Directing Effects
| Group Type | Examples | Effect on Rate | Directs to | Mechanism |
|---|---|---|---|---|
| Activating, o/p directors | –OH, –NH₂, –OR, –NHR, –alkyl | Faster than benzene | ortho and para | Donate electrons into ring (+M or +I); increase electron density at o and p positions |
| Deactivating, o/p directors | –F, –Cl, –Br, –I (halogens) | Slower than benzene | ortho and para | –I effect deactivates; +M of lone pair on halogen directs to o/p |
| Deactivating, m directors | –NO₂, –CN, –COOH, –CHO, –SO₃H, –COR | Much slower | meta | Withdraw electrons by –M; deplete o and p positions; meta positions least affected |
Why Meta Directors Direct to Meta
Electron-withdrawing groups (EWG) stabilise negative charge and destabilise positive charge. In the EAS arenium ion, the positive charge is delocalized at ortho and para positions. An EWG at these positions is particularly destabilizing — ortho and para attack is even less favorable. Meta attack puts the positive charge away from the EWG → meta product forms preferentially (by elimination, not because meta is activated).
Phenol (C₆H₅OH)
Properties and Reactions
Phenol is a weak acid (pKa 9.95) — weaker than carboxylic acids (~5) but stronger than alcohols (~16). The O–H bond is weakened by delocalisation of O lone pair into the ring, making H⁺ more easily lost. Phenoxide ion (C₆H₅O⁻) is stabilised by resonance.
Uses of Phenol
Antiseptic (original Lister disinfectant); manufacture of Bakelite (phenol-formaldehyde resin); aspirin (acetylsalicylic acid) synthesis; nylon-6,6; dyes; pharmaceuticals.
Toluene (Methylbenzene, C₆H₅CH₃)
Reactions
Methyl group is an o/p director and activator. Nitration gives mixture of ortho and para mononitrotoluene. Further nitration gives 2,4,6-trinitrotoluene (TNT) — a high explosive.
Benzaldehyde (C₆H₅CHO)
Properties and Reactions
Aromatic aldehyde — has properties of both benzene ring and aldehyde group. –CHO is a meta director and deactivator.
Uses: almond flavouring, perfumes, pharmaceutical synthesis.
Benzoic Acid (C₆H₅COOH)
Properties
pKa = 4.19. Stronger acid than aliphatic carboxylic acids (pKa ~5) due to resonance stabilisation of benzoate anion (negative charge spread over ring and carboxylate). Sparingly soluble in cold water; soluble in hot water, ethanol.
Uses: food preservative (E210); sodium benzoate (E211) in soft drinks; manufacture of plasticisers, dyes.
Aniline (Aminobenzene, C₆H₅NH₂) & Azo Dyes
Aniline Properties
Aniline is a weaker base than aliphatic amines (pKb ~9 vs ~4) because the lone pair on N is delocalised into the benzene ring, making it less available to accept H⁺.
Azo Dye Synthesis
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Exercises
- Predict the major product when toluene is nitrated. Explain your answer in terms of directing effects.Major products: 2-nitrotoluene (ortho) and 4-nitrotoluene (para). The –CH₃ methyl group is an ortho/para director and activator (donates electron density by +I effect, increasing ring electron density at ortho and para positions). Electrophile NO₂⁺ preferentially attacks these positions. The para isomer usually predominates over ortho (less steric hindrance).
- How would you distinguish between phenol and benzoic acid using simple chemical tests?Test 1 — NaHCO₃: benzoic acid reacts, giving CO₂ gas: C₆H₅COOH + NaHCO₃ → C₆H₅COO⁻Na⁺ + H₂O + CO₂↑. Phenol does NOT react (pKa 9.95 — too weak to react with HCO₃⁻, pKa 6.35).
Test 2 — FeCl₃(aq): phenol gives violet/purple colour; benzoic acid gives no colour (or pale brown iron complex).
Test 3 — Bromine water: phenol gives immediate white precipitate (2,4,6-tribromophenol); benzoic acid gives no reaction with Br₂(aq) without catalyst. - Explain why aniline is a weaker base than methylamine (CH₃NH₂).In methylamine, the lone pair on N is fully available to donate to H⁺ (accept proton). In aniline, the lone pair on N is delocalised into the benzene π-system (conjugation: N lone pair overlaps with ring p-orbitals). This reduces its availability to act as a proton acceptor. The resulting aniline is a much weaker base (pKb ~9) than methylamine (pKb ~3.4). The anilinium ion C₆H₅NH₃⁺ is also stabilised by resonance.
- Describe the synthesis of an azo dye from aniline and phenol. Write equations for both steps.Step 1 — Diazotisation (0–5°C, ice bath): C₆H₅NH₂ + NaNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + NaCl + H₂O. Must keep below 5°C — diazonium salt decomposes above this temperature.
Step 2 — Azo coupling (alkaline solution, 0–5°C): C₆H₅N₂⁺ + C₆H₅O⁻ → C₆H₅–N=N–C₆H₄O⁻ + H⁺ (orange azo compound). The diazonium ion acts as electrophile; phenoxide (alkaline) acts as nucleophile/coupling component at the para position. The N=N group (chromophore) absorbs visible light causing the colour. - Why does nitrobenzene give mainly meta-dinitrobenzene on further nitration?The –NO₂ group is a deactivating meta director. It withdraws electrons from the ring by –M (resonance) and –I (induction). In the EAS arenium ion intermediate, the positive charge is delocalised at ortho and para positions. An EWG (–NO₂) at these positions is additionally destabilising. Meta attack places the positive charge away from the existing –NO₂ group (least destabilised intermediate) → meta product is formed preferentially, giving 1,3-dinitrobenzene.
- Phenol reacts with bromine water to give a white precipitate but benzene does not. Explain.The –OH group in phenol is a powerful o/p activating group (lone pair on O donates into ring by +M effect, greatly increasing electron density at ortho and para positions). This activates the ring so much that Br₂ alone (without a Lewis acid catalyst) is electrophilic enough to attack. Three substitutions occur: 2,4,6-tribromophenol precipitates. Benzene's ring is not activated — Br₂ alone cannot overcome benzene's delocalisation energy; FeBr₃ catalyst is needed. The reaction with phenol is a sensitive and specific test for phenol.
Quiz — 25 Questions
Unit 5: Benzene Derivatives
25 QsAn ortho/para director:
The –NO₂ group is a meta director because:
Phenol reacts with NaHCO₃:
Phenol gives a white precipitate with bromine water because:
Toluene is oxidised by hot acidic KMnO₄ to give:
In the diazotisation step of azo dye synthesis, the temperature must be kept at 0–5°C because:
Aniline is less basic than methylamine because:
The chromophore in an azo dye is:
Which test distinguishes phenol from benzoic acid?
What major product forms when aniline is treated with excess Br₂(aq)?
Halogens (–F, –Cl, –Br, –I) are unusual directing groups because they are:
Benzaldehyde gives a positive Tollens' test but not a positive Fehling's test because:
The product of coupling benzenediazonium chloride with phenol in alkaline solution is:
Benzoic acid is a stronger acid than phenol because:
The side-chain halogenation of toluene in UV light gives primarily:
Which compound reacts with FeCl₃(aq) to give a violet colour?
Phenol is a weaker acid than benzoic acid but stronger than ethanol. This is because:
Why does –COOH group direct incoming electrophiles to the meta position?
TNT (2,4,6-trinitrotoluene) is made by:
The acetylation of aniline (with ethanoic anhydride) is useful in synthesis because:
Azo dyes are used in textiles because:
The correct order of acid strength is:
What is the structural feature common to all azo dyes?
When toluene undergoes side-chain halogenation with Cl₂/UV, the mechanism is:
Sodium benzoate is used as a food preservative because:
Unit 5 Quiz — Benzene Derivatives (25 Questions)
Select one answer eachThe -OH group attached to a benzene ring makes the ring:
Phenol is more acidic than ethanol because:
Phenol reacts with bromine water (no catalyst needed) to give:
Nitration of methylbenzene (toluene) gives mainly:
The -NO₂ group on a benzene ring is a:
TNT (2,4,6-trinitrotoluene) is produced by:
Chlorobenzene undergoes electrophilic substitution less readily than benzene because:
In the Friedel-Crafts acylation of benzene, the electrophile is:
Which reagents convert benzene into phenylamine (aniline)?
The amino group (-NH₂) attached to benzene ring is:
Phenylamine reacts with HCl to form:
Diazonium salts are formed by reacting phenylamine with:
Coupling of diazonium salt with phenol in alkaline solution gives:
The sulfonation of benzene uses:
Which product forms when benzene reacts with CH₃Cl in the presence of AlCl₃?
Aspirin is produced by reacting salicylic acid with:
Which compound is used as an antiseptic and is a chlorinated phenol?
The reaction of phenol with concentrated HNO₃/H₂SO₄ gives:
Halogenation of benzene requires a halogen carrier catalyst (e.g. FeBr₃) because:
What is the role of concentrated H₂SO₄ in the nitration of benzene?
Paracetamol is synthesised from:
Which group on benzene ring directs electrophile to meta position?
The reaction of a diazonium salt with CuCN gives:
Phenyl ethanoate (phenyl acetate) is formed by reacting phenol with:
Sulfonamide drugs work because they structurally resemble:
Unit Test — 50 marks
Section A — Short Answer
30 marksClassify the following groups as ortho/para directors or meta directors, and as activators or deactivators: (a) –OH; (b) –NO₂; (c) –Cl; (d) –COOH; (e) –CH₃. [5]
Describe three chemical tests to identify phenol, giving observations for each. [6]
Describe the synthesis of an orange azo dye from aniline and phenol. Give equations for both steps, state the conditions, and name the chromophore. [5]
Explain, with equations, why aniline is a weaker base than ethylamine but a stronger acid than cyclohexylamine. Use pKb values where relevant. [4]
Benzaldehyde is treated with: (a) Tollens' reagent [2]; (b) 2,4-DNPH [1]; (c) NaBH₄ [1]; (d) HCN [1]. Describe the observation and/or product for each. [5]
Compare the acidity of phenol, benzoic acid, and ethanol. Explain the differences in terms of the stability of their conjugate bases. Include pKa values. [5]
Section B — Extended Response
20 marks(a) Explain the concept of directing effects in electrophilic aromatic substitution. Distinguish clearly between ortho/para directors and meta directors in terms of their electron-donating/withdrawing ability. [4]
(b) Predict and explain the major product(s) of: (i) nitration of phenol; (ii) nitration of nitrobenzene; (iii) nitration of chlorobenzene. [6]
(b) (i) Phenol + HNO₃/H₂SO⃨: −OH is o/p director and activator. Expected products: 2-nitrophenol and 4-nitrophenol (ortho and para). In dilute conditions, 4-nitrophenol is major (para preferred, less steric hindrance). [2] (ii) Nitrobenzene + HNO₃/H₂SO⃨: −NO₂ is meta director. Major product: 1,3-dinitrobenzene. Ring is also deactivated → harsh conditions (fuming acids, high T) needed. [2] (iii) Chlorobenzene + HNO₃/H₂SO⃨: −Cl is o/p director (deactivator). Products: 2-chloronitrobenzene and 4-chloronitrobenzene. Para predominates (less steric crowding; also 4 gives more stable arenium ion). Rate slower than benzene due to −I deactivation by Cl. [2]
(a) Describe the physical and chemical properties of phenol that demonstrate both its acidic and aromatic character. [5]
(b) Compare the chemistry of the following four benzene derivatives, highlighting one distinctive reaction for each: phenol, toluene, benzaldehyde, aniline. [5]
(b) Phenol: distinctive reaction — bromination with Br₂(aq) (no catalyst) giving white 2,4,6-tribromophenol ppt. −OH activates ring so strongly that no Lewis acid is needed; all three o/p positions substituted. [~1.25] Toluene: distinctive — KMnO⃨/H⁺/heat: side chain oxidised to COOH regardless of chain length → benzoic acid. Also: UV/Cl₂ gives benzyl chloride (free radical, not EAS). [~1.25] Benzaldehyde: distinctive — positive Tollens' test (silver mirror) but negative Fehling's. Aromatic aldehydes are oxidised by Ag+ but not by Cu²+. Also 2,4-DNPH gives orange ppt. [~1.25] Aniline: distinctive — diazotisation + azo coupling to form azo dyes. C₆H₅NH₂ + NaNO₂/HCl (0–5°C) → C₆H₅N₂+Cl⁻; coupling with phenol in alkaline solution → orange azo compound. Also: acetylation to protect −NH₂ group in synthesis. [~1.25]