldehyde, ketone and carboxylic acid chapter.(i) Hot alkaline permanganate chops a whole side chain down to one \(\displaystyle \mathrm{-COOH} \). The rule: an alkyl group on a benzene ring is oxidised to a single carboxyl group provided it carries at least one benzylic hydrogen (a hydrogen on the carbon joined straight to the ring). Chain length does not matter — the extra carbons leave as \(\displaystyle \mathrm{CO_2} \). Ethylbenzene's \(\displaystyle \mathrm{-CH_2CH_3} \) has two benzylic hydrogens, so the entire two-carbon chain becomes one \(\displaystyle \mathrm{-COOH} \). In \(\displaystyle \mathrm{KOH} \) the product sits as the potassium salt; the free acid is released on acid work-up.\[\mathrm{C_6H_5CH_2CH_3 \;\xrightarrow[\Delta]{KMnO_4,\;KOH}\; C_6H_5COO^-K^+ \;\xrightarrow{H_3O^+}\; C_6H_5COOH} \]Missing product: benzoic acid, \(\displaystyle \mathrm{C_6H_5COOH} \).(ii) Thionyl chloride swaps the \(\displaystyle \mathrm{-OH} \) of a carboxyl group for \(\displaystyle \mathrm{-Cl} \). The rule: \(\displaystyle \mathrm{RCOOH + SOCl_2 \rightarrow RCOCl + SO_2\uparrow + HCl\uparrow} \). Both by-products are gases, which is why \(\displaystyle \mathrm{SOCl_2} \) is the preferred way to make an acid chloride. The starting material is benzene-$\displaystyle 1,2$-dicarboxylic acid (phthalic acid), and it has two \(\displaystyle \mathrm{-COOH} \) groups on neighbouring ring carbons, so both are converted.Missing product: benzene-$\displaystyle 1,2$-dicarbonyl dichloride (phthaloyl chloride), the ring carrying \(\displaystyle \mathrm{-COCl} \) at C-$\displaystyle 1$ and C-2.Worth separating the two heats you may have seen: phthalic acid warmed on its own loses a molecule of water between the two ortho groups and gives phthalic anhydride. Here \(\displaystyle \mathrm{SOCl_2} \) is written over the arrow, so chlorination is what happens.(iii) Semicarbazide is an ammonia derivative, so it condenses with the carbonyl. The rule: \(\displaystyle \mathrm{\!>\!C{=}O + H_2N{-}G \rightarrow \;>\!C{=}N{-}G + H_2O} \), a nucleophilic addition of the \(\displaystyle \mathrm{-NH_2} \) nitrogen followed by elimination of water. In semicarbazide, \(\displaystyle \mathrm{H_2N\!-\!CO\!-\!NH\!-\!NH_2} \), it is the \(\displaystyle \mathrm{-NH_2} \) of the hydrazine end that attacks, because the nitrogen next to the \(\displaystyle \mathrm{C{=}O} \) has its lone pair tied up in delocalisation.\[\mathrm{C_6H_5CHO + H_2NNHCONH_2 \rightarrow C_6H_5CH{=}N{-}NHCONH_2 + H_2O} \]Missing product: benzaldehyde semicarbazone.(iv) Benzene gains an acyl group only by Friedel–Crafts acylation. The product drawn is \(\displaystyle \mathrm{C_6H_5\!-\!CO\!-\!C_6H_5} \) (benzophenone), so one benzene ring supplied is the substrate and the other arrives with the carbonyl carbon already attached to it. The rule: an acid chloride plus anhydrous \(\displaystyle \mathrm{AlCl_3} \) generates the acylium ion \(\displaystyle \mathrm{R\!-\!\overset{+}{C}{=}O} \), which is the electrophile the ring attacks.Missing reagent: benzoyl chloride, \(\displaystyle \mathrm{C_6H_5COCl} \), with anhydrous \(\displaystyle \mathrm{AlCl_3} \).(v) Tollens' reagent oxidises an aldehyde and leaves a ketone alone. The starting material is $\displaystyle 4$-oxocyclohexane-$\displaystyle 1$-carbaldehyde: a saturated ring with \(\displaystyle \mathrm{{=}O} \) on one carbon and \(\displaystyle \mathrm{-CHO} \) on the carbon directly opposite. The rule: \(\displaystyle \mathrm{[Ag(NH_3)_2]^+} \) is a mild oxidant, and only the aldehyde — which still has a hydrogen on the carbonyl carbon to give up — is attacked. The ketone carbon has no such hydrogen and survives untouched, so the ring \(\displaystyle \mathrm{C{=}O} \) is carried through unchanged. Silver is deposited as the mirror.\[\mathrm{{-}CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow {-}COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O} \]Missing product: $\displaystyle 4$-oxocyclohexane-$\displaystyle 1$-carboxylic acid (obtained as its ammonium carboxylate, freed by acid), plus a silver mirror.(vi) \(\displaystyle \mathrm{NaCN} \) with \(\displaystyle \mathrm{HCl} \) makes \(\displaystyle \mathrm{HCN} \), which adds across the aldehyde to give a cyanohydrin. The rule: \(\displaystyle \mathrm{CN^-} \) adds to the carbonyl carbon, then the alkoxide picks up a proton, giving \(\displaystyle \mathrm{{>}C(OH)CN} \). The acid is added slowly so that a controlled amount of \(\displaystyle \mathrm{CN^-} \) stays available — pure \(\displaystyle \mathrm{HCN} \) alone is too poor a source of the nucleophile. Of the two groups on the ring, only the aldehyde carbon is electrophilic enough; the \(\displaystyle \mathrm{-COOH} \) carbon is deactivated by the lone pairs of its own \(\displaystyle \mathrm{-OH} \), so it merely gets protonated back after donating a proton.Missing product: the cyanohydrin, $\displaystyle 2$-[cyano(hydroxy)methyl]benzoic acid, \(\displaystyle \mathrm{2\text{-}(HO)(NC)CH\text{-}C_6H_4\text{-}COOH} \). (Because the new \(\displaystyle \mathrm{-OH} \) and the \(\displaystyle \mathrm{-COOH} \) are ortho to each other, this product readily loses water internally to a five-membered lactone, $\displaystyle 3$-oxo-$\displaystyle 1,3$-dihydro-$\displaystyle 2$-benzofuran-$\displaystyle 1$-carbonitrile; the addition product above is the answer being asked for.)(vii) One partner has no \(\displaystyle \mathrm{\alpha} \)-hydrogen, so this is a crossed aldol condensation with only one possible outcome. The rule: dilute \(\displaystyle \mathrm{NaOH} \) removes an \(\displaystyle \mathrm{\alpha} \)-hydrogen to build a carbanion, which attacks another carbonyl carbon; on heating, the \(\displaystyle \mathrm{\beta} \)-hydroxy carbonyl loses water. Benzaldehyde, \(\displaystyle \mathrm{C_6H_5CHO} \), has no \(\displaystyle \mathrm{\alpha} \)-hydrogen at all, so it can only be the carbonyl that is attacked. Propanal, \(\displaystyle \mathrm{CH_3CH_2CHO} \), supplies the carbanion at its \(\displaystyle \mathrm{\alpha} \)-carbon, \(\displaystyle \mathrm{-CH_2-} \).Addition first: \[\mathrm{C_6H_5CHO + \;^-CH(CH_3)CHO \rightarrow C_6H_5CH(OH)CH(CH_3)CHO} \] Then \(\displaystyle \Delta \) removes water, and the new double bond is kept because it is conjugated with both the ring and the \(\displaystyle \mathrm{C{=}O} \): \[\mathrm{C_6H_5CH(OH)CH(CH_3)CHO \xrightarrow{\Delta} C_6H_5CH{=}C(CH_3)CHO + H_2O} \]Missing product: $\displaystyle 2$-methyl-$\displaystyle 3$-phenylprop-$\displaystyle 2$-enal (\(\displaystyle \mathrm{\alpha} \)-methylcinnamaldehyde).(viii) \(\displaystyle \mathrm{NaBH_4} \) reduces aldehydes and ketones but not esters or acids. That selectivity is the whole point of the part. In ethyl $\displaystyle 3$-oxobutanoate, \(\displaystyle \mathrm{CH_3COCH_2COOC_2H_5} \), there are two carbonyls: a ketone and an ester. The hydride adds to the ketone carbon only, because the ester carbonyl is already stabilised by the lone pair of its \(\displaystyle \mathrm{-OC_2H_5} \) oxygen and is far less electrophilic. Step (ii), \(\displaystyle \mathrm{H^+} \), simply protonates the alkoxide that step (i) produced.\[\mathrm{CH_3COCH_2COOC_2H_5 \xrightarrow[(ii)\;H^+]{(i)\;NaBH_4} CH_3CH(OH)CH_2COOC_2H_5} \]Missing product: ethyl $\displaystyle 3$-hydroxybutanoate.(ix) \(\displaystyle \mathrm{CrO_3} \) takes a secondary alcohol to a ketone. The rule: a secondary alcohol has one hydrogen left on the carbinol carbon, so chromium(VI) removes that hydrogen along with the \(\displaystyle \mathrm{O\!-\!H} \) and stops at the ketone — there is no further oxidation without breaking a C–C bond. The starting material is cyclohexanol.Missing product: cyclohexanone.(x) The \(\displaystyle \mathrm{CH_2} \) hanging off the ring must end up as the \(\displaystyle \mathrm{-CHO} \) carbon, so the oxygen has to be delivered to the terminal carbon — that is anti-Markovnikov. Compare the two carbons of the exocyclic double bond: the product has \(\displaystyle \mathrm{-CHO} \) on the ring carbon, meaning the former \(\displaystyle \mathrm{{=}CH_2} \) is now the carbonyl carbon bonded to the ring. Acid-catalysed hydration would put the \(\displaystyle \mathrm{-OH} \) on the more substituted ring carbon, which is the wrong carbon. Hydroboration–oxidation puts boron, and then \(\displaystyle \mathrm{-OH} \), on the less substituted carbon:\[\mathrm{C_6H_{10}{=}CH_2 \xrightarrow[(ii)\;H_2O_2,\;OH^-]{(i)\;B_2H_6} C_6H_{11}{-}CH_2OH} \]That is a primary alcohol, cyclohexylmethanol. A primary alcohol goes to the aldehyde only with a mild, anhydrous oxidant — PCC — because \(\displaystyle \mathrm{KMnO_4} \) or \(\displaystyle \mathrm{K_2Cr_2O_7} \) would carry it on to the carboxylic acid:\[\mathrm{C_6H_{11}CH_2OH \xrightarrow{PCC} C_6H_{11}CHO} \]Missing reagents: (i) \(\displaystyle \mathrm{B_2H_6} \); (ii) \(\displaystyle \mathrm{H_2O_2/OH^-} \); (iii) PCC.(xi) Run the ozonolysis backwards: erase both carbonyl oxygens and join the two carbons with a double bond. The rule: \(\displaystyle \mathrm{O_3} \) followed by \(\displaystyle \mathrm{Zn/H_2O} \) cuts a \(\displaystyle \mathrm{C{=}C} \) into two carbonyl compounds, one from each end (the zinc is there to destroy \(\displaystyle \mathrm{H_2O_2} \), which would otherwise oxidise any aldehyde formed). Here both fragments are the same — two molecules of cyclohexanone — so each carbonyl carbon was a ring carbon of a cyclohexane ring, and the two rings were joined to each other by that double bond.\[\mathrm{C_6H_{10}{=}C_6H_{10} \xrightarrow[(ii)\;Zn/H_2O]{(i)\;O_3} 2\;C_6H_{10}{=}O} \]Missing starting material: cyclohexylidenecyclohexane (two cyclohexane rings sharing one \(\displaystyle \mathrm{C{=}C} \) between their carbons).Answer: (i) benzoic acid, \(\displaystyle \mathrm{C_6H_5COOH} \); (ii) phthaloyl chloride (benzene-$\displaystyle 1,2$-dicarbonyl dichloride); (iii) benzaldehyde semicarbazone, \(\displaystyle \mathrm{C_6H_5CH{=}N\text{-}NHCONH_2} \); (iv) \(\displaystyle \mathrm{C_6H_5COCl} \) with anhydrous \(\displaystyle \mathrm{AlCl_3} \); (v) $\displaystyle 4$-oxocyclohexane-$\displaystyle 1$-carboxylic acid with a silver mirror; (vi) the cyanohydrin $\displaystyle 2$-[cyano(hydroxy)methyl]benzoic acid; (vii) $\displaystyle 2$-methyl-$\displaystyle 3$-phenylprop-$\displaystyle 2$-enal, \(\displaystyle \mathrm{C_6H_5CH{=}C(CH_3)CHO} \); (viii) ethyl $\displaystyle 3$-hydroxybutanoate, \(\displaystyle \mathrm{CH_3CH(OH)CH_2COOC_2H_5} \); (ix) cyclohexanone; (x) (i) \(\displaystyle \mathrm{B_2H_6} \), (ii) \(\displaystyle \mathrm{H_2O_2/OH^-} \), (iii) PCC; (xi) cyclohexylidenecyclohexane.

A crowded carbonyl carbon cannot accept a bulky nucleophile as easily as an open one, and turning a flat, three-coordinate carbon into a crowded four-coordinate one costs even more when the neighbouring carbons are already loaded with methyl groups.Cyanohydrin formation is nucleophilic addition to a ketone: the cyanide ion, \(\displaystyle CN^-\), attacks the carbonyl carbon of \(\displaystyle R_2C=O\). As the new \(\displaystyle C-CN\) bond forms, the \(\displaystyle C=O\) \(\displaystyle \pi\) bond breaks and its electron pair moves fully onto oxygen, giving an alkoxide, \(\displaystyle R_2C(O^-)(CN)\), which is then protonated to the cyanohydrin \(\displaystyle R_2C(OH)(CN)\). This step converts the carbonyl carbon from planar \(\displaystyle sp^2\) (three groups around it) to tetrahedral \(\displaystyle sp^3\) (four groups around it), and the incoming \(\displaystyle CN^-\) has to approach along a fairly narrow trajectory close to the plane of the ring.In cyclohexanone the carbons flanking the carbonyl, C-$\displaystyle 2$ and C-$\displaystyle 6$, carry only hydrogens. Nothing blocks the approach of \(\displaystyle CN^-\) to the carbonyl carbon, and the tetrahedral product, \(\displaystyle 1\text{-hydroxycyclohexane-1-carbonitrile}\) (cyclohexanone cyanohydrin), is no more crowded than the ring itself, so the equilibrium lies well over to the addition product and the cyanohydrin forms in good yield.In $\displaystyle 2,2,6$-trimethylcyclohexanone the two carbons flanking the carbonyl carbon are themselves substituted: C-$\displaystyle 2$ carries two methyl groups (gem-dimethyl) and C-$\displaystyle 6$ carries one. These three methyl groups sit right next to the carbonyl carbon on both sides, so they physically block the path the cyanide ion must take to reach that carbon, and they also crowd the tetrahedral product that would result, \(\displaystyle 1\text{-hydroxy-2,2,6-trimethylcyclohexane-1-carbonitrile}\), far more than they crowd the planar starting ketone. Both effects -- a hindered approach and a strained product -- push the addition equilibrium back toward the ketone and \(\displaystyle HCN\), so only a small amount of cyanohydrin is obtained.Only the nitrogen whose lone pair is not being pulled into the carbonyl group is free to act as a nucleophile.Semicarbazide is \(\displaystyle H_2N-NH-CO-NH_2\): a carbonyl group, \(\displaystyle C=O\), with an \(\displaystyle -NH_2\) attached directly to it on one side (an amide-type nitrogen) and an \(\displaystyle -NH-NH_2\) chain attached to it on the other side (a hydrazine-type nitrogen bearing the second \(\displaystyle -NH_2\)).The \(\displaystyle -NH_2\) that sits directly on the carbonyl carbon behaves like the nitrogen of an amide: its lone pair delocalises into the adjacent \(\displaystyle C=O\) group, so semicarbazide has a resonance contributor \(\displaystyle H_2N-NH-C(-O^-)=NH_2^+\) alongside the normal structure. Because that nitrogen's lone pair is partly tied up in this way, it is a poor nucleophile and does not attack an aldehyde or ketone carbonyl carbon.The other \(\displaystyle -NH_2\), attached to the \(\displaystyle -NH-\) nitrogen and one bond further from the carbonyl group, is not part of this delocalised system. Its lone pair stays fully on nitrogen, so it is the nucleophile: it attacks the carbonyl carbon of an aldehyde or ketone, \(\displaystyle R_2C=O\); the \(\displaystyle C=O\) \(\displaystyle \pi\) electrons move onto oxygen to give a tetrahedral carbinolamine intermediate, \(\displaystyle R_2C(OH)(NH-NH-CO-NH_2)\); this intermediate then loses a molecule of water (the oxygen leaves as \(\displaystyle H_2O\) while a new \(\displaystyle C=N\) bond forms) to give the semicarbazone, \(\displaystyle R_2C=N-NH-CO-NH_2\). Only the terminal, unconjugated \(\displaystyle -NH_2\) takes part in this sequence.Fischer esterification is an equilibrium, so the product has to be removed from the mixture to keep the reaction moving forward instead of settling at a partial conversion.Esterification of a carboxylic acid with an alcohol under acid catalysis is a reversible reaction: \[RCOOH + R'OH \underset{H^+}{\rightleftharpoons} RCOOR' + H_2O \] The acid catalyst works by protonating the carbonyl oxygen of \(\displaystyle RCOOH\), which makes the carbonyl carbon more electrophilic; the alcohol oxygen of \(\displaystyle R'OH\) then attacks that carbon, and after a couple of proton transfers a molecule of water is lost to give the ester, \(\displaystyle RCOOR'\). Every one of these steps is reversible in the presence of \(\displaystyle H^+\) and water, so the same acid catalyst that makes the ester also hydrolyses it back to the acid and alcohol.Because an equilibrium constant fixes the ratio of products to reactants at equilibrium, simply mixing the acid and alcohol only ever reaches a partial conversion -- some \(\displaystyle RCOOH\) and \(\displaystyle R'OH\) remain unreacted, sitting alongside \(\displaystyle RCOOR'\) and \(\displaystyle H_2O\). Removing one of the products, water (for example using a drying agent or a Dean–Stark-type setup) or the ester itself (by distilling it off as it forms, for esters low-boiling enough to do this), keeps the product-side concentration low. By Le Chatelier's principle, the system responds to this constant removal by continuing to convert more \(\displaystyle RCOOH\) and \(\displaystyle R'OH\) into \(\displaystyle RCOOR'\) and \(\displaystyle H_2O\), since the reverse (hydrolysis) reaction cannot proceed once its own product has been taken away. This is why the ester or the water is removed as soon as it forms -- it pushes the reversible esterification toward completion instead of letting it stall at equilibrium.Answer: (i) Cyclohexanone's carbonyl carbon has only hydrogens on the flanking C-$\displaystyle 2$/C-$\displaystyle 6$ carbons, so \(\displaystyle CN^-\) adds easily to give $\displaystyle 1$-hydroxycyclohexane-$\displaystyle 1$-carbonitrile in good yield; in $\displaystyle 2,2,6$-trimethylcyclohexanone the three methyl groups on C-$\displaystyle 2$ and C-$\displaystyle 6$ block the approach of \(\displaystyle CN^-\) and overcrowd the tetrahedral product, so cyanohydrin formation is poor. (ii) In semicarbazide, \(\displaystyle H_2N-NH-CO-NH_2\), the \(\displaystyle -NH_2\) bonded directly to the carbonyl carbon has its lone pair delocalised into the \(\displaystyle C=O\) group (amide-type resonance) and so is a poor nucleophile; the other, terminal \(\displaystyle -NH_2\) is unconjugated and attacks the aldehyde/ketone carbonyl to form the semicarbazone, \(\displaystyle R_2C=N-NH-CO-NH_2\), after loss of water. (iii) Esterification, \(\displaystyle RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O\), is a reversible, acid-catalysed equilibrium; removing the water or the ester as it forms lowers the product concentration and, by Le Chatelier's principle, drives the equilibrium toward more ester instead of letting hydrolysis pull it back.

A negative Tollens' test rules out an aldehyde, and a positive iodoform test says there is a \(\displaystyle \text{CH}_3\text{-CO-} \) group in the molecule — so this is a methyl ketone, and the vigorous-oxidation clue is what pins down exactly which one.Step $\displaystyle 1$ — get the molecular formula from the percentage composition.Oxygen is "the rest": \(\displaystyle 100 - 69.77 - 11.63 = 18.60\% \).Convert each percentage to a mole ratio by dividing by the atomic mass:\[\text{C}: \frac{69.77}{12.00} = 5.814, \qquad \text{H}: \frac{11.63}{1.00} = 11.63, \qquad \text{O}: \frac{18.60}{16.00} = 1.1625 \]Divide every value by the smallest one (\(\displaystyle 1.1625\)) to get the simplest whole-number ratio:\[\text{C}: \frac{5.814}{1.1625} \approx 5.00, \qquad \text{H}: \frac{11.63}{1.1625} \approx 10.00, \qquad \text{O}: \frac{1.1625}{1.1625} = 1.00 \]Empirical formula: \(\displaystyle \text{C}_5\text{H}_{10}\text{O} \). Its formula mass is \(\displaystyle 5(12) + 10(1) + 16 = 86\ \text{g mol}^{-1} \), which already equals the given molecular mass of \(\displaystyle 86\). So \(\displaystyle n = \dfrac{86}{86} = 1\) — the molecular formula is exactly \(\displaystyle \text{C}_5\text{H}_{10}\text{O} \), no multiplication needed. (The step people skip: always check whether the empirical mass already matches the molecular mass before multiplying by some assumed \(\displaystyle n\).)Step $\displaystyle 2$ — count the degree of unsaturation.\[\text{DoU} = \frac{2(5) + 2 - 10}{2} = \frac{2}{2} = 1 \]One degree of unsaturation with one oxygen and no reaction pointing to a ring means one \(\displaystyle \text{C=O} \) — a saturated, open-chain carbonyl compound, \(\displaystyle \text{C}_5\text{H}_{10}\text{O} \).Step $\displaystyle 3$ — read the tests. Putting these together, the compound is a methyl ketone \(\displaystyle \text{CH}_3\text{-CO-C}_3\text{H}_7 \). There are three ways to arrange five carbons as a ketone: Pentan-$\displaystyle 3$-one is eliminated immediately: it cannot give the iodoform test. The other two both survive so far, so the oxidation data is what decides between them.Step $\displaystyle 4$ — use the oxidation products to choose the structure.Vigorous oxidation (hot, strong \(\displaystyle \text{KMnO}_4 \)) cleaves a ketone at the carbon–carbon bond next to the carbonyl. The sequence is: the carbonyl carbon keeps whichever group sits on one side of it and becomes the \(\displaystyle -\text{COOH}\) of one acid, while the carbon on the other side of the broken bond is itself oxidised into a brand-new \(\displaystyle -\text{COOH}\), carrying the rest of that chain with it.For pentan-$\displaystyle 2$-one, \(\displaystyle \text{CH}_3\text{-CO-CH}_2\text{-CH}_2\text{-CH}_3 \), the bond that breaks is between the carbonyl carbon and the first \(\displaystyle \text{CH}_2 \) of the propyl chain: That reproduces exactly the ethanoic acid and propanoic acid named in the question, and the carbon count checks out: \(\displaystyle 2 + 3 = 5\), the same as the starting ketone.For $\displaystyle 3$-methylbutan-$\displaystyle 2$-one, \(\displaystyle \text{CH}_3\text{-CO-CH(CH}_3)_2 \), the carbon on the far side of the carbonyl already carries two methyl branches and one H. A carboxylic-acid carbon can only bond to a single carbon substituent (as in \(\displaystyle \text{R-COOH} \)); this carbon would need to hold onto two methyl groups at once, so it cannot be oxidised cleanly into a simple \(\displaystyle -\text{COOH}\). This structure therefore cannot give a clean pair of straight-chain acids the way pentan-$\displaystyle 2$-one does, so it is rejected even though it also passes the iodoform test.So the compound that satisfies every clue — molecular formula \(\displaystyle \text{C}_5\text{H}_{10}\text{O} \), no Tollens' reduction, bisulphite addition, positive iodoform, and oxidative cleavage to ethanoic acid plus propanoic acid — is pentan-$\displaystyle 2$-one, also called methyl n-propyl ketone:\[\text{CH}_3\text{-CO-CH}_2\text{-CH}_2\text{-CH}_3 \]Answer: Pentan-$\displaystyle 2$-one (methyl n-propyl ketone), CH3-CO-CH2-CH2-CH3 — molecular formula \(\displaystyle \mathrm{C_{5}H_{10}O}\), formed by vigorous oxidative cleavage into ethanoic acid (CH3-COOH) and propanoic acid (CH3-CH2-COOH).

The number of resonance structures is not what decides stability — the energy and equivalence of those structures is, and by that measure the carboxylate ion beats the phenoxide ion even though it has fewer contributors.Acid strength tracks the stability of the conjugate base: the more a negative charge is spread out (delocalized) without being pushed onto an unwilling atom, the more stable the anion, the more the equilibrium\[\text{R--COOH} \rightleftharpoons \text{R--COO}^- + \text{H}^+ \]or\[\text{C}_6\text{H}_5\text{--OH} \rightleftharpoons \text{C}_6\text{H}_5\text{--O}^- + \text{H}^+ \]sits to the right, and the stronger the acid.Resonance in the carboxylate ion. Deprotonating a carboxylic acid gives the carboxylate ion, R–COO⁻, which is described by exactly two resonance structures:Structure $\displaystyle 1$: \(\displaystyle \text{R--C(=O)--O}^- \) (the negative charge and the lone C–O single bond sit on one oxygen, the C=O double bond on the other)Structure $\displaystyle 2$: \(\displaystyle \text{R--C(--O}^-\text{)=O} \) (the roles of the two oxygens are swapped)These two structures are strictly equivalent — same skeleton, same bond pattern, identical energy, differing only in which of the two chemically identical oxygen atoms is drawn holding the negative charge. Because they are equal in energy, the true structure (the resonance hybrid) is an exact $\displaystyle 50$:$\displaystyle 50$ blend: both C–O bonds become equal in length (bond order $\displaystyle 1.5$, intermediate between single and double), and each oxygen atom carries half a unit of negative charge, i.e. \(\displaystyle -\tfrac{1}{2}\) each. Two low-energy, equal-energy structures that place the charge on the most electronegative atom available (oxygen) give very effective delocalization — resonance stabilization is greatest precisely when the contributing structures are close in energy to each other.Resonance in the phenoxide ion. Deprotonating phenol gives \(\displaystyle \text{C}_6\text{H}_5\text{--O}^- \). Sending the oxygen lone pair into the ring generates more canonical structures — one with the charge purely on oxygen, and others (via the ortho and para positions) in which the negative charge is pushed onto ring carbons, with the ring itself thrown into a cross-conjugated (quinonoid) arrangement of double bonds. Counting these gives five resonance structures in total, so on a naive head count phenoxide looks "more delocalized" than carboxylate.But look at what those extra structures actually are: So phenoxide's extra resonance structures are numerous but weak: unequal in energy, charge-on-carbon, and partially de-aromatized. They add some stabilization, but far less per structure than carboxylate's two structures, which are equal-energy and keep the charge on oxygen throughout.Net comparison. Carboxylate is stabilized by two strong, energetically identical contributors; phenoxide is stabilized by one strong contributor (charge on O) plus several weak, high-energy ones (charge on ring carbon, ring de-aromatized). The carboxylate ion therefore ends up more stable — lower in energy relative to its acid — than the phenoxide ion is relative to phenol. A more stable conjugate base means the parent acid gives up its proton more readily, so the carboxylic acid is the stronger acid.This is exactly what the measured acidities show: for a typical carboxylic acid such as acetic acid , \(\displaystyle K_a \approx 1.8\times10^{-5} \) (\(\displaystyle \text{p}K_a \approx 4.76\)), while for phenol, \(\displaystyle K_a \approx 1.0\times10^{-10} \) (\(\displaystyle \text{p}K_a \approx 10\)) — the carboxylic acid is roughly a hundred-thousand times more dissociated than phenol under the same conditions, consistent with its far better-stabilized conjugate base.Answer: Carboxylate resonance uses only two structures, but they are exactly equivalent and keep the negative charge on the highly electronegative oxygen throughout, giving strong, symmetric delocalization. Phenoxide has more structures, but most of them force the negative charge onto ring carbons (less electronegative, energetically unfavourable) and partially destroy the ring's aromaticity, so they contribute weakly. Because the carboxylate ion is therefore the more stable, more effectively delocalized conjugate base, carboxylic acid loses its proton more readily than phenol does — carboxylic acid is the stronger acid.