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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 Ladder: Who Can Deprotonate Whom?Picric acid (2,4,6-trinitrophenol)pKa 0.4RCOOH (CH3COOH 4.8, HCOOH 3.8)pKa 4-5H2CO3 (CO2 + H2O)pKa 6.4p-NitrophenolpKa 7.2Phenol C6H5OHpKa 10.0H2OpKa 15.7ROH: CH3OH > 1° > 2° > 3°pKa 16-18HC≡CHpKa 25NH3pKa 35Alkanes R-HpKa ~50Dissolve in NaHCO3, CO2 fizzStronger than carbonic acidDissolve in NaOH, NOT in NaHCO3Phenols are weaker acids than H2CO3Too weak for aq. NaOHROH needs Na metal (H2 given off) or NaHNeed NaNH2 / RMgX / R-LiStrong base neededRule: a base deprotonates an acid only if the base's conjugate acid is weaker. Arrow = increasing acidity.
1° vs 2° vs 3° Alcohols: One Table, Six Exam HandlesProperty / test1° (RCH2OH)2° (R2CHOH)3° (R3COH)Lucas test (anhyd. ZnCl2 +conc. HCl, RT)No turbidity at RT (needsheat)Turbid in ~5 minTurbid immediatelyCu, 573 K (dehydrogenation)AldehydeKetoneAlkene (dehydrates)K2Cr2O7 / H+ (or KMnO4)Aldehyde, then carboxylicacidKetoneResists (only forced C-Ccleavage)Ease of acid dehydrationHardest (~443 K, conc. H2SO4)Medium (~440 K, 85% H3PO4)Easiest (~358 K, 20% H3PO4)Acidity in solutionMost acidic (CH3OH strongest)IntermediateLeast acidic (+I + poorsolvation)Rate with HX / via SN1SlowestMediumFastest (stable 3° cation)
Alcohol + Conc. H2SO4: Temperature Decides Ether vs Alkene2 CH3CH2OH + conc. H2SO4413 K, excess ALCOHOL443 K, excess ACIDCH3CH2-O-CH2CH3 (diethyl ether). SN2: a 2ndalcohol attacks the protonated one. Works for 1°only.CH2=CH2 (ethene). E1 / E2 loss of H2O. High Tfavours elimination (entropy).Ease of dehydration: 3° (358 K) > 2° (440 K) > 1° (443 K). Rate tracks carbocation stability.3° and most 2° alcohols give ALKENES, not ethers (hindered SN2, easy E1).Carbocation intermediate means rearrangement: (CH3)3C-CH(OH)CH3 → 2,3-dimethylbut-2-ene (1,2-CH3 shift,then Zaitsev), not 3,3-dimethylbut-1-ene.
⚠️ 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

Phenol: Reagent → Product Cheat-TableReagent / conditionsMajor productJEE catchBr2 in water (no catalyst)2,4,6-Tribromophenol (white ppt)All o/p sites brominated; ArOH is so activatedno Lewis acid is neededBr2 in CS2 or CHCl3, ~273 Ko- + p-Bromophenol (p major)Non-polar solvent suppresses phenoxideformation, so reaction stops at monoDil. HNO3, 298 Ko- + p-Nitrophenolo-isomer is steam volatile (intramolecularH-bond); p is not: separation by steamdistillationConc. HNO3 (best via sulphonationfirst)2,4,6-Trinitrophenol (picric acid)Direct conc. HNO3 oxidises phenol and gives tar;picric acid pKa ~0.4, dissolves in NaHCO3Conc. H2SO4, 298 Ko-Hydroxybenzenesulphonic acidKinetic productConc. H2SO4, 373 Kp-Hydroxybenzenesulphonic acidThermodynamic product (less crowded)Neutral FeCl3 (aq.)Violet / blue-violet complexPhenol test; also salicylic acid; carboxylicacids do not give itZn dust, heatBenzeneRemoves the -OH (reduction)Na2Cr2O7 / H2SO4p-BenzoquinoneOxidation of phenol; ring becomes a quinone
Two Ortho-Functionalisations of Phenol: Kolbe vs Reimer–TiemannFeatureKolbe (Kolbe–Schmitt)Reimer–TiemannReagentsC6H5ONa + CO2 (4–7 atm, ~400 K), then H+C6H5OH + CHCl3 + aq. NaOH (~340 K), then H+Attacking electrophileCO2 (weak electrophile; needs the veryelectron-rich phenoxide)Dichlorocarbene :CCl2 (from CHCl3 + OH-, via CCl3-and loss of Cl-)Major productSalicylic acid (o-hydroxybenzoic acid)Salicylaldehyde (o-hydroxybenzaldehyde)VariantsK-phenoxide / higher T shifts towardp-hydroxybenzoic acidCCl4 in place of CHCl3 gives salicylic acidFollow-upSalicylic acid + (CH3CO)2O → aspirinSalicylaldehyde → coumarin chemistry (Perkin),chelating ligandWhy ortho?Na+ chelates phenoxide O and incoming CO2 (cyclicTS)Ion-pair / chelation guides :CCl2 to the orthoposition
⚠️ 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

Ethers: Which Bond Breaks (HI) and How to Build Them (Williamson)HX reactivity: HI > HBr > HCl. Williamson rule: the HALIDE must be methyl or 1°; put the bulky group on the alkoxide.CaseReagentProductsReasoningCH3-O-CH2CH3 (1° / methyl)HI, 1 equivalentCH3I + CH3CH2OHSN2: I- attacks the LESS hindered alkylcarbon (methyl)(CH3)3C-O-CH3HI(CH3)3C-I + CH3OHSN1: protonated ether ionises to the stable3° cation; I- goes to the 3° carbonC6H5-O-CH3 (anisole)HIC6H5OH + CH3Isp2 C-O never cleaves; phenol does notconvert to ArIMake (CH3)3C-O-CH3Williamson(CH3)3CONa + CH3INever CH3ONa + (CH3)3CCl: 3° halide + strongbase gives isobutene (E2)Make C6H5-O-CH2CH3WilliamsonC6H5ONa + CH3CH2BrNever C2H5ONa + C6H5Br: aryl halide is inertto SN2

4. Making Aldehydes and Ketones

Named Routes to Aldehydes: Reagent → Reaction NameReactionReagentChangeThe catchRosenmundH2, Pd-BaSO4 (poisoned), boilingxyleneRCOCl → RCHOHCOCl is unstable, so no HCHO; poison stopsover-reduction to RCH2OHStephenSnCl2 + HCl (dry), then H2ORCN → RCHOVia aldimine hydrochloride; aldehyde keeps thenitrile's carbon countDIBAL-HDIBAL-H, -78 °C, then H2ORCOOR' or RCN → RCHOPartial reduction only at low temperature; LiAlH4would overshootEtardCrO2Cl2 in CS2, then H3O+ArCH3 → ArCHOInsoluble Etard complex protects the aldehydefrom over-oxidationGattermann–KochCO + HCl, anhyd. AlCl3 / CuClC6H6 → C6H5CHOHCOCl formed in situ; fails on stronglydeactivated ringsWackerPdCl2 / CuCl2, H2O, O2CH2=CH2 → CH3CHOOnly ethene gives an aldehyde; every higher1-alkene gives a methyl ketoneAlkynehydroborationB2H6 (or R2BH), then H2O2 / OH-RC≡CH → RCH2CHOAnti-Markovnikov; Hg2+/H2SO4 hydration gives theketone instead

5. Nucleophilic Addition to C=O

Reactivity Toward Nucleophilic Addition (Left = Most Reactive)HCHOno alkyl, leasthinderedRCHOone +I alkylArCHOring donates byresonanceR2C=Otwo +I alkyls, 2groups hinderR-CO-Aralkyl + ringAr2C=Otwo rings: leastreactiveTwo reasons, always give both:1. Electronic: alkyl (+I) and aryl (+M) groups reduce the partial positive charge on the carbonyl C.2. Steric: bigger groups block backside-of-plane attack and crowd the sp3 tetrahedral intermediate.Exception: Cl3C-CHO, ninhydrin, glyoxal form stable hydrates (gem-diols) because -I groups raise C=O reactivity.
Condensation with Ammonia Derivatives (H2N-Z): Products and the pH WindowReagentNameProductH2N-OHHydroxylamineOxime C=N-OHH2N-NH2HydrazineHydrazone C=N-NH2H2N-NH-C6H5PhenylhydrazinePhenylhydrazoneH2N-NH-C6H3(NO2)22,4-DNP (Brady's reagent)2,4-DNP derivative (orange-red ppt)H2N-NH-CO-NH2SemicarbazideSemicarbazone (only the NH2 on N-1 reacts)C=O + H2N-Z → C=N-Z + H2O (nucleophilic addition, then loss of water)pH (acidic → basic)rateoptimum pH ~ 3-5Too acidic: H2N-Z is protonated to -NH3+, soit cannot attack C=OToo basic: C=O is not protonated (lesselectrophilic) and the carbinolamine loseswater slowlySemicarbazide: the second NH2 is amide-like(resonance), non-nucleophilic
⚠️ 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

Aldehyde or Ketone + Base: Aldol, Cannizzaro or Crossed?Carbonyl compound + baseDoes it have an α-H?YESNO (HCHO, ArCHO, R3C-CHO)DILUTE NaOH / Ba(OH)2: ALDOL RCH2CHO → RCH(OH)CH(R)CHOHeat: dehydration (E1cB) → α,β-unsaturated carbonylCONC. (50%) NaOH / KOH: CANNIZZARO 2 ArCHO → ArCOO- +ArCH2OHRate-determining step: hydride (H-) transfer to a 2ndaldehyde moleculeCrossed versions (the high-yield single-product cases JEE favours)Claisen-Schmidt: ArCHO (no α-H) + CH3COR (has α-H) → ArCH=CH-COR, one product.Crossed Cannizzaro: HCHO + ArCHO → HCOO- (HCHO always oxidised) + ArCH2OH.Two DIFFERENT aldehydes that BOTH have α-H → FOUR aldol products (2 self + 2 cross).One aldehyde without α-H → only two products; neither has α-H → no aldol at all.
⚠️ 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

Choosing the Reducing AgentReagentConvertsLeaves aloneFails / avoid whenNaBH4CHO → CH2OH, ketone → CHOHC=C, COOH, COOR, CONH2, CN,NO2Cannot touch acids / estersLiAlH4All C=O incl. COOH, COOR, RCOCl →alcohols; CONH2, CN → aminesIsolated C=CNever in water / alcohols(violent)B2H6 (diborane)COOH → CH2OH (fast, selective)COOR, NO2, halidesAlso reduces C=C (hydroboration)H2 / Ni, Pt or PdC=O → CHOH and C=C → C-C-If a C=C must surviveClemmensen: Zn-Hg / conc.HClC=O → CH2Base-sensitive groups(esters, halides)Acid-sensitive groups: acetal,3° OH, some C=CWolff–Kishner: N2H4, KOH,glycol, ΔC=O → CH2 (via hydrazone, loses N2)Acid-sensitive groupsBase-sensitive groups: esters,amides, alkyl halides
Which Test Does Which Compound Pass?Fehling excludes benzaldehyde; NaHSO3 excludes bulky ketones (acetophenone, 3-pentanone); iodoform needs CH3CO- or CH3CH(OH)-; HCOOH is theodd acid that passes Tollens/Fehling.CompoundTollensFehlingSchiff2,4-DNPNaHSO3I2 / NaOHNaHCO3HCHO✓✓✓✓✓✗✗CH3CHO✓✓✓✓✓✓✗C6H5CHO✓✗✓✓✓✗✗CH3COCH3✗✗✗✓✓✓✗C6H5COCH3✗✗✗✓✗✓✗CH3CH2COCH2CH3✗✗✗✓✗✗✗HCOOH✓✓✗✗✗✗✓

8. Named Rearrangements and the Wittig Reaction

Four Rearrangements: Who Migrates?ReactionChangeReagentMigration rulePinacol1,2-diol → ketone (pinacol → pinacolone,3,3-dimethylbutan-2-one)conc. H2SO4 / H+OH leaves from the carbon giving themore stable cation; then aryl ≳ H > 3°> 2° > 1° > CH3 shifts to itBeckmannKetoxime → N-substituted amide (cyclohexanoneoxime → caprolactam → nylon-6)H2SO4, PCl5 or SOCl2The group ANTI to the leaving OHmigrates to nitrogenBaeyer–VilligerKetone → ester (O inserted next to C=O)RCO3H, m-CPBA, CF3CO3HH > 3° > cyclohexyl > 2° > Ar > 1° >CH3 (better cation-stabiliser migrates)Benzil–benzilicacidPhCO-COPh → Ph2C(OH)COO-NaOH, then H+OH- adds to one C=O; the aryl groupshifts to the adjacent C=O carbon

9. Grignard Reagent as a Synthetic Tool

Grignard Reagent (RMgX) + Electrophile, then H3O+SubstrateProductCarbon count / noteHCHO1° alcohol RCH2OH+1 CR'CHO2° alcohol R'CH(OH)RSecondary alcohols need an aldehyde otherthan HCHOR'COR''3° alcohol R'R''C(OH)RAny ketone gives a 3° alcoholCO2 (dry ice)Carboxylic acid RCOOH+1 C; key chain-extension routeEthylene oxide1° alcohol RCH2CH2OH+2 C; attack at the less hindered ringcarbon (SN2)R'CNKetone R'COR (via imine salt)Stops at ketone because the imine anion isnot electrophilicRCOCl or RCOOR' (excess RMgX)3° alcohol with TWO identical new R groupsKetone intermediate is more reactive thanthe starting acid derivativeHCOOR'2° alcohol with two identical R groupsFormate esters give 2°, not 3°H-Z: H2O, ROH, RCOOH, RNH2, RC≡CHAlkane R-H (RMgX is destroyed)Acidic H always wins: dry, aproticconditions are mandatory
⚠️ 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

Acid Strength: pKa Values You Should Be Able to RankAcidpKa (approx.)WhyHCOOH3.75No +I alkyl group to destabilise the carboxylateCH3COOH4.76+I of CH3 intensifies the negative charge on the anionCH3CH2COOH4.87More +I than CH3: weaker stillFCH2COOH2.59Strongest -I of the halogensClCH2COOH2.86-I falls as electronegativity falls: F > Cl > Br > ICl2CHCOOH1.29-I effects are additiveCl3CCOOH0.65Three Cl: one of the strongest simple organic acidsC6H5COOH4.20sp2 ring carbon is slightly electron-withdrawingp-O2N-C6H4COOH3.44-M and -I of NO2 stabilise the anionm-CH3O-C6H4COOH4.09Meta: only -I operates, so STRONGER than benzoicp-CH3O-C6H4COOH4.47Para: +M dominates, so WEAKER than benzoico-O2N-C6H4COOH2.17Ortho effect: any ortho substituent strengthens the acid
⚠️ 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

Nucleophilic Acyl Substitution: Reactivity Follows Leaving-Group QualityAcyl chloride RCOClLeaving group Cl- (conj. acid HCl, pKa -7)Acid anhydride (RCO)2OLeaving group RCOO- (RCOOH, pKa ~5)Ester RCOOR'Leaving group R'O- (R'OH, pKa ~16)Amide RCONH2Leaving group NH2- (NH3, pKa ~35)reactivity fallsDownhill is direct: RCOCl → anhydride / ester / amide with Nu-H. Uphill needs activation:RCOOH → RCOCl needs SOCl2 / PCl5; an ester cannot be turned back into an acid chloride directly.
Heating Effects: Ring Size Decides the ProductCompoundOn heatingRule of thumbOxalic acid (C2)CO2 + CO + H2O (with conc. H2SO4); HCOOH + CO2 at ~373 K inglycerolTwo COOH directly bondedMalonic acid (C3)CH3COOH + CO2 (~410 K)1,3-diacid decarboxylates via6-membered TSSuccinic acid (C4)Succinic anhydride (5-ring) + H2OAnhydride when a 5- or 6-ring can formGlutaric acid (C5)Glutaric anhydride (6-ring) + H2OSame ruleAdipic acid (C6)Cyclopentanone + CO2 + H2OKetone ring forms: 5-memberedPimelic acid (C7)Cyclohexanone + CO2 + H2OKetone ring forms: 6-memberedα-Hydroxy acidLactide (6-membered cyclic diester) + 2 H2OTwo molecules esterify each otherβ-Hydroxy acidα,β-Unsaturated acid + H2OConjugated alkene; 4-ring lactone toostrainedγ- / δ-Hydroxy acidγ-Lactone (5-ring) / δ-lactone (6-ring)Intramolecular esterification,spontaneous
⚠️ 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

⚠️ 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: