Unit II Model-Questions & Answers : Phenols, Aromatic Amines, Aryl Diazonium Salts and Aromatic Acids
Practice important short and long university questions with concise model answers, examiner expectations, essential keywords, and common mistakes.
How to Practice Theory Questions
Read the question and first prepare your own answer. Then open the model answer, compare the structure and keywords, and rewrite the answer within the suggested word limit.
Reaction-Writing and Examination Checklist
- Write the complete chemical formula of the starting material and major product.
- Place the reagent and essential condition on the reaction arrow.
- Name the major product, especially when ortho/para/meta orientation is involved.
- For named reactions, write both the equation and the reaction name.
- Explain the electronic reason after the reaction: +R activation for –OH/–NH₂ and –I/–R deactivation for –COOH.
- For diazonium chemistry, always show N₂↑ wherever appropriate because nitrogen evolution is a key driving force.
- Use structural formulas such as C₆H₅OH, C₆H₅NH₂, C₆H₅N₂⁺Cl⁻, C₆H₅COOH and o-HOC₆H₄COOH rather than relying only on reaction names.
Part A: Important 2-Mark Questions and Model Answers
Write each answer in approximately 40–60 words. Include the essential definition, scientific reason, equation, or example required by the question.
1. What is phenol? Give its molecular formula and structure.
Phenol is an aromatic hydroxy compound in which the hydroxyl group is directly attached to a benzene ring. Its molecular formula is C₆H₆O and its condensed structural formula is C₆H₅OH. The –OH group strongly activates the aromatic ring toward electrophilic substitution and makes phenol weakly acidic.
Examination Keywords: Formula: C₆H₅OH | Molecular formula: C₆H₆O
2. Why is phenol more acidic than ethanol?
Phenol ionises to form phenoxide ion. The negative charge in phenoxide is delocalised by resonance into the aromatic ring, which stabilizes the conjugate base. Ethanol forms ethoxide ion, where the negative charge remains mainly localized on oxygen. Therefore, phenol is more acidic than ethanol.
Examination Keywords: Reaction: C₆H₅OH ⇌ C₆H₅O⁻ + H⁺
3. Why does phenol react with NaOH but not normally with NaHCO₃?
Phenol is sufficiently acidic to react with the strong base NaOH, forming sodium phenoxide. However, phenol is weaker than carbonic acid, so it cannot normally displace carbonic acid from NaHCO₃ and liberate CO₂.
Examination Keywords: Reactions: C₆H₅OH + NaOH → C₆H₅ONa + H₂O; C₆H₅OH + NaHCO₃ → no appreciable reaction
4. Why is phenol ortho–para directing?
The oxygen atom of –OH has a lone pair that can donate electron density into the benzene ring by resonance. This increases electron density at the ortho and para positions and stabilizes the corresponding σ-complexes. Consequently, –OH is strongly activating and ortho–para directing.
Examination Keywords: Electronic effect: –OH shows a strong +R effect; it activates the ring.
5. What is the Kolbe–Schmitt reaction?
In the Kolbe–Schmitt reaction, sodium phenoxide reacts with carbon dioxide under heat and pressure to introduce a carboxyl group mainly at the ortho position. Acidification of sodium salicylate gives salicylic acid.
Examination Keywords: Reaction: C₆H₅ONa + CO₂ → o-HOC₆H₄COONa → H⁺ → o-HOC₆H₄COOH
6. What is the Reimer–Tiemann reaction?
The Reimer–Tiemann reaction introduces a formyl group (–CHO) mainly at the ortho position of phenol. Phenol is treated with CHCl₃ and aqueous NaOH, followed by hydrolysis. The major product is salicylaldehyde (o-hydroxybenzaldehyde).
Examination Keywords: Reaction: C₆H₅OH + CHCl₃ + NaOH → o-HOC₆H₄CHO (major)
7. What happens when phenol reacts with bromine water?
Phenol reacts rapidly with bromine water because the –OH group strongly activates the aromatic ring. Bromination occurs at both ortho positions and the para position, producing 2,4,6-tribromophenol as a white precipitate.
Examination Keywords: Reaction: C₆H₅OH + 3Br₂ → 2,4,6-C₆H₂Br₃OH + 3HBr
8. What is picric acid?
Picric acid is 2,4,6-trinitrophenol. It is formed by vigorous nitration of phenol. The three electron-withdrawing –NO₂ groups strongly stabilize the phenoxide ion, making picric acid considerably more acidic than phenol.
Examination Keywords: Reaction: C₆H₅OH + 3HNO₃ → 2,4,6-(NO₂)₃C₆H₂OH + 3H₂O
9. How is phenol converted into benzene?
Phenol is reduced by heating with zinc dust. The oxygen-containing functionality is removed and benzene is formed.
Examination Keywords: Reaction: C₆H₅OH + Zn → C₆H₆ + ZnO
10. Give two important synthetic reactions of phenol.
Two important synthetic reactions are the Kolbe–Schmitt reaction and the Reimer–Tiemann reaction. Kolbe–Schmitt reaction introduces –COOH to give salicylic acid, whereas Reimer–Tiemann reaction introduces –CHO mainly at the ortho position to give salicylaldehyde.
Examination Keywords: Kolbe–Schmitt: C₆H₅ONa + CO₂ → o-HOC₆H₄COOH; Reimer–Tiemann: C₆H₅OH + CHCl₃/NaOH → o-HOC₆H₄CHO
11. What is an aromatic amine? Give the formula of aniline.
An aromatic amine contains an amino group attached directly to an aromatic ring. Aniline is the simplest primary aromatic amine and has the formula C₆H₅NH₂. The nitrogen lone pair is conjugated with the benzene ring, influencing both its basicity and electrophilic substitution reactions.
Examination Keywords: Formula: C₆H₅NH₂
12. Why is aniline less basic than ethylamine?
In aniline, the nitrogen lone pair is delocalised into the benzene ring by resonance. This decreases its availability for protonation. In ethylamine, the lone pair remains more localized on nitrogen and is more available to accept H⁺. Hence aniline is less basic.
Examination Keywords: Electronic effect: lone-pair delocalisation into the aromatic ring.
13. Why is aniline ortho–para directing?
The –NH₂ group donates its lone pair into the aromatic ring by resonance. This increases electron density at the ortho and para positions and stabilizes the corresponding σ-complexes. Therefore, –NH₂ is strongly activating and ortho–para directing under conditions where the free amino group remains available.
Examination Keywords: Keywords: +R effect, activation, ortho–para direction.
14. What is diazotization of aniline?
Diazotization converts aniline into benzenediazonium chloride using sodium nitrite and hydrochloric acid at 0–5 °C. Nitrous acid is generated in situ and reacts with the primary aromatic amine. Low temperature is essential to minimize decomposition of the diazonium salt.
Examination Keywords: Reaction: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
15. Why is aniline protected before nitration?
The free –NH₂ group strongly activates the ring and can give uncontrolled reactions under strongly acidic nitration conditions. Acetylation converts –NH₂ into –NHCOCH₃, producing acetanilide. This moderates ring activation and permits more controlled electrophilic substitution. The amino group can later be regenerated by hydrolysis.
Examination Keywords: Reaction: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
16. What is acetanilide?
Acetanilide is the acetyl derivative of aniline, C₆H₅NHCOCH₃. It is prepared by acetylation of aniline and is commonly used as a protected amino compound during electrophilic aromatic substitution, particularly nitration.
Examination Keywords: Formula: C₆H₅NHCOCH₃
17. What happens when aniline reacts with bromine water?
Aniline reacts rapidly with bromine water because –NH₂ strongly activates the ring. Bromination occurs at both ortho positions and the para position, giving 2,4,6-tribromoaniline as a white precipitate.
Examination Keywords: Reaction: C₆H₅NH₂ + 3Br₂ → 2,4,6-C₆H₂Br₃NH₂ + 3HBr
18. What is formed when aniline is treated with nitrous acid at 0–5 °C?
Aniline undergoes diazotization to form benzenediazonium salt. Sodium nitrite and hydrochloric acid generate nitrous acid in situ. The reaction is kept at 0–5 °C because the diazonium product is unstable at higher temperature.
Examination Keywords: Reaction: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
19. Give two important reactions of aniline.
Aniline undergoes acetylation to form acetanilide and diazotization to form benzenediazonium chloride. It also undergoes rapid bromination to form 2,4,6-tribromoaniline. These reactions demonstrate the protection, basicity and strong activating effect of the amino group.
Examination Keywords: Key equations: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃; C₆H₅NH₂ → C₆H₅N₂⁺Cl⁻
20. What is the role of acetylation in aromatic amine chemistry?
Acetylation temporarily protects the amino group as an amide. The resulting acetanilide is less strongly activating than aniline, so electrophilic substitution becomes more controlled. After the required substitution, hydrolysis removes the acetyl group and regenerates the –NH₂ group.
Examination Keywords: Protection: –NH₂ → –NHCOCH₃ → –NH₂ after hydrolysis
21. What are aryl diazonium salts?
Aryl diazonium salts contain the functional group Ar–N₂⁺. Benzenediazonium chloride, C₆H₅N₂⁺Cl⁻, is a common example. They are prepared from primary aromatic amines at low temperature and are valuable intermediates because the diazonium group can be replaced by several other groups.
Examination Keywords: Formula: Ar–N₂⁺; example: C₆H₅N₂⁺Cl⁻
22. What is diazotization?
Diazotization is the conversion of a primary aromatic amine into an aryl diazonium salt using nitrous acid generated in situ from sodium nitrite and mineral acid. The reaction is carried out at 0–5 °C.
Examination Keywords: ArNH₂ + NaNO₂ + 2HCl → ArN₂⁺Cl⁻ + NaCl + 2H₂O
23. Why are diazonium salts prepared at 0–5 °C?
Aryl diazonium salts are relatively unstable at higher temperatures and may decompose with loss of nitrogen. Therefore, diazotization is performed at 0–5 °C to stabilize the intermediate and minimize side reactions.
Examination Keywords: Key condition: 0–5 °C
24. What is the Sandmeyer reaction?
The Sandmeyer reaction replaces the diazonium group by Cl, Br or CN using copper(I) salts. For example, benzenediazonium chloride reacts with CuCl to give chlorobenzene with evolution of nitrogen.
Examination Keywords: C₆H₅N₂⁺Cl⁻ + CuCl → C₆H₅Cl + N₂↑
25. What happens when a diazonium salt reacts with KI?
Iodide replaces the diazonium group to form an aryl iodide. Benzenediazonium chloride therefore gives iodobenzene on treatment with potassium iodide, with liberation of nitrogen gas.
Examination Keywords: C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂↑ + KCl
26. What is azo coupling?
Azo coupling is the reaction of an aryl diazonium ion with an activated aromatic compound such as phenol or aniline. The product contains an –N=N– linkage. Many azo compounds are intensely coloured and are used as dyes.
Examination Keywords: ArN₂⁺ + Ar′H → Ar–N=N–Ar′ + H⁺
27. Why are aryl diazonium salts important synthetic intermediates?
The diazonium group can be replaced by Cl, Br, I, F, CN, OH or H. Diazonium salts also undergo azo coupling. Formation of stable gaseous N₂ provides a strong driving force for many of these transformations.
Examination Keywords: Key transformations: ArN₂⁺ → ArCl, ArBr, ArI, ArF, ArCN, ArOH, ArH
28. What are aromatic carboxylic acids? Give an example.
Aromatic carboxylic acids contain a –COOH group directly attached to an aromatic ring. Benzoic acid, C₆H₅COOH, is the simplest example. Other important compounds include salicylic acid and substituted benzoic acids.
Examination Keywords: Benzoic acid formula: C₆H₅COOH
29. Why is benzoic acid acidic?
Benzoic acid loses H⁺ from its carboxyl group to form benzoate ion. The carboxylate ion is strongly resonance-stabilized because the negative charge is delocalised over two oxygen atoms. This stabilization favours ionization.
Examination Keywords: C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺
30. Why is –COOH meta-directing?
The –COOH group withdraws electron density from the benzene ring through –I and –R effects. It deactivates the ring and makes the σ-complexes from ortho and para attack less stable. Therefore, electrophilic substitution occurs preferentially at the meta position.
Examination Keywords: –COOH: deactivating, meta-directing
31. What happens when benzoic acid reacts with NaHCO₃?
Benzoic acid reacts with sodium bicarbonate to form sodium benzoate, water and carbon dioxide. Effervescence due to CO₂ is a characteristic test for carboxylic acids.
Examination Keywords: C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑
32. What is salicylic acid?
Salicylic acid is 2-hydroxybenzoic acid, o-HOC₆H₄COOH. It contains both phenolic –OH and carboxylic –COOH groups. It can be prepared by the Kolbe–Schmitt reaction and is the starting material for acetylsalicylic acid.
Examination Keywords: Formula: C₇H₆O₃; structure: o-HOC₆H₄COOH
33. What is aspirin chemically?
Aspirin is acetylsalicylic acid, chemically 2-acetoxybenzoic acid. It is prepared by acetylation of the phenolic –OH group of salicylic acid, commonly with acetic anhydride under acid catalysis.
Examination Keywords: o-HOC₆H₄COOH + (CH₃CO)₂O → o-CH₃COOC₆H₄COOH + CH₃COOH
34. Why does benzoic acid react with bases?
The –COOH group contains an acidic proton. Bases remove this proton to form resonance-stabilized benzoate salts. NaOH gives sodium benzoate, while NaHCO₃ gives sodium benzoate with CO₂ evolution.
Examination Keywords: C₆H₅COOH + NaOH → C₆H₅COONa + H₂O
35. Give pharmaceutical/formulation importance of aromatic acids.
Benzoic acid and benzoates are used as antimicrobial preservatives in suitable formulations. Salicylic acid is used in topical preparations for its keratolytic activity. Acetylsalicylic acid (aspirin), derived from salicylic acid, is an important medicinal compound.
Examination Keywords: Keywords: preservative, benzoate, keratolytic, aspirin
Part B: Three-Mark Questions and Model Answers
Practice these important three-mark questions using brief explanations, essential scientific terms, and suitable examples. Each model answer presents the points generally expected in a university examination.
1. Explain the acidic character of phenol with chemical equations.
Phenol is a weak acid because the O–H bond can ionise to give a phenoxide ion. The phenoxide ion is resonance-stabilized: the negative charge is delocalised from oxygen into the aromatic ring. Greater stabilization of the conjugate base favours ionization.
Phenol reacts with sodium hydroxide:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
However, phenol does not normally react with sodium bicarbonate because it is weaker than carbonic acid:
C₆H₅OH + NaHCO₃ → No appreciable reaction
Thus, phenol is acidic enough to react with NaOH but not strong enough to liberate CO₂ from NaHCO₃.
Examination Keywords: phenol acidity, +R effect, ortho–para direction, named reactions, major products
2. Explain the Kolbe–Schmitt reaction with equation and product.
Sodium phenoxide is obtained by treating phenol with NaOH. When sodium phenoxide is heated with CO₂ under pressure, carboxylation occurs predominantly at the ortho position. The resulting sodium salicylate is acidified to give salicylic acid.
Step 1: C₆H₅OH + NaOH → C₆H₅ONa + H₂O
Step 2: C₆H₅ONa + CO₂ → o-HOC₆H₄COONa
Step 3: o-HOC₆H₄COONa + H⁺ → o-HOC₆H₄COOH
Product: salicylic acid (2-hydroxybenzoic acid).
Examination Keywords: phenol acidity, +R effect, ortho–para direction, named reactions, major products
3. Explain the Reimer–Tiemann reaction with equation.
Phenol undergoes ortho-formylation with chloroform and aqueous alkali. Under the reaction conditions, chloroform generates a reactive carbon species that ultimately provides the formyl group. The major product after hydrolysis is salicylaldehyde.
C₆H₅OH + CHCl₃ + NaOH → o-HOC₆H₄CHO
The –OH group activates the ring and directs the incoming group mainly to the ortho position. The reaction is therefore an important method for preparing o-hydroxybenzaldehyde from phenol.
Examination Keywords: phenol acidity, +R effect, ortho–para direction, named reactions, major products
4. Explain bromination and nitration of phenol.
The –OH group strongly activates phenol toward electrophilic substitution and directs substitution to ortho and para positions.
Bromination:
C₆H₅OH + 3Br₂ → 2,4,6-C₆H₂Br₃OH + 3HBr
With bromine water, 2,4,6-tribromophenol is formed as a white precipitate.
Nitration under vigorous conditions gives picric acid:
C₆H₅OH + 3HNO₃ → 2,4,6-(NO₂)₃C₆H₂OH + 3H₂O
These reactions demonstrate the strong activating effect of –OH.
Examination Keywords: phenol acidity, +R effect, ortho–para direction, named reactions, major products
5. Explain the major chemical reactions of phenol in a concise format.
Phenol undergoes several important reactions. With NaOH it forms sodium phenoxide. Bromine water gives 2,4,6-tribromophenol. Vigorous nitration gives picric acid. Kolbe–Schmitt reaction gives salicylic acid after acidification, while Reimer–Tiemann reaction gives salicylaldehyde. Zinc dust reduces phenol to benzene.
Key equations:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
C₆H₅OH + 3Br₂ → 2,4,6-C₆H₂Br₃OH + 3HBr
C₆H₅OH + Zn → C₆H₆ + ZnO
Examination Keywords: phenol acidity, +R effect, ortho–para direction, named reactions, major products
6. Explain why aniline is less basic than aliphatic amines.
Model Answer: Aniline is less basic than aliphatic amines because the nitrogen lone pair is delocalised into the benzene ring by resonance. Therefore, the lone pair is less available for protonation. In aliphatic amines, the lone pair remains comparatively localized on nitrogen and alkyl groups increase electron density through the +I effect.
Reaction: C₆H₅NH₂ + H⁺ ⇌ C₆H₅NH₃⁺
Examination Keywords: resonance, lone-pair delocalisation, protonation, lower basicity
7. Explain the effect of substituents on the basicity of aromatic amines.
Model Answer: Electron-donating groups generally increase the basicity of aniline by increasing electron density and making the nitrogen lone pair more available. Electron-withdrawing groups generally decrease basicity by reducing electron density and lone-pair availability. Resonance effects are especially important at the ortho and para positions.
Reaction: p-CH₃C₆H₄NH₂ + H⁺ ⇌ p-CH₃C₆H₄NH₃⁺; p-NO₂C₆H₄NH₂ + H⁺ ⇌ p-NO₂C₆H₄NH₃⁺
Examination Keywords: EDG increase basicity; EWG decrease basicity; position affects resonance contribution
8. Explain bromination of aniline.
Model Answer: The –NH₂ group strongly activates the benzene ring through resonance donation and directs electrophilic substitution to the ortho and para positions. Therefore, aniline reacts rapidly with bromine water without a Lewis-acid catalyst and forms 2,4,6-tribromoaniline as a white precipitate.
Reaction: C₆H₅NH₂ + 3Br₂ → 2,4,6-C₆H₂Br₃NH₂ + 3HBr
Examination Keywords: strong activation, ortho–para direction, bromine water, white precipitate
9. Why is aniline protected by acetylation before nitration?
Model Answer: In strongly acidic nitrating medium, aniline can be protonated to the anilinium ion, –NH₃⁺, which is deactivating and meta directing. Moreover, the free –NH₂ group is strongly activating. Therefore, aniline is first converted into acetanilide. The –NHCOCH₃ group moderates activation while retaining ortho–para direction, so nitration becomes more controlled and predominantly gives the para product.
Reaction: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
C₆H₅NHCOCH₃ + HNO₃/H₂SO₄ → mainly p-NO₂C₆H₄NHCOCH₃ + H₂O
Examination Keywords: protection, acetanilide, controlled nitration, para product
10. Describe diazotization of aniline with reaction and condition.
Model Answer: Aniline is treated with nitrous acid generated in situ from sodium nitrite and hydrochloric acid. The reaction is maintained at 0–5 °C to form benzenediazonium chloride and minimize decomposition of the diazonium salt.
Reaction: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (0–5 °C)
Examination Keywords: diazotization, NaNO₂/HCl, 0–5 °C, benzenediazonium chloride
11. Explain acetylation and deacetylation of aniline.
Model Answer: Aniline reacts with acetic anhydride to form acetanilide. Acetylation temporarily protects the amino group and reduces excessive ring activation. After the required substitution, hydrolysis removes the acetyl group and regenerates the free –NH₂ group.
Reaction: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
C₆H₅NHCOCH₃ + H₂O → C₆H₅NH₂ + CH₃COOH
Examination Keywords: acetylation, protection, acetanilide, hydrolysis, deprotection
12. Explain why –NH₂ is an ortho–para directing group.
Model Answer: The nitrogen lone pair is donated into the aromatic ring through the +R effect. This resonance donation increases electron density at the ortho and para positions and stabilizes the corresponding sigma complexes more effectively than the meta sigma complex. Hence, the free –NH₂ group is strongly activating and ortho–para directing.
Reaction: C₆H₅NH₂ + E⁺ → mainly o-E-C₆H₄NH₂ + p-E-C₆H₄NH₂
Examination Keywords: +R effect, sigma-complex stability, strong activation, ortho–para direction
13. Describe the preparation of benzenediazonium chloride.
Aniline is dissolved in dilute HCl and cooled to 0–5 °C. A cold aqueous solution of NaNO₂ is added slowly. Nitrous acid is generated in situ and converts aniline into benzenediazonium chloride.
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
The low temperature prevents decomposition of the diazonium salt.
Examination Keywords: aniline, NaNO₂, HCl, 0–5 °C, diazotization
14. Explain the synthetic applications of aryl diazonium salts.
Aryl diazonium salts give aryl chlorides and bromides by Sandmeyer reactions, aryl iodides with KI, aryl fluorides by Balz–Schiemann, phenols by hydrolysis and arenes by reduction with H₃PO₂. They also form azo compounds by coupling with phenols and anilines.
Examination Keywords: Cl, Br, I, F, CN, OH, H, azo coupling
15. Explain the importance of nitrogen loss in diazonium reactions.
The diazonium group is an excellent leaving group because it can leave as molecular nitrogen, N₂. Nitrogen is highly stable and gaseous, so its formation and escape provide a strong driving force for substitution, hydrolysis and reduction reactions.
Examination Keywords: N₂, gaseous leaving group, driving force
16. Explain the acidity of benzoic acid.
Benzoic acid ionizes to form benzoate ion. The carboxylate ion is stabilized by resonance, with the negative charge delocalized over both oxygen atoms.
C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺
Electron-withdrawing substituents generally increase acidity by stabilizing the carboxylate ion, whereas electron-donating substituents generally decrease acidity.
Examination Keywords: carboxylate resonance, conjugate base, electron effects
17. Explain why –COOH is meta-directing.
The –COOH group withdraws electron density from the aromatic ring by inductive and resonance effects. Ortho and para σ-complexes contain unfavorable resonance contributors involving positive charge near the electron-withdrawing group. Meta substitution avoids this particularly unfavorable contributor and is therefore favoured.
Examination Keywords: –I, –R, deactivation, σ-complex, meta direction
18. Explain the reaction of benzoic acid with sodium bicarbonate.
Benzoic acid reacts with NaHCO₃ in an acid–base reaction. Carbonic acid formed initially decomposes into CO₂ and water.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑
The brisk effervescence of CO₂ is a useful qualitative test for carboxylic acids.
Examination Keywords: acid–base reaction, NaHCO₃, CO₂, effervescence
19. Explain the Kolbe–Schmitt preparation of salicylic acid.
Sodium phenoxide reacts with CO₂ under heat and pressure to form sodium salicylate mainly through ortho carboxylation. Acidification gives salicylic acid.
C₆H₅ONa + CO₂ → o-HOC₆H₄COONa → H⁺ → o-HOC₆H₄COOH
Examination Keywords: Kolbe–Schmitt, ortho carboxylation, salicylic acid
20. Explain the conversion of salicylic acid into aspirin.
Salicylic acid contains a phenolic –OH that undergoes acetylation with acetic anhydride in the presence of an acid catalyst.
o-HOC₆H₄COOH + (CH₃CO)₂O → o-CH₃COOC₆H₄COOH + CH₃COOH
The product is acetylsalicylic acid (aspirin).
Examination Keywords: acetylation, acetic anhydride, aspirin, phenolic ester
Part C: Long-Answer Questions and Model Answers
Practice these important long-answer questions using brief explanations, essential scientific terms, and suitable examples. Each model answer presents the points generally expected in a university examination.
1. Discuss the preparation, properties and important chemical reactions of phenol.
Phenol (C₆H₅OH) is an aromatic hydroxy compound in which –OH is directly attached to the benzene ring.
A. Preparation — Cumene process:
Cumene (isopropylbenzene) is oxidized by air to cumene hydroperoxide, which on acid cleavage gives phenol and acetone.
C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH → H⁺ → C₆H₅OH + (CH₃)₂CO
B. Acidic character:
Phenol is weakly acidic because the phenoxide ion is resonance-stabilized.
C₆H₅OH ⇌ C₆H₅O⁻ + H⁺
It reacts with NaOH:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
It does not normally react with NaHCO₃.
C. Electrophilic substitution:
The –OH group has a strong +R effect and is strongly activating and ortho–para directing.
Bromination:
C₆H₅OH + 3Br₂ → 2,4,6-C₆H₂Br₃OH + 3HBr
Nitration:
C₆H₅OH + 3HNO₃ → 2,4,6-(NO₂)₃C₆H₂OH + 3H₂O
D. Kolbe–Schmitt reaction:
C₆H₅ONa + CO₂ → o-HOC₆H₄COONa → H⁺ → o-HOC₆H₄COOH
Product: salicylic acid.
E. Reimer–Tiemann reaction:
C₆H₅OH + CHCl₃/NaOH → o-HOC₆H₄CHO
Product: salicylaldehyde.
F. Reduction:
C₆H₅OH + Zn → C₆H₆ + ZnO
Conclusion: Phenol is an important aromatic compound whose chemistry is governed by the resonance donation of –OH. Its acidity, ortho–para direction and named reactions make it highly important in pharmaceutical organic chemistry.
Examination Keywords: phenoxide resonance, acidity, +R effect, electrophilic substitution, Kolbe–Schmitt, Reimer–Tiemann
2. Explain the acidic character of phenol and discuss its important electrophilic substitution reactions.
Phenol is more acidic than aliphatic alcohols because its conjugate base, phenoxide ion, is resonance-stabilized. The negative charge is delocalized from oxygen into the aromatic ring. The greater stability of phenoxide favours ionization.
C₆H₅OH ⇌ C₆H₅O⁻ + H⁺
Phenol reacts with NaOH:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
It does not normally react with NaHCO₃ because phenol is weaker than carbonic acid.
Electrophilic substitution reactions:
1. Bromination — Bromine water gives 2,4,6-tribromophenol.
C₆H₅OH + 3Br₂ → 2,4,6-C₆H₂Br₃OH + 3HBr
2. Nitration — Vigorous nitration gives picric acid.
C₆H₅OH + 3HNO₃ → 2,4,6-(NO₂)₃C₆H₂OH + 3H₂O
3. Kolbe–Schmitt reaction — Carboxylation of sodium phenoxide gives salicylic acid after acidification.
C₆H₅ONa + CO₂ → o-HOC₆H₄COONa → H⁺ → o-HOC₆H₄COOH
4. Reimer–Tiemann reaction — Formylation gives salicylaldehyde.
C₆H₅OH + CHCl₃/NaOH → o-HOC₆H₄CHO
The –OH group therefore controls both the reactivity and orientation of phenol through resonance donation.
Examination Keywords: phenoxide resonance, acidity, +R effect, electrophilic substitution, Kolbe–Schmitt, Reimer–Tiemann
3. Discuss the preparation, basicity and important chemical reactions of aniline.
Aniline (C₆H₅NH₂) is the simplest primary aromatic amine.
A. Preparation by reduction of nitrobenzene:
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
B. Basicity:
Aniline is less basic than aliphatic amines because the nitrogen lone pair is delocalised into the benzene ring by resonance. Therefore, the lone pair is less available for protonation.
C. Acetylation:
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
Product: acetanilide.
D. Bromination:
C₆H₅NH₂ + 3Br₂ → 2,4,6-C₆H₂Br₃NH₂ + 3HBr
Product: 2,4,6-tribromoaniline.
E. Diazotization:
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Condition: 0–5 °C.
F. Electrophilic substitution:
The free –NH₂ group is strongly activating and ortho–para directing because of resonance donation. In strongly acidic medium, however, protonation gives –NH₃⁺, which is deactivating and meta directing.
Thus, the chemistry of aniline is governed by the availability of the nitrogen lone pair.
Examination Keywords: aniline, resonance, basicity, acetylation, acetanilide, nitration, para product, hydrolysis
4. Explain the basicity of aniline and the importance of amino-group protection.
Aniline is less basic than aliphatic amines because the nitrogen lone pair is conjugated with the benzene ring.
C₆H₅NH₂ + H⁺ ⇌ C₆H₅NH₃⁺
For controlled electrophilic substitution, aniline is commonly protected by acetylation:
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
The product, acetanilide, is less strongly activating than aniline and permits more controlled nitration. Nitration gives predominantly p-nitroacetanilide:
C₆H₅NHCOCH₃ + HNO₃ → p-NO₂C₆H₄NHCOCH₃ + H₂O
Hydrolysis regenerates the amino group:
p-NO₂C₆H₄NHCOCH₃ + H₂O → p-NO₂C₆H₄NH₂ + CH₃COOH
Therefore, protection is an important synthetic strategy for controlling reactivity and regioselectivity.
Examination Keywords: aniline, resonance, basicity, acetylation, acetanilide, nitration, para product, hydrolysis
5. Discuss the preparation and synthetic applications of aryl diazonium salts.
Aryl diazonium salts contain the functional group Ar–N₂⁺ and are among the most versatile intermediates in aromatic organic chemistry.
A. Preparation — Diazotization:
Aniline is treated with NaNO₂ and HCl at 0–5 °C.
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
The reaction is kept cold because the diazonium salt can decompose at higher temperature.
B. Important replacement reactions:
1. Sandmeyer chlorination:
C₆H₅N₂⁺Cl⁻ + CuCl → C₆H₅Cl + N₂↑
2. Sandmeyer bromination:
C₆H₅N₂⁺Cl⁻ + CuBr → C₆H₅Br + N₂↑
3. Sandmeyer cyanation:
C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂↑
4. Iodination:
C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂↑ + KCl
5. Hydrolysis:
C₆H₅N₂⁺Cl⁻ + H₂O → Δ → C₆H₅OH + N₂↑ + HCl
6. Reduction/deamination:
C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂↑ + H₃PO₃ + HCl
7. Balz–Schiemann fluorination:
C₆H₅N₂⁺BF₄⁻ → Δ → C₆H₅F + BF₃ + N₂↑
C. Azo coupling:
ArN₂⁺ + Ar′H → Ar–N=N–Ar′ + H⁺
The loss of stable N₂ makes diazonium substitutions favourable.
Conclusion: Aryl diazonium salts allow the preparation of aromatic derivatives that may be difficult to obtain by direct substitution and are therefore highly important in pharmaceutical organic chemistry.
Examination Keywords: diazonium salt, substitution, named reaction, major product, N₂ evolution
6. Discuss the important reactions of benzenediazonium chloride with equations.
Benzenediazonium chloride, C₆H₅N₂⁺Cl⁻, is a versatile intermediate.
1. Chlorination — CuCl:
C₆H₅N₂⁺Cl⁻ + CuCl → C₆H₅Cl + N₂↑
2. Bromination — CuBr:
C₆H₅N₂⁺Cl⁻ + CuBr → C₆H₅Br + N₂↑
3. Cyanation — CuCN:
C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂↑
4. Iodination — KI:
C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂↑ + KCl
5. Hydrolysis:
C₆H₅N₂⁺Cl⁻ + H₂O → Δ → C₆H₅OH + N₂↑ + HCl
6. Reduction — H₃PO₂:
C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂↑ + H₃PO₃ + HCl
7. Fluorination — Balz–Schiemann:
C₆H₅N₂⁺BF₄⁻ → Δ → C₆H₅F + BF₃ + N₂↑
8. Azo coupling:
C₆H₅N₂⁺ + C₆H₅OH → azo compound + H⁺
These reactions illustrate why the diazonium group is considered an exceptionally useful synthetic handle in aromatic chemistry.
Examination Keywords: diazonium salt, substitution, named reaction, major product, N₂ evolution
7. Discuss benzoic acid with respect to preparation, acidity, reactions and directing effect.
Benzoic acid (C₆H₅COOH) is the simplest aromatic carboxylic acid.
A. Preparation by side-chain oxidation:
Suitable alkylbenzenes containing a benzylic hydrogen undergo strong oxidation. Toluene, for example, gives benzoic acid:
C₆H₅CH₃ + 3[O] → C₆H₅COOH + H₂O
B. Acidity:
C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺
The benzoate ion is resonance-stabilized over two oxygen atoms. Electron-withdrawing substituents generally increase acidity, while electron-donating groups generally decrease it.
C. Reaction with NaOH:
C₆H₅COOH + NaOH → C₆H₅COONa + H₂O
D. Reaction with NaHCO₃:
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑
The effervescence of CO₂ is a characteristic test for carboxylic acids.
E. Electrophilic substitution:
The –COOH group is strongly deactivating and meta-directing because it withdraws electron density from the aromatic ring through –I and –R effects. Therefore, nitration and halogenation of benzoic acid generally favour the meta position.
F. Pharmaceutical/formulation importance:
Benzoic acid and benzoates have antimicrobial preservative applications in suitable formulations.
Conclusion: The carboxyl group controls the acidity, salt formation, deactivation and meta-directing behaviour of benzoic acid.
Examination Keywords: aromatic acid, preparation, balanced reaction, directing effect, pharmaceutical relevance
8. Discuss salicylic acid, its preparation, reactions and conversion into aspirin.
Salicylic acid is 2-hydroxybenzoic acid, o-HOC₆H₄COOH.
A. Preparation — Kolbe–Schmitt reaction:
Sodium phenoxide reacts with CO₂ under heat and pressure:
C₆H₅ONa + CO₂ → o-HOC₆H₄COONa
Acidification:
o-HOC₆H₄COONa + H⁺ → o-HOC₆H₄COOH
B. Structural features:
Salicylic acid contains a phenolic –OH and a carboxylic –COOH group. Both functional groups contribute to its chemical reactivity.
C. Reaction with NaOH:
o-HOC₆H₄COOH + NaOH → o-HOC₆H₄COONa + H₂O
D. Conversion into aspirin:
The phenolic hydroxyl group is acetylated with acetic anhydride:
o-HOC₆H₄COOH + (CH₃CO)₂O → o-CH₃COOC₆H₄COOH + CH₃COOH
Product: acetylsalicylic acid (aspirin).
E. Pharmaceutical relevance:
Salicylic acid is important in topical dermatological preparations, particularly for its keratolytic activity. Aspirin is a major medicinal derivative of salicylic acid.
Conclusion: Salicylic acid is an important example connecting aromatic acid chemistry, phenolic chemistry and pharmaceutical chemistry.
Examination Keywords: aromatic acid, preparation, balanced reaction, directing effect, pharmaceutical relevance