Phenols
Class 12 Chemistry Notes • NEB Syllabus 2083
Syllabus, 19 solved multiple choice questions, 4 short and 4 long answer questions with structures drawn for every reaction, handwritten and typed PDF notes and frequently asked questions on the preparation, properties, tests and uses of phenol.
1. Syllabus
Unit 11: Phenols
These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved short and long answer questions, and finish with the PDF notes for quick revision.
2. Multiple Choice Questions
Figure: Structures a, b, c and d of the question
- a) a benzene ring bonded directly to \(-OH\)
- b) a benzene ring bonded to \(-CH_2OH\)
- c) a saturated six-membered ring bonded to \(-OH\)
- d) a six-membered ring with one double bond bonded to \(-OH\)
Correct Answer: a) a benzene ring bonded directly to \(-OH\)
Reason: Phenol is a compound in which the \(-OH\) group is attached directly to a carbon atom of a benzene ring (\(C_6H_5OH\)). Structure b is benzyl alcohol, because the \(-OH\) is on the side chain \(-CH_2\), and structures c and d are cyclohexanol and cyclohexenol, because their rings are not aromatic.
- a) Benzenol
- b) Hydroxybenzene
- c) Carbolic acid
- d) All of the above
Correct Answer: a) Benzenol
Reason: Phenol is a retained IUPAC name, and the systematic form of it is benzenol (benzene + ol). Hydroxybenzene is only a descriptive way of reading the structure, and carbolic acid is the old trivial name of phenol, so it is not an IUPAC name. The textbook notes call phenol also monohydroxybenzene, so some teachers accept option d, but only benzenol is a purely systematic IUPAC name.
- a) Friedel-Crafts reaction
- b) Dow process
- c) Raschig process
- d) Kolbe-Schmitt reaction
Correct Answer: b) Dow process
Reason: In the Dow process chlorobenzene is heated with aqueous \(NaOH\) at about \(623\,K\) (\(350^\circ C\)) and \(300\) atm. Sodium phenoxide is formed, and acidification with \(HCl\) gives phenol. Friedel-Crafts reaction forms \(C-C\) bonds, the Raschig process makes chlorobenzene from benzene, and the Kolbe-Schmitt reaction converts phenol to salicylic acid.
Figure: Dow process
- a) \(H_2O\ (r.t.)\)
- b) \(H_2O/H^+\ (warm)\)
- c) \(NaOH\ (Cold)\)
- d) \(HNO_3\)
Correct Answer: b) \(H_2O/H^+\ (warm)\)
Reason: When an aqueous solution of benzene diazonium chloride is warmed with dilute acid (or boiled with water), the diazonium group is replaced by \(-OH\) and nitrogen is evolved: \(C_6H_5N_2^+Cl^-+H_2O\xrightarrow{warm}C_6H_5OH+N_2+HCl\). At room temperature the salt is stable, and cold alkali gives a diazotate or a coupling product, not phenol.
Figure: Hydrolysis of benzene diazonium chloride
- a) Phenol forms hydrogen bonding.
- b) Higher molecular weight of phenol.
- c) Presence of benzene ring
- d) Presence of ionic bond
Correct Answer: a) Phenol forms hydrogen bonding.
Reason: The polar \(O-H\) group of phenol forms intermolecular hydrogen bonds between phenol molecules, and this raises its boiling point (\(182^\circ C\)) above that of toluene (\(111^\circ C\)). It also forms hydrogen bonds with water, so phenol is more soluble than toluene. Toluene has only weak dispersion forces and cannot form hydrogen bonds. Molecular weights are nearly the same (\(94\) and \(92\)), so option b is not the reason. (The question in the book says alcohol in place of phenol, but the options all refer to phenol.)
- a) \(Water<Phenol<Ethanol\)
- b) \(Ethanol<Phenol<Water\)
- c) \(Phenol<Ethanol<Water\)
- d) \(Ethanol<Water<Phenol\)
Correct Answer: d) \(Ethanol<Water<Phenol\)
Reason: Phenol (\(pK_a\approx 10\)) is the strongest acid, because the phenoxide ion is stabilised by resonance. Water (\(pK_a=15.7\)) is a little more acidic than ethanol (\(pK_a\approx 16\)), because the \(+I\) effect of the ethyl group makes the ethoxide ion less stable. So the order is ethanol < water < phenol, which is also given in the textbook notes as phenol > water > alcohol. (In the printed question option d appears as water < ethanol < phenol, which is a typing error for this order, and it is the only option with phenol at the top.)
- a) Sodium phenoxide \(+H_2\)
- b) Sodium benzoate \(+CO_2\)
- c) Benzene \(+H_2O\)
- d) No reaction
Correct Answer: a) Sodium phenoxide \(+H_2\)
Reason: Phenol is acidic enough to react with active metals like sodium, which replace the hydrogen of the \(-OH\) group and liberate hydrogen gas: \(2C_6H_5OH+2Na\rightarrow 2C_6H_5ONa+H_2\uparrow\).
Figure: Phenol with sodium
- a) Oxidation
- b) Reduction
- c) Nucleophilic substitution
- d) Elimination
Correct Answer: b) Reduction
Reason: When phenol is distilled with zinc dust, zinc takes the oxygen as \(ZnO\) and benzene is formed: \(C_6H_5OH+Zn\xrightarrow{\Delta}C_6H_6+ZnO\). The \(-OH\) group is removed (loss of oxygen), so phenol is reduced to benzene and zinc is oxidised.
Figure: Phenol with zinc dust
- a) Azo dye
- b) Nitrophenol
- c) Benzene
- d) Salicylic acid
Correct Answer: a) Azo dye
Reason: In a slightly alkaline medium the phenoxide ion couples with the benzene diazonium ion at the para position. The product is \(p\)-hydroxyazobenzene (4-hydroxyazobenzene), an orange-yellow azo dye that contains the \(-N=N-\) group.
Figure: Coupling of phenol with benzene diazonium chloride
- a) \(CO_2\) and \(NaOH\)
- b) \(CH_3COCl\)
- c) \(HNO_3\)
- d) \(SOCl_2\)
Correct Answer: a) \(CO_2\) and \(NaOH\)
Reason: In Kolbe's reaction phenol is first changed to sodium phenoxide by \(NaOH\), and the phenoxide is heated with carbon dioxide at about \(400\,K\) and \(4-7\) atm. Acidification of sodium salicylate gives salicylic acid (2-hydroxybenzoic acid).
Figure: Kolbe's reaction, step 2 (carboxylation and acidification)
- a) Salicylic acid
- b) Salicylaldehyde
- c) Picric acid
- d) Aspirin
Correct Answer: b) Salicylaldehyde
Reason: In the Reimer-Tiemann reaction phenol is formylated (an \(-CHO\) group is introduced) by chloroform and aqueous alkali at about \(340\,K\). The reactive species is dichlorocarbene (\(:CCl_2\)), which attacks the ortho position of the phenoxide ion, and the product after acidification is salicylaldehyde (2-hydroxybenzaldehyde) as the major product, with a little \(p\)-hydroxybenzaldehyde.
Figure: Reimer-Tiemann reaction
- a) Reduction
- b) Oxidation
- c) Alkylation
- d) Acylation
Correct Answer: d) Acylation
Reason: The acetyl group \(CH_3CO-\) is introduced on the oxygen of phenol, so the reaction is acylation (here the product is the ester, phenyl acetate): \(C_6H_5OH+(CH_3CO)_2O\rightarrow C_6H_5OCOCH_3+CH_3COOH\). No alkyl group is added and no oxidation number changes in the ring. (The book prints the product as acetophenone, which is a ketone, \(C_6H_5COCH_3\); the product of this reaction is the ester phenyl acetate.)
Figure: Phenol with acetic anhydride
- a) \(o\)-nitrophenol and \(p\)-nitrophenol
- b) \(m\)-nitrophenol
- c) 2,4,6-trinitrophenol
- d) Only \(p\)-nitrophenol
Correct Answer: a) \(o\)-nitrophenol and \(p\)-nitrophenol
Reason: The \(-OH\) group is ortho and para directing, so the nitronium ion \(NO_2^+\) attacks these positions, and nothing goes to the meta position. With dilute nitric acid at about \(298\,K\) a mixture of \(o\)-nitrophenol and \(p\)-nitrophenol is formed. Trinitrophenol (picric acid) needs concentrated nitric acid with concentrated sulphuric acid.
Figure: Nitration of phenol with dilute nitric acid
- a) 2,4,6-tribromophenol
- b) \(o\)-bromophenol
- c) \(p\)-bromophenol
- d) Both b and c
Correct Answer: d) Both b and c
Reason: In a non-polar solvent like \(CS_2\) or \(CCl_4\) phenol does not ionise, so the ring is only moderately activated and one bromine enters, at the ortho and para positions. A mixture of \(o\)-bromophenol and \(p\)-bromophenol is obtained (the para isomer is the major one). 2,4,6-Tribromophenol is formed only in bromine water.
Figure: Bromination of phenol in a non-polar solvent
- a) Phenol is strongly deactivating
- b) It forms a stable complex with \(AlCl_3\)
- c) Over-alkylation occurs
- d) The \(-OH\) group is meta-directing
Correct Answer: b) It forms a stable complex with \(AlCl_3\)
Reason: The oxygen of phenol has lone pairs, and it donates them to the Lewis acid \(AlCl_3\) to form a stable coordinate complex. The catalyst is used up, and the oxygen carries a positive charge which deactivates the ring, so the reaction does not go smoothly (only a poor yield of cresols is obtained). Phenol is not deactivated in itself (it is strongly activated), and the \(-OH\) group is ortho and para directing.
Figure: Friedel-Crafts alkylation of phenol
- a) Formation of a phenoxide ion
- b) Formation of a complex with \(Fe^{3+}\)
- c) Oxidation of phenol
- d) Reduction of \(Fe^{3+}\)
Correct Answer: b) Formation of a complex with \(Fe^{3+}\)
Reason: The oxygen of phenol (as phenoxide) donates a lone pair to the \(Fe^{3+}\) ion and a coloured coordination complex, the hexaphenoxoferrate(III) ion, is formed: \(6C_6H_5OH+FeCl_3\rightarrow[Fe(OC_6H_5)_6]^{3-}+3HCl+3H^+\). It is a complex formation, not an oxidation or a reduction, and the iron stays as \(Fe^{3+}\). This is the neutral ferric chloride test of phenol.
- a) \(FeCl_3\)
- b) \(NaNO_2+H_2SO_4\)
- c) \(Br_2/H_2O\)
- d) Phthalic anhydride
Correct Answer: b) \(NaNO_2+H_2SO_4\)
Reason: In Liebermann's nitroso test phenol is warmed with sodium nitrite and concentrated sulphuric acid. A deep blue (or green) colour is formed, which turns red on dilution with water and changes again to deep blue on adding \(NaOH\). The coloured substance is the sodium salt of indophenol. \(FeCl_3\) gives a violet colour, \(Br_2/H_2O\) gives a white precipitate, and phthalic anhydride gives phenolphthalein.
Figure: Colour changes in Liebermann's test
- a) 2,4,6-tribromophenol
- b) Bromobenzene
- c) \(o\)- and \(p\)-bromophenol
- d) Only \(p\)-bromophenol
Correct Answer: a) 2,4,6-tribromophenol
Reason: In water phenol partly ionises to the phenoxide ion, which is extremely reactive, so all three activated positions (two ortho and one para) are brominated at once. The orange-brown colour of bromine disappears and a white precipitate of 2,4,6-tribromophenol is formed: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr\). It is used as a test for phenol.
Figure: Bromination of phenol in water
- a) Nylon-66
- b) Bakelite
- c) Teflon
- d) Polyethene
Correct Answer: b) Bakelite
Reason: Phenol condenses with formaldehyde (\(HCHO\)) in the presence of an acid or a base. First \(o\)- and \(p\)-hydroxybenzyl alcohols are formed, and these lose water repeatedly through \(-CH_2-\) links to give the cross-linked thermosetting polymer Bakelite (phenol-formaldehyde resin). Nylon-66 is a polyamide, Teflon is polytetrafluoroethene and polyethene is made from ethene.
Figure: First step in the formation of Bakelite
3. Numerical and Subjective Questions
Short Answer Questions
i) What is the IUPAC name of the compound \(C_6H_4(OH)CH_3\), with following structure?
Figure: The three cresols, C₆H₄(OH)CH₃
The compound \(C_6H_4(OH)CH_3\) is a methyl derivative of phenol. The carbon bearing \(-OH\) is numbered 1, so the structure with the methyl group on the next carbon (the first structure of the question) is 2-methylphenol (o-cresol), and the structure with the methyl group opposite to \(-OH\) (the second structure) is 4-methylphenol (p-cresol). The third isomer, with the methyl group at the meta position, is 3-methylphenol (m-cresol).
ii) Compare the polarity of phenol and toluene.
Phenol is much more polar than toluene. In phenol the oxygen is highly electronegative, so the \(C-O\) and \(O-H\) bonds are strongly polar, and the molecule has a dipole moment of about \(1.4\,D\). In toluene only the \(C-H\) bonds and the weak \(C(sp^2)-C(sp^3)\) bond are present, and the dipole moment is only about \(0.36\,D\), so toluene is almost non-polar.
| Property | Phenol | Toluene |
|---|---|---|
| Polar group | \(-OH\) | \(-CH_3\) |
| Dipole moment (approx.) | \(1.4\,D\) | \(0.36\,D\) |
| Nature | Polar | Almost non-polar |
| Hydrogen bonding | Yes | No |
iii) Draw the structural formula of \(C_6H_4(OH)_2\) and write IUPAC name.
Figure: The three isomers of C₆H₄(OH)₂
\(C_6H_4(OH)_2\) is dihydroxybenzene, and it has three positional isomers according to the positions of the two \(-OH\) groups: benzene-1,2-diol (catechol, also called 2-hydroxyphenol), benzene-1,3-diol (resorcinol, 3-hydroxyphenol) and benzene-1,4-diol (quinol or hydroquinone, 4-hydroxyphenol).
i) Compare the solubility of phenol and benzene.
Phenol is slightly soluble in water at room temperature (about \(8\) g in \(100\) mL), and it becomes completely miscible above \(65^\circ C\). Its \(-OH\) group forms hydrogen bonds with water molecules. Benzene is insoluble in water, because it is non-polar and cannot form hydrogen bonds. Both dissolve well in organic solvents. Phenol is less soluble than the small alcohols, because its large non-polar benzene ring reduces the interaction with water.
ii) Boiling point of phenol is greater than that of benzene and toluene. Explain.
Phenol molecules are held together by strong intermolecular hydrogen bonds (\(O-H\cdots O\)), because of the polar \(-OH\) group. Benzene and toluene are non-polar and have only weak van der Waals (London) forces, so less energy is needed to separate their molecules. So the boiling point of phenol (\(182^\circ C\)) is higher than that of toluene (\(111^\circ C\)) and benzene (\(80^\circ C\)), although their molecular masses are nearly the same.
iii) Arrange the following in the increasing order of boiling point. Benzene, toluene, phenol and cresol.
Benzene and toluene have only van der Waals forces, and toluene is higher because of its larger mass. Phenol and cresol are hydrogen bonded, so they boil higher, and cresol is the highest because of the larger molecular mass of the extra \(-CH_3\) group.
iv) Suggest the reason behind the lower boiling point and solubility of ortho-nitrophenol in comparison of m- and p- isomers.
Figure: o-Nitrophenol and p-Nitrophenol
In \(o\)-nitrophenol the \(-OH\) group and the \(-NO_2\) group are close to each other, and the hydrogen of \(-OH\) is attached to an oxygen of \(-NO_2\) by intramolecular hydrogen bonding (a six-membered chelate ring). So its \(-OH\) group is not free to bond with other molecules or with water: the molecules are held together only by weak forces, and the compound is volatile with steam (boiling point about \(216^\circ C\)) and is less soluble in water.
In \(m\)- and \(p\)-nitrophenol the groups are too far apart for this, so the \(-OH\) group forms intermolecular hydrogen bonds with other molecules and with water. This gives a higher boiling point, a higher melting point and a greater solubility in water. The difference is used to separate \(o\)-nitrophenol from \(p\)-nitrophenol by steam distillation.
i) How does the presence of an electron-withdrawing group at the meta position affect the reactivity of phenol towards electrophilic substitution?
The \(-OH\) group pushes electrons into the ring by its \(+M\) effect and makes the ortho and para positions rich in electrons. An electron-withdrawing group such as \(-NO_2\), \(-CN\) or \(-COOH\) at the meta position pulls electrons out of the ring by its \(-I\) effect (and cannot give any resonance help to the ortho and para carbons, because it is meta). So the net electron density of the whole ring falls, and the ring becomes less reactive (deactivated) than phenol itself. Electrophilic substitution becomes slower, and more drastic conditions are needed. The \(-OH\) group still decides the positions, so the electrophile goes to the positions ortho and para to \(-OH\) that are not blocked.
Benzene has a uniform \(\pi\)-electron cloud and no group to increase its electron density. The non-polar \(Br_2\) molecule is not polarised enough by it, so a Lewis acid like \(FeBr_3\) or \(AlCl_3\) is needed to generate the electrophile: \(Br_2+FeBr_3\rightarrow Br^{\delta+}\cdots Br\cdots FeBr_3^{\delta-}\) (that is, \(Br^++FeBr_4^-\)).
Phenol has a lone pair on oxygen which is delocalised into the ring (\(+M\) effect), as the resonating structures show.
Figure: Resonating structures of phenol (electron density rises at the ortho and para carbons)
This increases the electron density of the ring, especially at the ortho and para positions. The ring is so rich in electrons that it polarises the \(Br-Br\) bond by itself (the ring acts as the nucleophile) and no catalyst is needed. In water the phenoxide ion is even more reactive, and 2,4,6-tribromophenol is obtained at once. The \(\sigma\)-complex (arenium ion) formed from phenol is also extra stabilised by the oxygen lone pair, so the activation energy is lower.
Long Answer Questions
i) Phenoxide ion is more stable than ethoxide ion? Explain
Figure: Resonating structures of the phenoxide ion (the negative charge is spread over oxygen and the ortho and para carbons)
When phenol loses \(H^+\), the negative charge on the oxygen of the phenoxide ion is delocalised into the benzene ring through resonance. The lone pair on oxygen conjugates with the \(\pi\)-system, so the charge is shared by the oxygen and the ortho and para carbon atoms (structures I to IV, and the resonance hybrid). The spreading of charge over several atoms lowers the energy of the ion and makes it more stable.
In the ethoxide ion (\(CH_3CH_2O^-\)) the negative charge stays on the oxygen only (there is no resonance). The ethyl group also pushes electrons towards oxygen (\(+I\) effect), which increases the charge density and makes the ion less stable. So the phenoxide ion is more stable, and phenol gives up its proton more easily than ethanol.
ii) Compare the acidic strength of phenol, ethanol and water.
Phenol is the strongest acid among the three because its conjugate base is resonance stabilised. Water is a little more acidic than ethanol, since the \(+I\) effect of the ethyl group destabilises the ethoxide ion compared with the hydroxide ion. Phenol turns blue litmus red slightly, reacts with \(Na\) and with \(NaOH\), but it is a weaker acid than carbonic acid and does not liberate \(CO_2\) from sodium bicarbonate. Ethanol and water do not react with \(NaOH\) in this way.
Figure: Phenol dissolves in sodium hydroxide, which alcohols do not do
iii) Discuss the effect of substituent on acidic strength of phenolic group and compare the acidic strength of followings.
Figure: The four phenols compared in the question
A substituent changes the acidity by changing the stability of the phenoxide ion.
Electron-withdrawing groups (\(-NO_2\), \(-Cl\)) pull electrons away from the ring and the oxygen (\(-I\) effect, and \(-M\) effect for \(-NO_2\)). They spread the negative charge of the phenoxide ion further, so it is stabilised and the acidity increases. \(-NO_2\) is very effective, and \(-Cl\) works only by the \(-I\) effect, so it is less effective.
Electron-donating groups (\(-OCH_3\), \(-CH_3\)) push electrons into the ring (\(+M\) or \(+I\)), which increases the charge on oxygen and destabilises the phenoxide ion, so the acidity decreases.
| Compound | Substituent effect | \(pK_a\) (approx.) |
|---|---|---|
| p-Nitrophenol | \(-M, -I\) (strongly stabilises) | 7.15 |
| p-Chlorophenol | \(-I\) (stabilises) | 9.4 |
| Phenol | Reference | 9.95 |
| p-Methoxyphenol | \(+M\) (destabilises) | 10.2 |
This is the order of increasing acidic strength (the lower the \(pK_a\) the stronger the acid).
i) Suggest the reason behind ortho-, para- directing nature of \(-OH\) group in phenol.
Figure: Resonating structures of phenol (electron density rises at the ortho and para carbons)
The oxygen of \(-OH\) has two lone pairs. One of them is delocalised into the \(\pi\)-system of the ring (\(+M\) effect), so phenol is a resonance hybrid of the structures above. In structures II, III and IV the negative charge sits on the carbons at the ortho (two positions) and para positions, and no negative charge ever appears at the meta positions. So the electron density is highest at ortho and para, and the electrophile (\(E^+\)) attacks there. The arenium ion formed by ortho or para attack has an extra resonance structure in which oxygen bears the positive charge and every atom has an octet, and this makes it more stable than the ion from meta attack.
ii) Electrophilic aromatic substitution reaction in phenol is faster than benzene. Explain
The \(-OH\) group is a strong activating group. Through the \(+M\) effect the lone pair of oxygen increases the electron density of the whole ring (more at ortho and para), so the ring is a better nucleophile and attacks the electrophile more readily than benzene. The intermediate carbocation (\(\sigma\)-complex) is also stabilised by the lone pair of oxygen, so the activation energy is lower and the reaction is faster. That is why phenol is brominated and nitrated under milder conditions than benzene: it is nitrated by dilute nitric acid at room temperature, brominated without a catalyst and gives tribromophenol with bromine water, while benzene needs concentrated acids or a Lewis acid.
| Reaction | Benzene | Phenol |
|---|---|---|
| Nitration | conc. \(HNO_3\) + conc. \(H_2SO_4\), 330 K | dilute \(HNO_3\), 298 K |
| Bromination | \(Br_2\) + \(FeBr_3\) | \(Br_2\) water, no catalyst |
iii) With aqueous \(Br_2\), phenol produces tri-substituted product but with \(Br_2\) in \(CCl_4\) or \(CS_2\), it gives mono substituted product. Explain the reason.
Figure: Bromination of phenol in water
Figure: Bromination of phenol in a non-polar solvent
In water (a polar solvent) phenol ionises to a small extent to give the phenoxide ion, \(C_6H_5OH+H_2O\rightleftharpoons C_6H_5O^-+H_3O^+\). The \(O^-\) group is a much stronger electron donor than \(-OH\), so the ring is very strongly activated and all three positions (2, 4 and 6) are brominated at once. The product is the white precipitate of 2,4,6-tribromophenol.
In \(CS_2\) or \(CCl_4\) (non-polar solvents) phenol hardly ionises, and polar \(Br_2\) is not formed. The ring is only activated by the neutral \(-OH\) group, which is not enough for a second or third substitution at low temperature, so a monobromo product is obtained: \(p\)-bromophenol as the major and \(o\)-bromophenol as the minor product.
i) Phenol (\(-OH\)) \(\xrightarrow{?}\) (benzene ring) \(\xrightarrow{?}\) ? \(\xrightarrow{?}\) ?
The question gives the structure of the first product (a benzene ring) and leaves the reagents and the other two products open. The sequence that fits the preparation methods of the unit (phenol, then benzene, then benzenesulphonic acid, then sodium phenoxide, which returns to phenol on acidification) is:
Figure: Phenol to benzene to benzenesulphonic acid
Figure: Benzenesulphonic acid back to phenol
Step 1. Phenol is heated with zinc dust. The \(-OH\) group is removed as \(ZnO\) and benzene is formed: \(C_6H_5OH+Zn\xrightarrow{\Delta}C_6H_6+ZnO\).
Step 2. Benzene is heated with concentrated sulphuric acid (electrophilic substitution), and benzenesulphonic acid is formed: \(C_6H_6+H_2SO_4\xrightarrow{\Delta}C_6H_5SO_3H+H_2O\).
Step 3. Benzenesulphonic acid is fused with sodium hydroxide, and sodium phenoxide is formed (with sodium sulphite and water): \(C_6H_5SO_3H+2NaOH\xrightarrow{fusion}C_6H_5ONa+Na_2SO_3+2H_2O\). Acidification with \(HCl\) then gives phenol back: \(C_6H_5ONa+HCl\rightarrow C_6H_5OH+NaCl\).
i) Dilute \(HNO_3\)
Figure: Nitration of phenol with dilute nitric acid
Phenol is nitrated by dilute nitric acid at about \(298\,K\) to give a mixture of 2-nitrophenol (o-nitrophenol) and 4-nitrophenol (p-nitrophenol). The textbook notes take 2-nitrophenol as the major product and 4-nitrophenol as the minor product. The ortho isomer is volatile with steam (intramolecular hydrogen bonding), so the mixture is separated by steam distillation: \(C_6H_5OH+HNO_3\rightarrow C_6H_4(OH)NO_2+H_2O\).
ii) Conc. \(HNO_3\)
Figure: Nitration of phenol with concentrated nitric acid
With concentrated nitric acid in the presence of concentrated sulphuric acid all three activated positions (2, 4, 6) are nitrated, and 2,4,6-trinitrophenol (picric acid) is formed: \(C_6H_5OH+3HNO_3\xrightarrow{conc.\ H_2SO_4}C_6H_2(NO_2)_3OH+3H_2O\). The yield by this direct method is low, because nitric acid also oxidises phenol and gives tarry products.
iii) \(CH_3COCl\) in presence of pyridine
Figure: Acylation of phenol (esterification)
Phenol is acylated on its oxygen, and the ester phenyl acetate (phenyl ethanoate) is formed: \(C_6H_5OH+CH_3COCl\xrightarrow{pyridine}C_6H_5OCOCH_3+HCl\). Pyridine is a base and takes up the \(HCl\) liberated, so the reaction goes to completion. It is a nucleophilic acyl substitution in which the oxygen of phenol attacks the carbonyl carbon.
4. View PDF Notes
i. Handwritten Notes
ii. Typed PDF Notes
5. Frequently Asked Questions
What is phenol? (Phenols class 12)
What topics are included in Unit 11 Phenols of NEB Class 12 Chemistry?
How are phenols named? (Nomenclature of phenol Class 12)
What are the classes of phenols?
What are the three isomers of cresol?
How is phenol prepared from chlorobenzene (Dow process)?
How is phenol prepared from benzene diazonium chloride?
How is phenol prepared from benzenesulphonic acid and from anisole?
What are the physical properties of phenol?
Why is the boiling point of phenol higher than that of toluene?
Why is o-nitrophenol more volatile and less soluble than p-nitrophenol?
Why is phenol acidic? (Acidic nature of phenol)
What is the order of acidic strength of phenol, water and ethanol?
How do substituents affect the acidity of phenol?
What happens when phenol reacts with sodium and with sodium hydroxide?
What happens when phenol is heated with zinc dust, ammonia, or phthalic anhydride?
What is the coupling reaction of phenol?
What is Kolbe's reaction?
What is Reimer-Tiemann reaction?
What are the products of nitration of phenol?
What is the product of sulphonation of phenol?
Why does phenol give 2,4,6-tribromophenol with bromine water but a mono-bromo product in CS2?
Why is Friedel-Crafts alkylation of phenol not satisfactory?
What is the acylation of phenol?
What is Bakelite and how is it formed?
How do you test for phenol? (Test of phenol)
What are the uses of phenol? (Phenols class 12 uses)
Where can I get Phenols Class 12 notes PDF for NEB?
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