Alcohols and phenols: preparation and reactions

How are alcohols and phenols made, and how do they react?

In plain words: the O–H bond makes these compounds weak acids and lets them hydrogen bond, which raises boiling points and solubility. In alcohols the C–O bond can also break, so the –OH can be swapped for a halogen or removed with an H as water. In phenols the ring holds the C–O bond firmly, but the –OH makes the ring very reactive towards electrophiles.

Preparation of alcohols

1. From alkenes (NCERT pages 199–200)

• Acid catalysed hydration. Alkenes add water with an acid catalyst; with unsymmetrical alkenes the addition follows Markovnikov’s rule: CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃. NCERT’s three steps: (1) the alkene is protonated by H₃O⁺ to a carbocation; (2) water attacks the carbocation; (3) loss of a proton gives the alcohol.
• Hydroboration–oxidation. Diborane, (BH₃)₂, adds to the alkene to give a trialkylborane, which H₂O₂ in aqueous NaOH oxidises to the alcohol. Boron goes to the sp² carbon with more hydrogens, so the product looks like water added against Markovnikov’s rule; the yield is excellent. H. C. Brown first reported it in 1959 and shared the 1979 Nobel Prize in Chemistry with G. Wittig.

2. From carbonyl compounds (NCERT page 200)

• Aldehydes and ketones are reduced by H₂ over finely divided Pt, Pd or Ni, or by NaBH₄ or LiAlH₄: aldehydes give primary alcohols, ketones secondary alcohols.
• Carboxylic acids are reduced to primary alcohols in excellent yield by LiAlH₄, followed by water. LiAlH₄ is expensive, so commercially acids are first made into esters, which are reduced by H₂ with a catalyst.

3. From Grignard reagents (NCERT page 201). A Grignard reagent, RMgX (NCERT Unit 6 prepares it from a haloalkane and magnesium), adds to the C=O group; hydrolysis of the adduct gives the alcohol. Methanal gives a primary alcohol, other aldehydes secondary alcohols, and ketones tertiary alcohols. NCERT Example 7.2 (page 201): butanal on catalytic reduction gives butan-1-ol; propene with dilute H₂SO₄ gives propan-2-ol; propanone with CH₃MgBr and hydrolysis gives 2-methylpropan-2-ol.

Preparation of phenols

Phenol, also called carbolic acid, was first isolated from coal tar in the early nineteenth century; today it is made synthetically (NCERT page 201). Laboratory routes (NCERT page 202):

1. From haloarenes. Chlorobenzene fused with NaOH at 623 K and 320 atm gives sodium phenoxide, and acidification gives phenol.
2. From benzenesulphonic acid. Benzene is sulphonated with oleum; the benzenesulphonic acid, heated with molten NaOH, gives sodium phenoxide, and acidification gives phenol.
3. From diazonium salts. An aromatic primary amine with nitrous acid (NaNO₂ + HCl) at 273–278 K gives a diazonium salt, which is hydrolysed to phenol by warming with water or dilute acid.

• NCERT slip (pages 182 and 202): for chlorobenzene and NaOH at 623 K, the text of page 202 says 320 atmospheres, but the equation printed just below it says 300 atm, as Unit 6 does in both its text and its equation (page 182). Unit 6 also heats chlorobenzene in aqueous NaOH solution, where page 202 says “fused”. In an exam, use the pressure the question gives; otherwise write “NaOH, 623 K, 300 atm, then H⁺”, as both NCERT equations do.

Phenol from cumene

Most of the world’s phenol comes from cumene, isopropylbenzene (NCERT page 202):

• Cumene is oxidised by air to cumene hydroperoxide.
• Dilute acid converts this to phenol and acetone. Acetone, the by-product, is also obtained in large quantities this way.
• C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH, then C₆H₅C(CH₃)₂OOH → C₆H₅OH + CH₃COCH₃ (H⁺) (reasoned from NCERT page 202).

Physical properties

• Boiling points rise with the number of carbon atoms (stronger van der Waals forces) and, for alcohols, fall with branching (smaller surface area) (NCERT page 203).
• The –OH group forms intermolecular hydrogen bonds, so alcohols and phenols boil higher than hydrocarbons, ethers, haloalkanes and haloarenes of comparable molecular mass (NCERT pages 203–204). NCERT’s comparison (page 204): ethanol (46, 351 K), methoxymethane (46, 248 K) and propane (44, 231 K).
• Solubility in water comes from hydrogen bonds with water; it falls as the hydrophobic alkyl or aryl part grows, and several lower alcohols mix with water in all proportions (NCERT page 204).

Chemical reactions

• As nucleophiles alcohols react through the O–H bond: the oxygen’s lone pair attacks an electron-poor carbon, and H⁺ is lost. As electrophiles, once protonated (R–OH₂⁺), they react through the C–O bond, which breaks as a nucleophile attacks: Br⁻ + RCH₂–OH₂⁺ → RCH₂Br + H₂O (NCERT page 204).
• So the reactions fall into two groups: those that break O–H (acidity, esterification) and those that break C–O (with HX and PX₃, dehydration), which happen only in alcohols; phenols break the C–O bond only with zinc (NCERT pages 204–208). Oxidation forms a C=O bond by breaking O–H and C–H bonds (NCERT page 209).
• Esterification. Alcohols and phenols give esters with carboxylic acids or anhydrides (with a little concentrated H₂SO₄; with an acid the reaction is reversible, so the water is removed as it forms) and with acid chlorides (with pyridine to neutralise the HCl). Introducing an acetyl group, CH₃CO, is acetylation; acetylation of salicylic acid gives aspirin (NCERT pages 207–208).

Acidity of alcohols and phenols

• With metals. Alcohols and phenols react with sodium, potassium and aluminium to give alkoxides or phenoxides and hydrogen: 2R–OH + 2Na → 2R–ONa + H₂. Phenols also react with aqueous NaOH: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. Both are Brönsted acids, proton donors (NCERT page 205).
• Alcohols. The O–H bond is polar; electron-releasing alkyl groups increase the electron density on oxygen and lower the polarity, so the acid strength falls: primary > secondary ≫ tertiary (NCERT page 205). Alcohols are weaker acids than water: an alkoxide takes a proton from water, R–O⁻ + H₂O → R–OH + OH⁻, so alkoxides are stronger bases than hydroxide (sodium ethoxide is a stronger base than NaOH). With the lone pairs on oxygen, alcohols are also Brönsted bases (NCERT page 206).
• Phenols are more acidic, as their reaction with NaOH shows: the sp² carbon withdraws electron density from oxygen, raising the O–H polarity, and the negative charge of the phenoxide ion is delocalised over the ring, which stabilises it; the phenol molecule’s own resonance structures separate charge and stabilise it less (NCERT pages 206–207; see resonance).
• Substituents. Electron-withdrawing groups such as –NO₂ raise the acidity, most at ortho and para; electron-releasing alkyl groups lower it, so cresols are less acidic than phenol (NCERT page 207).

NCERT Table 7.3 (page 207) gives these pK_a values; the greater the pK_a, the weaker the acid:

• “A million times”. NCERT says phenol is a million times more acidic than ethanol (page 207). The pK_a values differ by 15.9 − 10.0 = 5.9, so the ratio of K_a values is 10^5.9 ≈ 8 × 10⁵: close to a million, as NCERT’s round figure says (reasoned).
• NCERT’s two books differ on phenol. Table 7.3 gives its pK_a as 10.0 (K_a = 1.0 × 10⁻¹⁰), but NCERT Class 11 Chemistry, Unit 6, Table 6.6 (page 195) gives K_a = 1.3 × 10⁻¹⁰, a pK_a of 9.89. In a problem, use the value the question gives.

Lucas test for primary, secondary and tertiary alcohols

• Alcohols react with hydrogen halides: ROH + HX → R–X + H₂O, and with PBr₃ to give alkyl bromides (NCERT page 208, as in NCERT Unit 6).
• The three classes react with HCl at very different rates, which is the Lucas test. Alcohols dissolve in Lucas reagent (concentrated HCl and ZnCl₂), but their chlorides do not and make the solution turbid. A tertiary alcohol gives turbidity immediately, as it forms the halide easily; a primary alcohol gives no turbidity at room temperature (NCERT page 208). A secondary alcohol comes in between: NCERT Unit 6 (page 164) gives the order of reactivity of alcohols with a haloacid as 3° > 2° > 1°.

Mechanism of dehydration of alcohols

• Alcohols lose water to give alkenes with a protic acid (concentrated H₂SO₄ or H₃PO₄) or a catalyst such as anhydrous ZnCl₂ or alumina. Ethanol needs concentrated H₂SO₄ at 443 K; secondary and tertiary alcohols react under milder conditions: propan-2-ol with 85% H₃PO₄ at 440 K gives propene, and 2-methylpropan-2-ol with 20% H₃PO₄ at 358 K gives 2-methylpropene (NCERT pages 208–209).
• Ease of dehydration: tertiary > secondary > primary, because tertiary carbocations are the most stable and easiest to form (NCERT page 209).
• NCERT’s mechanism for ethanol (page 209): (1) fast protonation gives the ethyl oxonium ion, CH₃CH₂OH₂⁺; (2) loss of water gives the carbocation, CH₃CH₂⁺, the slowest, rate-determining step; (3) loss of a proton gives ethene. The acid used in step 1 is released in step 3, and ethene is removed as it forms to drive the equilibrium.

Oxidation of alcohols

• Oxidation makes a C=O bond by breaking an O–H and a C–H bond: it is also called dehydrogenation (NCERT page 209).
• NCERT slip (pages 208–209): NCERT lists oxidation as item 4 under “Reactions involving cleavage of carbon – oxygen (C–O) bond in alcohols”, yet its own first sentence on oxidation says that an O–H and a C–H bond break while the C–O bond becomes C=O. In an exam, for the bonds broken in oxidation, write O–H and C–H.
• Primary alcohols give aldehydes, which can be oxidised further to carboxylic acids. Acidified KMnO₄ goes straight to the acid; CrO₃ in an anhydrous medium, or better pyridinium chlorochromate (PCC), stops at the aldehyde (NCERT pages 209–210).
• Secondary alcohols give ketones with CrO₃. Tertiary alcohols are not oxidised; under strong conditions (KMnO₄, heat) C–C bonds break and a mixture of acids with fewer carbons forms (NCERT page 210).
• Over heated copper at 573 K, primary and secondary alcohols are dehydrogenated to an aldehyde or ketone, while tertiary alcohols are dehydrated to alkenes (NCERT page 210).
• In the body, methanol and ethanol are oxidised to the aldehyde and then the acid. Methanol gives methanal and then methanoic acid, which can cause blindness and death; a patient is treated with intravenous dilute ethanol, which keeps the enzyme busy while the kidneys excrete the methanol (NCERT page 210).
• NCERT slip (page 210): NCERT says the ethanol swamps the enzyme that oxidises the aldehyde (HCHO) to the acid. Beyond the NCERT text, the ethanol competes for alcohol dehydrogenase, the enzyme of the first step, methanol to methanal, so little methanol is turned into the toxic products and the kidneys can excrete it (Wikipedia, “Methanol toxicity”). In a written answer, “ethanol slows the oxidation of methanol, giving the kidneys time to excrete it” is safe, and if you name the enzyme, name alcohol dehydrogenase; in an objective question that repeats NCERT’s sentence, recognise it as NCERT’s wording.

Nitration, sulphonation and halogenation of phenol

The –OH group activates the ring towards electrophilic substitution and directs to ortho and para, which resonance makes electron-rich (NCERT page 211; directive influence, Class 11).

• Nitration. Dilute HNO₃ at 298 K gives o- and p-nitrophenol, separated by steam distillation: o-nitrophenol is steam volatile because of intramolecular hydrogen bonding, while p-nitrophenol is held by intermolecular hydrogen bonds. Concentrated HNO₃ gives 2,4,6-trinitrophenol (picric acid), but in poor yield (NCERT page 211).
• Sulphonation then nitration. Picric acid is now made by treating phenol with concentrated H₂SO₄, which gives phenol-2,4-disulphonic acid, and then with concentrated HNO₃ (NCERT page 211). Picric acid is a strong acid because of its three electron-withdrawing –NO₂ groups.
• Halogenation. Br₂ in a solvent of low polarity (CS₂ or CHCl₃) at 273 K gives monobromophenols, mainly p-bromophenol with some o-bromophenol; no Lewis acid is needed, because –OH is so strongly activating. Bromine water gives a white precipitate of 2,4,6-tribromophenol: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr (NCERT page 212).
• NCERT slip (page 212): Example 7.5 (c) asks for the mononitration of “phenyl methanoate”, but the answer drawn has an –OCOCH₃ group, which is phenyl ethanoate (phenyl acetate); the product shown is 4-nitrophenyl ethanoate. In an exam, answer for the ester the question names: phenyl methanoate, HCOOC₆H₅, gives 4-nitrophenyl methanoate, and phenyl ethanoate gives 4-nitrophenyl ethanoate; in both, the ring O directs the –NO₂ mainly to the para position (reasoned from NCERT’s answer).

Kolbe’s and Reimer–Tiemann reactions

• Kolbe’s reaction. Phenol with NaOH gives the phenoxide ion, which is even more reactive than phenol and is substituted even by the weak electrophile CO₂; acidification gives ortho-hydroxybenzoic acid, salicylic acid (NCERT pages 212–213).
• Reimer–Tiemann reaction. Phenol with chloroform and aqueous NaOH puts a –CHO group at the ortho position: the intermediate substituted benzal chloride is hydrolysed by the alkali, and acidification gives salicylaldehyde (NCERT page 213).
• With zinc dust, phenol is reduced to benzene: C₆H₅OH + Zn → C₆H₆ + ZnO. With chromic acid (Na₂Cr₂O₇ and H₂SO₄) it gives benzoquinone, a conjugated diketone; in air, phenols slowly darken to mixtures containing quinones (NCERT page 213).

Solved examples

Common mistakes to avoid

Quick revision summary

• Alcohols from alkenes (H_₂O/H^⁺, Markovnikov; hydroboration–oxidation, anti-Markovnikov-looking), from aldehydes (1°) and ketones (2°) by reduction, from acids by LiAlH_₄, and from Grignard reagents (HCHO 1°, other aldehydes 2°, ketones 3°).
• Phenol from chlorobenzene (NaOH, 623 K, high pressure), benzenesulphonic acid, diazonium salts, and cumene (with acetone as by-product).
• Hydrogen bonding raises boiling points and solubility; branching lowers boiling points.
• Acidity: 1° > 2° ≫ 3° alcohols; alcohols < water < phenol; –NO_₂ raises and –CH_₃ lowers the acidity of phenol.
• Lucas test (conc. HCl + ZnCl_₂): 3° turbid at once, 1° not at room temperature. Dehydration: 3° > 2° > 1°, via a carbocation.
• Oxidation: 1° to aldehyde (PCC) or acid (KMnO_₄); 2° to ketone; 3° resists; Cu at 573 K dehydrogenates 1° and 2°, dehydrates 3°.
• Phenol: o,p-directing and activating; bromine water gives 2,4,6-tribromophenol; Kolbe gives salicylic acid, Reimer–Tiemann salicylaldehyde.

Questions students ask

Is phenol a strong acid or a weak acid?

Phenol is a weak acid: its pK_a is 10.0 (NCERT Table 7.3, page 207), and NCERT Unit 8 calls acids with pK_a between 5 and 15 weak, keeping “strong” for pK_a below 1 (NCERT page 250). It is far weaker than ethanoic acid, pK_a 4.76 (NCERT page 250), so it reacts with sodium hydroxide to give sodium phenoxide but, unlike carboxylic acids, gives no carbon dioxide with sodium hydrogencarbonate (NCERT pages 205 and 249). Three –NO₂ groups change that: 2,4,6-trinitrophenol, picric acid, is a strong acid (NCERT page 211).

Why are cresols less acidic than phenol, and which cresol is the most acidic?

Cresols are slightly weaker acids than phenol because the electron-releasing methyl group makes the phenoxide ion less stable (NCERT page 207); in NCERT Table 7.3 m-cresol is the most acidic, pK_a 10.1, against 10.2 for the other two and 10.0 for phenol. Beyond the NCERT text, finer values agree, 10.09 (m), 10.26 (p) and 10.29 (o) against 9.99 for phenol, with o and p too close to rank (PubChem, Hazardous Substances Data Bank). The reason (reasoned from NCERT page 206): the phenoxide ion’s negative charge spreads onto the ortho and para carbons, so a methyl group there destabilises the ion most, while a meta methyl sits on a carbon with none of that charge.

Is the dehydration of alcohols E1 or E2?

Acid-catalysed dehydration of secondary and tertiary alcohols follows the E1 path, a name beyond the NCERT text: the protonated alcohol slowly loses water to a carbocation, which then loses a proton, so the ease is 3° > 2° > 1° (NCERT page 209). Primary alcohols are often given an E2 path, a base removing a β-hydrogen as water leaves, in one step, since a primary carbocation is so unstable (Ian Hunt, University of Calgary course notes). NCERT, though, draws even ethanol’s dehydration through a carbocation (page 209), so in JEE and NEET answers write its three steps.

Why do tertiary alcohols resist oxidation?

Tertiary alcohols resist oxidation because the carbon carrying their –OH has no hydrogen: oxidation forms the C=O bond by removing the H of the O–H and an H from that carbon (NCERT page 209), and R₃C–OH has none to lose. A C=O bond could form there only by breaking a C–C bond (reasoned), so under the usual conditions tertiary alcohols are not oxidised, while strong agents such as KMnO₄ at high temperature break C–C bonds and give a mixture of acids with fewer carbons (NCERT page 210). Primary and secondary alcohols, which have that hydrogen, give aldehydes and ketones.

What is the electrophile in the Reimer–Tiemann reaction?

The electrophile in the Reimer–Tiemann reaction is dichlorocarbene, :CCl₂, a point beyond the NCERT text: the hydroxide ion takes a proton from chloroform, the CCl₃⁻ ion left behind loses a chloride ion, and the electron-poor carbene is attacked by the electron-rich phenoxide ion, mainly at the ortho position (Wikipedia, “Reimer–Tiemann reaction”). The –CHCl₂ group this puts on the ring is the substituted benzal chloride that NCERT names as the intermediate, and the alkali hydrolyses it to –CHO, giving salicylaldehyde after acidification (NCERT page 213). So chloroform supplies one carbon, which ends up as the aldehyde group (reasoned).

Why is the cumene process preferred for making phenol?

The cumene process is preferred because one process gives two bulk products, phenol and acetone: NCERT notes that most of the world’s phenol comes from cumene and that acetone, the by-product, is also obtained in large quantities this way (NCERT page 202). Beyond the NCERT text, cumene is made from benzene and propene by Friedel–Crafts alkylation, and the route pays only while acetone is wanted as well as phenol (Wikipedia, “Cumene process”). The chlorobenzene route needs fused NaOH at 623 K and very high pressure and gives no such by-product (NCERT page 202; reasoned).

Are Kolbe’s and Reimer–Tiemann reactions in the JEE Main 2027 and NEET (UG) 2026 syllabuses?

The Reimer–Tiemann reaction is named in Unit 17, Organic Compounds Containing Oxygen, of both JEE Main 2027 and NEET (UG) 2026, but Kolbe’s reaction is not named separately: it falls under the unit’s opening line, “General methods of preparation, properties, reactions and uses” (JEE Main 2027 wording). In JEE Main 2027’s words, the same unit names, for alcohols, the “Identification of primary, secondary and tertiary alcohols” and the “mechanism of dehydration”, and, for phenols, their “Acidic nature” and their halogenation, nitration and sulphonation. Read with its opening line, Unit 17 covers the whole of NCERT Section 7.4 (pages 199–213), Kolbe’s reaction included (reasoned).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Arrange water, ethanol and phenol in order of increasing acid strength, and give one reaction that shows phenol is the strongest.

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   Ethanol < water < phenol. Phenol dissolves in aqueous NaOH to give sodium phenoxide, C₆H₅OH + NaOH → C₆H₅ONa + H₂O, which alcohols do not (NCERT pages 205–206).

2. Q2. How would you distinguish pentan-1-ol from 2-methylbutan-2-ol by the Lucas test?

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   With Lucas reagent (conc. HCl + ZnCl₂), the tertiary 2-methylbutan-2-ol turns turbid at once, as its chloride forms easily; the primary pentan-1-ol gives no turbidity at room temperature (NCERT page 208).

3. Q3. o-Nitrophenol and p-nitrophenol have the same molar mass. Which has the higher boiling point, and why?

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   p-Nitrophenol. Its molecules are joined by intermolecular hydrogen bonds, which must be broken before it boils; in o-nitrophenol the hydrogen bond lies within each molecule, so the molecules are not associated (NCERT page 211; reasoned).

4. Q4. Name the products of (i) phenol with CHCl₃ and aqueous NaOH, then acid; (ii) sodium phenoxide with CO₂, then acid.

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   (i) Salicylaldehyde, 2-hydroxybenzaldehyde (Reimer–Tiemann). (ii) Salicylic acid, 2-hydroxybenzoic acid (Kolbe) (NCERT pages 212–213).