CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 08

Very Short Answer Questions (1 Mark)

Question 1. Name the phenol with molecular formula C₇H₈O which on treatment with Br₂ water readily gives a precipitate of C₇H₅OBr₃.
Answer:
m-Cresol forms a tribromo derivative because it has two o- and one p-position free w.r.t. OH group.

Teacher's Note:
1. In o- and p-cresol one of the activated positions is blocked by CH₃, so they can take up only two Br atoms.

Question 2. Which is steam volatile : ortho- or p-nitrophenol ? Give reason also. (Hr. Board 2011)
or Which of the following isomer is more volatile ? o-nitrophenol or p-nitrophenol ? (CBSE 2014)
Answer:
o-Nitrophenol undergoes intramolecular H-bonding (chelation) and hence exists as discrete molecules. As a result, less energy is needed to convert into vapours and hence o-nitrophenol is more volatile than p-nitrophenol which undergoes intermolecular H-bonding and exists as an associated molecule.

Teacher's Note:
1. This difference is used in practice: the two isomers formed on nitration of phenol are separated by steam distillation.

Question 3. Name the major product formed when sodium phenoxide is heated with CO₂ at 400 K under 4-7 atm pressure. What is the name of the reaction ?
Answer:
2-Hydroxybenzoic acid (salicylic acid). Kolbe’s reaction.
C₆H₅ONa + CO₂ \( \xrightarrow{400\text{ K},\ 4\text{-}7\text{ atm}} \) o-HO–C₆H₄–COONa \( \xrightarrow{H^+} \) o-HO–C₆H₄–COOH (Salicylic acid)

Teacher's Note:
1. The phenoxide ion, not phenol itself, is used because it is reactive enough to be attacked by the weak electrophile CO₂.

Question 4. Phenol is heated with chloroform and NaOH at 340 K. What is the main product formed ? Also give the name of the reaction involved.
Answer:
2-Hydroxybenzaldehyde (salicylaldehyde) ; Reimer-Tiemann reaction.
C₆H₅OH + CHCl₃ + aq. NaOH \( \xrightarrow{340\text{ K}} \) o-HO–C₆H₄–CHCl₂ (intermediate, as sodium salt) \( \xrightarrow{NaOH} \) o-NaO–C₆H₄–CHO \( \xrightarrow{H^+} \) o-HO–C₆H₄–CHO (Salicylaldehyde)

Teacher's Note:
1. The electrophile in the Reimer-Tiemann reaction is dichlorocarbene (:CCl₂), formed from CHCl₃ and NaOH.

Question 5. How will you synthesise salicylic acid from phenol ?
Answer:
Two methods are available :
(i) By Kolbe’s reaction. Sodium phenoxide (from phenol and NaOH) is heated with CO₂ at 400 K under 4-7 atm pressure and the product is acidified:
C₆H₅OH \( \xrightarrow{NaOH} \) C₆H₅ONa \( \xrightarrow[(ii)\ H^+]{(i)\ CO_2,\ 400\text{ K},\ 4\text{-}7\text{ atm}} \) o-HO–C₆H₄–COOH (Salicylic acid)
(ii) By Reimer-Tiemann reaction. Phenol is heated with carbon tetrachloride in presence of aq. NaOH at 340 K and the product so obtained on hydrolysis with dil. HCl gives salicylic acid.
C₆H₅OH + CCl₄ + aq. NaOH \( \xrightarrow{340\text{ K}} \) o-NaO–C₆H₄–COONa \( \xrightarrow{\text{dil. HCl}} \) o-HO–C₆H₄–COOH (Salicylic acid)

Teacher's Note:
1. With CHCl₃ the Reimer-Tiemann reaction gives salicylaldehyde; with CCl₄ it gives salicylic acid.

Question 6. Write equation involved in the acetylation of salicylic acid. (CBSE 2015)
Answer:
Salicylic acid on acetylation with acetic anhydride (in presence of a little conc. H₂SO₄) gives acetylsalicylic acid (aspirin):
o-HO–C₆H₄–COOH (Salicylic acid) + (CH₃CO)₂O \( \xrightarrow{H^+} \) o-CH₃COO–C₆H₄–COOH (Acetylsalicylic acid, Aspirin) + CH₃COOH

Teacher's Note:
1. Only the phenolic –OH is acetylated; the –COOH group stays unchanged in aspirin.

Question 7. How is phenolphthalein obtained ?
Answer:
By heating phenol with phthalic anhydride in presence of conc. H₂SO₄. Two molecules of phenol condense with one molecule of phthalic anhydride at the para positions:
Phthalic anhydride + 2 C₆H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ \Delta} \) Phenolphthalein (C₂₀H₁₄O₄) + H₂O

Teacher's Note:
1. Phenolphthalein is colourless in acidic medium and pink in basic medium, which is why it is used as an indicator.

Question 8. Give the mechanism of the following reaction,
2 CH₃CH₂OH \( \xrightarrow[413\text{K}]{\text{Dil. }H_2SO_4} \) CH₃CH₂OCH₂CH₃ + H₂O
Does this reaction follow S_N1 or S_N2 pathway?
Answer:
The reaction occurs by S_N2 mechanism. It takes place in three steps :
Step 1. Protonation of ethanol : CH₃CH₂–OH + H⁺ ⇄ CH₃CH₂–O⁺H₂
Step 2. Nucleophilic attack (S_N2) by an unprotonated ethanol molecule on the protonated ethanol with loss of water : CH₃CH₂–OH + CH₃CH₂–O⁺H₂ → CH₃CH₂–O⁺(H)–CH₂CH₃ + H₂O
Step 3. Loss of a proton : CH₃CH₂–O⁺(H)–CH₂CH₃ ⇄ CH₃CH₂–O–CH₂CH₃ + H⁺

Teacher's Note:
1. A 1° carbocation is too unstable to form, so the second alcohol molecule displaces water in one step; that is why this is S_N2.

Question 9. How are simple and mixed ethers prepared using diazomethane ? (Maharashtra Board 2012)
Answer:
Dimethyl ether is the only simple ether which can be prepared by the action of diazomethane on methanol in presence of tetrafluoroboric acid (HBF₄).
⁻:CH₂–N⁺≡N (Diazomethane) + HBF₄ → CH₃–N⁺≡N + BF₄⁻
CH₃–O(H) + CH₃–N⁺≡N → CH₃–O–CH₃ (Dimethyl ether) + N₂ + H⁺
Reaction of alcohols, other than methanol, with CH₂N₂ in presence HBF₄ gives mixed ethers.
CH₃CH₂OH (Ethanol) + ⁻:CH₂–N⁺≡N \( \xrightarrow{HBF_4} \) CH₃CH₂–O–CH₃ (Ethyl methyl ether) + N₂

Teacher's Note:
1. Diazomethane always delivers a CH₃ group, so the ether obtained always contains at least one methyl group.

Question 10. How do you account for the miscibility of ethoxyethane with water ?
Answer:
Solubility or miscibility of ethoxyethane with water is due to H- bonding between H-atoms of H₂O and O atom of ethoxyethane as shown below :
(CH₃CH₂)₂O : ···· H–O–H (the O atom of the ether accepts H-bonds from two water molecules)

Teacher's Note:
1. Ether can only accept H-bonds (it has no O–H), so it is far less soluble in water than an alcohol of similar size; it is only partly miscible.

Question 11. A little water is there on the inner surface of a conical flask. Can this water be removed by shaking with ether? Answer with reason. (West Bengal Board 2012)
Answer:
No, because water does not dissolve in ether. Instead alcohol can be used to remove water since water dissolves in alcohol.

Teacher's Note:
1. Water and alcohol can both donate and accept H-bonds, so they mix freely; ether cannot donate one.

Question 12. Why are higher ethers insoluble in water ?
Answer:
Higher ethers are insoluble in water because due to the bigger size of the alkyl groups, the oxygen atom in ethers fails to form intermolecular H-bonds with water.

Teacher's Note:
1. The larger the hydrocarbon part, the more it shields the oxygen and the more the molecule behaves like an alkane.

Question 13. Suggest a simple method to remove peroxides from ethers.
Answer:
By shaking the ether with an aqueous solution of KI or a ferrous salt. The peroxides oxidise iodide ions to iodine (or Fe²⁺ to Fe³⁺) and are themselves reduced, so they are destroyed. The ether is then distilled.

Teacher's Note:
1. Peroxides form when ethers stand in air and light; they are explosive on heating, so old ether must be tested before distillation.

Question 14. Why are Grignard reagents prepared in ether but not in benzene ?
Answer:
Due to presence of lone pairs of electrons on the oxygen atom, ethers form coordinate complexes with Grignard reagents but not in benzene. As a result, Grignard reagents dissolve in ether.
2 (CH₃CH₂)₂O (Diethyl ether) + RMgX (Grignard reagent) → RMgX·2O(CH₂CH₃)₂ (Mg coordinated to the two ether oxygens)

Teacher's Note:
1. The ether must also be dry: Grignard reagents are destroyed by water, giving alkanes.

Question 15. Write the products obtained when benzyl phenyl ether is heated with HI.
Answer:
Protonation of benzyl phenyl ether gives oxononium ion in which O–C₆H₅ has some double bond character due to resonance between the lone pairs of electrons on the oxygen atom and the sp²-hybridized carbon of the phenyl ring. Therefore, attack by I⁻ ion exclusively breaks the weaker O–CH₂C₆H₅ bond forming phenol and benzyl iodide.
C₆H₅–O–CH₂–C₆H₅ (Benzyl phenyl ether) \( \xrightarrow{H^+} \) C₆H₅–O⁺(H)–CH₂–C₆H₅ \( \xrightarrow{I^-} \) C₆H₅–OH (Phenol) + I–CH₂–C₆H₅ (Benzyl iodide)

Teacher's Note:
1. Aryl–O bonds are never cleaved by HI; so an aryl alkyl ether always gives a phenol and an alkyl iodide, never an aryl iodide.

Question 16. Write the structure of main product of the following reaction.
C₆H₅–OC₂H₅ + HI → (CBSE 2018)
Answer:
In phenetole (C₆H₅–O–C₂H₅), C₆H₅–O bond has some double bond character due to resonance and hence more difficult to break. As a result, nucleophilic attack by I⁻ ion occurs on C₂H₅ group forming ethyl iodide and phenol.
C₆H₅–O–C₂H₅ (Phenetole) \( \xrightarrow[\Delta]{HI} \) [C₆H₅–O⁺(H)–C₂H₅] \( \xrightarrow{I^-} \) C₆H₅–OH (Phenol) + C₂H₅–I (Ethyl iodide)

Teacher's Note:
1. Even excess HI does not convert the phenol formed into iodobenzene.

Short Answer Questions (2 Marks)

Question 17. How is ethanol converted into diethyl ether? (J & K Board 2012)
Answer:
Ethanol is heated with conc. H₂SO₄ at 413 K, with ethanol in excess (dehydration of alcohol) :
2 C₂H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 413\text{ K}} \) C₂H₅–O–C₂H₅ (Diethyl ether) + H₂O

Teacher's Note:
1. Industrially, the same dehydration is done by passing ethanol vapour over alumina at about 513 K.

Question 18. How can diethyl ether be prepared from the following ?
(i) ethyl iodide (ii) ethyl alcohol.
Write chemical equation in each case. (Raj. Board, 2011)
Answer:
(i) By heating ethyl iodide with sodium ethoxide (Williamson synthesis) or with dry silver oxide :
C₂H₅I + C₂H₅ONa → C₂H₅OC₂H₅ + NaI ;   2C₂H₅I + Ag₂O \( \xrightarrow{\Delta} \) C₂H₅OC₂H₅ + 2AgI
(ii) By heating excess of ethyl alcohol with conc. H₂SO₄ at 413 K :
2C₂H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 413\text{ K}} \) C₂H₅OC₂H₅ + H₂O

Teacher's Note:
1. Dry silver oxide is a neat alternative for symmetrical ethers; moist silver oxide would give the alcohol instead.

Question 19. How will you obtain anisole from phenol ?
or How will you convert phenol to anisole ? (CBSE 2015, 2019)
Answer:
Phenol is converted into sodium phenoxide by treating with NaOH, and sodium phenoxide is then heated with methyl iodide (Williamson synthesis) :
C₆H₅OH + NaOH → C₆H₅ONa + H₂O ;   C₆H₅ONa + CH₃I \( \xrightarrow{\Delta} \) C₆H₅–O–CH₃ (Anisole) + NaI

Teacher's Note:
1. Methyl iodide (or dimethyl sulphate) must supply the CH₃ group; iodobenzene and sodium methoxide would not react.

Question 20. What are symmetrical and unsymmetrical ethers? How can diethyl ether be prepared from ethyl alcohol? (Uttarakhand Board 2012)
Answer:
Ethers in which both the groups attached to the oxygen atom are the same are called symmetrical or simple ethers, e.g., C₂H₅–O–C₂H₅. Ethers in which the two groups are different are called unsymmetrical or mixed ethers, e.g., CH₃–O–C₂H₅, C₆H₅–O–CH₃.
Preparation of diethyl ether from ethyl alcohol : by heating excess of ethyl alcohol with conc. H₂SO₄ at 413 K :
2C₂H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 413\text{ K}} \) C₂H₅–O–C₂H₅ + H₂O

Teacher's Note:
1. Acid dehydration gives only symmetrical ethers; a mixture of two alcohols would give three different ethers.

Question 21. Preparation of unsymmetrical ethers by Williamson synthesis requires proper choice of reactants. Comment.
Answer:
Williamson synthesis is an S_N2 reaction, while alkoxide ions are also strong bases. If the alkyl halide is 3° (or 2°), elimination predominates and an alkene is formed. Hence, the alkyl halide should be primary, and the bulky group should be taken as the alkoxide.
For example, tert-butyl methyl ether is prepared from sodium tert-butoxide and methyl bromide : (CH₃)₃C–ONa + CH₃Br → (CH₃)₃C–O–CH₃ + NaBr ; whereas (CH₃)₃C–Br + CH₃ONa gives mainly 2-methylpropene.
Similarly, for aryl alkyl ethers, sodium phenoxide and alkyl halide are used, since aryl halides do not undergo nucleophilic substitution.

Teacher's Note:
1. Rule of thumb: halide primary, alkoxide may be bulky.

Question 22. tert-Butyl methyl ether, in principle, can be prepared by treating either (i) sodium methoxide and tert- butyl chloride or (ii) sodium tert-butoxide and methyl chloride. Which one is preferred and why ?
Answer:
Method (ii) is preferred. In method (i), the 3° alkyl halide with the strongly basic methoxide ion undergoes elimination (E2) rather than substitution, and 2-methylpropene is the main product :
(CH₃)₃C–Cl + CH₃ONa → (CH₃)₂C=CH₂ + CH₃OH + NaCl
In method (ii), methyl chloride readily undergoes S_N2 attack by the tert-butoxide ion and gives the ether :
(CH₃)₃C–ONa + CH₃Cl → (CH₃)₃C–O–CH₃ + NaCl

Teacher's Note:
1. Methyl halides cannot undergo elimination at all (no β-H), which makes them ideal partners in Williamson synthesis.

Question 23. How will you account for the following ?
Ethers possess a net dipole moment even if they are symmetrical in structure.
Answer:
In ethers, the oxygen atom is sp³-hybridised and the molecule is bent (V-shaped) with the C–O–C bond angle about 110°. The two C–O bond dipoles therefore do not cancel each other, and the molecule has a net dipole moment (diethyl ether 1.18 D) even when the two alkyl groups are identical.

Teacher's Note:
1. Symmetry cancels dipoles only when the bond moments are opposite (180°), as in CO₂; a bent molecule always keeps a resultant.

Question 24. Write a note on nitration of anisole (HP Board 2011)
Answer:
Anisole undergoes nitration with a mixture of conc. HNO₃ and conc. H₂SO₄. The –OCH₃ group activates the ring (+R effect) and directs the electrophile (NO₂⁺) to the ortho and para positions. A mixture of 2-nitroanisole (minor) and 4-nitroanisole (major) is formed :
C₆H₅OCH₃ \( \xrightarrow{\text{Conc. }H_2SO_4 + \text{conc. }HNO_3} \) o-O₂N–C₆H₄–OCH₃ + p-O₂N–C₆H₄–OCH₃
The para isomer predominates because attack at the ortho position is sterically hindered by the –OCH₃ group.

Teacher's Note:
1. Always name both products and say which is major; the para isomer is favoured on steric grounds.

Question 25. Explain why cleavage of alkyl phenyl ethers with HBr always produces phenol and alkyl bromides and not bromobenzene and alkanols.
Answer:
On protonation, alkyl phenyl ether forms C₆H₅–O⁺(H)–R. The O–aryl bond has partial double bond character (resonance of oxygen lone pair with the ring) and so it is stronger than the O–alkyl bond. Moreover, Br⁻ cannot attack the sp²-hybridised carbon of the benzene ring. Therefore, Br⁻ attacks the alkyl carbon and the weaker O–R bond breaks, giving phenol and alkyl bromide : C₆H₅–O–R + HBr → C₆H₅OH + R–Br. Phenol does not react further with HBr to give bromobenzene.

Teacher's Note:
1. This is true even with excess HBr or HI: the phenol never becomes a halobenzene.

Question 26. Write the reagent and equation in the conversion of anisole into p-methoxyacetophenone.
Answer:
Reagent : acetyl chloride (CH₃COCl) in presence of anhydrous AlCl₃ (Friedel-Crafts acylation).
C₆H₅OCH₃ + CH₃COCl \( \xrightarrow{\text{Anhyd. }AlCl_3} \) p-CH₃O–C₆H₄–COCH₃ (4-Methoxyacetophenone, major) + o-CH₃O–C₆H₄–COCH₃ (2-Methoxyacetophenone, minor) + HCl

Teacher's Note:
1. Acetic anhydride with anhydrous AlCl₃ can be used in place of acetyl chloride.

Question 27. Describe simple chemical tests to distinguish between (a) diethyl ether and propanol.
(b) Ethanol and dimethyl ether.
Answer:
(a) (i) Add a few pieces of sodium metal. Whereas propanol evolves H₂ gas, diethyl ether does not. 2CH₃CH₂CH₂OH + 2Na → 2CH₃CH₂CH₂ONa + H₂↑
(ii) Add SOCl₂ or PCl₅ ; propanol evolves HCl gas at r.t. but diethyl ether does not.
(b) (i) Ethanol evolves H₂ gas with sodium metal, while dimethyl ether does not react.
(ii) Iodoform test : Ethanol gives a yellow ppt. of iodoform on warming with I₂ and NaOH, while dimethyl ether does not.

Teacher's Note:
1. Ethers have no O–H group, so every test based on the acidic O–H (Na, PCl₅, SOCl₂) is negative for them.

Question 28. Why are ethers used as solvents ?
Answer:
(i) Ethers are chemically quite inert : they do not react with bases, dilute acids, oxidising and reducing agents or active metals like sodium under ordinary conditions.
(ii) They dissolve a large number of organic compounds (fats, oils, resins, gums), and due to lone pairs on oxygen they can solvate (coordinate with) species like Grignard reagents.
(iii) Low boiling ethers (like diethyl ether) can be easily removed from the product by evaporation.
Hence, ethers are used as solvents for many reactions (e.g., Grignard reactions, LiAlH₄ reductions) and for extraction of organic compounds.

Teacher's Note:
1. Diethyl ether is very volatile and highly inflammable, so it is handled away from flames.

Short Answer Questions (3 Marks)

Question 29. How will you distinguish between the following pairs by chemical reactions ?
(i) Phenol and methanol
(ii) Ethanol and phenol (CBSE 2017)
(iii) Phenol and 1-propanol (CBSE 2019)
(iv) Propanol and 2-methylpropan-2-ol (CBSE 2017, 2019)
(v) C₆H₅OH and C₆H₅CH₂OH
(vi) Phenol and acetic acid
(vii) Phenol and benzoic acid. (CBSE 2013)
(viii) Phenol and chlorobenzene
Answer:
(i) FeCl₃ test : Phenol gives a violet colour with neutral FeCl₃ solution while methanol does not.
(ii) FeCl₃ test : Phenol gives a violet colour with neutral FeCl₃; ethanol does not. Alternatively, ethanol gives a yellow ppt. of iodoform with I₂/NaOH while phenol does not.
(iii) Bromine water test : Phenol decolourises bromine water giving a white ppt. of 2,4,6-tribromophenol; 1-propanol does not. (FeCl₃ test also works.)
(iv) Lucas test : 2-Methylpropan-2-ol (3°) gives turbidity immediately with Lucas reagent, while propan-1-ol (1°) gives no turbidity at room temperature.
(v) FeCl₃ test : C₆H₅OH gives a violet colour with neutral FeCl₃; benzyl alcohol does not.
(vi) NaHCO₃ test : Acetic acid gives brisk effervescence of CO₂ with NaHCO₃ solution; phenol does not. CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
(vii) NaHCO₃ test : Benzoic acid gives brisk effervescence of CO₂ with NaHCO₃; phenol does not. (Also, phenol gives a violet colour with neutral FeCl₃ while benzoic acid gives a buff coloured ppt.)
(viii) FeCl₃ test : Phenol gives a violet colour with neutral FeCl₃ while chlorobenzene does not. (Also, phenol dissolves in aq. NaOH while chlorobenzene does not.)

Teacher's Note:
1. Write the test, the observation for each compound, and an equation where possible; “observation” earns the mark.

Question 30. Give the structures of A and B.
(i) C₆H₅N₂Cl \( \xrightarrow[\Delta]{+H_2O} \) A \( \xrightarrow[\text{water}]{Br_2} \) B (Raj. Board 2012)
(ii) C₆H₅Cl \( \xrightarrow[\text{Pressure}]{NaOH,\ 623\text{K}} \) A \( \xrightarrow{H^+/H_2O} \) B (Jharkhand Board 2012)
(iii) A \( \xleftarrow[(ii)\ H^+]{(i)\ CHCl_3 + NaOH,\ 340\text{K}} \) Phenol \( \xrightarrow[(ii)\ H^+]{(i)\ NaOH/CO_2,\ 410\text{K}} \) B (Kerala Board 2012)
(iv) Phenol \( \xrightarrow[\Delta]{Zn\ \text{dust}} \) A \( \xrightarrow[+\text{conc. }H_2SO_4]{\text{conc. }HNO_3} \) B (Uttarakhand Board 2012)
Answer:
(i) A = C₆H₅OH (Phenol) ; B = 2,4,6-Br₃C₆H₂OH (2,4,6-Tribromophenol)
(ii) A = C₆H₅ONa (Sodium phenoxide) ; B = C₆H₅OH (Phenol)
(iii) A = o-HO–C₆H₄–CHO (Salicylaldehyde) ; B = o-HO–C₆H₄–COOH (Salicylic acid)
(iv) A = C₆H₆ (Benzene) ; B = C₆H₅NO₂ (Nitrobenzene)

Teacher's Note:
1. In (ii), NaOH is in excess, so the phenol formed is at once converted into its sodium salt; acid is needed to free phenol.

Question 31. Outline a scheme for the synthesis of each of the following :
(i) Ethyl methyl ether (ii) tert-Butyl methyl ether (iii) Anisole and (iv) Phenetole
Answer:
(i) C₂H₅OH \( \xrightarrow{Na} \) C₂H₅ONa \( \xrightarrow{CH_3I} \) C₂H₅–O–CH₃ (Ethyl methyl ether) + NaI
(ii) (CH₃)₃COH \( \xrightarrow{Na} \) (CH₃)₃CONa \( \xrightarrow{CH_3Br} \) (CH₃)₃C–O–CH₃ (tert-Butyl methyl ether) + NaBr
(iii) C₆H₅OH \( \xrightarrow{NaOH} \) C₆H₅ONa \( \xrightarrow{CH_3I} \) C₆H₅–O–CH₃ (Anisole) + NaI
(iv) C₆H₅OH \( \xrightarrow{NaOH} \) C₆H₅ONa \( \xrightarrow{C_2H_5I} \) C₆H₅–O–C₂H₅ (Phenetole) + NaI

Teacher's Note:
1. All four are Williamson syntheses; in each case the halide is a simple primary halide.

Question 32. Give reasons for the following : (i) The boiling points of ethers are lower than their isomeric alcohols.
(ii) The boiling points of ethers are much lower than those of alcohols of comparable molecular masses. (CBSE 2007)
(iii) Dimethyl ether is completely miscible with water but diethyl ether is soluble in water to a small extent (only 8% by weight). (Pb Board 2012)
(iv) Ethyl alcohol (mol. mass = 46) boils at 351 K while diethyl ether (mol. mass = 74) boils at 308 K.
Or The boiling point of ethanol is higher than that of methoxymethane. (CBSE 2009)
(v) The C–O–C bond angle in ethers is higher than the H–O–H angle in water though oxygen is sp³-hybridized in both these cases.
(vi) The dipole moment of diethyl ether (1.18 D) is lower than that of water (1.84 D).
Answer:
(i) and (ii) Alcohols have an O–H bond and are associated by intermolecular H-bonding, while ethers have no H attached to oxygen and cannot form intermolecular H-bonds; their molecules are held only by weak dipole-dipole forces. Hence ethers boil much lower than isomeric alcohols or alcohols of comparable molecular masses.
(iii) The oxygen of ethers can form H-bonds with water. In dimethyl ether the small methyl groups do not hinder this, so it is completely miscible. In diethyl ether the larger ethyl groups (hydrophobic) hinder H-bond formation with water, so it is only slightly soluble.
(iv) Ethanol molecules are associated by intermolecular H-bonding, which needs extra energy to break, while ether molecules are not H-bonded. Hence ethanol boils higher than diethyl ether (or methoxymethane) in spite of its lower molecular mass.
(v) In ethers, the two bulky alkyl groups repel each other (steric repulsion), which widens the C–O–C angle to about 111.7°; in water the small H atoms cause little repulsion and the lone pair-lone pair repulsion reduces the H–O–H angle to 104.5°.
(vi) The C–O bond is less polar than the O–H bond (electronegativity difference O–C < O–H), and the larger C–O–C angle makes the two bond moments cancel to a greater extent. Hence the resultant dipole moment of diethyl ether is lower than that of water.

Teacher's Note:
1. Ethers can accept but not donate H-bonds: that single fact explains their low boiling points and their partial solubility in water.

Question 33. How does diethyl ether react with :
(i) HBr at 373 K
(ii) HI at 373 K
(iii) Dil. H₂SO₄ under pressure and (iv) BF₃ ?
Answer:
(i) C₂H₅–O–C₂H₅ + HBr \( \xrightarrow{373\text{ K}} \) C₂H₅Br (Bromoethane) + C₂H₅OH
(ii) C₂H₅–O–C₂H₅ + HI \( \xrightarrow{373\text{ K}} \) C₂H₅I (Iodoethane) + C₂H₅OH ; with excess HI : C₂H₅–O–C₂H₅ + 2HI → 2C₂H₅I + H₂O
(iii) Hydrolysis occurs to give ethanol : C₂H₅–O–C₂H₅ + H₂O \( \xrightarrow{\text{Dil. }H_2SO_4,\ \text{pressure}} \) 2C₂H₅OH
(iv) Being a Lewis base, diethyl ether forms a coordination complex (boron trifluoride etherate) with the Lewis acid BF₃ : (C₂H₅)₂O: + BF₃ → (C₂H₅)₂O→BF₃

Teacher's Note:
1. Ether oxygen behaves as a Lewis base in all four reactions: it is first protonated by the acid, or coordinates to BF₃.

Question 34. (a) How does anisole react with :
(i) Br₂ in CS₂
(ii) Conc. HNO₃–H₂SO₄ mixture (HP Board 2011)
(iii) HI at 373 K ?
(iv) Anisole is treated with CH₃Cl/anhydrous AlCl₃. (CBSE (Foreign) 2017)
(b) How will you convert anisole to phenol ? (CBSE 2007)
Answer:
(a) (i) Bromination occurs at o- and p-positions, even without a catalyst : C₆H₅OCH₃ + Br₂ \( \xrightarrow{CS_2} \) p-Br–C₆H₄–OCH₃ (4-Bromoanisole, major) + o-Br–C₆H₄–OCH₃ (2-Bromoanisole, minor) + HBr
(ii) Nitration gives a mixture of o- and p-nitroanisole : C₆H₅OCH₃ \( \xrightarrow{\text{Conc. }H_2SO_4 + \text{conc. }HNO_3} \) p-O₂N–C₆H₄–OCH₃ (4-Nitroanisole, major) + o-O₂N–C₆H₄–OCH₃ (2-Nitroanisole, minor)
(iii) C₆H₅–O–CH₃ + HI \( \xrightarrow{373\text{ K}} \) C₆H₅OH (Phenol) + CH₃I (Methyl iodide)
(iv) Friedel-Crafts alkylation : C₆H₅OCH₃ + CH₃Cl \( \xrightarrow{\text{Anhyd. }AlCl_3} \) p-CH₃–C₆H₄–OCH₃ (4-Methoxytoluene, major) + o-CH₃–C₆H₄–OCH₃ (2-Methoxytoluene, minor) + HCl
(b) By heating anisole with HI : C₆H₅OCH₃ + HI \( \xrightarrow{373\text{ K}} \) C₆H₅OH + CH₃I

Teacher's Note:
1. –OCH₃ is an activating, o,p-directing group like –OH, though a little less activating.

Question 35. Discuss the following :
(i) Cleavage of ethers by acids
(ii) Electrophilic substitution reactions in aromatic ethers.
Answer:
(i) Cleavage of ethers by acids : Ethers are cleaved by heating with HI or HBr (373 K). R–O–R + HX → RX + ROH; with excess HX the alcohol formed is also converted into alkyl halide. The order of reactivity is HI > HBr > HCl.
Mechanism : the ether is first protonated, R–O–R′ + HX → R–O⁺(H)–R′ + X⁻. Then the halide ion attacks : if the groups are 1° or 2°, X⁻ attacks the smaller (less hindered) alkyl group by S_N2, e.g., CH₃–O–CH₂CH₃ + HI → CH₃I + CH₃CH₂OH. If one group is 3°, the reaction is S_N1 and the halide goes to the 3° group, e.g., (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH. Aryl alkyl ethers always give phenol and alkyl halide : C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I.
(ii) Electrophilic substitution in aromatic ethers : The alkoxy group is o, p-directing and activates the ring (+R effect). Anisole undergoes
Halogenation : with Br₂ in ethanoic acid (no catalyst needed) giving mainly p-bromoanisole.
Nitration : with conc. H₂SO₄ + conc. HNO₃ giving o- and p-nitroanisole (para major).
Friedel-Crafts reaction : with CH₃Cl (alkylation) or CH₃COCl (acylation) in presence of anhyd. AlCl₃ giving o- and p-methylanisole or o- and p-methoxyacetophenone (para major).

Teacher's Note:
1. In cleavage questions first decide the mechanism (S_N1 if a 3° group is present, otherwise S_N2); the products follow from it.

Question 36. How will you account for the following ?
(i) Methyl phenyl ether reacts with HI to give phenol and methyl iodide and not iodobenzene and methyl alcohol.
(ii) The order of reactivity of halogen acids towards cleavage of ethers is : HI > HBr > HCl.
(iii) t-Butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of the butyl methyl ether. (CBSE 2019)
Answer:
(i) In the protonated ether C₆H₅–O⁺(H)–CH₃, the O–C₆H₅ bond has partial double bond character due to resonance and is stronger, while the O–CH₃ bond is weaker. Also, I⁻ cannot attack the sp² carbon of the ring. Hence I⁻ attacks the CH₃ group (S_N2), giving CH₃I and phenol.
(ii) The cleavage requires protonation of the ether and then nucleophilic attack by X⁻. HI is the strongest acid (weakest H–X bond) and I⁻ is the best nucleophile, followed by HBr and Br⁻, while HCl is the weakest acid and Cl⁻ the poorest nucleophile. Hence the reactivity order is HI > HBr > HCl.
(iii) Sodium methoxide is a strong base and t-butyl chloride is a 3° halide which is sterically hindered for S_N2 attack. So, elimination (E2) predominates over substitution : (CH₃)₃C–Cl + CH₃ONa → CH₂=C(CH₃)₂ (2-Methylpropene) + CH₃OH + NaCl

Teacher's Note:
1. Point (iii) is the reason why tert-butyl methyl ether is made from sodium tert-butoxide and a methyl halide.

Question 37. Under what conditions do ethers form oxonium salts ? How is the presence of peroxides in ethers detected and removed ?
Answer:
Oxonium salts : Ethers, being Lewis bases, dissolve in cold concentrated strong acids such as conc. H₂SO₄ or conc. HCl and form oxonium salts : R₂O + HCl (cold, conc.) → [R₂O⁺H]Cl⁻ ; R₂O + H₂SO₄ → [R₂O⁺H]HSO₄⁻. On dilution with water the ether is regenerated.
Detection of peroxides : The ether is shaken with an aqueous solution of ferrous ammonium sulphate and potassium thiocyanate. If peroxides are present, Fe²⁺ is oxidised to Fe³⁺, which gives a blood red colour with SCN⁻ : Fe³⁺ + 3SCN⁻ → Fe(SCN)₃ (Blood red). (Alternatively, peroxides liberate iodine from acidified KI, which turns starch blue.)
Removal of peroxides : by shaking the ether with an acidified solution of ferrous sulphate (or aq. KI), which reduces the peroxides, followed by distillation of the ether. Storing ether in dark coloured bottles with a little copper or iron wire reduces their formation.

Teacher's Note:
1. Ether must never be distilled to dryness; the peroxides concentrate in the residue and can explode.

Question 38. Write the chemical equations, when :
(i) Benzyl phenyl ether reacts with HI at 373 K.
(ii) tert-Butyl methyl ether reacts with HBr at 373 K.
(iii) Tetrahydrofuran reacts with excess of HBr at 373 K.
(iv) Phenetole reacts with HI at 373 K.
(v) 2-Ethoxybutane reacts with HI at 373 K.
(vi) 2-Methoxy-2-phenylpropane reacts with HI at 373 K.
(vii) Diethyl ether reacts with Cl₂ in dark.
(viii) Ethoxyethane is heated with excess of HI at 373 K.
(ix) Methoxyethane is heated with excess HI. (Assam Board 2013)
Answer:
(i) C₆H₅–O–CH₂C₆H₅ + HI \( \xrightarrow{373\text{ K}} \) C₆H₅OH (Phenol) + C₆H₅CH₂I (Benzyl iodide)
(ii) (CH₃)₃C–O–CH₃ + HBr \( \xrightarrow{373\text{ K}} \) (CH₃)₃C–Br (tert-Butyl bromide) + CH₃OH (S_N1 : the stable 3° carbocation is formed)
(iii) Tetrahydrofuran (cyclic C₄H₈O) + 2HBr \( \xrightarrow{373\text{ K}} \) Br–CH₂CH₂CH₂CH₂–Br (1,4-Dibromobutane) + H₂O
(iv) C₆H₅–O–C₂H₅ + HI \( \xrightarrow{373\text{ K}} \) C₆H₅OH (Phenol) + C₂H₅I (Ethyl iodide)
(v) CH₃CH₂–CH(OC₂H₅)–CH₃ + HI \( \xrightarrow{373\text{ K}} \) CH₃CH₂CH(OH)CH₃ (Butan-2-ol) + CH₃CH₂I (Iodoethane) (S_N2 : I⁻ attacks the less hindered ethyl group)
(vi) C₆H₅–C(CH₃)₂–OCH₃ + HI \( \xrightarrow{373\text{ K}} \) C₆H₅–C(CH₃)₂–I (2-Iodo-2-phenylpropane) + CH₃OH (S_N1 : the stable 3° benzylic carbocation is formed)
(vii) Substitution occurs at the α-carbon atoms : CH₃CH₂–O–CH₂CH₃ + 2Cl₂ \( \xrightarrow{\text{Dark}} \) CH₃CHCl–O–CHClCH₃ (α,α′-Dichlorodiethyl ether) + 2HCl
(viii) C₂H₅–O–C₂H₅ + 2HI \( \xrightarrow{373\text{ K}} \) 2C₂H₅I + H₂O
(ix) CH₃–O–C₂H₅ + 2HI \( \xrightarrow{\Delta} \) CH₃I + C₂H₅I + H₂O

Teacher's Note:
1. With excess HI both groups end up as alkyl iodides; with one mole only the less hindered group (S_N2) or the 3°/benzylic group (S_N1) becomes the iodide.

Question 39. Complete the following reactions :
(i) C₆H₅ONa + CH₃Br \( \xrightarrow{\Delta} \) (Hr. Board 2011)
(ii) CH₃–O–C₂H₅ + HI \( \xrightarrow{373\text{K}} \) (Hr. Board 2011)
(iii) CH₃OCH₂CH₃ + HBr \( \xrightarrow{373\text{K}} \) (Hr. Board 2011)
(iv) CH₃–C(CH₃)₂–OCH₃ + HI → (CBSE 2016)
(v) CH₃–CH(CH₃)–O–CH₂CH₃ \( \xrightarrow{HI} \) (CBSE 2017)
(vi) (CH₃)₃C–OCH₃ \( \xrightarrow{HI} \) (CBSE (Foreign) 2017)
(vii) C₆H₅–OC₂H₅ + HI → (CBSE 2018)
Answer:
(i) C₆H₅ONa + CH₃Br \( \xrightarrow{\Delta} \) C₆H₅–O–CH₃ (Anisole) + NaBr
(ii) CH₃–O–C₂H₅ + HI \( \xrightarrow{373\text{K}} \) CH₃I + C₂H₅OH
(iii) CH₃OCH₂CH₃ + HBr \( \xrightarrow{373\text{K}} \) CH₃Br + CH₃CH₂OH
(iv) (CH₃)₃C–OCH₃ + HI → (CH₃)₃C–I (2-Iodo-2-methylpropane) + CH₃OH
(v) (CH₃)₂CH–O–CH₂CH₃ + HI → (CH₃)₂CH–OH (Propan-2-ol) + CH₃CH₂I (Iodoethane)
(vi) (CH₃)₃C–OCH₃ + HI → (CH₃)₃C–I + CH₃OH
(vii) C₆H₅–OC₂H₅ + HI → C₆H₅OH (Phenol) + C₂H₅I (Ethyl iodide)

Teacher's Note:
1. Compare (ii) and (iv): the methyl group gets the iodide in an S_N2 cleavage, but the 3° group gets it in an S_N1 cleavage.

Question 40. Describe the following :
(i) Kolbe’s reaction (CBSE 2010)
(ii) Reimer-Tiemann reaction (HP Board 2011 ; Pb. Board 2012 ; Hr. Board 2012 ; CBSE 2010, 2010 S, 2013, 2014, 2019)
(iii) Coupling reaction (CBSE 2010 ; Hr. Board 2011, 2012 ; HP Board 2011, 2013)
(iv) Schotten-Baumann reaction.
(v) Esterification reaction. (Pb. Board 2008, 2009 ; Hr. Board 2012 ; West Bengal Board 2013)
(vi) Fischer esterification. (CBSE 2007)
(vii) Williamson’s synthesis. (J & K Board 2011 ; Hr. Board 2011 ; CBSE 2010, 2010 S, 2013)
(viii) Friedel-Crafts alkylation of phenol. (CBSE 2019)
Answer:
(i) Kolbe’s reaction : When sodium phenoxide is heated with CO₂ at 400 K under 4-7 atm pressure, sodium salicylate is formed, which on acidification gives salicylic acid : C₆H₅ONa + CO₂ \( \xrightarrow{400\text{ K},\ 4\text{-}7\text{ atm}} \) o-HO–C₆H₄–COONa \( \xrightarrow{H^+} \) o-HO–C₆H₄–COOH
(ii) Reimer-Tiemann reaction : When phenol is heated with chloroform and aqueous NaOH at 340 K followed by acidification, salicylaldehyde (2-hydroxybenzaldehyde) is formed : C₆H₅OH + CHCl₃ + 3NaOH \( \xrightarrow{340\text{ K}} \) o-NaO–C₆H₄–CHO \( \xrightarrow{H^+} \) o-HO–C₆H₄–CHO. The electrophile is dichlorocarbene (:CCl₂). With CCl₄ in place of CHCl₃, salicylic acid is formed.
(iii) Coupling reaction : Benzenediazonium chloride reacts with phenol in mildly alkaline medium (273-278 K) to form p-hydroxyazobenzene, an orange dye : C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{OH^-} \) p-HO–C₆H₄–N=N–C₆H₅ + HCl
(iv) Schotten-Baumann reaction : Benzoylation of phenols (or alcohols, amines) with benzoyl chloride in presence of aqueous NaOH : C₆H₅OH + C₆H₅COCl \( \xrightarrow{NaOH} \) C₆H₅COOC₆H₅ (Phenyl benzoate) + HCl
(v) Esterification reaction : Alcohols and phenols react with carboxylic acids, acid chlorides or acid anhydrides to form esters : R–OH + R′COOH \( \xrightarrow{H^+} \) R′COOR + H₂O ; R–OH + (R′CO)₂O \( \xrightarrow{H^+} \) R′COOR + R′COOH ; R–OH (or Ar–OH) + R′COCl \( \xrightarrow{\text{Pyridine}} \) R′COOR (or R′COOAr) + HCl
(vi) Fischer esterification : Heating a carboxylic acid with an alcohol in presence of a little conc. H₂SO₄ (or dry HCl) to form an ester. The reaction is reversible; water is removed as formed to shift the equilibrium : CH₃COOH + C₂H₅OH ⇄ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄)
(vii) Williamson’s synthesis : An alkyl halide reacts with sodium alkoxide (or phenoxide) to give an ether (S_N2) : R–X + R′ONa → R–O–R′ + NaX, e.g., CH₃I + C₂H₅ONa → CH₃OC₂H₅ + NaI. Primary alkyl halides give the best yields; 3° halides give alkenes.
(viii) Friedel-Crafts alkylation of phenol : Phenol reacts with an alkyl halide in presence of anhydrous AlCl₃ to give a mixture of o- and p-alkylphenols : C₆H₅OH + CH₃Cl \( \xrightarrow{\text{Anhyd. }AlCl_3} \) o-CH₃–C₆H₄–OH (o-Cresol) + p-CH₃–C₆H₄–OH (p-Cresol) + HCl. (Yields are low because phenol forms a complex with AlCl₃. Anisole undergoes the same reaction readily giving 2- and 4-methoxytoluene.)

Teacher's Note:
1. Named reactions are scored on two things: the correct reagents and conditions over the arrow, and the correct product name.

Long Answer Questions (5 Marks)

Question 41. Discuss briefly two methods by which primary, secondary and tertiary alcohols be distinguished.
Answer:
1. Lucas test : The alcohol is shaken with Lucas reagent (conc. HCl + anhydrous ZnCl₂) at room temperature. Alkyl chlorides formed are insoluble and appear as turbidity.
3° alcohols give turbidity immediately : (CH₃)₃COH + HCl \( \xrightarrow{ZnCl_2} \) (CH₃)₃CCl + H₂O
2° alcohols give turbidity within about 5 minutes : (CH₃)₂CHOH + HCl \( \xrightarrow{ZnCl_2} \) (CH₃)₂CHCl + H₂O
1° alcohols do not give turbidity at room temperature (only on heating).
2. Action of heated copper (573 K) : Vapours of the alcohol are passed over heated copper and the product is identified.
1° alcohols give aldehydes (which reduce Tollens’ reagent) : RCH₂OH \( \xrightarrow{Cu,\ 573\text{ K}} \) RCHO + H₂
2° alcohols give ketones (which do not reduce Tollens’ reagent) : R₂CHOH \( \xrightarrow{Cu,\ 573\text{ K}} \) R₂CO + H₂
3° alcohols give alkenes (which decolourise bromine water) : (CH₃)₃COH \( \xrightarrow{Cu,\ 573\text{ K}} \) (CH₃)₂C=CH₂ + H₂O
(Victor Meyer’s test is another method : the nitroalkanes obtained from 1°, 2° and 3° alcohols give blood red, blue and no colour respectively with nitrous acid followed by alkali.)

Teacher's Note:
1. In the Lucas test, report the time for turbidity; in the copper test, report how the product is identified, not only its name.

Question 42. What is the IUPAC name of tert-butyl methyl ether ? Write one method each for the synthesis and cleavage of ethers. Give chemical equations.
Answer:
IUPAC name of tert-butyl methyl ether, (CH₃)₃C–O–CH₃ : 2-Methoxy-2-methylpropane.
Synthesis (Williamson synthesis) : An alkyl halide is heated with sodium alkoxide. For tert-butyl methyl ether, sodium tert-butoxide is heated with methyl bromide (a 3° halide is not used because it would give an alkene) :
(CH₃)₃C–ONa + CH₃–Br \( \xrightarrow{\Delta} \) (CH₃)₃C–O–CH₃ + NaBr
Cleavage (with HI) : Ethers are cleaved by HI at 373 K. With tert-butyl methyl ether the reaction follows S_N1 mechanism because the stable 3° carbocation is formed, so tert-butyl iodide and methanol are obtained :
(CH₃)₃C–O–CH₃ + HI \( \xrightarrow{373\text{ K}} \) (CH₃)₃C–I + CH₃OH
For ethers with 1°/2° groups the cleavage is S_N2 and the iodide is formed from the smaller group, e.g., CH₃–O–C₂H₅ + HI → CH₃I + C₂H₅OH.

Teacher's Note:
1. Ether names put the larger group as the parent alkane and the smaller group as alkoxy: here propane is the parent and methoxy the prefix.