CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Write the IUPAC name of the following compound:
H₃C–C(CH₃)=C(Br)–CH₂–OH [CBSE (AI) 2017]
Answer:
2-Bromo-3-methylbut-2-en-1-ol

Teacher's Note:
1. Number the chain from the end nearer to –OH (the principal group), even though that gives the double bond and substituents their locants afterwards.

Question 2. Write the IUPAC name of the following compound:
C₆H₅–CH=CH–CH₂–OH [CBSE (AI) 2017]
Answer:
3-phenylprop-2-en-1-ol

Teacher's Note:
1. This is cinnamyl alcohol; the phenyl ring is named as a substituent on the three-carbon chain carrying the –OH.

Question 3. Write the IUPAC name of the following compound:
CH₃–O–CH₂–CH(OH)–CH₃ [CBSE Chennai 2015]
Answer:
1-methoxypropan-2-ol.

Teacher's Note:
1. When –OH and an ether linkage are both present, the alcohol is the principal group and –OCH₃ becomes the methoxy prefix.

Question 4. How would you convert ethanol to ethene? [CBSE (AI) 2011]
Answer:
CH₃CH₂OH (Ethanol) →^{conc. H₂SO₄, 443 K} CH₂=CH₂ (Ethene) + H₂O

Teacher's Note:
1. Temperature decides the product: at 443 K ethanol gives ethene, but at 413 K with excess ethanol it gives diethyl ether.

Question 5. How will you synthesise the following alcohol from appropriate alkene:
(CH₃)₂C(OH)CH₂CH₃ [CBSE 2019 (56/4/1)]
Answer:
CH₃–C(CH₃)=CHCH₃ →^{H₂O, H+} CH₃–C(OH)(CH₃)–CH₂CH₃

Teacher's Note:
1. Acid-catalysed hydration follows Markovnikov’s rule: the –OH goes to the carbon that forms the more stable (here 3°) carbocation.

Question 6. o-Nitrophenol has lower boiling point than p-nitrophenol. Explain.
Answer:
Due to intramolecular H-bonding o-nitrophenol exists as a discrete molecule whereas due to intermolecular H-bonding p-nitrophenol exists as associated molecules. As a result of this p-nitrophenol has higher boiling point than o-nitrophenol.

Teacher's Note:
1. The same intramolecular H-bond makes o-nitrophenol steam volatile, which is how the two isomers are separated.

Question 7. Of the two hydroxy organic compounds ROH and R′OH, the first one is basic and the other is acidic in behaviour. How is R different from R′?
Answer:
R is an alkyl group whereas R′ is an aryl group.

Teacher's Note:
1. An aryl group makes the –OH acidic because the phenoxide ion formed is resonance stabilised; an alkyl group (+I) does the opposite.

Question 8. Arrange the following compounds in the increasing order of their acid strengths:
4-nitrophenol, phenol, 2,4,6-trinitrophenol [CBSE (F) 2013]
Answer:
Phenol < 4-nitrophenol < 2,4,6-trinitrophenol

Teacher's Note:
1. Each extra –NO₂ group at an ortho/para position stabilises the phenoxide ion further, so picric acid is the strongest.

Question 9. Phenol is an acid but does not react with sodium bicarbonate solution. Why?
Answer:
Phenol is a weaker acid than carbonic acid (H₂CO₃) and hence does not liberate CO₂ from sodium bicarbonate.

Teacher's Note:
1. This gives a simple test: carboxylic acids give brisk effervescence with NaHCO₃, phenols do not.

Question 10. Suggest a reagent for conversion of ethanol to ethanoic acid.
Answer:
(i) CrO₃–H₂SO₄
(ii) KMnO₄/OH⁻, (iii) H₃O⁺
(iv) Acidified KMnO₄ or K₂Cr₂O₇

Teacher's Note:
1. A strong oxidising agent is needed to take the 1° alcohol past the aldehyde stage; PCC would stop at ethanal.

Question 11. How is toluene obtained from phenol? [CBSE Delhi 2013C]
Answer:
Phenol →^{Zn (dust), Δ} Benzene →^{+CH₃Cl/Anhyd. AlCl₃} Toluene

Teacher's Note:
1. Zinc dust removes the oxygen of phenol (C₆H₅OH + Zn → C₆H₆ + ZnO); the methyl group is then added by Friedel–Crafts alkylation.

Question 12. Suggest a reagent for the following conversion:
CH₃–CH=CH–CH(OH)–CH₂–CH₃ → CH₃–CH=CH–CO–CH₂–CH₃
Answer:
CrO₃, Pyridine and HCl (Pyridinium chlorochromate)

Teacher's Note:
1. PCC oxidises the alcohol without touching the C=C double bond; KMnO₄ would attack the double bond too.

Question 13. What is denatured alcohol?
Answer:
Alcohol is made unfit for drinking by mixing some copper sulphate and pyridine in it. This is called denatured alcohol.

Teacher's Note:
1. Copper sulphate gives a colour and pyridine a foul smell, so the alcohol cannot be mistaken for a drink.

Question 14. Explain why sodium metal can be used for drying diethyl ether but not ethyl alcohol.
Answer:
Due to presence of an active hydrogen atom, ethyl alcohol reacts with sodium metal.
2CH₃–CH₂–OH + 2Na → 2CH₃–CH₂–O⁻Na⁺ + H₂
Diethyl ether, on the other hand, does not have replaceable hydrogen atom therefore does not react with sodium metal hence can be dried by metallic sodium.

Teacher's Note:
1. Sodium removes traces of water from ether without reacting with the ether itself.

Question 15. What is the order of dehydration of primary, secondary and tertiary alcohols?
Answer:
Tertiary alcohols > Secondary alcohols > Primary alcohols.

Teacher's Note:
1. Dehydration goes through a carbocation, so the order follows carbocation stability.

Short Answer Questions (2 Marks)

Question 16. How will you convert:
(i) Propene to Propan-1-ol?
(ii) Ethanal to Propan-2-ol? [CBSE Delhi 2013]
Answer:
(i) CH₃CH=CH₂ (Propene) + HBr →^Peroxide CH₃CH₂CH₂Br →^{KOH (aq), Δ} CH₃CH₂CH₂OH (Propan-1-ol)
(ii) CH₃CHO (Ethanal) →^CH₃MgI (CH₃)₂CH–OMgI →^{H+, H₂O} (CH₃)₂CH–OH (Propan-2-ol)

Teacher's Note:
1. In (i) the peroxide effect (anti-Markovnikov addition) is needed so that Br, and later OH, ends up on the terminal carbon.

Question 17. Which is a stronger acid—phenol or cresol? Explain.
Answer:
All the cresols are weaker acids than phenols. Methyl group has +I effect (positive inductive effect) as well as hyperconjugation effect but the hyperconjugation effect predominates over the +I effect. Since both these effects increase the electron density on the O–H bond and hence all the cresols are weaker acids than phenols.
As hyperconjugation effect can operate only through ortho and para positions and not through meta positions, therefore, meta-cresol is stronger acid than ortho and para-cresols. However, due to stronger +I effect at ortho position than at para position (+I effect decreases with distance), ortho-cresol is a weaker acid than para-cresol. Thus, the order of acidic strength in increasing order is:
ortho-cresol < para-cresol < meta-cresol < phenol

Teacher's Note:
1. Electron-releasing groups destabilise the phenoxide ion, so any methyl-substituted phenol is a weaker acid than phenol itself.

Question 18. Arrange: water, ethanol and phenol in increasing order of acidity and give reason for your answer.
Answer:
Increasing order of acidity is ethanol < water < phenol. The phenoxide ion obtained after the removal of a proton is stabilised by resonance whereas the ethoxide ion obtained after the removal of a proton is destabilised by +I effect of –C₂H₅ group. Therefore, phenol is stronger acid than ethanol. On the other hand, ethanol is weaker acid than water because electron releasing –C₂H₅ group in ethanol increases the electron density on oxygen and hence the polarity of O–H bond in ethanol decreases which results in the decreasing acidic strength. Hence, acidic strength increases in the order given above.

Teacher's Note:
1. Always reason through the stability of the anion formed: resonance-stabilised phenoxide > hydroxide > ethoxide (destabilised by +I).

Question 19. Answer the following questions.
(i) Among HI, HBr, HCl, HI is most reactive towards alcohols. Why?
(ii) Of the two alcohols; (a) CH₂=CH–CH₂OH and (b) CH₂=CH–CH₂–CH₂OH, which one will react more easily with conc. HCl in the presence of ZnCl₂? [CBSE (F) 2012]
Answer:
(i) HI has the lowest bond dissociation energy due to longer bond length that is why it is most reactive.
(ii) CH₂=CH–CH₂OH

Teacher's Note:
1. In (ii) the allyl alcohol reacts faster because the allylic carbocation formed is stabilised by resonance; the other alcohol gives an unstabilised 1° carbocation.

Question 20. Account for the following:
(i) Phenol does not react with NaHCO₃ whereas carboxylic acids react.
(ii) Phenol is more easily nitrated than benzene.
Answer:
(i) RCOOH + NaHCO₃ → RCOONa + H₂CO₃ (weaker acid than RCOOH)
C₆H₅OH (Phenol) + NaHCO₃ ↛ C₆H₅O⁻Na⁺ (Sodium phenoxide ion) + H₂CO₃ (stronger acid than phenol)
Such acid–base reaction takes place only if the acid formed is weaker than the reacting acid. In other words, phenol is not a strong enough acid to react with NaHCO₃.
(ii) Nitration involves attack of electrophile nitronium ion (NO₂⁺) on benzene ring. Due to +R effect of –OH group electron density on benzene ring increases. Therefore, phenol is more easily nitrated as compared to benzene.

Teacher's Note:
1. Phenol is nitrated even by dilute HNO₃ at room temperature, whereas benzene needs a conc. HNO₃/H₂SO₄ mixture.

Question 21. Answer the following questions:
(i) Dipole moment of phenol is smaller than that of methanol. Why?
(ii) In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why?
Answer:
(i) In phenol, C–O bond is less polar due to electron-withdrawing effect of benzene ring whereas in methanol, C–O bond is more polar due to electron-releasing effect of –CH₃ group.
(ii) Phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution and hence undergoes electrophilic substitution with carbon dioxide which is a weak electrophile.

Teacher's Note:
1. The negative charge on oxygen in phenoxide pushes even more electron density into the ring than a neutral –OH, which is needed for the weak electrophile CO₂.

Question 22. Write the mechanism of the following reaction:
2CH₃CH₂OH →^{conc. H₂SO₄, 413 K} CH₃–CH₂–O–CH₂–CH₃ [CBSE Delhi 2016, 2019 (56/4/3), 2020 (56/3/2)]
Answer:
The formation of ether is a nucleophilic bimolecular reaction (S_N2) involving the attack of alcohol molecule on protonated alcohol as shown below:
(i) CH₃–CH₂–OöH#776;H (Ethanol) + H⁺ → CH₃–CH₂–O⁺H₂
(ii) CH₃–CH₂–OöH#776;H + CH₃–CH₂–O⁺H₂ → CH₃–CH₂–O⁺(H)–CH₂–CH₃ + H₂O
(the lone pair on O of the ethanol molecule attacks the CH₂ carbon of the protonated alcohol and water leaves)
(iii) CH₃–CH₂–O⁺(H)–CH₂–CH₃ → CH₃–CH₂–O–CH₂–CH₃ (Diethyl ether) + H⁺

Teacher's Note:
1. In curved-arrow answers, show the arrow from the O lone pair to the carbon, and from the C–O bond to the leaving O of the protonated alcohol.

Question 23. What happens when
(i) Phenol reacts with Conc. HNO₃?
(ii) Ethyl chloride reacts with NaOC₂H₅?
Write the chemical equations involved in the above reactions. [CBSE 2019 (56/4/3)]
Answer:
(i) C₆H₅OH (Phenol) →^{conc. HNO₃} 2,4,6-Trinitrophenol (picric acid, –NO₂ groups at both ortho positions and the para position to –OH)
(ii) CH₃CH₂–Cl (Ethyl chloride) + CH₃–CH₂O⁻Na⁺ (Sodium ethoxide) → CH₃CH₂–O–CH₂–CH₃ (Diethyl ether) + NaCl

Teacher's Note:
1. Reaction (ii) is Williamson synthesis; it works well here because ethyl chloride is a 1° halide.

Question 24. How would you obtain
(i) picric acid (2,4,6-trinitrophenol) from phenol?
(ii) 2-Methylpropene from 2-Methylpropanol? [CBSE Delhi 2011]
Answer:
(i) Phenol →^{H₂SO₄ (conc.)} Phenol-2,4-disulphonic acid →^{HNO₃ (conc.)} Picric acid (2,4,6-trinitrophenol)
(ii) CH₃–CH(CH₃)–CH₂OH (2-Methylpropanol) →^+SOCl₂ CH₃–CH(CH₃)–CH₂–Cl →^{+KOH (alc.)} CH₃–C(CH₃)=CH₂ (2-methylpropene)

Teacher's Note:
1. Sulphonating first protects the ring from oxidation by conc. HNO₃; the –SO₃H groups are then replaced by –NO₂ groups.

Question 25. (i) Predict the major product of acid catalysed dehydration of 1-Methylcyclohexanol.
(ii) You are given benzene, conc. H₂SO₄, NaOH and dil. HCl. Write the preparation of phenol using these reagents. [CBSE SQP 2015]
Answer:
(i) 1-Methylcyclohexene
(ii) C₆H₆ \( \xrightarrow{\text{conc. }H_2SO_4,\ \Delta} \) C₆H₅SO₃H \( \xrightarrow{NaOH,\ \text{fuse},\ 575\,K} \) C₆H₅ONa \( \xrightarrow{\text{dil. }HCl} \) C₆H₅OH

Teacher's Note:
1. In (i) the double bond forms towards the ring carbon carrying CH₃, giving the more substituted (more stable) alkene – Saytzeff’s rule.

Question 26. Draw the structures of any two isomeric alcohols (other than 1° alcohols) having molecular formula C₅H₁₂O and give their IUPAC names. [CBSE SQP 2015]
Answer:
Any two isomers out of the following:
(i) CH₃–CH₂–CH₂–CH(OH)–CH₃   Pentan-2-ol
(ii) CH₃–CH₂–CH(OH)–CH₂–CH₃   Pentan-3-ol
(iii) CH₃–CH(CH₃)CH(OH)–CH₃   3-Methylbutan-2-ol
(iv) CH₃–CH₂–C(CH₃)(OH)–CH₃   2-Methylbutan-2-ol

Teacher's Note:
1. Check each structure: the carbon carrying OH must be attached to two (2°) or three (3°) other carbons; (i)–(iii) are secondary and (iv) is tertiary.

Short Answer Questions (3 Marks)

Question 27. Explain the mechanism of the following reactions:
(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by hydrolysis.
(ii) Acid catalysed dehydration of an alcohol forming an alkene.
OR
Give the mechanism for the formation of ethene from ethanol. [CBSE 2020 (56/1/2)]
(iii) Acid catalysed hydration of an alkene forming an alcohol.
Answer:
(i) Step I: Nucleophilic addition of Grignard reagent to carbonyl group.
>C^δ+=O^δ− + R^δ−–Mg^δ+–X → [>C(R)–O⁻Mg⁺X] (Adduct)
(the alkyl group R attacks the carbonyl carbon and the π electrons of C=O shift to oxygen)
Step II: Hydrolysis
[>C(R)–O⁻Mg⁺X] →^H₂O >C(R)–OH + Mg(OH)X
(ii) The mechanism of dehydration of ethanol involves the following steps:
Step 1: Formation of protonated alcohol
CH₃–CH₂–OH (Ethanol) + H⁺ →^Fast CH₃–CH₂–O⁺H₂ (Ethyl oxonium ion)
Step 2: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction.
CH₃–CH₂–O⁺H₂ →^Slow CH₃–C⁺H₂ + H₂O
Step 3: Formation of ethene by elimination of a proton
H–CH₂–C⁺H₂ → CH₂=CH₂ (Ethene) + H⁺
The acid used in step 1 is released in step 3. To drive the equilibrium to the right, ethene is removed as it is formed.
(iii) >C=C< (Alkene) + H₂O ⇌^H+ >CH–C(OH)<
Mechanism:
Step I: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.
H₂O + H⁺ → H₃O⁺
>C=C< + H–O⁺H₂ ⇌ >CH–C⁺< + H₂O
Step II: Nucleophilic attack of water on carbocation.
>CH–C⁺< + H₂O ⇌ >CH–C(O⁺H₂)<
Step III: Deprotonation to form an alcohol
>CH–C(O⁺H₂)< + H₂O → >CH–C(OH)< + H₃O⁺

Teacher's Note:
1. Note that dehydration of an alcohol (ii) and hydration of an alkene (iii) are the reverse of each other and pass through the same carbocation.

Question 28. How do you convert the following:
(i) Phenol to anisole
(ii) Propan-2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol [CBSE Delhi 2015]
Answer:
(i) C₆H₅OH (Phenol) →^{+ NaOH} C₆H₅O⁻Na⁺ →^{+ CH₃Br} C₆H₅OCH₃ (Anisole)
(ii) CH₃–CH(OH)–CH₃ (Propan-2-ol) →^PCC CH₃–CO–CH₃ →^{(i) CH₃MgBr; (ii) H₂O} CH₃–C(OH)(CH₃)–CH₃ (2-methylpropan-2-ol)
(iii) C₆H₅NH₂ (Aniline) →^{NaNO₂/HCl, 273–278 K} C₆H₅N₂⁺Cl⁻ →^{H₂O, Warm} C₆H₅OH (Phenol)

Teacher's Note:
1. To add one carbon to a 2° alcohol and make it 3°, oxidise to the ketone first, then add CH₃MgBr.

Question 29. Arrange the following compounds in decreasing order of acidity.
(i) H₂O, ROH, HC≡CH
(ii) CH₃O–C₆H₄–OH (p), O₂N–C₆H₄–OH (p), C₆H₅OH
(iii) CH₃OH, H₂O, C₆H₅OH
Answer:
(i) H₂O > ROH > HC≡CH
(ii) p-O₂N–C₆H₄–OH > C₆H₅OH > p-CH₃O–C₆H₄–OH
(iii) C₆H₅OH > H₂O > CH₃OH

Teacher's Note:
1. Check the direction asked: this question wants decreasing order, so the strongest acid comes first.

Question 30. Give one chemical test to distinguish between the following:
(i) Phenol and propan-1-ol
(ii) Ethanol and dimethyl ether
(iii) propan-1-ol and 2-Methyl-propan-2-ol [CBSE 2019 (56/5/1)]
Answer:
(i) Phenol gives a violet coloration with neutral FeCl₃ solution while propan-1-ol does not.
6C₆H₅OH (Phenol) + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ + 3H⁺ + 3HCl
(ii) Ethanol gives yellow precipitate of iodoform when warmed with NaOI (NaOH/I₂) while dimethylether does not.
CH₃CH₂OH (Ethanol) + 4I₂ + 6NaOH →^Δ CHI₃↓ (Iodoform, yellow ppt) + HCOONa + 5NaI + 5H₂O
(iii) Propan-1-ol being a 1° alcohol does not produce turbidity at room temperature on treatment with Lucas reagent (anhyd. ZnCl₂ + conc. HCl) whereas 2-methylpropan-2-ol being a 3° alcohol gives turbidity immediately at room temperature on treatment with Lucas reagent.
(CH₃)₃C–OH (2-methyl-propan-2-ol) →^{Anhyd. ZnCl₂ + conc. HCl} (CH₃)₃C–Cl (2-Chloro-2-methylpropane, turbidity appears immediately)

Teacher's Note:
1. For any distinguishing test, state what each compound does: one gives the observation and the other does not.

Question 31. An organic compound ‘A’ having molecular formula C₃H₆ on treatment with aqueous H₂SO₄ gives ‘B’ which on treatment with HCl/ZnCl₂ gives ‘C’. The compound C on treatment with ethanolic KOH gives back the compound ‘A’. Identify the compounds A, B, C.
Answer:
A = CH₃–CH=CH₂ (Propene);   B = CH₃–CH(OH)–CH₃ (Propan-2-ol);   C = CH₃–CH(Cl)–CH₃ (2-chloropropane)
CH₃–CH=CH₂ (Propene, A) →^{aq. H₂SO₄ (Hydration)} CH₃–CH(OH)–CH₃ (Propan-2-ol, B) →^HCl/ZnCl₂ CH₃–CH(Cl)–CH₃ (2-Chloropropane, C)
CH₃–CH(Cl)–CH₃ (C) →^{Ethanolic KOH} CH₃–CH=CH₂ (Propene)

Teacher's Note:
1. Hydration of propene follows Markovnikov’s rule, so B is the 2° alcohol propan-2-ol, not propan-1-ol.

Question 32. A compound ‘A’ is optically active. On mild oxidation, it gives a compound ‘B’ but on vigorous oxidation gives another compound ‘C’. C along with D is also formed from B by reaction with iodine and alkali. Deduce the structures of A, B, C, D.
Answer:
CH₃–CH₂–CH(OH)–CH₃ (A, sec-Butyl alcohol) →^{(O), Mild oxidation} CH₃–CH₂–CO–CH₃ (B, Ethyl methyl ketone)
A →^{(O), Vigorous oxidation} CH₃–CH₂–COOH (C)
B →^I₂/NaOH CH₃–CH₂–COOH (Propanoic acid, C) + CHI₃↓ (D)
A = Butan-2-ol, B = Butan-2-one, C = Propanoic acid, D = Iodoform

Teacher's Note:
1. The iodoform reaction removes the CH₃CO– carbon as CHI₃, so the acid formed has one carbon less than the ketone (it is obtained as its sodium salt and then acidified).

Question 33. Account for the following:
(i) Rectified spirit cannot be converted into absolute alcohol by simple distillation.
(ii) Diethyl ether does not react with sodium.
(iii) Phenols do not undergo substitution of the –OH group like alcohols.
Answer:
(i) Rectified spirit containing 95% ethyl alcohol and 5% water forms an azeotropic mixture which distils at a constant temperature of 351.13 K.
(ii) Since diethyl ether does not contain an active hydrogen attached to oxygen like alcohols and phenols, it does not react with sodium.
(iii) The C–O bond in phenols has some double bond character due to resonance and hence cannot be easily cleaved by nucleophile. In contrast, the C–O bond in alcohols is a pure single bond and hence can be easily cleaved by nucleophile.

Teacher's Note:
1. An azeotrope boils at a constant temperature with the same composition in liquid and vapour, so distillation cannot separate its components.

Question 34. Write the main product(s) in each of the following reactions:
(i) CH₃–C(CH₃)₂–O–CH₃ + HI →
(ii) CH₃–CH=CH₂ →^{(i) B₂H₆; (ii) 3H₂O₂/OH−}
(iii) C₆H₅–OH →^{(i) aq. NaOH; (ii) CO₂, H+} [CBSE Delhi 2016]
Answer:
(i) (CH₃)₃C–O–CH₃ (Methyl tert. butyl ether) + H–I → (CH₃)₃C–I (Tert. butyl iodide) + CH₃–OH (Methyl alcohol)
(ii) CH₃–CH=CH₂ (Propene) →^{(i) B₂H₆; (ii) 3H₂O₂/OH−} CH₃–CH₂–CH₂–OH (n-propyl alcohol)
(iii) C₆H₅OH (Phenol) →^{(i) aq NaOH; (ii) CO₂/H+} o-HO–C₆H₄–COOH (Salicylic acid)

Teacher's Note:
1. Hydroboration–oxidation gives the anti-Markovnikov alcohol (propan-1-ol), whereas acid-catalysed hydration of propene gives propan-2-ol.

Question 35. How will you bring the following conversions?
(i) Phenol to benzyl alcohol
(ii) Phenol to m-bromophenol
(iii) Phenol to aspirin.
Answer:
(i) Phenol →^{Zn dust, Δ} Benzene →^{CH₃Cl/AlCl₃} Toluene →^{Cl₂, hν} C₆H₅CH₂–Cl (Benzyl chloride) →^{NaOH (aq)} C₆H₅CH₂–OH (Benzyl alcohol)
(ii) Phenol →^{Zn (dust), Δ} Benzene →^{HNO₃ (conc.), H₂SO₄ (conc.)} Nitrobenzene →^{Br₂/FeBr₃} m-Bromonitrobenzene
→^{Sn/HCl (conc.)} m-Bromoaniline →^{NaNO₂/HCl, 273–278 K} m-Bromobenzenediazonium chloride →^{H₂O, Δ} m-Bromophenol
(iii) Phenol →^NaOH Sodium phenoxide →^{(i) CO₂, 410 K/4–7 atm; (ii) H₂O/H+} Salicylic acid (o-HO–C₆H₄–COOH) →^{(CH₃CO)₂O, H₂SO₄ (conc.)} Aspirin (o-CH₃COO–C₆H₄–COOH)

Teacher's Note:
1. In (ii) the –NO₂ group is used as a meta director and is later converted to –OH via the amine and the diazonium salt.

Question 36. How would you obtain the following?
(i) Benzoquinone from phenol [CBSE 2020 (56/1/2)]
(ii) 2-Methylpropan-2-ol from methylmagnesium bromide
(iii) Propan-2-ol from propene [CBSE (AI) 2011, (F) 2011]
Answer:
(i) C₆H₅OH (Phenol) →^{Na₂Cr₂O₇/H₂SO₄} Benzoquinone (O=C₆H₄=O, the two C=O groups para to each other)
(ii) CH₃–CO–CH₃ (Propanone) + CH₃MgBr (Methyl magnesium bromide) → [(CH₃)₃C–O⁻Mg⁺Br] →^H₂O (CH₃)₃C–OH (2-methylpropan-2-ol)
(iii) CH₃–CH=CH₂ (Propene) + H₂O →^{H+, Markovnikov’s addition} CH₃–CH(OH)–CH₃ (Propan-2-ol)

Teacher's Note:
1. A 3° alcohol is obtained when a Grignard reagent adds to a ketone; with methanal it would give a 1° and with other aldehydes a 2° alcohol.

Question 37. Give reasons for the following:
(i) Phenol is more acidic than methanol.
(ii) The C–O–H bond angle in alcohols is slightly less than the tetrahedral angle (109°28′).
(iii) (CH₃)₃C–O–CH₃ on reaction with HI gives (CH₃)₃C–I and CH₃–OH as the main products and not (CH₃)₃C–OH and CH₃–I. [CBSE Allahabad 2015]
Answer:
(i) In phenol, the phenoxide ion obtained after the removal of a proton is stabilised by resonance whereas there is no resonance in the alkoxide ion of methanol.
Moreover, due to +I effect of CH₃ group the electron density in O–H bond increases which makes release of H⁺ difficult.
(ii) It is due to the repulsion between the lone pair of electrons on oxygen atoms (the two unshared electron pairs on O in >C–Ö–H).
(iii) The reaction between (CH₃)₃COCH₃ and HI follows S_N1 mechanism. For an S_N1 reaction, the formation of product is controlled by stability of the carbocation formed in the slowest step. Since tert.butyl carbonium ion (CH₃)₃C⁺ formed after the cleavage of C–O bond in the slowest step is more stable than methyl carbonium ion (CH₃⁺) therefore (CH₃)₃C–I and CH₃OH are the main products.

Teacher's Note:
1. The C–O–H angle in methanol is about 108.9°: lone pair–lone pair repulsion squeezes it below 109°28′.

Question 38. Give the major products that are formed by heating each of the following ethers with HI.
(i) CH₃–CH₂–CH(CH₃)–CH₂–O–CH₂–CH₃
(ii) CH₃–CH₂–CH₂–O–C(CH₃)₂–CH₂–CH₃
(iii) C₆H₅–CH₂–O–C₆H₅
Answer:
(i) CH₃–CH₂–CH(CH₃)–CH₂OH + CH₃CH₂I
(ii) CH₃CH₂CH₂OH + CH₃CH₂–C(CH₃)₂–I
(iii) C₆H₅–CH₂I + C₆H₅–OH

Teacher's Note:
1. Primary/primary ethers cleave by S_N2 (I⁻ attacks the smaller group); a 3° or benzyl group cleaves by S_N1 and takes the iodine. The aryl–O bond is never broken.

Long Answer Questions (5 Marks)

Question 39. (i) Write the formula of reagents used in the following reactions:
(a) Bromination of phenol to 2,4,6-tribromophenol
(b) Hydroboration of propene and then oxidation to propanol.
(ii) Arrange the following compound groups in the increasing order of their property indicated:
(a) p-nitrophenol, ethanol, phenol (acidic character)
(b) Propanol, propane, propanal (boiling point)
(iii) Write the mechanism (using curved arrow notation) of the following reaction:
CH₃–CH₂–O⁺H₂ →^CH₃CH₂OH CH₃–CH₂–O⁺(H)–CH₂–CH₃ + H₂O [CBSE Delhi 2017]
Answer:
(i) (a) Br₂(aq)
(b) B₂H₆, H₂O, H₂O₂ and OH⁻
(ii) (a) Ethanol < Phenol < p-nitrophenol
(b) Propane < Propanal < Propanol
(iii) CH₃–CH₂–ÖH + CH₃–CH₂–O⁺H₂ → CH₃–CH₂–O⁺(H)–CH₂–CH₃ + H₂O
Curved arrows: one arrow from the lone pair on O of the ethanol molecule to the CH₂ carbon of the protonated alcohol; a second arrow from the C–O bond of the protonated alcohol to its oxygen, which leaves as H₂O (S_N2).

Teacher's Note:
1. Boiling points: alcohols (H-bonding) > aldehydes (dipole–dipole) > alkanes (only weak dispersion forces) of similar mass.

Question 40. (i) How can you obtain phenol from (a) Cumene, (b) Benzene sulphonic acid, (c) Benzene diazonium chloride?
(ii) Write the structure of the major product obtained from dinitration of 3-methylphenol.
(iii) Write the reaction involved in Kolbe’s reaction. [CBSE 2019 (56/2/1)]
Answer:
(i) (a) C₆H₅–CH(CH₃)₂ (Cumene) →^{+ O₂} C₆H₅–C(CH₃)₂–O–O–H (Cumene hydroperoxide) →^{H+, H₂O} C₆H₅OH (Phenol) + CH₃–CO–CH₃ (Acetone)
(b) C₆H₅SO₃H (Benzene sulphonic acid) →^NaOH C₆H₅SO₃⁻Na⁺ →^{+NaOH, fuse, 573 K} C₆H₅O⁻Na⁺ →^H+ C₆H₅OH (Phenol)
(c) C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride) + H₂O →^warm C₆H₅OH (Phenol) + N₂ + HCl
(ii) 3-Methylphenol →^{HNO₃(Conc.)/H₂SO₄(conc.), Dinitration} 2,4-Dinitro-5-methylphenol
(the two –NO₂ groups enter ortho and para to –OH, at C-2 and C-4; CH₃ is at C-5)
(iii) Kolbe’s reaction:
C₆H₅OH (Phenol) →^+NaOH C₆H₅O⁻Na⁺ (Sodium phenoxide) →^{CO₂, 400 K, 4–7 atm} o-HO–C₆H₄–COO⁻Na⁺ →^H+ o-HO–C₆H₄–COOH (Salicylic acid)

Teacher's Note:
1. In (ii) the –OH group, the strongest activator, decides where the nitro groups go; acetone from the cumene process is a valuable by-product.