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Class XII ⚗️ Chemistry Ch 7 of 10

Alcohols, Phenols and Ethers

CUET unit: Alcohols, Phenols and Ethers

📌 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

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Q1. Which of the following converts propene into propan-1-ol?

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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:

▸ Show answer & explanation

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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