JEE Chemistry Alcohols Phenols and Ethers Notes Set 12

Phenols: Structure and Acidity

A phenol has an OH group attached directly to a benzene ring. The oxygen lone pair is delocalised into the ring, so the C-O bond is short and strong, the O-H bond is weakened, and the ring is strongly activated towards electrophiles. Loss of a proton gives the phenoxide ion, whose negative charge is spread over oxygen and the ortho and para carbons; this is why phenol (\( pK_a \approx 10 \)) is much more acidic than alcohols but weaker than carboxylic acids and carbonic acid, so it dissolves in NaOH but not in \( \ce{NaHCO3} \). Electron-withdrawing groups at ortho or para positions (for example nitro) increase acidity, and picric acid (2,4,6-trinitrophenol) is as strong as a mineral acid.

Phenol, o-, m- and p-cresol, catechol (benzene-1,2-diol), resorcinol (1,3-diol) and hydroquinone (1,4-diol).

Preparation

Industrial routes are (i) the cumene process, in which cumene is oxidised by air to cumene hydroperoxide and then treated with dilute acid to give phenol and acetone, and (ii) the Dow process, in which chlorobenzene is heated with NaOH at about 630 K and 300 atm. Laboratory routes include hydrolysis of a diazonium salt with warm water, alkali fusion of sodium benzenesulphonate, and decarboxylation of salicylic acid with soda lime.

Cumene process. The tertiary benzylic hydrogen of cumene is removed by a radical chain with oxygen. In the hydroperoxide, acid protonates the terminal oxygen; the phenyl group migrates from carbon to oxygen as water leaves, and hydrolysis of the resulting cation gives phenol and acetone. The aryl group with the greater electron-donating ability migrates better.

Electrophilic Substitution of Phenol

Because OH is a powerful ortho/para director, phenol reacts under mild conditions and often polysubstitutes.

Halogenation. Bromine water gives a white precipitate of 2,4,6-tribromophenol immediately. In a non-polar solvent (\( \ce{CS2} \) or \( \ce{CCl4} \)) at low temperature the mono-bromo compounds form, mainly the para isomer. Groups at the para position, such as \( \ce{-SO3H} \) and \( \ce{-COOH} \), can be replaced by bromine (ipso substitution), so p-hydroxybenzenesulphonic acid and salicylic acid with excess bromine water also give 2,4,6-tribromophenol.

Nitration. Dilute \( \ce{HNO3} \) at low temperature gives a mixture of o- and p-nitrophenol. The ortho isomer has intramolecular hydrogen bonding, so it is steam volatile, has a lower boiling point and is less soluble in water; it is separated by steam distillation. Concentrated nitric acid oxidises phenol extensively, so picric acid is made by sulphonating phenol first and then nitrating the sulphonic acid.

Sulphonation. At about 288 K (kinetic control) the ortho acid predominates; at about 373 K (thermodynamic control) the para acid does. Alkali fusion then gives catechol and hydroquinone respectively.

Friedel-Crafts and related reactions. Alkylation with an alcohol and acid gives mainly the para alkylphenol (for example 4-tert-butylphenol). With formaldehyde and base (Lederer-Manasse) hydroxymethylphenols form, and phenol with excess formaldehyde gives the polymer Bakelite.

Reactions Using Phenoxide: Reimer-Tiemann, Kolbe, Ethers and Esters

Reimer-Tiemann formylation. Phenol with chloroform and aqueous NaOH, followed by acid, gives salicylaldehyde (ortho, major) and 4-hydroxybenzaldehyde (para, minor). Hydroxide removes a proton from \( \ce{CHCl3} \) to form \( \ce{CCl3^-} \), which loses \( \ce{Cl^-} \) to give the neutral electrophile dichlorocarbene, \( \ce{:CCl2} \). The carbene attacks the phenoxide ring and the resulting \( \ce{-CHCl2} \) group is hydrolysed to CHO. With \( \ce{CCl4} \) in place of \( \ce{CHCl3} \) the product is salicylic acid. If both ortho positions are blocked, the para product forms; if only the para position is blocked (p-cresol), some abnormal dienone with \( \ce{CH3} \) and \( \ce{CHCl2} \) on the same carbon appears as a minor product.

Kolbe carboxylation. Sodium phenoxide takes up carbon dioxide at about 400 K and 5 atm; acidification gives salicylic acid. Acetylation of salicylic acid with acetic anhydride gives aspirin (acetylsalicylic acid). Salicylaldehyde and salicylic acid are more volatile and less soluble than their para isomers because of intramolecular hydrogen bonding.

Williamson synthesis and esterification. Phenoxide with an alkyl halide gives an aryl alkyl ether (anisole from \( \ce{CH3I} \)); HI cleaves it back to phenol and alkyl iodide, never to iodobenzene. Phenol is acylated at oxygen with acid chlorides or anhydrides in base to give phenyl esters (Schotten-Baumann). The Fries rearrangement of a phenyl ester with \( \ce{AlCl3} \) moves the acyl group to the ortho or para ring position.

Claisen rearrangement. Allyl phenyl ether on heating undergoes a [3,3] sigmatropic shift to give 2-allylphenol. When both ortho positions are blocked, a second [3,3] (Cope) shift carries the allyl group to the para position; each shift inverts the allyl unit, so after two shifts the original O-CH2 carbon is bonded to the ring.

Other reactions. Zinc dust distillation reduces phenol to benzene. Chromic acid or sodium dichromate oxidises phenol to p-benzoquinone. Catechol with \( \ce{CH2I2} \) and NaOH gives a methylenedioxy ring (1,3-benzodioxole).

Tests for Phenols

Phenols turn blue litmus red only faintly, dissolve in NaOH but not in \( \ce{NaHCO3} \), give a violet colour with neutral \( \ce{FeCl3} \) (a hexaphenoxoferrate(III) complex), give the Liebermann nitroso test (a red or brown colour that turns blue on adding strong alkali), couple with diazonium salts in weak alkali (azo dyes), and decolourise bromine water with a white precipitate.

Worked Examples

Example 1. A compound \( \ce{C7H8O} \) dissolves in NaOH, is insoluble in \( \ce{NaHCO3} \) and gives \( \ce{C7H5OBr3} \) with bromine water. It is a phenol (acidic), and three Br atoms require all of positions 2, 4 and 6 relative to OH to be free, so the methyl group must be meta: it is m-cresol.

Example 2. Phenol is treated with \( \ce{CHCl3}/NaOH \) to give P (major) and Q (minor). P has a lower boiling point, lower melting point and lower water solubility than Q because P is the ortho isomer with an intramolecular hydrogen bond (salicylaldehyde), whereas Q (the para isomer) forms intermolecular hydrogen bonds.

Example 3 (yield calculation). If 1.00 mol of salicylic acid is acetylated to aspirin in 80% yield, the amount of aspirin formed is 0.80 mol. The molar mass of aspirin \( \ce{C9H8O4} \) is 180 g/mol, so the mass is \( 0.80 \times 180 = 144\,\text{g} \).