Alcohols, Phenols, Ethers, Carbonyl Compounds and Carboxylic Acids
Organic
Weightage: 4-6 Questions (16-24 Marks)
JEE Unit 17
“Welcome back! You already know the NCERT chapters on alcohols, phenols, ethers, aldehydes, ketones and carboxylic acids, so this guide skips the textbook story and keeps what JEE Main asks. This is the biggest organic unit, and it rewards one habit above all: name the reactive site before you name the reagent. Every question is really asking which of four things is happening: an O–H being deprotonated (acidity), a C–O bond being broken (substitution, cleavage), a C=O being attacked (addition, condensation, reduction), or an α-hydrogen being removed (aldol, haloform). Decide which one applies and the product follows.”
— SCORECHEM ACADEMIC TEAM
1. Alcohols: Acidity, Tests, Dehydration and Oxidation
- Acidity order of the O–H family: RCOOH > H2CO3 > phenol > H2O > ROH, and within the alcohols CH3OH > 1° > 2° > 3°. Alkyl groups push electrons (+I) onto oxygen and also block solvation of the alkoxide, so 3° alcohols are the weakest acids. JEE keys treat water as more acidic than every alcohol (methanol is in fact nearly equal to water).
- What dissolves in what: carboxylic acids dissolve in NaHCO3 (CO2 fizz) and NaOH; ordinary phenols dissolve in NaOH only; alcohols dissolve in neither but react with sodium metal (H2 gas). This three-step ladder is the standard way a question separates an acid, a phenol and an alcohol in one mixture.
- Lucas test (anhydrous ZnCl2 + conc. HCl) is an SN1 test: turbidity is the insoluble alkyl chloride separating out. Allylic and benzylic 1° alcohols give cloudiness at once, like 3°, because their carbocations are resonance-stabilised. Methanol and lower 1° alcohols need heating and can fail it altogether.
- Oxidation: PCC (or Cu at 573 K) stops a 1° alcohol at the aldehyde; K2Cr2O7/H+, KMnO4 or Jones reagent runs it on to the acid. A 2° alcohol gives a ketone with any of these; a 3° alcohol resists, which is why oxidation separates 3° from 1°/2° in a mixture.
- Ether vs alkene: ethanol with conc. H2SO4 gives diethyl ether at 413 K (excess alcohol, SN2) and ethene at 443 K (excess acid, elimination). Because a carbocation is involved above that temperature, rearrangement is possible: 3,3-dimethylbutan-2-ol gives 2,3-dimethylbut-2-ene by a 1,2-methyl shift, not the unrearranged alkene.
- Diols and triols: ethylene glycol is cleaved by HIO4 or Pb(OAc)4 between the two carbinol carbons to give 2 HCHO (this cleavage works on any vicinal diol); glycerol dehydrates with KHSO4 to pungent acrolein (CH2=CH–CHO) and nitrates to nitroglycerine (dynamite when absorbed on kieselguhr).
⚠️ JEE Trap: "methanol and ethanol are both oxidised to carboxylic acids, so both pass the iodoform test" is false. Only ethanol (CH3CH(OH)–) gives iodoform. Methanol oxidises to HCHO then HCOOH, neither of which has the CH3CO– unit, so the pair is a textbook distinguishing question.
2. Phenols: Acidity and Ring Reactions
- Why phenol beats an alcohol by ~6 pKa units: the phenoxide ion delocalises its charge into the ring (resonance), while an alkoxide is localised on oxygen and destabilised by +I. Substituents then act exactly as in benzoic acid: −M/−I groups at ortho/para (NO2) strengthen it enormously, +I/+M groups (CH3, OCH3) weaken it slightly, and meta-NO2 helps only through −I (a smaller effect). Order: p-nitrophenol > o-nitrophenol > m-nitrophenol > phenol > p-cresol; picric acid sits far above all of them.
- Reactivity of the ring: −OH is a strong o/p director and activator, so phenol brominates in water without a catalyst to the tribromo compound. Friedel–Crafts is poor on phenol because AlCl3 coordinates to oxygen.
- Preparation of phenol: cumene hydroperoxide rearrangement (phenol + acetone, the industrial route); chlorobenzene with NaOH at 623 K and 300 atm (Dow); diazonium salt warmed with water; sodium benzenesulphonate fused with NaOH.
- Identification: neutral FeCl3 gives a violet colour; Br2/water gives a white precipitate; NaHCO3 gives no fizz (except picric acid).
⚠️ JEE Trap: o-nitrophenol is steam-volatile, p-nitrophenol is not — and the reason is H-bonding, not acidity. o-Nitrophenol forms an intramolecular H-bond that keeps molecules separate (lower b.p., volatile); p-nitrophenol forms intermolecular H-bonds (higher b.p., associated). Questions pair this with a b.p. or solubility ranking.
3. Ethers: Williamson Design and HI Cleavage
- Williamson = SN2. Choose the pair so the halide is methyl or 1° and the alkoxide carries any bulky group. A 3° halide with an alkoxide gives an alkene (E2); an aryl or vinyl halide gives nothing. Intramolecular Williamson on a halohydrin (X–CH2CH2–OH + base) closes an epoxide.
- HI cleavage decides by mechanism: primary/secondary alkyl groups go by SN2, so iodide attacks the smaller alkyl; a tertiary group goes by SN1, so iodide ends up on the tertiary carbon. An alkyl aryl ether always gives phenol + alkyl iodide. With excess HI both alkyl groups end up as RI.
- Physical and hazard facts: ethers have b.p. close to alkanes of equal mass (no H-bond donor) but dissolve in cold conc. H2SO4 (oxonium salt formation), which separates them from alkanes. On standing in air they form explosive hydroperoxides at the α-carbon; FeSO4/KSCN turns blood-red (Fe3+) to detect them, and shaking with FeSO4 or storing over Cu2O/Na removes them.
4. Making Aldehydes and Ketones
- Shared routes: oxidation or dehydrogenation of 1°/2° alcohols; ozonolysis with Zn/H2O (the Zn stops over-oxidation by H2O2); hydration of alkynes with Hg2+/H2SO4 (ethyne gives CH3CHO, every other alkyne a ketone); alkaline hydrolysis of gem-dihalides (terminal gives aldehyde, non-terminal gives ketone).
- Ketone-specific routes: Friedel–Crafts acylation (ArH + RCOCl/AlCl3); R2Cd + RCOCl (stops at the ketone, unlike RMgX); nitrile + RMgX then H3O+; dry distillation of calcium salts; decarboxylation of β-keto acids; Wacker oxidation of higher 1-alkenes.
- Counting from ozonolysis: one mole of a compound giving two carbonyls in one molecule means a ring or diene was present; the number of C=O groups produced tells you how many C=C were cleaved.
5. Nucleophilic Addition to C=O
- Mechanism: the nucleophile attacks the planar sp2 carbon perpendicular to the plane (slow step) to give a tetrahedral alkoxide, then protonation. Reactivity falls as steric crowding and electron donation rise, and is raised by −I groups (chloral hydrate, ninhydrin, glyoxal hydrates are stable gem-diols).
- HCN gives a cyanohydrin (base-catalysed, because CN⊖ is the real nucleophile); hydrolysis gives an α-hydroxy acid. NaHSO3 gives a crystalline adduct with aldehydes, methyl ketones and cyclic ketones, but not bulky ketones (acetophenone, benzophenone), and the carbonyl is regenerated with acid or alkali, so it is used for purification.
- Alcohols: aldehyde + 2 ROH/HCl(g) gives an acetal; ketone + ethylene glycol gives a cyclic ketal. Acetals are stable to base, Grignard reagents and LiAlH4, but hydrolysed by dilute aqueous acid, which makes them protecting groups (protect the aldehyde, reduce the ester elsewhere, deprotect).
⚠️ JEE Trap: semicarbazide has two –NH2 groups but only one reacts. The terminal –NH2 (attached to C=O) is delocalised by amide resonance and is not nucleophilic, so the –NH2 on the other nitrogen forms the semicarbazone. Equally, the reaction is fastest at pH ~3–5, not in strong acid or strong base.
6. α-Hydrogen Chemistry: Aldol, Cannizzaro, Haloform
- α-H acidity (pKa ~19–20) comes from the −I effect of C=O and resonance in the enolate. It drives aldol, haloform, HVZ and every condensation below.
- Aldol: enolate formation, attack on another carbonyl (rate-determining), protonation, then dehydration by E1cB on heating. Intramolecular aldol of 1,4- and 1,5-diketones closes 5- and 6-membered rings. Count the products: two different carbonyls both with α-H give four products; one without α-H gives two; neither gives none.
- Cannizzaro: no α-H, conc. alkali, hydride transfer is rate-determining. HCHO, ArCHO and R3C–CHO qualify; glyoxal gives glycolate by the intramolecular version. Remember that an aldehyde WITH α-H under the same conditions does aldol instead.
- Haloform (X2 + NaOH, i.e. NaOX): substrate must contain CH3CO– or CH3CH(OH)– (the latter is first oxidised by NaOX). Products: CHX3 + carboxylate with one carbon fewer. Iodoform is a yellow solid (m.p. 119 °C).
- Related condensations: Claisen–Schmidt (ArCHO + methyl ketone, base), Perkin (ArCHO + anhydride + sodium salt of the acid, cinnamic acid), Knoevenagel (active methylene + carbonyl, pyridine/piperidine), benzoin (2 PhCHO, CN⊖ catalyst).
⚠️ JEE Trap: "an aldehyde with α-H in conc. NaOH gives Cannizzaro" is false. Enolate formation is far faster than hydride transfer, so aldol wins. Cannizzaro is reserved for aldehydes with NO α-H, whatever the base strength, and a question will usually hide this by naming HCHO, benzaldehyde or pivaldehyde.
7. Reduction, Oxidation and Distinguishing Tests
- Pick Clemmensen or Wolff–Kishner by the other groups present: acid-sensitive (acetal, tertiary OH) means Wolff–Kishner; base-sensitive (ester, alkyl halide) means Clemmensen.
- Mild oxidants: Tollens' reagent (ammoniacal AgNO3, silver mirror) oxidises ALL aldehydes, including benzaldehyde, plus α-hydroxy ketones (fructose) and formic acid. Fehling's and Benedict's (Cu2+ to red Cu2O) oxidise aliphatic aldehydes only, not benzaldehyde. Schiff's test restores magenta colour with aldehydes, not ketones.
- Baeyer's test (cold dilute KMnO4, purple to brown) detects unsaturation (C=C, C≡C) and some oxidisable groups; carbonyl compounds with no C=C do not respond, which separates them from unsaturated alcohols.
- Formaldehyde and acetaldehyde polymers: HCHO gives paraformaldehyde (linear) and trioxane (cyclic trimer), and urotropine with NH3 (urinary antiseptic); CH3CHO gives paraldehyde (trimer, conc. H2SO4, sedative) and metaldehyde (tetramer, dry HCl at 0 °C). Acetone with conc. H2SO4 gives mesitylene (1,3,5-trimethylbenzene).
8. Named Rearrangements and the Wittig Reaction
- Pinacol: protonate one OH, lose water to the more stable cation, then a group from the neighbouring carbon shifts and a carbonyl forms. The exact aryl-versus-hydrogen ranking varies with the substrate, so JEE questions are built where the cation or the aryl group clearly decides.
- Beckmann stereospecificity: the group anti to the OH migrates, so the E or Z oxime geometry decides which amide forms. Acetophenone oxime (phenyl anti to OH) gives acetanilide, which hydrolyses to aniline and acetic acid.
- Wittig: Ph3P=CR2 + C=O gives C=C plus Ph3P=O through an oxaphosphetane; the strong P=O bond is the driving force, and the double bond forms exactly where the carbonyl was.
9. Grignard Reagent as a Synthetic Tool
- Preparation rules: R–X + Mg in dry ether (the ether lone pairs solvate Mg); reactivity of halide I > Br > Cl >> F. The substrate must not contain acidic H (–OH, –COOH, –NH2, terminal alkyne) or a group the reagent would attack (–CHO, –CN, –NO2).
- Reaction priority in a polyfunctional molecule: acidic H > aldehyde > ketone > acid chloride/ester > alkyl halide. A Grignard reagent plus terminal alkyne gives the alkyne's Grignard and an alkane (CH4 from CH3MgBr): this is how methane volumes get measured in numericals.
⚠️ JEE Trap: working back from a target alcohol has several valid answers, so a question asks for "a pair", not "the pair". To make (CH3)2CHCH2CH(OH)CH(CH3)2 you can combine isobutyraldehyde with isobutylMgBr or 3-methylbutanal with isopropylMgBr. Check each option's carbon skeleton rather than hunting for one memorised pair.
10. Carboxylic Acids: Acidity and Reactions
- Dimer and physical trends: acids exist as cyclic hydrogen-bonded dimers even in the vapour, giving b.p. above alcohols of similar mass; even-carbon acids melt higher than their odd neighbours (better packing).
- Dicarboxylic acids: oxalic pKa1 1.27 is far stronger than a monoacid because the second COOH is adjacent and electron-withdrawing; maleic (cis) is a stronger first acid than fumaric (trans) because the mono-anion is stabilised by an intramolecular H-bond, but its second dissociation is weaker for the same reason.
- Reactions of the acid: Fischer esterification (acid-catalysed, reversible; use excess alcohol or remove water; the oxygen of the alcohol ends up in the ester, so acyl–oxygen bond breaks); HVZ (X2/red P then H2O, needs an α-H, so HCOOH and benzoic acid cannot); LiAlH4 or B2H6 reduce COOH to CH2OH (NaBH4 cannot, and B2H6 leaves esters and NO2 untouched); Arndt–Eistert adds one CH2 (SOCl2, CH2N2, Ag2O/H2O).
- Decarboxylations: soda-lime on RCOONa gives R–H (one carbon fewer); Kolbe electrolysis of RCOO⊖ gives R–R at the anode (2(n−1) carbons); Hunsdiecker (RCOOAg + Br2, CCl4) gives R–Br with one carbon fewer by free-radical chain; β-keto acids and malonic acids lose CO2 through a six-membered cyclic transition state at mild temperature.
- Ring substitution: −COOH is a meta director and deactivator, so nitration or chlorination of benzoic acid gives the 3-substituted product.
⚠️ JEE Trap: tert-butylbenzene is NOT oxidised by KMnO4, even though toluene and ethylbenzene are. Side-chain oxidation of an alkylbenzene needs at least one benzylic hydrogen. Toluene and ethylbenzene both give benzoic acid (the whole chain is cut back to COOH), but tert-butylbenzene has none and is untouched.
11. Acid Derivatives and Heating Effects
- Why the order: a weaker conjugate base leaves faster. Acyl chlorides also lack effective Cl→C=O resonance donation (poor 3p–2p overlap), while amides have strong N→C=O donation, so they are least electrophilic. Water-sensitive acyl chlorides and anhydrides cannot be handled in protic solvents.
- Acyl chloride toolbox: + R2Cd gives a ketone (stops there); + RMgX (excess) gives a 3° alcohol; + LiAlH4 gives RCH2OH; + H2/Pd–BaSO4 gives RCHO (Rosenmund); + benzene/AlCl3 gives a ketone.
- Esters: acid hydrolysis is reversible, saponification is irreversible (carboxylate is deprotonated and unreactive); LiAlH4 or Na/EtOH (Bouveault–Blanc) gives two alcohols; DIBAL-H at −78 °C stops at the aldehyde; pyrolysis of an ester with a β-H on the alkyl side gives alkene + acid by syn elimination through a six-membered transition state.
- Amides: dehydration (P2O5) gives a nitrile; LiAlH4 reduces to an amine with the SAME carbon count; Hofmann bromamide degradation (Br2/KOH) gives an amine with ONE carbon fewer (details in Unit 18).
⚠️ JEE Trap: LiAlH4 on an amide keeps the carbon count; Hofmann degradation loses one. RCONH2 → RCH2NH2 (LiAlH4) and RCONH2 → RNH2 (Br2/KOH) are placed side by side in options because students merge the two.
12. How JEE Frames These Questions
- Match-the-list on reagent → name: Rosenmund (H2, Pd–BaSO4), Stephen (SnCl2, HCl), Etard (CrO2Cl2, CS2), Gattermann–Koch (CO, HCl, AlCl3), Wolff–Kishner (N2H4, KOH), Clemmensen (Zn–Hg, HCl), Tollens, Fehling. Learn each reagent→name pair as a flashcard, in both directions.
- Which pairs does a test distinguish? For each pair, ask whether exactly one member has the required group (CH3CO–, free CHO, aliphatic CHO, acidic COOH). Pairs like ethanol/methanol and anisole/acetone work for the iodoform test; acetic/propanoic acid and diethyl ether/pentan-3-one do not.
- Statement-I / Statement-II on aldol: the "always gives four products" statement is false whenever one partner lacks α-H; read for "always", "only", "never".
- Hidden-aldehyde identification: a compound that is iodoform-positive but Tollens-negative, and after dilute acid is positive in both, is an acetal of a keto-aldehyde, e.g. CH3COCH2CH(OCH3)2.
- Rank the acids from reaction sequences: identify each product first (diazonium → CuCN → hydrolysis gives benzoic acid; oxidation of an aldehyde gives its own acid; RMgX + CO2 gives RCOOH one carbon longer), then rank by pKa: HCOOH > PhCOOH > PhCH2COOH > CH3CH2COOH.
- Isomer-set statements (same formula, different functional groups): test each statement against EVERY isomer separately — 2,4-DNP positive for all carbonyls, Baeyer's for only those with C=C, Tollens for only the aldehyde, iodoform for only the methyl ketone. All aromatic isomers give a sooty flame, so a statement naming several of them can still be true.
⚠️ JEE Trap: sequences that end in an acid ranking hide a one-carbon change in the middle. Grignard + CO2 adds a carbon; Tollens oxidation keeps the same carbon count (propanal gives propanoic acid, not acetic acid); diazonium → CN → COOH adds one carbon to the ring. Count carbons at every arrow before assigning structures.
13. Quick Sheet and Checklist
| Concept | Key Fact |
|---|---|
| Acid strength | RCOOH > H2CO3 > phenol > H2O > ROH (CH3OH > 1° > 2° > 3°) |
| Lucas | 3° instant, 2° ~5 min, 1° none at RT; allylic/benzylic fast |
| Ethanol + H2SO4 | 413 K ether (SN2); 443 K ethene |
| Williamson | Halide methyl/1°; bulky group on alkoxide; never Ar–X |
| Ether + HI | SN2 at smaller alkyl; SN1 at 3°; Ar–O–R gives ArOH + RI |
| Kolbe / Reimer–Tiemann | PhONa + CO2 gives salicylic acid; PhOH + CHCl3/NaOH gives salicylaldehyde (:CCl2) |
| Aldehyde routes | Rosenmund, Stephen, DIBAL-H, Etard, Gattermann–Koch; Wacker gives CH3CHO only from ethene |
| Reactivity to Nu | HCHO > RCHO > ArCHO > R2CO > RCOAr > Ar2CO |
| H2N–Z reactions | Optimum pH 3–5; semicarbazone uses only one NH2 |
| Aldol / Cannizzaro | α-H + dilute base = aldol; no α-H + conc. base = Cannizzaro; HCHO oxidised in crossed |
| Haloform | CH3CO– or CH3CH(OH)–; product has one carbon fewer |
| Tollens / Fehling | Tollens: all CHO; Fehling: aliphatic CHO only |
| Reductions | NaBH4: CHO/ketone only; LiAlH4: everything polar; B2H6: COOH selective |
| Rearrangements | Beckmann anti-migration; B–V: H > 3° > 2° > Ar > 1° > CH3 |
| Grignard | HCHO → 1°; RCHO → 2°; ketone/ester → 3°; CO2 → RCOOH; acidic H destroys it |
| Acid derivatives | RCOCl > (RCO)2O > ester > amide |
| Heating diacids | Oxalic: CO2/CO; malonic: CH3COOH; C4/C5: anhydride; C6/C7: cyclic ketone |
Before the exam, check you can:
- Rank phenol, picric acid, acetic acid, carbonic acid, water and ethanol, and say which dissolve in NaHCO3 and NaOH.
- Predict ether versus alkene from ethanol and conc. H2SO4 by temperature, and spot when a carbocation rearrangement intervenes.
- Design a Williamson synthesis for any unsymmetrical ether, and predict which bond HI cleaves in a given ether.
- Write the Kolbe and Reimer–Tiemann products and name the electrophile in each.
- Match every named aldehyde/ketone preparation to its reagent and starting material.
- Give BOTH reasons (electronic and steric) for aldehydes reacting faster than ketones.
- Decide aldol, Cannizzaro or crossed variants from the α-H count alone, and count the products.
- State exactly which compounds pass iodoform, Tollens, Fehling, Schiff and 2,4-DNP.
- Choose Clemmensen or Wolff–Kishner, NaBH4 or LiAlH4 from the other groups present.
- Say which group migrates in Pinacol, Beckmann and Baeyer–Villiger.
- Predict Grignard products from HCHO, RCHO, ketones, esters, nitriles, CO2 and epoxides, and explain why acidic H wrecks it.
- Rank substituted benzoic and chloroacetic acids by pKa, including the meta/para methoxy and ortho effects.
- Predict the product of heating each dicarboxylic and hydroxy acid.