Quick Recap — Oxygen Functional Groups
- Alcohols (): monohydric/di/tri; classified 1/2/3 by the carbon bearing .
- Phenols ( on a benzene ring), ethers (), aldehydes (), ketones (), carboxylic acids ().
- Common names: methanol (wood spirit), ethane-1,2-diol (glycol), propane-1,2,3-triol (glycerol), methanal (formaldehyde), propanone (acetone), ethanoic acid (acetic acid).
- Alcohols and acids hydrogen-bond, giving high boiling points; carboxylic acids exist as dimers.
Beyond-NCERT JEE Essentials
High-yield named reactions, selectivity rules and the traps examiners build around them. This assumes the basics (functional groups, naming, hydrogen bonding) from the recap and pushes into product prediction and reasoning.
Alcohols, Phenols and Ethers
Lucas test (distinguishes 1/2/3 alcohols). Lucas reagent is conc. HCl + anhydrous ZnCl2 and it reacts by SN1, so the rate follows carbocation stability: a tertiary alcohol gives immediate turbidity (cloudiness) at room temperature, a secondary one clouds in about 5 minutes, and a primary one stays clear cold (it reacts only on heating). Reactivity 3 > 2 > 1. [JEE Tip] The turbidity is the insoluble alkyl chloride separating out; any question phrased as rate with conc. HCl/ZnCl2 is really asking for carbocation stability. Lucas only works for alcohols small enough to dissolve in the reagent (up to about 6 carbons).
Victor Meyer test also sorts 1/2/3 alcohols, but by colour. The alcohol is turned into R-I (red P + I2), then into the nitroalkane R-NO2 (AgNO2), then treated with nitrous acid (HNO2) and made alkaline: primary gives a RED colour (nitrolic acid), secondary a BLUE colour (pseudonitrole), tertiary stays COLOURLESS. [JEE Tip] Remember 1-red, 2-blue, 3-colourless. The tertiary case is colourless because the carbon carrying -NO2 has no hydrogen left to react.
Williamson ether synthesis. Alkoxide or phenoxide (R-ONa) + alkyl halide (R'-X) -> ether R-O-R' by SN2. For an unsymmetrical ether, pick the LESS hindered halide (methyl or primary) and the MORE hindered alkoxide; a tertiary halide only eliminates. So tert-butyl methyl ether must come from sodium tert-butoxide + CH3I, never from tert-butyl bromide + sodium methoxide (that gives isobutylene by E2). [JEE Tip] Aryl halides fail Williamson (no SN2 at an sp2 carbon), so to bond oxygen to a ring the ring must supply the phenoxide, not the halide (this is how anisole, C6H5-O-CH3, is made: sodium phenoxide + CH3I).
Reimer-Tiemann reaction. Phenol + CHCl3 + aqueous NaOH (about 340 K), then acidify -> salicylaldehyde (2-hydroxybenzaldehyde), the -CHO landing ORTHO to -OH. The true electrophile is dichlorocarbene (:CCl2), generated from CHCl3 by base. [JEE Tip] Replace CHCl3 by CCl4 under the same conditions and you get salicylic acid (2-hydroxybenzoic acid) instead, because CCl4 installs -CCl3 which hydrolyses to -COOH. Both are ortho-selective; mixing up CHCl3 (aldehyde) and CCl4 (acid) is a classic trap.
Kolbe (Kolbe-Schmitt) reaction. Sodium phenoxide + CO2 (about 400 K, 4-7 atm), then acidify -> salicylic acid (2-hydroxybenzoic acid), with -COOH ortho to -OH. This is the industrial precursor to aspirin. [JEE Tip] Do not confuse this with Kolbe electrolysis (electrolysis of a carboxylate salt, which couples the R groups to give an alkane R-R). Phenol + CO2 gives salicylic acid; carboxylate electrolysis gives R-R.
Phenol acidity and substituent effects. Phenol (pKa about 10) is far more acidic than alcohols (pKa about 16-18) because phenoxide delocalises its charge into the ring. Electron-withdrawing groups (-NO2, -X) at ortho or para RAISE acidity by stabilising phenoxide; electron-donating groups (-CH3, -OCH3) LOWER it. Order: picric acid (2,4,6-trinitrophenol) >> p-nitrophenol > phenol > p-cresol. [JEE Tip] ortho- and para-nitrophenol are much stronger acids than meta-nitrophenol, because only from ortho/para can -NO2 withdraw charge by resonance (-M); from meta it works by -I alone. Note phenol is still weaker than carbonic acid, so phenol does NOT liberate CO2 from NaHCO3 (whereas any carboxylic acid does) - a favourite distinguishing point.
Ether cleavage by HI. Ethers are cleaved by hot HX with reactivity HI > HBr > HCl (iodide is the best nucleophile and HI the strongest acid). The oxygen is protonated, then iodide attacks. In a simple mixed alkyl ether, iodide hits the SMALLER, less hindered carbon by SN2: CH3-O-C2H5 + HI -> CH3I + C2H5OH. If one group is tertiary, it instead leaves as the stable carbocation (SN1): CH3-O-C(CH3)3 + HI -> CH3OH + (CH3)3C-I. For anisole the aryl C-O bond never breaks, so C6H5-O-CH3 + HI -> C6H5OH (phenol) + CH3I. [JEE Tip] Rule of thumb: the alkyl-oxygen bond that gives the more stable carbocation, or the less hindered SN2 centre, becomes the iodide; the aryl-oxygen bond is untouched, always leaving phenol.
Aldehydes and Ketones
Nucleophilic addition is the master reaction: the carbonyl carbon is electrophilic (C=O is polarised, C carrying delta+ and O carrying delta-), so a nucleophile adds to carbon and oxygen takes up H. Reactivity is aldehydes > ketones (two electron-donating, bulky alkyl groups deactivate a ketone both electronically and sterically) and aliphatic > aromatic (ring conjugation feeds electron density into the carbonyl). HCHO is the most reactive carbonyl of all.
Aldol and Cross-Aldol (need alpha-H). Two carbonyl molecules that have an alpha-H, in dilute NaOH: the enolate of one adds to the carbonyl of the other -> a beta-hydroxy carbonyl (the aldol); warming dehydrates it to an alpha,beta-unsaturated carbonyl. Example: 2 CH3CHO -> CH3CH(OH)CH2CHO (3-hydroxybutanal) -> CH3CH=CHCHO (but-2-enal) on heating. A CROSS-aldol between two different alpha-H partners gives a messy four-product mixture; it becomes clean only when one partner has NO alpha-H (e.g., HCHO or C6H5CHO) and so can act only as the electrophile. [JEE Tip] Aldol absolutely requires an alpha-H; no alpha-H means no aldol.
Cannizzaro reaction (no alpha-H). An aldehyde that has NO alpha-H, warmed with conc. NaOH, disproportionates: one molecule is oxidised to the carboxylate salt, the other reduced to the alcohol. 2 HCHO + NaOH -> CH3OH + HCOONa; 2 C6H5CHO + NaOH -> C6H5CH2OH + C6H5COONa. [JEE Tip] In a CROSSED Cannizzaro, HCHO is always the one oxidised (to formate) because it is the easiest to oxidise, so it drives the other aldehyde entirely to the alcohol - a clean route to benzyl alcohol from benzaldehyde. Decision rule for an aldehyde in strong base: alpha-H present -> Aldol; alpha-H absent -> Cannizzaro.
Clemmensen vs Wolff-Kishner. Both take a carbonyl C=O all the way down to CH2 (complete deoxygenation), but under opposite conditions. Clemmensen uses Zn-Hg (zinc amalgam) with conc. HCl (ACIDIC) - choose it when the molecule is base-sensitive. Wolff-Kishner uses NH2NH2 (forming the hydrazone) then KOH or NaOH with heat (BASIC) - choose it when the molecule is acid-sensitive. Either one turns acetophenone (C6H5COCH3) into ethylbenzene (C6H5CH2CH3). [JEE Tip] Both reduce only aldehyde/ketone carbonyls to CH2; the carbonyl of an acid, ester or amide is left alone. Select the method by which reagent (acid or base) the rest of the molecule can survive.
Named reactions that STOP at an aldehyde.
- Rosenmund reduction: acyl chloride RCOCl + H2 over Pd-BaSO4 (a poisoned catalyst) -> RCHO; the poison prevents over-reduction to the alcohol.
- Etard reaction: toluene (C6H5CH3) + CrO2Cl2 (chromyl chloride), then hydrolysis -> benzaldehyde (C6H5CHO).
- Gattermann-Koch reaction: benzene + CO + HCl with anhydrous AlCl3 (and CuCl) -> benzaldehyde (C6H5CHO); a direct formylation of the ring.
- Stephen reduction: nitrile RCN + SnCl2/HCl -> an aldimine, then hydrolysis -> RCHO. [JEE Tip] Group them by starting material: Rosenmund from an acid chloride, Stephen from a nitrile, Etard and Gattermann-Koch build the -CHO directly onto an arene.
HVZ (Hell-Volhard-Zelinsky) reaction. A carboxylic acid that has an alpha-H + X2 with red P -> the alpha-halo acid: CH3COOH + Cl2/red P -> ClCH2COOH. [JEE Tip] No alpha-H, no reaction - so HCOOH and (CH3)3C-COOH (no alpha-hydrogen) do not undergo HVZ. The alpha-halo acid is the standard springboard to alpha-hydroxy and alpha-amino acids.
Haloform reaction. A methyl ketone (a CH3-CO- unit) OR an alcohol oxidisable to one (a CH3-CH(OH)- unit, i.e. a methyl carbinol), treated with X2/NaOH -> the haloform CHX3 + a carboxylate. With I2/NaOH this is the iodoform test, giving a yellow CHI3 precipitate. POSITIVE: ethanal (CH3CHO), every methyl ketone (acetone, acetophenone), ethanol, and secondary alcohols of the form CH3CH(OH)R such as isopropanol. NEGATIVE: HCHO, methanol, benzaldehyde, benzophenone. [JEE Tip] The test needs the acetyl-type fragment CH3-CO- or CH3-CH(OH)- specifically; count the carbons on the carbonyl (or carbinol) before answering.
Tollens vs Fehling (aldehydes, not ketones). Tollens reagent (ammoniacal AgNO3) gives a silver mirror with essentially every aldehyde; Fehling solution (alkaline Cu2+ with tartrate) gives a red Cu2O precipitate with ALIPHATIC aldehydes only. Ketones give neither. [JEE Tip] Benzaldehyde is Tollens-POSITIVE but Fehling-NEGATIVE - the standard way to separate an aromatic aldehyde from an aliphatic one. Also HCOOH and its formate esters are Tollens-positive, because their carbonyl carbon still carries a hydrogen (an aldehyde-like H).
Carboxylic Acids
Acidity and -I effects. Across families the order is carboxylic acid > carbonic acid > phenol > water > alcohol. Among the acids themselves, HCOOH (pKa 3.75) > CH3COOH (pKa 4.76): acetic acid is weaker because its methyl group is electron-donating (+I) and destabilises the carboxylate. Electron-withdrawing (-I) groups strengthen an acid, their effect is additive and it fades with distance: CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH, and FCH2COOH > ClCH2COOH > BrCH2COOH (fluorine is the strongest -I halogen). A chlorine on C-2 strengthens the acid far more than the same chlorine on C-4. [JEE Tip] For any acidity ranking: count the -I groups, weigh how close each sits to -COOH, and remember an alkyl group (+I) always weakens the acid.
Decarboxylation. Heating the sodium salt of an acid with soda lime (NaOH + CaO) gives the alkane with one fewer carbon plus Na2CO3: CH3COONa -> CH4. [JEE Tip] Acids with a carbonyl beta to -COOH (beta-keto acids) lose CO2 especially easily on gentle warming. (Kolbe electrolysis of the carboxylate is a separate decarboxylation that couples two R groups to R-R.)
Reactivity of acid derivatives toward nucleophilic acyl substitution: acid chloride > anhydride > ester > amide. The order tracks two linked factors - how strongly the attached group donates its lone pair back into the C=O (donation stabilises the derivative and slows attack) and how good a leaving group it is. Chloride is a poor donor but an excellent leaving group, so acyl chlorides are the most reactive; -NH2 is a strong donor and a terrible leaving group, so amides are the least reactive. [JEE Tip] The same order tells you the ease of hydrolysis and means you can move DOWN the series (chloride -> anhydride -> ester -> amide) easily but not back up.
Worked Examples — Beyond-NCERT Essentials
Example 1 - Aldol or Cannizzaro? Of ethanal (CH3CHO), benzaldehyde (C6H5CHO), 2,2-dimethylpropanal ((CH3)3C-CHO) and methanal (HCHO), which undergo aldol condensation and which undergo Cannizzaro with conc. NaOH? The only deciding factor is the alpha-hydrogen (a hydrogen on the carbon next to -CHO): aldol needs one, Cannizzaro needs none. Ethanal has three alpha-H (its CH3), so it does aldol. Benzaldehyde carries its -CHO on the ring with no alpha-carbon, so Cannizzaro. 2,2-dimethylpropanal has a quaternary alpha-carbon, (CH3)3C-, bearing no hydrogen, so Cannizzaro. Methanal has no carbon besides the carbonyl, hence no alpha-H, so Cannizzaro. Answer: Aldol - ethanal only. Cannizzaro - benzaldehyde, 2,2-dimethylpropanal and methanal.
Example 2 - Count the iodoform-positive compounds. How many of these give a yellow precipitate with I2/NaOH: ethanol, methanol, propan-2-ol, propan-1-ol, acetone, acetaldehyde, benzaldehyde, acetophenone? The iodoform test is positive only for a CH3-CO- group or a CH3-CH(OH)- group (the latter is first oxidised to CH3-CO-). Ethanol (CH3CH(OH)H, a methyl carbinol) positive; methanol negative; propan-2-ol (CH3CH(OH)CH3) positive; propan-1-ol (its carbinol carbon is CH2, not CH3CH(OH)-) negative; acetone (methyl ketone) positive; acetaldehyde (a CH3-CO- unit) positive; benzaldehyde (no methyl on the carbonyl) negative; acetophenone (C6H5COCH3, a methyl ketone) positive. Answer: 5 - ethanol, propan-2-ol, acetone, acetaldehyde and acetophenone.
Example 3 - Distinguish by Tollens and Fehling. Three unlabelled bottles hold benzaldehyde, propanal and propanone. Using only Tollens and Fehling, tell them apart. Ketones fail both tests, so propanone gives neither a silver mirror nor a red precipitate - it is the one negative to both. The two aldehydes both give a silver mirror with Tollens. Split them with Fehling, which responds to ALIPHATIC aldehydes only: propanal gives the red Cu2O precipitate while benzaldehyde (aromatic) does not. Answer: propanone negative to both; propanal Tollens-positive and Fehling-positive; benzaldehyde Tollens-positive but Fehling-negative.
Example 4 - Order the acids by strength. Rank in decreasing acidity: CH3COOH, FCH2COOH, ClCH2COOH, Cl2CHCOOH and HCOOH, and justify with -I effects. Acidity rises as the carboxylate is more stabilised by electron withdrawal. Cl2CHCOOH has two -I chlorines and is strongest (pKa about 1.3). Among the single-substituent acids, fluorine is a stronger -I atom than chlorine, so FCH2COOH (pKa about 2.6) beats ClCH2COOH (pKa about 2.9). HCOOH has no electron-donating alkyl group, so it is stronger than acetic acid but weaker than the halo-acids (pKa 3.75). CH3COOH is weakest (pKa 4.76) because its methyl is electron-donating (+I). Answer: Cl2CHCOOH > FCH2COOH > ClCH2COOH > HCOOH > CH3COOH.
Example 5 - Clemmensen vs Wolff-Kishner. Give the product when acetophenone (C6H5COCH3) meets (a) Zn-Hg/conc. HCl and (b) NH2NH2 then KOH/heat. Which would you pick for a substrate carrying an acid-labile group? Both reactions reduce the ketone C=O to CH2, so both give the same product: ethylbenzene, C6H5CH2CH3. They differ only in conditions. (a) Clemmensen is strongly acidic (conc. HCl) and is ruled out by an acid-sensitive group. (b) Wolff-Kishner is strongly basic (KOH, heat, via the hydrazone) and is ruled out by a base-sensitive group. For an acid-labile substrate, choose Wolff-Kishner. Answer: both give ethylbenzene (C6H5CH2CH3); use Wolff-Kishner when the molecule is acid-sensitive, Clemmensen when it is base-sensitive.
Example 6 - Reimer-Tiemann product (and the CCl4 trap). Phenol is heated with CHCl3 and aqueous NaOH at 340 K and the mixture is acidified. What is the major product, and how would it change with CCl4 in place of CHCl3? CHCl3 with NaOH generates dichlorocarbene (:CCl2), the Reimer-Tiemann electrophile. Phenoxide steers it ORTHO, and on hydrolysis and acidification the -CHCl2 becomes -CHO, giving salicylaldehyde (2-hydroxybenzaldehyde). With CCl4 the ring instead receives -CCl3, which hydrolyses to -COOH, giving salicylic acid (2-hydroxybenzoic acid) - the same product as the Kolbe reaction. Answer: with CHCl3, salicylaldehyde (2-hydroxybenzaldehyde); with CCl4, salicylic acid (2-hydroxybenzoic acid); both ortho.
Example 7 - Ether cleavage with HI. Predict the products of one mole of cold HI with (a) ethyl methyl ether CH3-O-C2H5, (b) tert-butyl methyl ether CH3-O-C(CH3)3, and (c) anisole C6H5-O-CH3. HI protonates the ether oxygen, then iodide follows the easiest path. (a) Both groups are small, so iodide attacks the LESS hindered carbon (the methyl) by SN2: products CH3I + C2H5OH. (b) The tert-butyl group forms a stable tertiary carbocation (SN1), leaving as the cation and grabbing iodide: products (CH3)3C-I + CH3OH. (c) The aryl C-O bond cannot break (sp2, partial double-bond character), so iodide takes the methyl: products C6H5OH (phenol) + CH3I. Answer: (a) CH3I + ethanol; (b) tert-butyl iodide + methanol; (c) phenol + CH3I.
Example 8 - Lucas and iodoform on isomeric alcohols. For butan-1-ol, butan-2-ol and 2-methylpropan-2-ol, predict (i) the Lucas test at room temperature and (ii) the iodoform test. Lucas reagent reacts by SN1, so turbidity appears fastest for the most stable carbocation: 2-methylpropan-2-ol (tertiary) clouds IMMEDIATELY, butan-2-ol (secondary) in about 5 minutes, butan-1-ol (primary) stays clear cold. For iodoform we need a CH3-CH(OH)- unit: only butan-2-ol, CH3CH(OH)CH2CH3, has it, so only butan-2-ol is iodoform-positive (butan-1-ol has a -CH2OH carbinol and the tertiary alcohol has three methyls but no CH3CH(OH)- unit). Answer: Lucas - tert-butyl alcohol immediate, butan-2-ol about 5 min, butan-1-ol no change cold; iodoform - only butan-2-ol positive.