Class 12 · Chemistry · Chapter 7

Alcohols, Phenols and Ethers

How alcohols, phenols and ethers are classified, named, prepared and reacted, and why phenol is acidic while ethanol is not. The board tests this chapter through mechanisms (hydration, dehydration, HI on ethers), give-reason questions on acidity and boiling points, name reactions (Kolbe, Reimer-Tiemann, Williamson) and the conversion puzzles of the organic unit.

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Before you startWhat the board asks from this chapter

Alcohols, Phenols and Ethers typically carries about 6 marks within the CBSE organic chemistry unit for the 70-mark theory paper: usually one or two 1-mark items, one 2- or 3-mark reasoning or mechanism question, and a share of a 4-mark case study or a 5-mark question drawn from the organic unit.

Question typeMarksUsually asked
MCQ1Order of acidity or boiling points, which alcohol gives immediate turbidity with Lucas reagent, the intermediate in dehydration, the product of phenol with bromine water or dilute HNO3, the name reaction that gives salicylic acid.
Assertion–Reason1Phenol versus ethanol acidity, o-nitrophenol being steam volatile, ethanol boiling higher than methoxymethane, –OH being ortho/para directing, alcohols as Bronsted acids and bases.
Very short answer2IUPAC names; give reasons (o-nitrophenol more acidic than o-methoxyphenol, phenols do not undergo C–OH cleavage, ether bond angle above tetrahedral); product of a given reaction; distinguish two alcohols by a test.
Short answer3Three name reactions with equations; reagents for a list of conversions; the mechanism of acid-catalysed hydration, dehydration or HI cleavage of an ether.
Long answer5An identify-A-to-E puzzle starting from ethanol or phenol, or a set of conversions with one mechanism, or five give-reason parts.
Case-based4A passage on the preparation and cleavage of ethers, or on the electrophilic substitution and acidity of phenol, with four sub-questions.

Hot topics, in order

  1. Acidity: phenol versus ethanol versus water, effect of nitro and methyl groups, resonance of the phenoxide ion
  2. Mechanisms: acid-catalysed hydration of ethene, acid dehydration of ethanol, HI cleavage of ethers (SN2 and SN1 cases)
  3. Name reactions with equations: Kolbe, Reimer-Tiemann, Williamson synthesis, cumene process, hydroboration-oxidation
  4. Give-reason questions on boiling points, solubility and o-nitrophenol steam volatility
  5. Reactions of phenol: nitration, bromination (CS2 versus bromine water), oxidation, zinc dust
  6. Oxidation of alcohols with the right reagent (PCC, CrO3, acidified KMnO4, Cu at 573 K) and why tertiary alcohols resist oxidation
  7. Lucas test and distinguishing primary, secondary and tertiary alcohols
  8. Reactions of ethers: HI on anisole, electrophilic substitution of anisole

The whole chapter, in orderNotes

01 7.1 Classification

An alcohol has –OH on an sp3 carbon of an aliphatic chain; a phenol has –OH bonded directly to an sp2 carbon of a benzene ring; an ether has an oxygen bonded to two carbon groups, R–O–R′. Everything about reactivity in this chapter follows from which carbon the oxygen is attached to.

Alcohols and phenols are classified first by the number of –OH groups: monohydric (ethanol, phenol), dihydric (ethane-1,2-diol, catechol) and trihydric (propane-1,2,3-triol, glycerol). Monohydric alcohols are then divided by the carbon carrying the –OH. If that carbon is sp3, the alcohol is primary (1°), secondary (2°) or tertiary (3°) according to whether the carbon is joined to one, two or three other carbons. Allylic alcohols have –OH on a carbon next to a C=C (CH2=CH–CH2OH) and benzylic alcohols have –OH on a carbon attached to a benzene ring (C6H5CH2OH); both may be 1°, 2° or 3°. If the –OH is on an sp2 carbon the compound is a vinylic alcohol (CH2=CH–OH, unstable) or a phenol.

Ethers are simple or symmetrical when both groups on oxygen are the same (C2H5OC2H5) and mixed or unsymmetrical when they differ (CH3OC2H5, anisole C6H5OCH3).

Exam tip Benzyl alcohol, C6H5CH2OH, is an alcohol (–OH on sp3 carbon), not a phenol. Calling it a phenol costs the 1 mark in classification questions.

02 7.2 Nomenclature

Alcohols. Replace the final -e of the alkane by -ol and number the chain so the –OH carbon gets the lowest locant: CH3CH(OH)CH3 is propan-2-ol, (CH3)3COH is 2-methylpropan-2-ol. With two or three –OH groups keep the -e: ethane-1,2-diol, propane-1,2,3-triol. A ring carrying –OH is a cycloalkanol; when the –OH is on a side chain the ring becomes a prefix, as in cyclohexylmethanol. Double bonds go before the -ol: prop-2-en-1-ol.

Phenols. The parent is phenol, numbered from the –OH carbon, or use ortho/meta/para: 2-methylphenol is o-cresol, 4-methylphenol is p-cresol. Benzene-1,2-diol is catechol, benzene-1,3-diol is resorcinol, benzene-1,4-diol is hydroquinone. 2-Hydroxybenzoic acid is salicylic acid and 2-hydroxybenzaldehyde is salicylaldehyde; both names recur in Kolbe and Reimer-Tiemann questions.

Ethers. Common names list both alkyl groups alphabetically: ethyl methyl ether. In IUPAC the ether is an alkoxyalkane: the larger group is the parent alkane and the smaller group with the oxygen is the alkoxy prefix. CH3OC2H5 is methoxyethane, CH3OCH(CH3)2 is 2-methoxypropane, C6H5OCH3 is methoxybenzene (anisole), C6H5OC2H5 is ethoxybenzene (phenetole), and C2H5OCH2CH(CH3)2 is 1-ethoxy-2-methylpropane.

Exam tip In an alkoxyalkane the parent chain is the larger carbon group. Write 1-ethoxy-2-methylpropane, with substituents in alphabetical order.

03 7.3 Structures of Functional Groups

In alcohols the oxygen is sp3 hybridised with two lone pairs, so C–O–H is bent. In methanol the C–O–H angle is 108.9°, slightly less than 109.5°, because the two lone pairs repel the bond pairs more strongly. The C–O bond length is 142 pm.

In phenol the C–O bond is shorter, 136 pm, for two reasons you must both state: the oxygen lone pair is conjugated with the ring, giving the bond partial double-bond character, and the sp2 carbon is more electronegative than an sp3 carbon so it holds the bond closer.

In ethers the oxygen is again sp3, but the C–O–C angle is larger than tetrahedral: 111.7° in dimethyl ether, with C–O bonds of 141 pm. The angle opens because the two bulky alkyl groups repel each other. Since the molecule is bent, the C–O dipoles do not cancel and ethers have a net dipole moment.

Two standard 1-mark questions come from here: why the C–O bond of phenol is shorter than in methanol, and why the C–O–C angle of an ether exceeds 109.5°.

Exam tip Give both reasons for the short C–O bond in phenol (conjugation and the sp2 carbon); one reason earns half the marks.

04 7.4.1 Preparation of Alcohols

From alkenes. (i) Acid-catalysed hydration: an alkene adds water with dilute H2SO4 following Markovnikov's rule, so propene gives propan-2-ol. Mechanism, three steps: (1) the alkene is protonated by H3O+ to give a carbocation (slow, rate determining); (2) water attacks the carbocation to give a protonated alcohol; (3) the protonated alcohol loses a proton to water, giving the alcohol and regenerating H3O+. (ii) Hydroboration-oxidation: diborane (BH3)2 adds to the alkene to give a trialkylborane, which is oxidised by H2O2 in aqueous NaOH to the alcohol. The result is anti-Markovnikov addition of water: propene gives propan-1-ol.

From carbonyl compounds. Aldehydes give primary and ketones give secondary alcohols with H2/Pd, NaBH4 or LiAlH4. Carboxylic acids are reduced to primary alcohols by LiAlH4; commercially the acid is first converted to an ester which is hydrogenated.

From Grignard reagents. RMgX adds to a C=O to give an adduct, which is hydrolysed to the alcohol. Methanal gives a primary alcohol, any other aldehyde a secondary alcohol, a ketone a tertiary alcohol. So CH3MgBr with propanone gives 2-methylpropan-2-ol; C2H5MgCl with methanal gives propan-1-ol.

CH3CH=CH2 → [(i) (BH3)2 (ii) H2O2/OH−] → CH3CH2CH2OH
(BH3)2diborane; anti-Markovnikov alcohol
RMgX + R′CHO → R′CH(OMgX)R → [H2O] → R′CH(OH)R + Mg(OH)X
RMgXGrignard reagent; methanal gives 1°, other aldehydes 2°, ketones 3°
Exam tip The hydration mechanism earns one mark per step: carbocation formation, attack by water, deprotonation. Show curved arrows and label the first step slow.

05 7.4.2 Preparation of Phenols

NCERT gives four routes, each of which has been asked as a 1- or 2-mark equation.

From haloarenes. Chlorobenzene is fused with NaOH at 623 K and 300 atm to give sodium phenoxide; acidification gives phenol. The harsh conditions are needed because the aryl C–Cl bond resists nucleophilic substitution.

From benzenesulphonic acid. Benzene is sulphonated with oleum; the acid is fused with molten NaOH to give sodium phenoxide, then acidified. This answers the NCERT question that gives you benzene, conc. H2SO4 and NaOH.

From diazonium salts. Aniline is diazotised with NaNO2/HCl at 273–278 K to benzenediazonium chloride; warming with water hydrolyses it to phenol with loss of N2.

From cumene (industrial). Cumene (isopropylbenzene, from benzene and propene by Friedel-Crafts alkylation) is oxidised by air to cumene hydroperoxide, C6H5C(CH3)2OOH, which with dilute acid gives phenol and propanone. Both products are valuable, which is why the process is preferred.

C6H5Cl + NaOH → [623 K, 300 atm] → C6H5ONa → [H+] → C6H5OH
623 K, 300 atmdrastic conditions; aryl C–Cl is unreactive
C6H5CH(CH3)2 → [O2] → C6H5C(CH3)2OOH → [H3O+] → C6H5OH + CH3COCH3
C6H5CH(CH3)2cumene; acetone is the by-product
Exam tip Name the intermediate cumene hydroperoxide and write acetone as the by-product; forgetting the by-product costs half a mark.

06 7.4.3 Physical Properties of Alcohols and Phenols

Boiling points. Alcohols and phenols boil much higher than hydrocarbons, ethers and haloalkanes of comparable molecular mass because of intermolecular hydrogen bonding through the polar O–H group. Propan-1-ol (molar mass 60, b.p. 370 K) boils far above butane (58, b.p. 273 K), and ethanol (351 K) above its isomer methoxymethane (249 K), which has no O–H. Within alcohols the boiling point rises with the number of carbons (larger van der Waals forces) and falls with branching (smaller surface area), so the order is propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol.

Solubility. Alcohols and phenols dissolve in water by hydrogen bonding with water. Solubility falls as the hydrophobic alkyl or aryl part grows: methanol, ethanol and propanol are miscible, higher alcohols are not, phenol is only sparingly soluble.

o- and p-Nitrophenol. In o-nitrophenol the –OH and –NO2 are close enough to form an intramolecular hydrogen bond, so the molecules are not associated with each other; it has a lower boiling point, lower water solubility and is steam volatile. p-Nitrophenol forms intermolecular hydrogen bonds and is not steam volatile. A mixture of the two is therefore separated by steam distillation, the ortho isomer distilling over.

Exam tip Always name the force: intermolecular hydrogen bonding for alcohols, van der Waals forces for chain length, intramolecular hydrogen bonding for o-nitrophenol.

07 7.4.4 Cleavage of the O–H Bond: Acidity and Esterification

Alcohols and phenols react with Na, K and Al to give alkoxides or phenoxides and hydrogen: 2C2H5OH + 2Na → 2C2H5ONa + H2. So they are Bronsted acids; they also act as Bronsted bases through the lone pairs on oxygen (protonated alcohols appear in many mechanisms). Phenol additionally reacts with aqueous NaOH to give sodium phenoxide; alcohols do not.

Acidity of alcohols. Order 1° > 2° > 3°, and all alcohols are weaker acids than water. An alkyl group is electron releasing (+I): it raises electron density on oxygen and destabilises the alkoxide ion by intensifying its negative charge. Water (pKa about 15.7) is a slightly stronger acid than ethanol (15.9), so an alkoxide is a stronger base than hydroxide.

Acidity of phenol. Phenol (pKa 10.0) is about a million times more acidic than ethanol. The oxygen is bonded to an electron-withdrawing sp2 carbon, which polarises the O–H bond, and the phenoxide ion is stabilised by resonance, the negative charge being delocalised over the ortho and para carbons; the resonance structures of phenol itself carry charge separation and are less stabilising. Ethoxide has no delocalisation. Order: phenol > water > ethanol.

Substituents. Electron-withdrawing groups at ortho or para stabilise the phenoxide and raise acidity: p-nitrophenol pKa 7.1, o-nitrophenol 7.2, m-nitrophenol 8.3 (weaker, no direct resonance with the O−). Electron-releasing groups (–CH3, –OCH3) lower acidity: cresols 10.1–10.2. Hence o-nitrophenol is more acidic than o-methoxyphenol.

Esterification. With carboxylic acids (conc. H2SO4, reversible, water removed), acid chlorides (in pyridine) or anhydrides, alcohols and phenols give esters. Acetylation of salicylic acid with ethanoic anhydride gives aspirin.

C6H5OH + NaOH → C6H5O−Na+ + H2O
C6H5O−Na+sodium phenoxide; alcohols do not react with NaOH
R–OH + R′COOH ⇌ [conc. H2SO4] ⇌ R′COOR + H2O
R′COORester; equilibrium shifted by removing water
Exam tip For 'compare acidity of phenol and ethanol' draw the resonance structures of the phenoxide ion (charge on O, two ortho and one para carbon) and state that ethoxide has none. The drawing is what the marking scheme looks for.

08 7.4.4 Cleavage of the C–O Bond in Alcohols

These reactions remove –OH as water. Only alcohols do this; phenols do not, because the C–O bond of phenol has partial double-bond character and the sp2 carbon cannot be attacked by a nucleophile.

With hydrogen halides. R–OH + HX → R–X + H2O. Tertiary alcohols react with conc. HCl at room temperature; primary and secondary need anhydrous ZnCl2. The Lucas test (conc. HCl + anhydrous ZnCl2) distinguishes the classes by the time taken for turbidity (the insoluble alkyl chloride): tertiary immediately, secondary within about five minutes, primary none at room temperature. Reactivity 3° > 2° > 1°, following carbocation stability. Bromides are made with NaBr/H2SO4 or PBr3.

With PCl5 and SOCl2. R–OH + PCl5 → R–Cl + POCl3 + HCl; R–OH + SOCl2 → R–Cl + SO2 + HCl. Thionyl chloride is preferred because both by-products are gases.

Dehydration. With conc. H2SO4, H3PO4 or hot alumina, alcohols lose water to give alkenes. Ethanol gives ethene with conc. H2SO4 at 443 K; propan-2-ol needs 85% H3PO4 at 440 K and 2-methylpropan-2-ol only 20% H3PO4 at 358 K, so ease is 3° > 2° > 1°. Mechanism: (1) protonation of ethanol to the protonated alcohol; (2) loss of water to give the ethyl carbocation (slow, rate determining); (3) loss of a proton from the adjacent carbon to give ethene and regenerate the acid. Where two alkenes are possible the more substituted one is major (Saytzeff rule).

R–OH + HCl → [anhyd. ZnCl2] → R–Cl + H2O
conc. HCl + anhyd. ZnCl2Lucas reagent; turbidity time distinguishes 1°, 2°, 3°
CH3CH2OH → [conc. H2SO4, 443 K] → CH2=CH2 + H2O
443 Kat 413 K the same reagents give diethyl ether
Exam tip Dehydration and hydration are the reverse of each other. Learn one thoroughly and write the carbocation step as the slow step in both.

09 7.4.4 Oxidation of Alcohols

Oxidation removes hydrogen from the carbon carrying –OH and from the –OH itself, so it is also called dehydrogenation. The product depends on the class of alcohol and the strength of the oxidising agent, and the examiner tests whether you pick the right reagent.

Primary alcohols. Acidified KMnO4 or acidified K2Cr2O7 takes a primary alcohol through the aldehyde to the carboxylic acid. To stop at the aldehyde use CrO3 in anhydrous medium or, better, PCC (pyridinium chlorochromate, a complex of chromium trioxide with pyridine and HCl). Propan-1-ol gives propanoic acid with alkaline KMnO4 and propanal with PCC.

Secondary alcohols give ketones with CrO3 or acidified dichromate: propan-2-ol → propanone.

Tertiary alcohols do not oxidise under normal conditions because the carbinol carbon has no hydrogen. Under vigorous conditions C–C bonds break, giving a mixture of carboxylic acids with fewer carbons.

Copper at 573 K. Alcohol vapour over heated copper gives an aldehyde from a primary alcohol, a ketone from a secondary alcohol, and an alkene (dehydration) from a tertiary alcohol; a good test to distinguish the three.

RCH2OH → [PCC] → RCHO; RCH2OH → [KMnO4/H+] → RCOOH
PCCmild, stops at the aldehyde; KMnO4 goes to the acid
(CH3)3C–OH → [Cu, 573 K] → (CH3)2C=CH2 + H2O
Cu, 573 K1° gives aldehyde, 2° gives ketone, 3° dehydrates
Exam tip Benzyl alcohol to benzoic acid needs acidified KMnO4; to benzaldehyde it needs PCC. Swapping these is the commonest loss in the reagent question.

10 7.4.4 Reactions of Phenols: Electrophilic Substitution, Kolbe, Reimer-Tiemann

The –OH group is strongly activating and ortho/para directing: the oxygen lone pair is delocalised into the ring and the resonance structures place negative charge at the ortho and para carbons, so electrophiles attack there under mild conditions.

Nitration. Dilute HNO3 at 298 K gives o- and p-nitrophenol, separated by steam distillation. Conc. HNO3 gives 2,4,6-trinitrophenol, picric acid; because the yield is poor, picric acid is made by sulphonating phenol to phenol-2,4-disulphonic acid and then treating with conc. HNO3.

Halogenation. In CS2 at 273 K, where phenol is not ionised, bromine gives o- and p-bromophenol (para major). In water phenol ionises to the strongly activated phenoxide, and bromine water gives a white precipitate of 2,4,6-tribromophenol at once.

Kolbe's reaction. Sodium phenoxide (a stronger nucleophile than phenol) with CO2 under pressure, then acidification, gives 2-hydroxybenzoic acid, salicylic acid.

Reimer-Tiemann reaction. Phenol heated with chloroform and aqueous NaOH (about 340 K): dichlorocarbene (:CCl2) from CHCl3 attacks the ortho position, the –CHCl2 group is hydrolysed by alkali, and acidification gives 2-hydroxybenzaldehyde, salicylaldehyde. With CCl4 the product is salicylic acid.

Zinc dust. Phenol distilled with zinc dust is reduced to benzene: C6H5OH + Zn → C6H6 + ZnO. Oxidation. Chromic acid oxidises phenol to benzoquinone; in air phenol slowly darkens.

C6H5OH + 3Br2 (aq) → 2,4,6-tribromophenol (white ppt) + 3HBr
Br2 (aq)bromine water; in CS2 at 273 K only monobromination
C6H5ONa + CO2 → [pressure, then H+] → 2-HO–C6H4–COOH
CO2Kolbe's reaction; salicylic acid
Exam tip Kolbe gives the acid (from CO2); Reimer-Tiemann gives the aldehyde (from CHCl3). Both substitute ortho. Write the sodium salt intermediate and the acidification step.

11 7.5 Some Commercially Important Alcohols

Methanol, CH3OH, is called wood spirit because it was once made by destructive distillation of wood. Today it is made from water gas over a ZnO–Cr2O3 catalyst at 573–673 K and 200–300 atm: CO + 2H2 → CH3OH. It is a colourless liquid, b.p. 337 K, and highly poisonous: small quantities cause blindness and large quantities death. It is used as a solvent in paints and varnishes and to make methanal.

Ethanol, C2H5OH, is made by fermentation of the sugars in molasses (the liquor left after crystallising cane sugar) or from starch. Yeast supplies two enzymes: invertase hydrolyses sucrose to glucose and fructose, and zymase converts these to ethanol and carbon dioxide: C6H12O6 → 2C2H5OH + 2CO2. Fermentation is carried out in the absence of air, because air oxidises the ethanol to ethanoic acid and the wine turns sour; once the alcohol content reaches about 14 per cent the yeast is killed and fermentation stops.

Ethanol (b.p. 351 K) is a solvent in paints and medicine. Commercial alcohol is made unfit for drinking by adding copper sulphate (colour) and pyridine (foul smell), usually with methanol: this is denatured alcohol or methylated spirit. Ethanol is a mild intoxicant; drinking methylated spirit causes blindness and death.

CO + 2H2 → [ZnO–Cr2O3, 573–673 K, 200–300 atm] → CH3OH
ZnO–Cr2O3catalyst for methanol from water gas
Exam tip Name the enzymes correctly: invertase for sucrose to glucose and fructose, zymase for glucose to ethanol, and say fermentation is anaerobic.

12 7.6.1 Preparation of Ethers

Dehydration of alcohols. With conc. H2SO4 at 413 K ethanol loses water between two molecules to give ethoxyethane; at 443 K the same reagents give ethene. The reaction is SN2: one ethanol is protonated, a second ethanol attacks its carbon and displaces water, and the protonated ether loses a proton. It works for primary alcohols only; secondary and tertiary alcohols eliminate instead, because the hindered carbon cannot be attacked from the back and the stable carbocation loses a proton to give an alkene. The ethers formed are symmetrical.

Williamson synthesis. An alkyl halide is heated with a sodium alkoxide: R–X + R′O−Na+ → R–O–R′ + NaX. The alkoxide displaces the halide by SN2. Symmetrical and unsymmetrical ethers can be made, including aryl alkyl ethers from sodium phenoxide and CH3Br (the halide must be the alkyl partner, since aryl halides do not undergo SN2).

The limitation is the halide. It should be primary; a tertiary halide gives an alkene because the alkoxide acts as a base and eliminates HX. To make tert-butyl ethyl ether use sodium tert-butoxide and ethyl bromide; sodium ethoxide with tert-butyl bromide gives 2-methylpropene. Likewise 2-methoxy-2-methylpropane comes from sodium tert-butoxide and methyl bromide.

2CH3CH2OH → [conc. H2SO4, 413 K] → C2H5OC2H5 + H2O
413 KSN2; only for 1° alcohols
R–X + R′–O−Na+ → R–O–R′ + Na+X−
R–Xprimary alkyl halide (SN2); tertiary halide gives alkene
Exam tip Put the bulky or tertiary group on the alkoxide and the small primary group on the halide, and write 'SN2' explicitly; it is a marking point.

13 7.6.2 Physical Properties of Ethers

Ethers are colourless, pleasant-smelling liquids (dimethyl and ethyl methyl ether are gases). Because C–O–C is bent, the two C–O dipoles do not cancel and ethers have a small net dipole moment.

Boiling points. Ethers have no O–H, so they cannot hydrogen-bond with each other. Their boiling points are close to those of alkanes of similar molecular mass and far below the isomeric alcohols: methoxymethane (molar mass 46) boils at 249 K while ethanol boils at 351 K; ethoxyethane (74) boils at 308 K while butan-1-ol boils at 391 K. Diethyl ether is volatile and inflammable and is used as a low-boiling solvent.

Solubility. The ether oxygen can accept a hydrogen bond from water, so lower ethers are slightly soluble: the miscibility of ethoxyethane with water resembles that of butan-1-ol. Solubility falls as the alkyl groups grow. Ethers are good solvents for organic compounds and for Grignard reagents because they are unreactive and aprotic.

The two comparisons the examiner expects: alcohols boil higher than isomeric ethers because of intermolecular hydrogen bonding; ethers dissolve in water about as much as the isomeric alcohol because the ether oxygen still hydrogen-bonds with water.

Exam tip Do not write that ethers are insoluble in water. NCERT compares the miscibility of ethoxyethane with that of butan-1-ol.

14 7.6.3 Chemical Reactions of Ethers

Cleavage of the C–O bond. Ethers are the least reactive of the three classes; the C–O bond breaks only with excess HI or HBr under drastic conditions (HI most effective). R–O–R + HX → R–X + R–OH, and with excess HX the alcohol is also converted: R–O–R + 2HX → 2R–X + H2O. Mechanism: (1) the ether oxygen is protonated to an oxonium ion; (2) the halide ion attacks the less hindered alkyl carbon by SN2 and displaces the alcohol. In a mixed ether the halide goes on the smaller group: CH3OC2H5 + HI → CH3I + C2H5OH. Two exceptions carry marks. With a tertiary group the path is SN1: the protonated ether loses the stable tertiary carbocation, so (CH3)3C–O–CH3 + HI gives (CH3)3C–I and CH3OH. With an aryl group the halide attacks the alkyl carbon only, because the aryl C–O bond has partial double-bond character and an sp2 carbon can neither undergo SN2 nor form a phenyl cation: anisole + HI → phenol + CH3I, never iodobenzene.

Electrophilic substitution in anisole. The –OCH3 group is activating and ortho/para directing. (i) Bromine in ethanoic acid gives mainly p-bromoanisole, with no Lewis acid needed. (ii) CH3Cl with anhydrous AlCl3 gives 2- and 4-methoxytoluene. (iii) CH3COCl with anhydrous AlCl3 gives 2- and 4-methoxyacetophenone (para major). (iv) Conc. HNO3 with conc. H2SO4 gives 2- and 4-nitroanisole.

R–O–R′ + HI → R–I + R′–OH (I− attacks the smaller group); excess HI → R–I + R′–I + H2O
HIHI > HBr > HCl; SN2 unless the group is tertiary (SN1)
C6H5OCH3 + HI → C6H5OH + CH3I
C6H5OHthe aryl–O bond survives; iodobenzene is never formed
Exam tip Three HI products to memorise: methoxyethane gives CH3I and ethanol; tert-butyl methyl ether gives tert-butyl iodide and methanol; anisole gives phenol and CH3I.

VocabularyKey terms

Monohydric, dihydric, trihydric alcoholAlcohols containing one, two or three hydroxyl groups, e.g. ethanol, ethane-1,2-diol and propane-1,2,3-triol.
Primary, secondary, tertiary alcoholClassification by the number of carbons (one, two or three) attached to the sp3 carbon carrying the –OH group.
Allylic and benzylic alcoholAlcohols with –OH on an sp3 carbon adjacent to a C=C (prop-2-en-1-ol) or to a benzene ring (benzyl alcohol).
PhenolA compound in which –OH is attached directly to an sp2 carbon of an aromatic ring; C6H5OH is the simplest.
Symmetrical and unsymmetrical etherR–O–R′ is symmetrical (simple) when R = R′, e.g. diethyl ether, and unsymmetrical (mixed) when they differ, e.g. anisole.
Hydroboration-oxidationAddition of diborane to an alkene followed by oxidation with H2O2/OH−, giving the anti-Markovnikov alcohol, e.g. propene to propan-1-ol.
CumeneIsopropylbenzene; its aerial oxidation to cumene hydroperoxide followed by acid treatment is the industrial route to phenol and acetone.
Lucas reagentConc. HCl with anhydrous ZnCl2; distinguishes alcohols by how quickly turbidity appears (3° immediate, 2° in about five minutes, 1° none at room temperature).
PCCPyridinium chlorochromate, a complex of chromium trioxide with pyridine and HCl; a mild oxidant that takes primary alcohols to aldehydes without going on to acids.
Kolbe's reactionSodium phenoxide with carbon dioxide under pressure, then acidification, gives 2-hydroxybenzoic acid (salicylic acid).
Reimer-Tiemann reactionPhenol heated with chloroform and aqueous NaOH gives, after acidification, 2-hydroxybenzaldehyde (salicylaldehyde); the electrophile is dichlorocarbene.
Williamson synthesisPreparation of an ether by the SN2 reaction of a primary alkyl halide with a sodium alkoxide or phenoxide.
Picric acid2,4,6-Trinitrophenol, formed from phenol and concentrated nitric acid; strongly acidic because of three electron-withdrawing nitro groups.
Denatured alcoholCommercial ethanol made unfit for drinking by adding copper sulphate and pyridine, usually with methanol; also called methylated spirit.

Keep this on one pageFormula sheet

NameExpressionNotes
Acid-catalysed hydration of alkeneR–CH=CH2 + H2O → [dil. H2SO4] → R–CH(OH)–CH3Markovnikov; carbocation (slow) → attack by water → deprotonation
Hydroboration-oxidationR–CH=CH2 → [(i) (BH3)2 (ii) H2O2, OH−] → R–CH2–CH2OHanti-Markovnikov primary alcohol
Grignard on carbonylR–MgX + C=O → adduct → [H2O] → alcohol + Mg(OH)XHCHO gives 1°, other RCHO gives 2°, ketone gives 3°
Cumene processC6H5CH(CH3)2 → [O2] → C6H5C(CH3)2OOH → [H+] → C6H5OH + CH3COCH3industrial phenol; acetone by-product
Dehydration of ethanolC2H5OH → [conc. H2SO4, 443 K] → CH2=CH2; at 413 K → C2H5OC2H5ease 3° > 2° > 1°; carbocation is the slow step
Oxidation of alcoholsRCH2OH → [PCC] → RCHO; → [KMnO4/H+] → RCOOH; R2CHOH → [CrO3] → R2CO3° resist oxidation; Cu at 573 K dehydrogenates 1° and 2°, dehydrates 3°
Lucas testR–OH + HCl → [anhyd. ZnCl2] → R–Cl + H2Oturbidity: 3° at once, 2° in ~5 min, 1° none at room temperature
Kolbe's reactionC6H5ONa + CO2 → [pressure, H+] → 2-HO–C6H4–COOHsalicylic acid; acetylation with (CH3CO)2O gives aspirin
Reimer-Tiemann reactionC6H5OH + CHCl3 + 4NaOH → [340 K] → 2-NaO–C6H4–CHO + 3NaCl + 3H2O → [H+] → 2-HO–C6H4–CHOsalicylaldehyde via dichlorocarbene; CCl4 gives salicylic acid
Williamson synthesisR–X + R′O−Na+ → R–O–R′ + NaXSN2; primary halide; tertiary halide gives alkene
HI cleavage of ethersR–O–R′ + HI → R–I + R′OH (smaller R takes I); C6H5OCH3 + HI → C6H5OH + CH3ISN2 on the less hindered carbon; SN1 when a group is tertiary
pK<sub>a</sub> values (NCERT)p-nitrophenol 7.1, o-nitrophenol 7.2, m-nitrophenol 8.3, phenol 10.0, m-cresol 10.1, o- and p-cresol 10.2, ethanol 15.9lower pKa means stronger acid; phenol > water > ethanol

Practise with a pencilDiagrams that earn marks

Mechanism of acid-catalysed dehydration of ethanol to ethene

Three steps in a row. Step 1: CH3CH2–OH with a curved arrow from an oxygen lone pair to H+, giving CH3CH2–O+H2 (protonated alcohol, fast). Step 2: a curved arrow from the C–O bond onto oxygen, releasing H2O and leaving CH3CH2+; label 'slow, rate determining'. Step 3: a curved arrow from a C–H bond of the CH3 onto C+ forming CH2=CH2, the proton taken by HSO4− to regenerate H2SO4. Name each intermediate.

Resonance structures of phenol and the phenoxide ion

Two rows. Top, phenol: the Kekulé structure, then three structures with O+H on oxygen and a negative charge at ortho, para and the other ortho carbon (charge separation). Bottom, phenoxide: O− on the ring, then three structures with C=O and the negative charge at ortho, para, ortho. Note under the drawing that the phenoxide has no charge separation, so phenol loses a proton readily, and that the ortho/para charge explains why –OH directs electrophiles there.

Cumene route to phenol

A three-box flow. Box 1: benzene + CH3CH=CH2 with anhydrous AlCl3 giving cumene, C6H5CH(CH3)2. Box 2: air (O2) giving cumene hydroperoxide with the O–O–H shown explicitly. Box 3: dilute H2SO4 giving phenol and propanone side by side. Put the reagent above each arrow and label 'by-product: acetone'.

TextbookNCERT questions the board keeps asking

7.3 (i) Write the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. (ii) Classify the isomers as primary, secondary and tertiary alcohols.3 m

Eight isomers. Primary: CH3CH2CH2CH2CH2OH, pentan-1-ol; CH3CH2CH(CH3)CH2OH, 2-methylbutan-1-ol; (CH3)2CHCH2CH2OH, 3-methylbutan-1-ol; (CH3)3CCH2OH, 2,2-dimethylpropan-1-ol. Secondary: CH3CH2CH2CH(OH)CH3, pentan-2-ol; CH3CH2CH(OH)CH2CH3, pentan-3-ol; (CH3)2CHCH(OH)CH3, 3-methylbutan-2-ol. Tertiary: CH3CH2C(CH3)2OH, 2-methylbutan-2-ol.

7.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.2 m

Diborane adds across the C=C of an alkene (boron to the carbon with more hydrogens) to give a trialkylborane, which is oxidised by H2O2 in aqueous NaOH to the alcohol. The net result is anti-Markovnikov addition of water. Example: 6CH3CH=CH2 + (BH3)2 → 2(CH3CH2CH2)3B; (CH3CH2CH2)3B + 3H2O2 → [OH−] → 3CH3CH2CH2OH + H3BO3. Propene gives propan-1-ol, whereas acid-catalysed hydration gives propan-2-ol.

7.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.2 m

o-Nitrophenol. Its –OH and –NO2 groups form an intramolecular hydrogen bond, so the molecules are not associated with each other and it has a low boiling point and low water solubility. In p-nitrophenol the groups are far apart; intermolecular hydrogen bonds associate the molecules, so it is less volatile and stays behind.

7.9 Give the equations of reactions for the preparation of phenol from cumene.2 m

Aerial oxidation: C6H5CH(CH3)2 + O2 → C6H5C(CH3)2–O–O–H (cumene hydroperoxide). Acid treatment: C6H5C(CH3)2OOH → [dil. H2SO4] → C6H5OH + CH3COCH3. Cumene itself comes from benzene and propene by Friedel-Crafts alkylation.

7.11 Write the mechanism of hydration of ethene to yield ethanol.3 m

Step 1 (slow): protonation of ethene by H3O+ gives the ethyl carbocation: CH2=CH2 + H3O+ → CH3CH2+ + H2O.

Step 2: water attacks the carbocation: CH3CH2+ + H2O → CH3CH2–O+H2.

Step 3: another water molecule removes the proton: CH3CH2–O+H2 + H2O → CH3CH2OH + H3O+. Show curved arrows in each step.

7.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.2 m

(i) Sulphonation: C6H6 + H2SO4 (oleum) → C6H5SO3H + H2O. (ii) Neutralisation and alkali fusion: C6H5SO3H + NaOH → C6H5SO3Na + H2O; C6H5SO3Na + 2NaOH → [fuse] → C6H5ONa + Na2SO3 + H2O. (iii) Acidification: C6H5ONa + H+ → C6H5OH + Na+.

7.14 Give two reactions that show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.3 m

(i) 2C6H5OH + 2Na → 2C6H5ONa + H2. (ii) C6H5OH + NaOH → C6H5ONa + H2O.

Phenol (pKa 10.0) is far more acidic than ethanol (15.9). In phenol the oxygen is on an electron-withdrawing sp2 carbon, and the phenoxide ion is resonance stabilised with the negative charge spread over the ortho and para carbons. In ethanol the electron-releasing ethyl group intensifies the charge on the ethoxide oxygen and there is no resonance. Ethanol does not react with NaOH.

7.15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?2 m

The –NO2 group is electron withdrawing (−I, −R): it weakens the O–H bond and disperses the negative charge of the nitrophenoxide ion by resonance, stabilising it. The –OCH3 group is electron releasing (+R): it pushes electron density onto the phenoxide oxygen and destabilises the anion. A more stable conjugate base means a stronger acid, so o-nitrophenol (pKa 7.2) is far more acidic than o-methoxyphenol (about 10).

7.17 Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO4 solution. (ii) Bromine in CS2 with phenol. (iii) Dilute HNO3 with phenol. (iv) Treating phenol with chloroform in presence of aqueous NaOH.4 m

(i) CH3CH2CH2OH → [alk. KMnO4, then H+] → CH3CH2COOH. (ii) C6H5OH + Br2 → [CS2, 273 K] → o-bromophenol (minor) + p-bromophenol (major) + HBr. (iii) C6H5OH + dil. HNO3 → [298 K] → o-nitrophenol + p-nitrophenol + H2O. (iv) C6H5OH + CHCl3 + 4NaOH → [340 K] → sodium 2-formylphenoxide + 3NaCl + 3H2O; acidification gives salicylaldehyde (Reimer-Tiemann reaction).

7.19 Write the mechanism of acid dehydration of ethanol to yield ethene.3 m

Step 1 (fast): CH3CH2OH + H+ → CH3CH2–O+H2 (protonated alcohol).

Step 2 (slow, rate determining): CH3CH2–O+H2 → CH3CH2+ + H2O (ethyl carbocation).

Step 3: CH3CH2+ → CH2=CH2 + H+; the proton is taken by HSO4−, regenerating H2SO4. Conditions: conc. H2SO4, 443 K.

7.21 Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol to 2,4,6-tribromophenol. (iv) Benzyl alcohol to benzoic acid. (v) Dehydration of propan-2-ol to propene. (vi) Butan-2-one to butan-2-ol.3 m

(i) Acidified KMnO4 or acidified K2Cr2O7. (ii) PCC or CrO3 in anhydrous medium. (iii) Bromine water. (iv) Acidified KMnO4. (v) 85% H3PO4 at 440 K (or conc. H2SO4). (vi) NaBH4 or LiAlH4.

7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.3 m

The reaction is SN2, so the halide must be primary. (i) A tertiary halide gives an alkene by elimination, the alkoxide acting as a base: (CH3)3C–Br + C2H5ONa → (CH3)2C=CH2 + C2H5OH + NaBr. tert-Butyl ethyl ether must be made from (CH3)3CONa and C2H5Br. (ii) Aryl halides cannot be the halide partner because they do not undergo SN2; anisole is made from sodium phenoxide and CH3Br, and diaryl ethers cannot be prepared this way.

7.28 Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene (iii) benzyl ethyl ether.3 m

(i) CH3CH2CH2OCH2CH2CH3 + HI → CH3CH2CH2I + CH3CH2CH2OH. (ii) C6H5OCH3 + HI → C6H5OH + CH3I; the aryl C–O bond is not cleaved. (iii) C6H5CH2OC2H5 + HI → C6H5CH2I + C2H5OH; the resonance-stabilised benzyl carbocation forms (SN1), so iodide goes to the benzyl group.

7.33 When 3-methylbutan-2-ol is treated with HBr, the product is 2-bromo-2-methylbutane. Give a mechanism for this reaction.3 m

(1) Protonation: (CH3)2CH–CH(OH)–CH3 + HBr → (CH3)2CH–CH(O+H2)–CH3 + Br−. (2) Loss of water gives the secondary carbocation (CH3)2CH–C+H–CH3. (3) A 1,2-hydride shift from C-3 to C-2 gives the more stable tertiary carbocation (CH3)2C+–CH2CH3. (4) Br− attacks: (CH3)2C(Br)–CH2CH3, 2-bromo-2-methylbutane. State the rearrangement explicitly; it is the key marking point.

Previous years and frequently askedBoard questions with model answers

MCQ The correct order of acid strength is:
  1. C6H5OH > H2O > ROH
  2. C6H5OH > ROH > H2O
  3. ROH > C6H5OH > H2O
  4. H2O > C6H5OH > ROH
Frequently asked1 m
Answer: (a)
Phenoxide is resonance stabilised, so phenol is the strongest acid. Alcohols are weaker than water because the +I alkyl group destabilises the alkoxide.
MCQ Identify the correct increasing order of boiling points:
  1. Propan-1-ol < Butan-1-ol < Butan-2-ol < Pentan-1-ol
  2. Pentan-1-ol < Butan-1-ol < Butan-2-ol < Propan-1-ol
  3. Propan-1-ol < Butan-2-ol < Butan-1-ol < Pentan-1-ol
  4. Butan-1-ol < Butan-2-ol < Propan-1-ol < Pentan-1-ol
Frequently asked1 m
Answer: (c)
Boiling point rises with carbon number and falls with branching, so butan-2-ol boils below butan-1-ol but both lie between propan-1-ol and pentan-1-ol.
MCQ Lucas reagent produces cloudiness immediately with:
  1. 2-Methylpropan-1-ol
  2. Butan-2-ol
  3. 2-Methylpropan-2-ol
  4. Butan-1-ol
Frequently asked1 m
Answer: (c)
Only a tertiary alcohol reacts with conc. HCl/ZnCl2 at once, through a stable tertiary carbocation. Butan-2-ol takes about five minutes; the primary alcohols show no turbidity at room temperature.
MCQ Phenol can be converted to salicylic acid by:
  1. Reimer-Tiemann reaction
  2. Friedel-Crafts reaction
  3. Kolbe's reaction
  4. Coupling reaction
Frequently asked1 m
Answer: (c)
Kolbe's reaction: sodium phenoxide with CO2 under pressure, then acidification, gives 2-hydroxybenzoic acid. Reimer-Tiemann gives salicylaldehyde.
MCQ Williamson synthesis of ethers is an example of:
  1. SN1 reaction
  2. SN2 reaction
  3. Electrophilic addition
  4. Elimination reaction
Frequently asked1 m
Answer: (b)
The alkoxide ion attacks the primary alkyl halide from the back side and displaces the halide in one bimolecular step.
MCQ The intermediate formed in the slowest step of the acid-catalysed dehydration of an alcohol is a:
  1. Protonated alcohol
  2. Carbanion
  3. Free radical
  4. Carbocation
Frequently asked1 m
Answer: (d)
The rate-determining step is loss of water from the protonated alcohol to form a carbocation; the protonated alcohol forms in the fast first step.
Assertion–Reason Assertion (A): Phenol is less acidic than 4-methylphenol. Reason (R): Electron-releasing groups decrease the acidity of phenol.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (d)
The methyl group is electron releasing and destabilises the phenoxide ion, so 4-methylphenol (pKa 10.2) is weaker than phenol (10.0). A is false, R is true.
Assertion–Reason Assertion (A): A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation. Reason (R): o-Nitrophenol is steam volatile because of intermolecular hydrogen bonding.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (c)
The ortho isomer is steam volatile because of intramolecular hydrogen bonding, which prevents association between molecules. The reason as stated is wrong.
Assertion–Reason Assertion (A): The –OH group in phenol directs the incoming group to the meta position of the ring. Reason (R): The –OH group activates the aromatic ring towards electrophilic substitution.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (d)
The oxygen lone pair raises electron density at the ortho and para positions, so –OH is ortho/para directing and activating. A is false; R is true.
Very short answer Give reasons: (i) o-Nitrophenol is more acidic than o-methoxyphenol. (ii) Phenols do not undergo substitution of the –OH group like alcohols.Frequently asked2 m

(i) –NO2 is electron withdrawing: it weakens the O–H bond and stabilises the nitrophenoxide ion by dispersing its charge. –OCH3 is electron releasing: it intensifies the charge on the methoxyphenoxide ion and destabilises it. A more stable conjugate base means a stronger acid.

(ii) The C–O bond of phenol has partial double-bond character (resonance with the ring) and the carbon is sp2. Neither SN1 (unstable phenyl cation) nor SN2 (no back-side attack on a ring carbon) is possible, so the –OH is not replaced.

Very short answer How will you distinguish between butan-1-ol and 2-methylpropan-2-ol using Lucas reagent? Write the reaction involved.Frequently asked2 m

Add Lucas reagent (conc. HCl + anhydrous ZnCl2) to each at room temperature. 2-Methylpropan-2-ol, a tertiary alcohol, gives turbidity immediately as insoluble tert-butyl chloride separates: (CH3)3C–OH + HCl → [ZnCl2] → (CH3)3C–Cl + H2O. Butan-1-ol, a primary alcohol, gives no turbidity at room temperature because the primary carbocation does not form readily.

Very short answer Write the structures of the major products: (i) Anisole is treated with HI. (ii) Phenol is treated with conc. HNO3.Frequently asked2 m

(i) C6H5OCH3 + HI → C6H5OH (phenol) + CH3I; iodide attacks the methyl carbon because the aryl C–O bond is not cleaved. (ii) C6H5OH + 3HNO3 (conc.) → 2,4,6-trinitrophenol (picric acid) + 3H2O; draw –OH at C-1 and –NO2 at C-2, C-4 and C-6.

Very short answer (i) Why is the C–O–C bond angle in ethers slightly greater than the tetrahedral angle? (ii) Why is the C–O bond length in phenol shorter than that in methanol?Frequently asked2 m

(i) The two bulky alkyl groups on the sp3 oxygen repel each other, opening the angle to 111.7° in dimethyl ether. (ii) In phenol the oxygen lone pair is delocalised into the ring, giving the C–O bond partial double-bond character, and the oxygen is bonded to a more electronegative sp2 carbon; the bond is 136 pm against 142 pm in methanol.

Short answer Write the chemical equations for the following: (a) Reimer-Tiemann reaction (b) Kolbe's reaction (c) Friedel-Crafts acylation of anisole.Frequently asked3 m

(a) C6H5OH + CHCl3 + 4NaOH → [340 K] → sodium 2-formylphenoxide + 3NaCl + 3H2O; then H+ gives 2-hydroxybenzaldehyde (salicylaldehyde).

(b) C6H5OH + NaOH → C6H5ONa; C6H5ONa + CO2 → [pressure] → sodium salicylate → [H+] → 2-hydroxybenzoic acid (salicylic acid).

(c) C6H5OCH3 + CH3COCl → [anhyd. AlCl3] → 4-methoxyacetophenone (major) + 2-methoxyacetophenone (minor) + HCl.

Short answer (a) Name the reagents used to convert a primary alcohol to (i) an aldehyde (ii) a carboxylic acid. (b) Why do tertiary alcohols resist oxidation under normal conditions? (c) What happens when the vapours of 2-methylpropan-2-ol are passed over heated copper at 573 K?Frequently asked3 m

(a) (i) PCC or CrO3 in anhydrous medium. (ii) Acidified KMnO4 or acidified K2Cr2O7.

(b) Oxidation removes a hydrogen from the carbon bearing –OH along with the –OH hydrogen. A tertiary alcohol has no hydrogen on that carbon, so oxidation would need a C–C bond to break, which requires vigorous conditions.

(c) Dehydration, not dehydrogenation: (CH3)3C–OH → [Cu, 573 K] → (CH3)2C=CH2 + H2O.

Short answer Write the mechanism of the acid-catalysed hydration of an alkene to form an alcohol.Frequently asked3 m

Take propene with dilute H2SO4. Step 1 (slow): protonation gives the more stable secondary carbocation (Markovnikov): CH3CH=CH2 + H3O+ → CH3C+HCH3 + H2O. Step 2: water attacks the carbocation: CH3C+HCH3 + H2O → CH3CH(O+H2)CH3. Step 3: deprotonation by water gives propan-2-ol and regenerates H3O+, so the acid is a catalyst.

Short answer What happens when: (i) cumene hydroperoxide is treated with dilute acid, (ii) phenol is distilled with zinc dust, (iii) ethanol is heated with conc. H2SO4 at 413 K? Write the equations.Frequently asked3 m

(i) C6H5C(CH3)2OOH → [dil. H2SO4] → C6H5OH + CH3COCH3; phenol and propanone (cumene process). (ii) C6H5OH + Zn → C6H6 + ZnO; phenol is reduced to benzene. (iii) 2C2H5OH → [conc. H2SO4, 413 K] → C2H5OC2H5 + H2O; diethyl ether by intermolecular dehydration (at 443 K ethene forms instead).

Long answer An organic compound A (C2H6O) reacts with sodium metal to give B and hydrogen gas. A reacts with iodine and NaOH to give a yellow precipitate C. When A is heated with conc. H2SO4 at 413 K it gives D (C4H10O). D reacts with excess HI to give E. Identify A, B, C, D and E and write the reactions involved.Frequently asked5 m

A is ethanol, CH3CH2OH: formula C2H6O, reacts with Na, and gives the iodoform test because it has a CH3CH(OH)– unit.

B: 2CH3CH2OH + 2Na → 2CH3CH2ONa (sodium ethoxide) + H2.

C: CH3CH2OH + 4I2 + 6NaOH → CHI3 (iodoform, yellow) + HCOONa + 5NaI + 5H2O.

D: 2CH3CH2OH → [conc. H2SO4, 413 K] → C2H5OC2H5 (ethoxyethane, C4H10O) + H2O.

E: C2H5OC2H5 + 2HI (excess) → 2C2H5I (iodoethane) + H2O. With one equivalent of HI the products would be iodoethane and ethanol; excess HI converts the ethanol too.

Long answer (a) Write the reagents used in the following conversions: (i) Phenol to 2,4,6-tribromophenol (ii) Propene to propan-1-ol (iii) Butan-2-one to butan-2-ol. (b) Write the mechanism of the acid-catalysed dehydration of ethanol to ethene.Frequently asked5 m

(a) (i) Bromine water. (ii) Diborane (BH3)2 followed by H2O2/NaOH (hydroboration-oxidation). (iii) NaBH4 or LiAlH4, or H2/Ni.

(b) Conc. H2SO4, 443 K. Step 1 (fast): CH3CH2OH + H+ → CH3CH2O+H2. Step 2 (slow, rate determining): CH3CH2O+H2 → CH3CH2+ + H2O. Step 3: HSO4− removes a β-hydrogen, CH3CH2+ → CH2=CH2 + H+, regenerating the acid. Draw curved arrows in each step.

Long answer Account for the following: (i) Phenol is more acidic than ethanol. (ii) Tertiary alcohols are dehydrated more easily than primary alcohols. (iii) Anisole with HI gives phenol and iodomethane, not iodobenzene and methanol. (iv) Sodium ethoxide with tert-butyl bromide does not give tert-butyl ethyl ether. (v) Alcohols have higher boiling points than ethers of comparable molecular mass.Frequently asked5 m

(i) The phenoxide ion is resonance stabilised with the charge delocalised over the ortho and para carbons; ethoxide has no delocalisation and is destabilised by the +I ethyl group. The sp2 carbon in phenol also polarises the O–H bond.

(ii) Dehydration goes through a carbocation in the slow step. A tertiary carbocation is more stable than a primary one, so it forms under milder conditions (20% H3PO4 at 358 K against conc. H2SO4 at 443 K).

(iii) After protonation, iodide attacks the methyl carbon by SN2. The aryl C–O bond has partial double-bond character, the ring carbon cannot be attacked from the back, and a phenyl cation is too unstable for SN1.

(iv) tert-Butyl bromide is tertiary; the ethoxide ion acts as a base and eliminates HBr to give 2-methylpropene. Williamson synthesis is SN2 and needs a primary halide, so use sodium tert-butoxide with ethyl bromide.

(v) Alcohols associate by intermolecular hydrogen bonding through O–H; ethers have no O–H and are held only by weak dipole and van der Waals forces. Ethanol boils at 351 K, methoxymethane at 249 K.

Case-based Read the passage and answer the questions. Ethers are prepared by dehydration of primary alcohols with conc. H2SO4 at 413 K or by Williamson synthesis, in which an alkyl halide reacts with a sodium alkoxide by an SN2 mechanism. The C–O bond of an ether is cleaved only under drastic conditions with excess HI or HBr; in aryl alkyl ethers the alkyl–oxygen bond breaks, and when one alkyl group is tertiary the cleavage follows an SN1 path.
(i) Complete the equation: CH3OCH2CH3 + HI → ?
(ii) Name the alkyl halide and the sodium alkoxide needed to prepare tert-butyl ethyl ether by Williamson synthesis.
(iii) Anisole with HI gives phenol and iodomethane. Justify.
(iv) Why is the C–O–C bond angle in ethers greater than the tetrahedral angle?
Frequently asked4 m

(i) CH3OCH2CH3 + HI → CH3I + CH3CH2OH; iodide attacks the smaller methyl carbon by SN2.

(ii) Bromoethane and sodium tert-butoxide, (CH3)3CONa. tert-Butyl bromide with sodium ethoxide would give 2-methylpropene.

(iii) The O–C6H5 bond has partial double-bond character from resonance with the ring; the sp2 ring carbon cannot undergo SN2 and a phenyl cation is unstable. So I− attacks the methyl carbon, giving CH3I and phenol.

(iv) The two bulky alkyl groups on the sp3 oxygen repel each other, opening the angle to 111.7° in dimethyl ether.

Case-based Read the passage and answer the questions. The –OH group of phenol is strongly activating and ortho/para directing in electrophilic substitution because the oxygen lone pair is delocalised into the ring. Phenol is therefore nitrated by dilute nitric acid at room temperature and brominated by bromine water without a catalyst. Phenol is also a weak acid whose acidity is altered by substituents on the ring.
(i) Write the products when phenol reacts with (a) Br2 in CS2 at 273 K (b) conc. HNO3.
(ii) Why does phenol not undergo protonation readily like alcohols?
(iii) Arrange in increasing order of acidity: phenol, p-cresol, p-nitrophenol.
(iv) Name the isomer of nitrophenol that is steam volatile.
Frequently asked4 m

(i) (a) o-Bromophenol and p-bromophenol (para major). (b) 2,4,6-Trinitrophenol (picric acid).

(ii) The lone pair on the phenolic oxygen is delocalised into the ring by resonance, so it is less available to bond to a proton; in alcohols the lone pairs are localised on oxygen.

(iii) p-Cresol < phenol < p-nitrophenol (pKa 10.2, 10.0, 7.1). –CH3 destabilises the phenoxide; –NO2 stabilises it.

(iv) o-Nitrophenol, because intramolecular hydrogen bonding prevents association between molecules.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Explaining phenol's acidity by 'oxygen is more electronegative'. The examiner wants the resonance stabilisation of the phenoxide ion (charge on ortho and para carbons) and the sp2 carbon argument; draw the resonance structures.
  • Mixing up Kolbe's reaction and the Reimer-Tiemann reaction. Kolbe uses CO2 and gives salicylic acid; Reimer-Tiemann uses CHCl3/NaOH and gives salicylaldehyde. Write the sodium salt intermediate and the acidification step.
  • Giving iodobenzene and methanol as the products of anisole with HI. The aryl C–O bond is never cleaved; the products are phenol and iodomethane.
  • Using a tertiary alkyl halide in Williamson synthesis. The reaction is SN2 and a tertiary halide gives an alkene; put the bulky group on the alkoxide.
  • Forgetting the temperature in the ethanol plus conc. H2SO4 reaction: 413 K gives diethyl ether, 443 K gives ethene.
  • Choosing acidified KMnO4 to convert a primary alcohol to an aldehyde. It goes on to the acid; the aldehyde needs PCC, CrO3 in anhydrous medium or Cu at 573 K.
  • Saying o-nitrophenol is steam volatile because of intermolecular hydrogen bonding. It is intramolecular bonding within the ortho isomer.
  • Skipping the 'slow step' label and curved arrows in mechanism questions; the carbocation-forming step must be marked rate determining.

How to prepareStrategy for this chapter

Divide the chapter into three blocks. Block one is alcohols: the preparation routes (hydration with its mechanism, hydroboration-oxidation, Grignard with the 1°/2°/3° rule), the physical property arguments, and the reactions at the O–H bond, at the C–O bond (HX and the Lucas test, SOCl2, dehydration with its mechanism) and oxidation with the right reagent for each product. Block two is phenol: the four preparations (cumene first), the acidity argument with the phenoxide resonance structures and the pKa table, and the six reactions of phenol. Block three is ethers: dehydration at 413 K, Williamson synthesis and its limitation, HI cleavage with its three cases, and the four electrophilic substitutions of anisole.

Memorise the numbers the board asks for: 108.9°, 111.7°, 136 pm and 142 pm, 413 K and 443 K, 623 K and 300 atm, 573 K for copper, and the pKa values. Write the two mechanisms (hydration and dehydration) and the HI mechanism from memory until each takes under three minutes; they are the 2- and 3-mark questions that recur most.

Practise the give-reason set: phenol versus ethanol acidity, o-nitrophenol versus o-methoxyphenol, ethanol versus methoxymethane boiling point, steam volatility, C–O bond length in phenol, C–O–C angle in ethers, why phenols do not undergo C–OH cleavage. Then solve two or three identify-A-to-E puzzles starting from ethanol or phenol, since the 5-mark question is usually of that type. In the exam give this chapter about 12 minutes and always write reagents and conditions above the arrow.

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