Intext Questions
1. Classify the following as primary, secondary and tertiary alcohols: (i) (CH₃)₃C–CH₂OH (ii) H₂C=CH–CH₂OH (iii) CH₃–CH₂–CH₂–OH (iv) C₆H₅–CH(OH)–CH₃ (v) C₆H₅–CH₂–CH(OH)–CH₃ (vi) C₆H₅–CH=CH–C(OH)(CH₃)₂
Answer
Primary: (i), (ii), (iii) – the –OH carbon is attached to one other carbon.
Secondary: (iv), (v) – the –OH carbon is attached to two carbons.
Tertiary: (vi) – the –OH carbon is attached to three carbons.
2. Identify allylic alcohols in the above examples.
Answer
(ii) H₂C=CH–CH₂OH and (vi) C₆H₅–CH=CH–C(OH)(CH₃)₂, because in both the –OH group is on an sp³ carbon next to a C=C double bond. ((iv) is a benzylic alcohol, not allylic.)
3. Name the following compounds according to IUPAC system. (i) CH₃–CH₂–CH(CH₂Cl)–CH(CH₂OH)–CH(CH₃)–CH₃ (ii) CH₃–CH(CH₃)–CH₂–CH(OH)–CH(CH₂OH)–CH₃ (iii) cyclohexanol with Br at C-3 (iv) H₂C=CH–CH(OH)–CH₂–CH₂–CH₃ (v) CH₃–C(CH₃)=C(Br)–CH₂OH
Answer
(i) 4-Chloro-3-ethyl-2-(1-methylethyl)butan-1-ol
(ii) 2,5-Dimethylhexane-1,3-diol
(iii) 3-Bromocyclohexanol
(iv) Hex-1-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-1-ol
4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal? (i) CH₃–CH(CH₃)–CH₂OH (ii) cyclohexylmethanol, C₆H₁₁–CH₂OH
Answer
Methanal with any RMgX gives a primary alcohol RCH₂OH.
(i) (CH₃)₂CH–MgBr + HCHO
⇒ (CH₃)₂CH–CH₂–OMgBr —H₂O
⇒ (CH₃)₂CHCH₂OH + Mg(OH)Br
(ii) C₆H₁₁MgBr (cyclohexylmagnesium bromide) + HCHO
⇒ C₆H₁₁CH₂OMgBr —H₂O
⇒ C₆H₁₁CH₂OH + Mg(OH)Br
5. Write structures of the products of the following reactions: (i) CH₃–CH=CH₂ —H₂O/H⁺→ (ii) 2-oxocyclohexyl–CH₂–COOCH₃ (methyl (2-oxocyclohexyl)ethanoate) —NaBH₄→ (iii) CH₃–CH₂–CH(CH₃)–CHO —NaBH₄→
Answer
(i) CH₃–CH(OH)–CH₃ (propan-2-ol; Markovnikov addition of water).
(ii) The ring C=O is reduced to CH–OH but the ester group is untouched (NaBH₄ does not reduce esters): 2-hydroxycyclohexyl–CH₂–COOCH₃.
(iii) CH₃–CH₂–CH(CH₃)–CH₂OH (2-methylbutan-1-ol).
6. Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl–ZnCl₂ (b) HBr and (c) SOCl₂. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol
Answer
(i) Butan-1-ol: (a) CH₃CH₂CH₂CH₂Cl (1-chlorobutane; primary alcohols react only on heating with Lucas reagent) (b) CH₃CH₂CH₂CH₂Br (c) CH₃CH₂CH₂CH₂Cl + SO₂ + HCl
(ii) 2-Methylbutan-2-ol, CH₃CH₂C(OH)(CH₃)₂: (a) CH₃CH₂C(Cl)(CH₃)₂ (2-chloro-2-methylbutane, formed at once – turbidity immediately) (b) CH₃CH₂C(Br)(CH₃)₂ (c) CH₃CH₂C(Cl)(CH₃)₂
7. Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol
Answer
(i) 1-Methylcyclohexene (the more substituted alkene, Saytzeff product).
(ii) A mixture of but-1-ene and but-2-ene; but-2-ene (CH₃CH=CHCH₃) is the major product because the primary carbocation formed first rearranges by hydride shift to the more stable secondary carbocation, CH₃CH₂C⁺HCH₃, which loses H⁺ to give the more substituted alkene.
8. Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer
In o- and p-nitrophenoxide ions the negative charge on oxygen spreads into the ring (onto the ortho and para carbons) and further onto the –NO₂ group.
p-Nitrophenoxide: ⁻O–C₆H₄–NO₂ ↔ structures with negative charge on C-2, C-4 and C-6 ↔ O=C₆H₄=N⁺(–O⁻)₂, a quinonoid form in which the charge sits on the oxygen atoms of the nitro group.
o-Nitrophenoxide: similarly ⁻O–C₆H₄–NO₂(o) ↔ ring-charged forms ↔ O=C₆H₄=N⁺(O⁻)₂ with charge on the nitro oxygens.
This extra delocalisation onto the electronegative nitro oxygens stabilises these phenoxide ions much more than the phenoxide ion itself (whose charge is spread only over O and ring carbons), so the nitrophenols are stronger acids.
9. Write the equations involved in the following reactions: (i) Reimer–Tiemann reaction (ii) Kolbe's reaction
Answer
(i) C₆H₅OH + CHCl₃ + aq. NaOH (340 K)
⇒ intermediate o-(dichloromethyl)sodium phenoxide
⇒ NaOH
⇒ sodium salicylaldehyde
⇒ H⁺
⇒ o-HOC₆H₄CHO (salicylaldehyde, 2-hydroxybenzaldehyde).
(ii) C₆H₅OH + NaOH
⇒ C₆H₅ONa; C₆H₅ONa + CO₂ (400 K, 4–7 atm)
⇒ o-HOC₆H₄COONa —H⁺
⇒ o-HOC₆H₄COOH (salicylic acid, 2-hydroxybenzoic acid).
10. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer
The alkyl halide must be primary (from ethanol) and the alkoxide secondary (from the other alcohol), so that substitution and not elimination occurs.
C₂H₅OH + HBr → C₂H₅Br + H₂O
CH₃CH₂CH(CH₃)CH(OH)CH₃ + Na
⇒ CH₃CH₂CH(CH₃)CH(ONa)CH₃ + ½H₂
CH₃CH₂CH(CH₃)CH(ONa)CH₃ + C₂H₅Br
⇒ CH₃CH₂CH(CH₃)CH(OC₂H₅)CH₃ (2-ethoxy-3-methylpentane) + NaBr
11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? (i) 4-bromonitrobenzene + CH₃ONa (ii) sodium 4-nitrophenoxide + CH₃Br
Answer
Set (ii): p-O₂NC₆H₄ONa + CH₃Br
⇒ p-O₂NC₆H₄OCH₃ + NaBr. In set (i) the bromine is on an aryl carbon; the C–Br bond has partial double-bond character and aryl halides do not readily undergo nucleophilic substitution with alkoxide, so (ii), an SN2 attack of phenoxide on the unhindered CH₃Br, is the right choice.
12. Predict the products of the following reactions: (i) CH₃–CH₂–CH₂–O–CH₃ + HBr → (ii) C₆H₅–OC₂H₅ + HBr → (iii) C₆H₅–OC₂H₅ —conc. H₂SO₄/conc. HNO₃→ (iv) (CH₃)₃C–OC₂H₅ —HI→
Answer
(i) CH₃CH₂CH₂OH + CH₃Br (Br⁻ attacks the less hindered methyl group by SN2).
(ii) C₆H₅OH + C₂H₅Br (the O–aryl bond is not broken).
(iii) o-Nitrophenetole (minor) and p-nitrophenetole, p-O₂NC₆H₄OC₂H₅ (major).
(iv) (CH₃)₃C–I + C₂H₅OH (cleavage via the stable tert-butyl carbocation, SN1).
Exercises
1. Write IUPAC names of the following compounds: (i) CH₃–CH(CH₃)–CH(OH)–C(CH₃)₂–CH₃ (ii) CH₃–CH(OH)–CH₂–CH(OH)–CH(C₂H₅)–CH₂–CH₃ (iii) CH₃–CH(OH)–CH(OH)–CH₃ (iv) HO–CH₂–CH(OH)–CH₂–OH (v) phenol with CH₃ at ortho position (vi) phenol with CH₃ at para position (vii) phenol with CH₃ groups at positions 2 and 5 (viii) phenol with CH₃ groups at both ortho positions (ix) CH₃–O–CH₂–CH(CH₃)–CH₃ (x) C₆H₅–O–C₂H₅ (xi) C₆H₅–O–C₇H₁₅(n–) (xii) CH₃–CH₂–O–CH(CH₃)–CH₂–CH₃
Answer
(i) 2,2,4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-Dimethylphenol
(viii) 2,6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane
2. Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane-1,3,5-triol (iv) 2,3-Diethylphenol (v) 1-Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 4-Chloro-3-ethylbutan-1-ol.
Answer
(i) CH₃–C(OH)(CH₃)–CH₂–CH₃
(ii) C₆H₅–CH₂–CH(OH)–CH₃
(iii) HOCH₂–CH₂–C(OH)(CH₃)–CH₂–C(OH)(CH₃)–CH₃
(iv) Benzene ring with OH at C-1 and C₂H₅ groups at C-2 and C-3
(v) CH₃CH₂–O–CH₂CH₂CH₃
(vi) CH₃–CH(OC₂H₅)–CH(CH₃)–CH₂–CH₃
(vii) C₆H₁₁–CH₂OH (a CH₂OH group on a cyclohexane ring)
(viii) CH₃CH₂–C(OH)(C₆H₁₁)–CH₂CH₃
(ix) Five-membered ring: OH on C-1 and a C=C between C-3 and C-4
(x) HO–CH₂–CH₂–CH(C₂H₅)–CH₂Cl
3. (i) Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names. (ii) Classify the isomers of alcohols in question (i) as primary, secondary and tertiary alcohols.
Answer
There are eight isomeric alcohols:
1. CH₃CH₂CH₂CH₂CH₂OH – Pentan-1-ol – primary
2. CH₃CH(CH₃)CH₂CH₂OH – 3-Methylbutan-1-ol – primary
3. CH₃CH₂CH(CH₃)CH₂OH – 2-Methylbutan-1-ol – primary
4. (CH₃)₃CCH₂OH – 2,2-Dimethylpropan-1-ol – primary
5. CH₃CH(OH)CH₂CH₂CH₃ – Pentan-2-ol – secondary
6. CH₃CH₂CH(OH)CH₂CH₃ – Pentan-3-ol – secondary
7. CH₃CH(OH)CH(CH₃)CH₃ – 3-Methylbutan-2-ol – secondary
8. CH₃CH₂C(OH)(CH₃)₂ – 2-Methylbutan-2-ol – tertiary
4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer
The two have similar molecular masses (60 and 58 u), but propanol molecules are associated through intermolecular hydrogen bonds via the –OH group. Extra energy is needed to break these, so propanol boils higher (370 K) than butane (273 K), whose molecules are held only by weak van der Waals forces.
5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer
The –OH group of an alcohol forms hydrogen bonds with water molecules, which releases enough energy to break the water–water hydrogen bonds. Hydrocarbons cannot form hydrogen bonds with water, so they are insoluble.
6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer
Addition of diborane to an alkene to give a trialkylborane, followed by oxidation with alkaline H₂O₂, which converts it to an alcohol. The net result is addition of water against Markovnikov's rule (–OH goes to the less substituted carbon), with good yield.
Example: CH₃CH=CH₂ + (H–BH₂)₂
⇒ CH₃CH₂CH₂–BH₂
⇒ (CH₃CH₂CH₂)₃B; (CH₃CH₂CH₂)₃B + 3H₂O₂ —OH⁻
⇒ 3CH₃CH₂CH₂OH (propan-1-ol) + B(OH)₃
7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer
o-CH₃C₆H₄OH – 2-Methylphenol (o-cresol); m-CH₃C₆H₄OH – 3-Methylphenol (m-cresol); p-CH₃C₆H₄OH – 4-Methylphenol (p-cresol).
8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer
o-Nitrophenol is steam volatile. Its –OH and –NO₂ groups are close enough to form an intramolecular hydrogen bond, so its molecules do not associate with each other. p-Nitrophenol forms intermolecular hydrogen bonds, so its molecules are associated and it is far less volatile.
9. Give the equations of reactions for the preparation of phenol from cumene.
Answer
C₆H₅CH=CH₂… (cumene is made from benzene and propene) C₆H₆ + CH₂=CHCH₃ —H⁺
⇒ C₆H₅CH(CH₃)₂ (cumene).
C₆H₅CH(CH₃)₂ + O₂ (air)
⇒ C₆H₅C(CH₃)₂–O–OH (cumene hydroperoxide)
C₆H₅C(CH₃)₂OOH —H⁺/H₂O
⇒ C₆H₅OH (phenol) + CH₃COCH₃ (acetone)
10. Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer
C₆H₅Cl + NaOH —623 K, 300 atm
⇒ C₆H₅ONa + HCl(neutralised); C₆H₅ONa + HCl
⇒ C₆H₅OH + NaCl (Dow's process).
11. Write the mechanism of hydration of ethene to yield ethanol.
Answer
Step 1: Protonation of the alkene by electrophilic attack of H₃O⁺: CH₂=CH₂ + H₃O⁺
⇒ CH₃–CH₂⁺ + H₂O
Step 2: Nucleophilic attack of water on the carbocation: CH₃CH₂⁺ + H₂O
⇒ CH₃CH₂–O⁺H₂
Step 3: Deprotonation: CH₃CH₂–O⁺H₂ + H₂O
⇒ CH₃CH₂OH + H₃O⁺
12. You are given benzene, conc. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer
C₆H₆ + H₂SO₄ (oleum/conc.)
⇒ C₆H₅SO₃H (benzenesulphonic acid) + H₂O
C₆H₅SO₃H + NaOH
⇒ C₆H₅SO₃Na + H₂O; C₆H₅SO₃Na + 2NaOH (fused, ~573 K)
⇒ C₆H₅ONa + Na₂SO₃ + H₂O
C₆H₅ONa + H⁺ (dil. H₂SO₄)
⇒ C₆H₅OH (phenol)
13. Show how will you synthesise: (i) 1-phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction. (iii) pentan-1-ol using a suitable alkyl halide?
Answer
(i) C₆H₅CH=CH₂ (styrene) + H₂O —H⁺
⇒ C₆H₅CH(OH)CH₃ (Markovnikov addition).
(ii) C₆H₁₁CH₂Br + NaOH (aq)
⇒ C₆H₁₁CH₂OH + NaBr (primary halide, SN2).
(iii) CH₃CH₂CH₂CH₂CH₂Cl + KOH (aq)
⇒ CH₃CH₂CH₂CH₂CH₂OH + KCl.
14. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer
(1) 2C₆H₅OH + 2Na
⇒ 2C₆H₅ONa + H₂ (2) C₆H₅OH + NaOH
⇒ C₆H₅ONa + H₂O (phenol reacts with aqueous NaOH, alcohols do not).
Phenol (pKa ≈ 10) is far more acidic than ethanol (pKa ≈ 15.9). In the phenoxide ion the negative charge is delocalised over the ring by resonance, stabilising it; the ethoxide ion has no resonance and its charge is intensified by the electron-releasing ethyl group.
15. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Answer
–NO₂ is electron-withdrawing (–I and –R); it helps disperse the negative charge of the phenoxide ion, stabilising it and releasing H⁺ easily. –OCH₃ is electron-donating (+R), which intensifies the negative charge on the phenoxide ion and destabilises it, so o-methoxyphenol is less acidic.
16. Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Answer
A lone pair on the oxygen of –OH is delocalised into the ring (+R effect). Resonance structures place a negative charge at the ortho and para positions, increasing electron density there. The ring becomes more attractive to electrophiles (activated), and substitution occurs mainly at ortho and para positions.
17. Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO₄ solution. (ii) Bromine in CS₂ with phenol. (iii) Dilute HNO₃ with phenol. (iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer
(i) CH₃CH₂CH₂OH —alk. KMnO₄, [O]
⇒ CH₃CH₂COOH (propanoic acid)
(ii) C₆H₅OH + Br₂ —CS₂, 273 K
⇒ o-BrC₆H₄OH (minor) + p-BrC₆H₄OH (major) + HBr
(iii) C₆H₅OH + dil. HNO₃ (298 K)
⇒ o-nitrophenol + p-nitrophenol
(iv) C₆H₅OH + CHCl₃ + aq. NaOH
⇒ (then H⁺) o-HOC₆H₄CHO (salicylaldehyde) – Reimer–Tiemann reaction
18. Explain the following with an example. (i) Kolbe's reaction. (ii) Reimer-Tiemann reaction. (iii) Williamson ether synthesis. (iv) Unsymmetrical ether.
Answer
(i) Kolbe's reaction: sodium phenoxide (more reactive than phenol towards electrophiles) reacts with the weak electrophile CO₂ to give o-hydroxybenzoic acid. C₆H₅ONa + CO₂ (400 K, 4–7 atm)
⇒ o-HOC₆H₄COONa —H⁺
⇒ salicylic acid.
(ii) Reimer–Tiemann reaction: phenol with chloroform and aqueous NaOH introduces a –CHO group at the ortho position. C₆H₅OH + CHCl₃ + 3NaOH
⇒ (then H⁺) salicylaldehyde.
(iii) Williamson synthesis: an alkyl halide (preferably primary) reacts with a sodium alkoxide or phenoxide by SN2 to give an ether. CH₃I + C₂H₅ONa
⇒ CH₃OC₂H₅ + NaI.
(iv) Unsymmetrical (mixed) ether: an ether in which the two groups on oxygen are different, e.g. CH₃OC₂H₅ (methoxyethane), C₆H₅OCH₃ (anisole).
19. Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer
(Conc. H₂SO₄, 443 K)
Step 1: Protonation of –OH: CH₃CH₂–OH + H⁺
⇒ CH₃CH₂–O⁺H₂ (fast)
Step 2: Loss of water to form carbocation: CH₃CH₂–O⁺H₂
⇒ CH₃CH₂⁺ + H₂O (slow, rate-determining)
Step 3: Loss of a β-proton: CH₃–CH₂⁺
⇒ CH₂=CH₂ + H⁺
20. How are the following conversions carried out? (i) Propene → Propan-2-ol. (ii) Benzyl chloride → Benzyl alcohol. (iii) Ethyl magnesium chloride → Propan-1-ol. (iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer
(i) CH₃CH=CH₂ + H₂O —H⁺→ CH₃CH(OH)CH₃
(ii) C₆H₅CH₂Cl + aq. NaOH
⇒ C₆H₅CH₂OH + NaCl
(iii) C₂H₅MgCl + HCHO
⇒ C₂H₅CH₂OMgCl —H₂O
⇒ CH₃CH₂CH₂OH + Mg(OH)Cl
(iv) CH₃MgBr + CH₃COCH₃
⇒ (CH₃)₃C–OMgBr —H₂O
⇒ (CH₃)₃C–OH + Mg(OH)Br
21. Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol to 2,4,6-tribromophenol. (iv) Benzyl alcohol to benzoic acid. (v) Dehydration of propan-2-ol to propene. (vi) Butan-2-one to butan-2-ol.
Answer
(i) Acidified KMnO₄ or acidified K₂Cr₂O₇ (strong oxidant)
(ii) Pyridinium chlorochromate (PCC) in CH₂Cl₂ (or CrO₃ in anhydrous medium)
(iii) Bromine water
(iv) Acidified (or alkaline) KMnO₄
(v) 85% H₃PO₄ at 440 K (or conc. H₂SO₄ with heating)
(vi) NaBH₄ or LiAlH₄ (or H₂ with Ni/Pt/Pd catalyst)
22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer
Both have formula C₂H₆O, but ethanol has an O–H bond and its molecules are associated by intermolecular hydrogen bonding. Methoxymethane has no H on oxygen and cannot hydrogen-bond with itself, so it boils far lower (248 K vs 351 K).
23. Give IUPAC names of the following ethers: (i) C₂H₅OCH₂–CH(CH₃)–CH₃ (ii) CH₃OCH₂CH₂Cl (iii) O₂N–C₆H₄–OCH₃(p) (iv) CH₃CH₂CH₂OCH₃ (v) cyclohexane carrying two CH₃ groups on one carbon and OC₂H₅ on the opposite (C-4) carbon (vi) C₆H₅–OC₂H₅
Answer
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 4-Nitroanisole (1-methoxy-4-nitrobenzene)
(iv) 1-Methoxypropane
(v) 4-Ethoxy-1,1-dimethylcyclohexane
(vi) Ethoxybenzene
24. Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane
Answer
(i) Sodium propoxide + 1-bromopropane: CH₃CH₂CH₂ONa + CH₃CH₂CH₂Br
⇒ CH₃CH₂CH₂OCH₂CH₂CH₃ + NaBr
(ii) Sodium phenoxide + bromoethane: C₆H₅ONa + C₂H₅Br
⇒ C₆H₅OC₂H₅ + NaBr
(iii) Sodium tert-butoxide + bromomethane: (CH₃)₃CONa + CH₃Br
⇒ (CH₃)₃COCH₃ + NaBr
(iv) Sodium ethoxide + bromomethane (or iodomethane): C₂H₅ONa + CH₃Br
⇒ CH₃OC₂H₅ + NaBr
25. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer
Alkoxides are strong bases as well as nucleophiles. With secondary and especially tertiary alkyl halides, elimination competes with or wins over substitution, giving alkenes instead of ethers. E.g. for tert-butyl methyl ether, (CH₃)₃CBr + CH₃ONa
⇒ (CH₃)₂C=CH₂ + CH₃OH + NaBr (alkene, not ether). The correct choice is CH₃Br + (CH₃)₃CONa. Also, aryl halides cannot be used as the halide component (e.g. C₆H₅Br + C₂H₅ONa gives no ether), because aryl halides resist nucleophilic substitution.
26. How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Answer
By heating propan-1-ol with conc. H₂SO₄ at about 413 K (excess alcohol): 2CH₃CH₂CH₂OH
⇒ CH₃CH₂CH₂–O–CH₂CH₂CH₃ + H₂O.
Step 1: CH₃CH₂CH₂–OH + H⁺
⇒ CH₃CH₂CH₂–O⁺H₂
Step 2 (SN2): a second alcohol molecule attacks: CH₃CH₂CH₂–OH + CH₃CH₂CH₂–O⁺H₂
⇒ CH₃CH₂CH₂–O⁺(H)–CH₂CH₂CH₃ + H₂O
Step 3: loss of proton: CH₃CH₂CH₂–O⁺(H)–CH₂CH₂CH₃
⇒ CH₃CH₂CH₂–O–CH₂CH₂CH₃ + H⁺
27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer
Ether formation by this route is an SN2 attack of an alcohol on a protonated alcohol, which is hindered for 2° and 3° alkyl groups. These alcohols instead readily form stable carbocations that lose a proton, so elimination dominates and alkenes are the main product. E.g. (CH₃)₃COH —H⁺
⇒ (CH₃)₂C=CH₂.
28. Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.
Answer
(i) CH₃CH₂CH₂OCH₂CH₂CH₃ + HI
⇒ CH₃CH₂CH₂I + CH₃CH₂CH₂OH (with excess HI, the alcohol also gives CH₃CH₂CH₂I)
(ii) C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I
(iii) C₆H₅CH₂OC₂H₅ + HI
⇒ C₆H₅CH₂I + C₂H₅OH (cleavage at the benzyl side, via the stable benzyl carbocation / easy substitution at benzylic carbon)
29. Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer
(i) The oxygen of –OR donates a lone pair into the ring (+R effect), increasing the electron density of the ring, so it is attacked by electrophiles more easily than benzene.
(ii) In the resonance structures of anisole, the negative charge (extra electron density) appears at the ortho and para positions (C-2, C-4, C-6), so electrophiles attack mainly at these positions.
30. Write the mechanism of the reaction of HI with methoxymethane.
Answer
Step 1: Protonation: CH₃–O–CH₃ + HI
⇒ CH₃–O⁺(H)–CH₃ + I⁻
Step 2: I⁻ attacks a methyl carbon by SN2: I⁻ + CH₃–O⁺(H)–CH₃
⇒ CH₃I + CH₃OH
Step 3 (excess HI): CH₃OH + HI
⇒ CH₃–O⁺H₂ + I⁻
⇒ CH₃I + H₂O
31. Write equations of the following reactions: (i) Friedel-Crafts reaction – alkylation of anisole. (ii) Nitration of anisole. (iii) Bromination of anisole in ethanoic acid medium. (iv) Friedel-Craft's acetylation of anisole.
Answer
(i) C₆H₅OCH₃ + CH₃Cl —anhyd. AlCl₃, CS₂
⇒ 2-methoxytoluene (minor) + 4-methoxytoluene (major) + HCl
(ii) C₆H₅OCH₃ —conc. H₂SO₄ + conc. HNO₃
⇒ 2-nitroanisole (minor) + 4-nitroanisole (major)
(iii) C₆H₅OCH₃ + Br₂ —CH₃COOH
⇒ p-bromoanisole (major, ~90%) + o-bromoanisole (minor) + HBr (no Lewis acid needed, ring is activated)
(iv) C₆H₅OCH₃ + CH₃COCl —anhyd. AlCl₃
⇒ 2-methoxyacetophenone (minor) + 4-methoxyacetophenone (major) + HCl
32. Show how would you synthesise the following alcohols from appropriate alkenes? (i) 1-methylcyclohexanol (ii) 4-methylheptan-4-ol (iii) pentan-2-ol (iv) 2-cyclohexylbutan-2-ol
Answer
Acid-catalysed hydration (H₂O/H⁺) follows Markovnikov's rule, so the –OH ends up on the more substituted carbon.
(i) 1-Methylcyclohexene (or methylenecyclohexane) + H₂O —H⁺
⇒ 1-methylcyclohexanol
(ii) CH₃CH₂CH₂C(CH₃)=CHCH₂CH₃ (4-methylhept-3-ene) + H₂O —H⁺
⇒ CH₃CH₂CH₂C(OH)(CH₃)CH₂CH₂CH₃
(iii) CH₂=CHCH₂CH₂CH₃ (pent-1-ene) + H₂O —H⁺
⇒ CH₃CH(OH)CH₂CH₂CH₃ (pent-2-ene would also give pentan-3-ol, so pent-1-ene is preferred)
(iv) C₆H₁₁–C(CH₃)=CHCH₃ (2-cyclohexylbut-2-ene) or C₆H₁₁–C(=CH₂)CH₂CH₃ + H₂O —H⁺
⇒ C₆H₁₁–C(OH)(CH₃)CH₂CH₃
33. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: CH₃–CH(CH₃)–CH(OH)–CH₃ —HBr→ CH₃–C(Br)(CH₃)–CH₂–CH₃. Give a mechanism for this reaction. (Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Answer
Step I (protonation): (CH₃)₂CH–CH(OH)–CH₃ + H⁺
⇒ (CH₃)₂CH–CH(O⁺H₂)–CH₃
Step II (loss of water): (CH₃)₂CH–CH(O⁺H₂)–CH₃
⇒ (CH₃)₂CH–C⁺H–CH₃ (2° carbocation) + H₂O
Step III (1,2-hydride shift from C-3): (CH₃)₂CH–C⁺H–CH₃
⇒ (CH₃)₂C⁺–CH₂–CH₃ (more stable 3° carbocation)
Step IV (attack of Br⁻): (CH₃)₂C⁺–CH₂CH₃ + Br⁻
⇒ (CH₃)₂C(Br)–CH₂CH₃ (2-bromo-2-methylbutane)
Important Extra Questions
How does the Lucas test distinguish primary, secondary and tertiary alcohols?
Answer
With conc. HCl + ZnCl₂, tertiary alcohols give turbidity (insoluble alkyl chloride) immediately, secondary in about five minutes, and primary not at room temperature (only on heating).
Why is phenol more reactive than benzene towards bromination? Give the product with bromine water.
Answer
–OH strongly activates the ring by +R effect; with bromine water phenol gives a white precipitate of 2,4,6-tribromophenol without any catalyst.
What is denatured alcohol?
Answer
Commercial ethanol made unfit for drinking by adding copper sulphate (colour) and pyridine (bad smell).
Why do ethers have lower boiling points than isomeric alcohols but similar solubility in water?
Answer
Ethers cannot hydrogen-bond with themselves (no O–H), so boil low; but the ether oxygen can accept hydrogen bonds from water, so solubility is similar to alcohols of the same mass.
What happens when a tertiary alcohol is passed over heated copper at 573 K?
Answer
It undergoes dehydration (not dehydrogenation) to an alkene, e.g. (CH₃)₃COH
⇒ (CH₃)₂C=CH₂ + H₂O; primary and secondary alcohols give aldehydes and ketones instead.