📌 Snapshot
- Why ethanol boils far above its isomer dimethyl ether, and why phenol is a million times more acidic.
- Preparation, acidity, dehydration, oxidation and named reactions of alcohols, phenols and ethers in clear tables.
- Lucas test, Williamson synthesis, ether cleavage with HI and the phenol reactions that decide CUET questions.
- Exam traps, a reaction sheet and twelve NTA style practice MCQs with every wrong option explained.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 7.
Two compounds, one formula, a hundred degrees apart
Ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃) both have the formula C₂H₆O and the same mass. Yet ethanol is a liquid that boils at 351 K, while methoxymethane is a gas that boils at 249 K. The only difference is where the hydrogen sits. In ethanol it is bonded to oxygen, so molecules can grip each other by hydrogen bonds. In the ether it is bonded only to carbon, so they cannot. That one fact, the –OH group makes hydrogen bonds, explains the boiling points, the solubility and much of the reactivity of this chapter. The second big idea is that the oxygen is bonded to a different kind of carbon in each family, which decides how acidic and how reactive the compound is.
The three families and their structures
| Family | General idea | Example |
|---|---|---|
| Alcohol | –OH on an sp³ carbon of an alkyl group | CH₃CH₂OH |
| Phenol | –OH directly on an sp² carbon of a benzene ring | C₆H₅OH |
| Ether | R–O–R′, with both hydrogens of water replaced by alkyl or aryl groups | CH₃OCH₃, C₆H₅OCH₃ |
Alcohols by the number of –OH groups: monohydric (one), dihydric (two, such as ethane-1,2-diol), trihydric (three, such as propane-1,2,3-triol or glycerol) and polyhydric. Monohydric alcohols by the carbon: primary (1°), secondary (2°) and tertiary (3°) alcohols have the –OH on a carbon with one, two or three other carbons attached. An allylic alcohol has –OH on an sp³ carbon next to C=C, a benzylic alcohol has it next to a benzene ring, and a vinylic alcohol (CH₂=CH–OH) has it on an sp² carbon of a double bond. Ethers are simple (symmetrical) if both groups are the same (CH₃CH₂OCH₂CH₃) and mixed (unsymmetrical) if not (C₂H₅OC₆H₅).
| Structural feature | Value | Why |
|---|---|---|
| C–O–H angle in methanol | 108.9°, slightly less than tetrahedral | Lone pairs on oxygen repel the bonding pairs more strongly |
| C–O–C angle in methoxymethane | 111.7°, slightly more than tetrahedral | The bulky alkyl groups repel each other |
| C–O bond length | Methanol 142 pm, ether 141 pm, phenol 136 pm | In phenol the oxygen lone pair is conjugated with the ring and the carbon is sp², so the bond has some double bond character |
Naming them
In IUPAC names an alcohol takes the ending -ol, replacing the final "e" of the alkane. The chain is numbered to give –OH the lowest locant. In polyhydric alcohols the "e" is kept (ethane-1,2-diol). Phenols are named as substituted phenols. Ethers are named as alkoxy alkanes: the smaller group becomes the alkoxy prefix on the larger parent chain.
| Structure | Name |
|---|---|
| CH₃CH(OH)CH₃ | Propan-2-ol |
| HOCH₂CH₂OH | Ethane-1,2-diol |
| HOCH₂CH(OH)CH₂OH | Propane-1,2,3-triol |
| o-CH₃C₆H₄OH | 2-Methylphenol (o-cresol) |
| Benzene with OH at 1,2 / 1,3 / 1,4 | Benzene-1,2-diol (catechol), benzene-1,3-diol (resorcinol), benzene-1,4-diol (hydroquinone) |
| CH₃OC₂H₅ | Methoxyethane |
| C₂H₅OCH₂CH₂CH₃ | 1-Ethoxypropane |
| C₆H₅OCH₃ | Methoxybenzene (anisole) |
| C₆H₅OC₂H₅ | Ethoxybenzene |
Worked example: naming. Name CH₃CH(CH₃)CH₂CH₂OH and (CH₃)₂CHOCH₂CH₃.
| Step | What we do | CH₃CH(CH₃)CH₂CH₂OH | (CH₃)₂CHOCH₂CH₃ |
|---|---|---|---|
| 1 | Alcohol or ether? | Alcohol: parent chain must contain the carbon bearing –OH | Ether: find the two groups on oxygen |
| 2 | Choose and number | Four carbon chain, numbered from the –OH end, so –OH is on C1 and a methyl is on C3 | Groups are ethyl and isopropyl. Smaller one is the alkoxy group |
| 3 | Name | 3-Methylbutan-1-ol | 2-Ethoxypropane |
Making alcohols
| Route | Reagent | Product |
|---|---|---|
| Acid catalysed hydration of an alkene | Dilute H₂SO₄ or H₃PO₄, water | Markovnikov alcohol: OH goes to the more substituted carbon. Propene gives propan-2-ol |
| Hydroboration–oxidation | B₂H₆ (diborane), then H₂O₂ with NaOH | Anti Markovnikov alcohol: propene gives propan-1-ol. Boron adds to the less substituted carbon |
| Reduction of an aldehyde | H₂ with Ni, Pd or Pt, or NaBH₄, or LiAlH₄ | Primary alcohol |
| Reduction of a ketone | Same reagents | Secondary alcohol |
| Reduction of a carboxylic acid or ester | LiAlH₄ (NaBH₄ is too weak) | Primary alcohol |
| Grignard reagent with HCHO | RMgX, then H₂O | Primary alcohol |
| Grignard with another aldehyde | RMgX, then H₂O | Secondary alcohol |
| Grignard with a ketone | RMgX, then H₂O | Tertiary alcohol |
Acid catalysed hydration happens in three steps. The key is the middle one, which is why the rule is Markovnikov's.
| Step | What happens | Why |
|---|---|---|
| 1 | The alkene is protonated: CH₃CH=CH₂ + H⁺ → CH₃C⁺HCH₃ | H adds to the carbon that already has more H, giving the more stable secondary carbocation |
| 2 | Water attacks the carbocation | The positive carbon is electron poor |
| 3 | The oxonium ion loses H⁺ to give propan-2-ol | The catalyst H⁺ is regenerated |
Examples from Grignard reagents: CH₃MgBr with HCHO gives ethanol, with CH₃CHO gives propan-2-ol, and with CH₃COCH₃ gives 2-methylpropan-2-ol (a tertiary alcohol) after hydrolysis.
Making phenols
| Route | Steps |
|---|---|
| From chlorobenzene (Dow process) | C₆H₅Cl is heated with NaOH at 623 K and high pressure to give sodium phenoxide, then acidified |
| From benzenesulphonic acid | Benzene with oleum gives C₆H₅SO₃H, which is fused with NaOH and acidified |
| From a diazonium salt | Aniline with NaNO₂ and HCl at 273–278 K, then warm with water: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl |
| Cumene process (industrial) | Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, which is treated with dilute acid to give phenol and acetone |
Beyond the textbook. For the exam: the cumene process gives phenol and acetone together. In reality: this is the main industrial route to both, and its economics depend on acetone, which is a by-product that is always made in a fixed amount alongside the phenol: C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH, followed by C₆H₅C(CH₃)₂OOH → C₆H₅OH + CH₃COCH₃. Why it matters: it is a good example of a cheap starting material, benzene and propene, converted into two valuable products without waste.
Physical properties: it all comes back to hydrogen bonding
| Compound (similar mass) | Boiling point (K) | Reason |
|---|---|---|
| Butane | 273 | Only weak London forces |
| Ethoxyethane | 308 | Polar C–O–C, but no O–H, so no H-bonding between molecules |
| Pentan-1-ol | 411 | O–H bonds link molecules by H-bonds |
| Pentane | 309 | Same as ethoxyethane, so an ether boils like an alkane of similar mass |
| Butan-1-ol | 391 | Alcohols boil far above alkanes, ethers and haloalkanes |
For the PYQ ordering: butane < ethoxyethane < pentanal < pentan-1-ol. Pentanal is polar (dipole–dipole forces) but cannot hydrogen bond to itself, so it sits between the ether and the alcohol.
Boiling points rise with chain length and fall with branching: butan-1-ol (391 K) > butan-2-ol (373 K) > 2-methylpropan-2-ol (356 K).
Solubility in water. The lower alcohols mix with water in all proportions because they hydrogen bond to it. As the non-polar alkyl part grows, solubility falls. Ethers are slightly soluble, since the oxygen can accept H-bonds from water but has none to give. Phenol is moderately soluble because the large non-polar ring offsets the –OH.
Acidity: why phenol is acidic and alcohols barely are
Alcohols and phenols are both weak Brønsted acids. They react with active metals to give hydrogen:
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂ and 2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂
Only phenol also reacts with aqueous sodium hydroxide: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. Alcohols do not.
Why phenol is more acidic than ethanol. After losing H⁺, an alkoxide (RO⁻) keeps its negative charge on oxygen alone. A phenoxide ion can spread the charge over the oxygen and the ring carbons through five resonance structures, so it is much more stable. A more stable conjugate base means a stronger acid. The difference is large: phenol has pKa about 10 and ethanol 15.9, so phenol is about a million times more acidic.
Why 1° alcohols are more acidic than 3°. Alkyl groups push electrons towards oxygen. In a tertiary alkoxide there are more such groups, so the negative charge is less stable and the alcohol is less willing to lose H⁺. The order is phenol > water > 1° > 2° > 3° alcohols. Alcohols are weaker acids than water, so alkoxides are stronger bases than hydroxide.
Substituents on phenol. An electron withdrawing group helps spread the negative charge on the phenoxide and makes the phenol more acidic. An electron releasing group does the opposite.
| Compound | pKa | Reading |
|---|---|---|
| Ethanol | 15.9 | Weaker acid than water |
| Phenol | 10.0 | Reference |
| o-Cresol, p-Cresol | 10.2 | CH₃ releases electrons, so slightly weaker |
| m-Nitrophenol | 8.3 | NO₂ withdraws electrons |
| o-Nitrophenol | 7.2 | NO₂ at ortho strongly withdraws |
| p-Nitrophenol | 7.1 | NO₂ at para is the most effective |
| 2,4,6-Trinitrophenol (picric acid) | Strong acid | Three NO₂ groups at o and p positions |
Worked example: ranking acidity. Arrange phenol, 3-nitrophenol, 3,5-dinitrophenol and 2,4,6-trinitrophenol in increasing acid strength.
| Step | What we do | Result |
|---|---|---|
| 1 | Acidity depends on how well the phenoxide is stabilised | More electron withdrawing groups give a stronger acid |
| 2 | Compare the number and position of NO₂ groups | None, one (meta), two (both meta), three (two ortho and one para) |
| 3 | Ortho and para NO₂ groups help most. Meta groups help less, but two beat one | Phenol < 3-nitrophenol < 3,5-dinitrophenol < 2,4,6-trinitrophenol |
The common wrong step is to think that a meta nitro group has no effect. It still withdraws electrons through the sigma framework, only less than at ortho or para.
Pause and predict: will ethanol dissolve in aqueous NaOH to form sodium ethoxide?
Check your answer: no. Ethanol is a weaker acid than water, so hydroxide cannot deprotonate it appreciably. Sodium ethoxide must be made with metallic sodium.
Reactions of alcohols that break the O–H bond
Esterification. An alcohol (or a phenol) reacts with a carboxylic acid, an acid anhydride or an acid chloride to give an ester.
| Reactant | Conditions | Note |
|---|---|---|
| Carboxylic acid | Conc. H₂SO₄ catalyst | Reversible: RCOOH + R′OH ⇌ RCOOR′ + H₂O |
| Acid anhydride | H₂SO₄ or pyridine | Not reversible |
| Acid chloride | Pyridine | Pyridine takes up the HCl, which keeps the reaction going |
Acetylation of salicylic acid with acetic anhydride gives aspirin: C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH.
Reactions that break the C–O bond
With hydrogen halides. R–OH + HX → R–X + H₂O. Reactivity is 3° > 2° > 1°, and HI > HBr > HCl. The Lucas test uses this to tell the classes of alcohol apart. Lucas reagent is concentrated HCl with ZnCl₂. The alkyl chloride formed is insoluble, so the clear solution turns cloudy.
| Alcohol class | Observation at room temperature | Reason |
|---|---|---|
| Tertiary | Turbidity immediately | Forms a stable 3° carbocation at once |
| Secondary | Turbidity in about 5 minutes | Slower: a less stable 2° carbocation |
| Primary | No turbidity at room temperature | A 1° carbocation is too unstable. Needs heat |
Other reagents: PBr₃ gives R–Br, and SOCl₂ gives R–Cl cleanly.
Dehydration. The same alcohol gives different products depending on temperature and which reagent is in excess.
| Conditions | Reaction | Product |
|---|---|---|
| Conc. H₂SO₄, 443 K, excess acid | C₂H₅OH → CH₂=CH₂ + H₂O | Ethene (an alkene) |
| Conc. H₂SO₄, 413 K, excess alcohol | 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O | Ethoxyethane (an ether) |
Ease of dehydration to alkenes is 3° > 2° > 1°. The temperature needed falls from 443 K for primary alcohols to 440 K for secondary and 358 K for tertiary alcohols. The mechanism is:
| Step | What happens | Why |
|---|---|---|
| 1 | The –OH is protonated to –OH₂⁺ | Water is a far better leaving group than OH⁻ |
| 2 (slow) | Water leaves, giving a carbocation | The rate determining step. More substituted carbocations form faster |
| 3 | A neighbouring H⁺ is lost to form C=C | The acid is regenerated |
Oxidation. The product depends on the class of alcohol and the reagent.
| Alcohol | Reagent | Product |
|---|---|---|
| Primary | PCC (pyridinium chlorochromate), or CrO₃ in anhydrous medium | Aldehyde, stops there |
| Primary | Acidified KMnO₄ or K₂Cr₂O₇ | Carboxylic acid (through the aldehyde) |
| Secondary | CrO₃ or K₂Cr₂O₇ | Ketone |
| Tertiary | Mild oxidants | No reaction, since there is no H on the carbinol carbon. Harsh conditions break C–C bonds |
Passing alcohol vapour over copper at 573 K removes hydrogen instead: CH₃CH₂OH → CH₃CHO + H₂ (1°), (CH₃)₂CHOH → (CH₃)₂CO + H₂ (2°). A tertiary alcohol dehydrates to an alkene instead, (CH₃)₃COH → (CH₃)₂C=CH₂ + H₂O.
Beyond the textbook. For the exam: methanol is toxic and causes blindness, and ethanol is made by fermentation. In reality: methanol is poisonous because the body's liver enzyme converts it into formaldehyde and then formic acid. Ethanol is used as an antidote, because the enzyme prefers ethanol, which slows the conversion of methanol and lets it be excreted. Why it matters: it explains why denatured alcohol is made undrinkable, with CuSO₄ (colour) and pyridine (foul smell), rather than harmless.
Phenols: reactions of the ring
The –OH group is strongly activating and ortho and para directing, because its lone pair feeds electron density into the ring. So phenol reacts under milder conditions than benzene.
| Reaction | Reagent and conditions | Product |
|---|---|---|
| Nitration, dilute | Dilute HNO₃, 298 K | Mixture of o-nitrophenol and p-nitrophenol |
| Nitration, concentrated | Conc. HNO₃ (with H₂SO₄) | 2,4,6-Trinitrophenol (picric acid) |
| Bromination, non-polar solvent | Br₂ in CS₂ or CHCl₃, low temperature | Mainly p-bromophenol, with some ortho |
| Bromination, water | Bromine water | 2,4,6-Tribromophenol, a white precipitate: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr |
| Kolbe's reaction | Sodium phenoxide, CO₂ (high pressure, heat), then H⁺ | Salicylic acid (2-hydroxybenzoic acid) |
| Reimer–Tiemann | CHCl₃ with aqueous NaOH, then H⁺ | Salicylaldehyde (–CHO at ortho) |
| Reduction | Zn dust, heat | Benzene: C₆H₅OH + Zn → C₆H₆ + ZnO |
| Oxidation | Chromic acid | Benzoquinone |
The two nitrophenols can be separated by steam distillation. o-Nitrophenol has an intramolecular hydrogen bond between the –OH and the neighbouring –NO₂, so it is a single, less polar molecule that evaporates easily and is steam volatile. p-Nitrophenol can only hydrogen bond to other molecules, so it forms associated molecules, is less volatile and stays behind.
Pause and predict: phenol is treated with bromine water and a white precipitate forms. What is it, and why is a Lewis acid not needed?
Check your answer: 2,4,6-tribromophenol. The –OH group activates the ring so strongly that it brominates all three positions with no catalyst, in a polar solvent that also helps ionise Br₂.
Worked example: a reaction chain. Phenol is treated in turn with (1) Zn dust, (2) CH₃Cl and anhydrous AlCl₃, (3) K₂Cr₂O₇ with H₂SO₄, and (4) HNO₃ with H₂SO₄. Find the final product.
| Step | Reagent | Reaction | Product |
|---|---|---|---|
| 1 | Zn dust | Removes the oxygen of phenol | Benzene |
| 2 | CH₃Cl, AlCl₃ | Friedel–Crafts methylation | Toluene |
| 3 | K₂Cr₂O₇ / H₂SO₄ | Side chain CH₃ is oxidised | Benzoic acid |
| 4 | HNO₃ / H₂SO₄ | Nitration. –COOH withdraws electrons and directs to the meta position | m-Nitrobenzoic acid |
The common wrong step is to give an ortho or para product at the end. A –COOH group is meta directing.
Commercial alcohols
Methanol (wood spirit) is now made from synthesis gas: CO + 2H₂ → CH₃OH over a ZnO–Cr₂O₃ catalyst at high temperature and pressure. It boils at 337 K and is highly toxic.
Ethanol is made by fermentation of sugars. Invertase turns sucrose into glucose and fructose, and zymase turns these into ethanol and carbon dioxide, without air:
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆, then C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Zymase stops working when the ethanol content reaches about 14 percent. Ethanol is also made industrially by hydrating ethene.
Ethers: making them
| Method | Reaction | Limits |
|---|---|---|
| Dehydration of alcohols | 2ROH → R–O–R + H₂O with H₂SO₄ at 413 K | Works for primary alcohols only. Secondary and tertiary alcohols give alkenes. Gives symmetrical ethers |
| Williamson synthesis | R–X + R′ONa → R–O–R′ + NaX | An SN2 reaction, so the halide must be primary. Gives mixed ethers |
Williamson with a tertiary halide fails. An alkoxide is also a strong base, and with a bulky halide it takes a proton and makes an alkene instead.
Worked example: choosing the Williamson route. Prepare (a) tert-butyl methyl ether and (b) anisole.
| Target | Wrong route | Right route | Reasoning |
|---|---|---|---|
| (CH₃)₃C–O–CH₃ | CH₃ONa + (CH₃)₃CBr gives 2-methylpropene by elimination | (CH₃)₃CONa + CH₃Br → (CH₃)₃COCH₃ + NaBr | The halide must be the primary (methyl) one |
| C₆H₅OCH₃ | C₆H₅Br + CH₃ONa (aryl halides do not undergo SN2) | C₆H₅ONa + CH₃Br → C₆H₅OCH₃ + NaBr | Use the phenoxide and a primary alkyl halide |
Ethers: reactions
Ethers are quite unreactive, which is why they are good solvents. The C–O bond is cleaved only by strong acids.
Cleavage by HX. The order of reactivity is HI > HBr > HCl. Excess HI converts a dialkyl ether to two alkyl iodides: C₂H₅OC₂H₅ + 2HI → 2C₂H₅I + H₂O. The mechanism is: the ether oxygen is protonated, and then iodide attacks one carbon.
| Ether | Mechanism | Products | Why |
|---|---|---|---|
| CH₃–O–C₂H₅ (primary, primary) | SN2 | CH₃I + C₂H₅OH | I⁻ attacks the less hindered, smaller alkyl group |
| (CH₃)₃C–O–CH₃ | SN1 | (CH₃)₃C–I + CH₃OH | The tertiary C–O bond breaks to form a stable tertiary carbocation |
| C₆H₅–O–CH₃ (anisole) | SN2 at the methyl | C₆H₅OH + CH₃I | The aryl C–O bond is stronger and an sp² carbon cannot be attacked in SN2. The phenol formed does not react further with HI |
Electrophilic substitution of anisole. The –OR group is activating and ortho, para directing, like –OH, so anisole is more reactive than benzene.
| Reaction | Conditions | Major product |
|---|---|---|
| Bromination | Br₂ in ethanoic acid (no FeBr₃ needed) | p-Bromoanisole (about 90 percent) |
| Friedel–Crafts alkylation or acylation | CH₃Cl or CH₃COCl with AlCl₃ | Mixture of o and p products, mainly para |
| Nitration | Conc. HNO₃ and H₂SO₄ | Mixture of o- and p-nitroanisole |
Beyond the textbook. For the exam: ethers are inert and are good solvents. In reality: diethyl ether slowly reacts with oxygen in air on standing and forms explosive peroxides, so laboratory bottles are stored in the dark, with stabilisers, and old bottles must not be distilled to dryness. Why it matters: inertness to acids and bases does not mean inertness to oxygen.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "Tertiary alcohols are the most acidic." | They are the most reactive in other reactions | Acidity runs 1° > 2° > 3°, because alkyl groups destabilise the alkoxide |
| "Alcohols dissolve in NaOH like phenols." | Both have –OH | Only phenol reacts with aqueous NaOH. Alcohols are weaker acids than water |
| "Hydration and hydroboration give the same alcohol." | Both add water across C=C | Hydration is Markovnikov (propan-2-ol). Hydroboration–oxidation is anti Markovnikov (propan-1-ol) |
| "Lucas test: primary reacts fastest." | Primary alcohols are least hindered | Tertiary immediately, secondary in about 5 minutes, primary not at room temperature |
| "Anisole + HI gives methanol and iodobenzene." | Either C–O bond could break | Always phenol + CH₃I. The aryl C–O bond is stronger and cannot undergo substitution |
| "Williamson synthesis with a tertiary halide gives the ether." | The reaction looks general | It gives an alkene. Use the tertiary alkoxide with a primary halide |
| "Br₂ in CS₂ and bromine water give the same product." | Same reagent, same phenol | Non-polar solvent gives monobromophenol. Water gives white 2,4,6-tribromophenol |
| "Both nitrophenols are steam volatile." | They are isomers | Only o-nitrophenol is, because of intramolecular H-bonding |
| "Ethanol at 443 K and 413 K gives the same product." | Same reagents | 443 K and excess acid gives ethene. 413 K and excess alcohol gives ethoxyethane |
Reaction sheet
| Reaction | Reagents and conditions | What changes | Example |
|---|---|---|---|
| Hydration | H₂O, H⁺ | Alkene to alcohol (Markovnikov) | Propene → propan-2-ol |
| Hydroboration | B₂H₆, then H₂O₂ / OH⁻ | Alkene to alcohol (anti Markovnikov) | Propene → propan-1-ol |
| Grignard | RMgX then H₂O | Carbonyl to alcohol | Propanone + CH₃MgBr → 2-methylpropan-2-ol |
| Esterification | RCOOH, H⁺ | ROH to ester | Salicylic acid → aspirin |
| Lucas test | Conc. HCl, ZnCl₂ | ROH to RCl | Tertiary: turbid at once |
| Dehydration | Conc. H₂SO₄, 443 K | Alcohol to alkene | Ethanol → ethene |
| Ether formation | Conc. H₂SO₄, 413 K | 2ROH to R–O–R | Ethanol → ethoxyethane |
| Oxidation | PCC, or K₂Cr₂O₇ / KMnO₄ | 1° to CHO or COOH, 2° to ketone | Propan-2-ol → propanone |
| Kolbe | C₆H₅ONa, CO₂, then H⁺ | Adds –COOH ortho | Phenol → salicylic acid |
| Reimer–Tiemann | CHCl₃, aq. NaOH | Adds –CHO ortho | Phenol → salicylaldehyde |
| Williamson | R–X (primary), R′ONa | Makes R–O–R′ | CH₃Br + C₆H₅ONa → anisole |
| Ether cleavage | Excess HI | R–O–R′ to RI and R′I | Anisole → phenol + CH₃I |
60-second revision
| Topic | Lock this in |
|---|---|
| Boiling point | Alcohols (H-bonding) ≫ aldehydes > ethers ≈ alkanes. Falls with branching |
| Hydration | Markovnikov. Hydroboration–oxidation is anti Markovnikov |
| Grignard | HCHO gives 1°, other aldehydes give 2°, ketones give 3° alcohols |
| Phenol manufacture | Cumene process gives phenol and acetone |
| Acidity | Phenol > water > 1° > 2° > 3°. EWG on o/p raises, EDG lowers. Picric acid is strong |
| Lucas | 3° immediate, 2° about 5 min, 1° none |
| Dehydration | 443 K gives alkene. 413 K gives ether. Ease 3° > 2° > 1° |
| Oxidation | 1° gives CHO (PCC) or COOH. 2° gives ketone. 3° resists |
| Phenol reactions | Dilute HNO₃ gives o + p. Conc. gives picric acid. Br₂ water gives white 2,4,6 tribromophenol. Kolbe, Reimer–Tiemann, Zn |
| o vs p nitrophenol | Ortho is steam volatile (intramolecular H-bond) |
| Williamson | SN2, primary halide only. Tertiary halide gives alkene |
| HI cleavage | 1° groups: smaller alkyl gives RI. Tertiary group gives tert-RI by SN1. Aryl alkyl ether gives phenol + RI |
| Anisole EAS | o,p directing. Bromination gives p-bromoanisole in ethanoic acid |
Where this chapter connects
- Before this: Haloalkanes and Haloarenes supplies SN1, SN2, the Williamson mechanism and the idea of a good leaving group, which are used again for the Lucas test and ether cleavage.
- Next: Aldehydes, Ketones and Carboxylic Acids is where alcohols oxidise to and where Grignard reagents add, so the products here become the starting materials there.
- Class XI link: Hydrocarbons gives the Markovnikov and anti Markovnikov addition to alkenes that start the preparation of alcohols.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. Which of the following converts propene into propan-1-ol?
▸ Show answer & explanation
Answer: B
Hydroboration–oxidation adds boron to the less substituted carbon and replaces it with OH, giving the anti Markovnikov alcohol, propan-1-ol. Acid catalysed hydration (A) and HBr followed by aqueous KOH (C) both follow Markovnikov's rule and give propan-2-ol. Hydrogenation (D) gives propane.
Q2. Which reagent oxidises a primary alcohol to an aldehyde and stops at that stage?
▸ Show answer & explanation
Answer: C
PCC is a mild oxidant that does not oxidise the aldehyde further. Acidified permanganate, hot nitric acid and excess acidified dichromate all carry the oxidation on to the carboxylic acid.
Q3. Methylmagnesium bromide is reacted with propanone and the product is hydrolysed. The alcohol obtained is:
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Answer: A
A Grignard reagent adds to a ketone to give a tertiary alcohol. CH₃ adds to the carbonyl carbon of (CH₃)₂C=O, which gives (CH₃)₃C–OH, 2-methylpropan-2-ol. Propan-2-ol would form from ethanal, and butan-2-ol and butan-1-ol would need other carbonyl compounds.
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Q4. The correct increasing order of acidic strength of phenol, 3-nitrophenol, 3,5-dinitrophenol and 2,4,6-trinitrophenol is:
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Answer: D
Electron withdrawing nitro groups stabilise the phenoxide ion, so more of them give a stronger acid. Ortho and para nitro groups are most effective, so picric acid is the strongest. Phenol, with none, is the weakest. Option A puts 3-nitrophenol below phenol. Option B swaps the mono and dinitro compounds. Option C is the reverse order.
Q5. The correct increasing order of boiling points of n-butane, pentanal, ethoxyethane and pentan-1-ol is:
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Answer: B
n-Butane has only weak London forces. Ethoxyethane is slightly polar. Pentanal has stronger dipole–dipole forces, and pentan-1-ol forms hydrogen bonds and boils highest. The other options place a more polar compound below a less polar one.
Q6. Which of the following gives turbidity immediately with Lucas reagent at room temperature?
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Answer: C
The tertiary alcohol forms a stable tertiary carbocation quickly and gives the insoluble alkyl chloride at once. Butan-2-ol, a secondary alcohol, takes about five minutes. The primary alcohols, butan-1-ol and ethanol, give no turbidity at room temperature.
Q7. When ethanol is heated with concentrated H₂SO₄ at 443 K and with excess ethanol at 413 K, the main products are respectively:
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Answer: A
At 443 K with excess acid, dehydration to the alkene (ethene) occurs. At 413 K with excess alcohol, one molecule of ethanol attacks a protonated ethanol and gives the ether, ethoxyethane. Option B swaps the conditions. Ethane and ethanal are not products of dehydration.
Q8. Match the reaction in List I with the organic product in List II. List I: (i) Kolbe's reaction of sodium phenoxide (ii) Reimer–Tiemann reaction of phenol (iii) Phenol heated with zinc dust (iv) Phenol with dilute HNO₃ at 298 K. List II: (p) Salicylaldehyde (q) Salicylic acid (r) Benzene (s) A mixture of o-nitrophenol and p-nitrophenol.
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Answer: D
Kolbe's reaction introduces a –COOH group ortho to the –OH to give salicylic acid. Reimer–Tiemann introduces –CHO to give salicylaldehyde. Zinc dust removes the oxygen and gives benzene, and dilute nitric acid at room temperature gives the o and p nitrophenols. The other options swap at least one pair.
Q9. Assertion (A): o-Nitrophenol is steam volatile but p-nitrophenol is not. Reason (R): o-Nitrophenol has intramolecular hydrogen bonding while p-nitrophenol has intermolecular hydrogen bonding.
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Answer: A
In o-nitrophenol the –OH bonds to the neighbouring nitro group inside the same molecule, so molecules are not linked to one another and the compound evaporates with steam. In p-nitrophenol the molecules link to each other, so it is far less volatile. This is the basis of their separation by steam distillation.
Q10. Phenol is treated successively with zinc dust, CH₃Cl with anhydrous AlCl₃, K₂Cr₂O₇ with H₂SO₄, and finally HNO₃ with H₂SO₄. The final product is:
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Answer: D
Zinc converts phenol to benzene, Friedel–Crafts methylation gives toluene, and dichromate oxidises the methyl group to –COOH to give benzoic acid. The carboxyl group withdraws electrons and directs the nitro group to the meta position. Options A, B and C all place the nitro group at positions that a –COOH group does not direct to.
Q11. The best route to prepare tert-butyl methyl ether by the Williamson synthesis is:
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Answer: B
The Williamson synthesis is SN2, so the halide must be primary. Sodium tert-butoxide with methyl bromide gives the ether. Option A uses a tertiary halide, which gives 2-methylpropene by elimination. Option C is dehydration, which gives mixtures and alkenes with a tertiary alcohol. Option D would give the symmetrical ether or an alkene.
Q12. In the reaction (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH, the products are formed because:
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Answer: C
When one alkyl group is tertiary, protonated ether ionises to a stable tertiary carbocation and methanol, and the iodide then combines with the carbocation. If SN2 operated, I⁻ would attack the methyl group and give CH₃I and tert-butyl alcohol, so option A describes the wrong products. Option B is not the reason, and HI does not reduce ethers.
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