CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. ortho-Nitrophenol has lower boiling point than p-nitrophenol. Why? [CBSE Comptt. Delhi 2012]
Answer:
p-Nitrophenol shows intermolecular hydrogen bonding. So it has higher boiling point. Whereas o-nitrophenol shows intramolecular H-bonding hence lower boiling point.

Teacher's Note:
1. Intramolecular H-bonds form within one molecule and do not hold molecules together, so they do not raise the boiling point.

Question 2. The C–O bond is much shorter in phenol than in ethanol. Give reason. [CBSE Comptt. Delhi 2012]
Answer:
Because in phenol, conjugation of unshared electron pair over oxygen with aromatic ring results in partial double bond character in carbon-oxygen bond. In phenol, oxygen is attached to sp² hybridised carbon atom while in ethanol it is attached to sp³ hybridised carbon atom. The bond formed between oxygen and sp² hybridised carbon is more stable i.e., shorter than that formed between oxygen and sp³ hybridised carbon.

Teacher's Note:
1. Two reasons earn full marks: partial double bond character from resonance, and the greater s-character of the sp² carbon.

Question 3. Out of CH₃OH and C₆H₅OH (phenol), which one is more acidic? [CBSE Comptt. Delhi 2016]
Answer:
C₆H₅OH (phenol), as phenoxide ion is more stable due to resonance which is absent in alkoxide ion.

Teacher's Note:
1. Compare the stability of the anions formed, not of the acids: the more stable the conjugate base, the stronger the acid.

Question 4. What happens when phenol is oxidized by Na₂Cr₂O₇/H₂SO₄? [CBSE Comptt. OD Set-1 2017]
Answer:
C₆H₅OH \( \xrightarrow{Na_2Cr_2O_7/H_2SO_4} \) O=C₆H₄=O (p-benzoquinone)
∴ Benzoquinone is formed.

Teacher's Note:
1. Benzoquinone is a conjugated diketone: the two C=O groups are para to each other on a six-membered ring with two C=C bonds.

Question 5. What happens when phenol is heated with zinc dust? [CBSE Comptt. OD Set-2 2017]
Answer:
C₆H₅OH + Zn → C₆H₆ + ZnO
Benzene is formed.

Teacher's Note:
1. Zinc dust removes the oxygen of phenol – remember this as the standard first step whenever a conversion from phenol to a benzene derivative (toluene, acetophenone) is asked.

Question 6. What happens when phenol is treated with bromine water? [CBSE Comptt. OD Set-3 2017]
Answer:
C₆H₅OH + 3Br₂ (aq) → 2,4,6-Br₃C₆H₂OH + 3HBr
2,4,6-Tribromophenol is formed (white precipitate).

Teacher's Note:
1. With bromine water all three o/p positions are substituted; to get only monobromophenols use Br₂ in a non-polar solvent such as CS₂ at low temperature.

Question 7. Write the IUPAC name of the given compound:
CH₃–CH(CH₃)–CH₂–O–CH₂–CH₃ [CBSE OD 2015]
Answer:
1-Ethoxy-2-methylpropane.

Teacher's Note:
1. Choose the larger group (the branched C₄ group) as the parent alkane and number it from the carbon attached to oxygen.

Question 8. Write the IUPAC name of the following compound:
CH₃–O–C(CH₃)₂–CH₃ [CBSE OD Set-3 2017]
Answer:
2-Methoxy-2-methylpropane

Teacher's Note:
1. This is methyl tert-butyl ether (MTBE); both the methoxy and the methyl substituents sit on C-2 of propane.

Question 9. Write the equation involved in the reaction Williamson’s ether synthesis. [CBSE OD 2013]
Answer:
R–X (Alkyl halide) + R′–ONa (Sodium alkoxide) → R–O–R′ (Ether) + NaX
e.g. CH₃I (Methyl iodide) + C₂H₅ONa (Sodium ethoxide) → CH₃–O–C₂H₅ (Ethyl methyl ether) + NaI

Teacher's Note:
1. Use a primary alkyl halide in Williamson synthesis; with a tertiary halide the alkoxide acts as a base and an alkene is formed instead.

Question 10. How would you obtain acetophenone from phenol? [CBSE Comptt. OD 2013]
Answer:
C₆H₅OH (Phenol) \( \xrightarrow[-ZnO]{\text{Zn dust},\ \Delta} \) C₆H₆ (Benzene) \( \xrightarrow{CH_3COCl/AlCl_3} \) C₆H₅COCH₃ (Acetophenone)

Teacher's Note:
1. Friedel-Crafts acylation cannot be done on phenol directly in good yield, so the OH is first removed by distilling with zinc dust.

Question 11. An alkoxide is a stronger base than hydroxide ion. Justify. [CBSE SQP 2013]
Answer:
Due to the presence of an alkyl group, higher electron density is found in alkoxide ion.

Teacher's Note:
1. The +I effect of the alkyl group makes the lone pairs on oxygen more available for accepting a proton.

Question 12. The number of alcohols giving iodoform test among the following is:
CH₃CH₂OH, CH₃OH, CH₃CH₂CH₂OH, (CH₃)₂CHOH, CH₃CH₂CH₂CH₂OH, CH₃CH₂CH(OH)CH₃, CH₃CH(OH)CH(CH₃)₂, (C₂H₅)₂CHOH, (CH₃)₃COH
Answer:
4
Alcohols having the CH₃CH(OH)– group give the iodoform test: CH₃CH₂OH, (CH₃)₂CHOH, CH₃CH₂CH(OH)CH₃ and CH₃CH(OH)CH(CH₃)₂.

Teacher's Note:
1. (C₂H₅)₂CHOH is a 2° alcohol but has no CH₃ on the carbinol carbon, and (CH₃)₃COH has no H on it – neither gives the test.

Question 13. Write IUPAC name of the following compound:
CH₃–CH(CH₃)–CH₂–CH(OH)–CH(CH₃)–CH₂OH
Answer:
2,5-Dimethylhexane-1,3-diol.

Teacher's Note:
1. Number the chain from the end nearer to the –OH groups; the –OH positions (1,3) get the lowest locants before the methyl groups.

Question 14. How is phenol obtained from aniline?
Answer:
C₆H₅NH₂ —(HONO, 0–5°C)→ C₆H₅N₂⁺Cl⁻ —(H₂O/H⁺, Δ)→ C₆H₅OH
Aniline is diazotised to benzenediazonium chloride, which on warming with water gives phenol.

Teacher's Note:
1. Keep the diazotisation at 0–5°C; the diazonium salt is hydrolysed to phenol only when it is warmed.

Question 15. Why is phenol acidic in nature?
Answer:
Due to stability of phenoxide ion by resonance.

Teacher's Note:
1. The negative charge of the phenoxide ion is delocalised over the ring, which is why phenol is more acidic than alcohols and water.

Short Answer Questions (2 Marks)

Question 16. Explain why is O=C=O nonpolar while R–O–R is polar. [NCERT Exemplar]
Answer:
CO₂ is a linear molecule. The dipole moment of two C=O bonds are equal and opposite. Hence, they cancel each other. So, the dipole moment of CO₂ is zero and it is a non-polar molecule (O=C=O, μ = 0).
While for ethers, two dipoles are pointing in the same direction. These two dipoles do not cancel the effect of each other. Therefore, there is a finite resultant dipole and hence, R—O—R is a polar molecule (bent molecule, C–O–C angle 111.7°, μ ≠ 0).

Teacher's Note:
1. Polarity depends on shape as well as on bond polarity: equal bond dipoles cancel only in a symmetric (linear) arrangement.

Question 17. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. [NCERT]
Answer:
The formation of ethers by dehydration of alcohol is a bimolecular reaction (S_N2) involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alkenes are formed.

Teacher's Note:
1. Secondary and tertiary alcohols also form fairly stable carbocations, which readily lose H⁺ to give alkenes.

Question 18. Write one chemical reaction to illustrate the following:
(i) Reimer-Tiemann reaction
(ii) Williamson’s synthesis
Answer:
(i) C₆H₅OH —(CHCl₃ + NaOH(aq))→ [o-Na⁺O⁻–C₆H₄–CHCl₂] —((i) NaOH, (ii) H⁺)→ o-HO–C₆H₄–CHO (salicylaldehyde)
(ii) C₆H₅ONa + CH₃Br → C₆H₅OCH₃ + NaBr

Teacher's Note:
1. In Williamson’s synthesis of an alkyl aryl ether, the phenoxide must be the nucleophile and the alkyl halide the substrate – aryl halides do not undergo S_N2.

Question 19. Account for the following:
(i) Phenol has a smaller dipole moment than methanol.
(ii) Phenol undergoes electrophilic substitution reactions.
Answer:
(i) Due to –ve charge on oxygen being delocalised by resonance (the C–O bond in phenol is less polar).
(ii) Due to greater electron density on the ring than benzene (the –OH group releases electrons into the ring by resonance).

Teacher's Note:
1. In methanol the C–O bond points towards a fully electron-rich O; in phenol the electron-withdrawing ring pulls the density back, lowering the dipole moment.

Question 20. Complete the following equations and name the products:
(i) Phenol + FeCl₃ →
(ii) C₆H₅OH + CHCl₃ + NaOH —(340 K)→
(iii) C₆H₅OH + Br₂(aq) →
(iv) C₆H₅ONa + CO₂ —(400 K, 4–7 atm)→
Answer:
(i) 3C₆H₅OH + FeCl₃ → [(C₆H₅O)₃Fe] + 3HCl   (ferric phenoxide, violet colour)
(ii) C₆H₅OH + CHCl₃ + NaOH → o-HO–C₆H₄–CHO + NaCl + H₂O   (salicylaldehyde)
(iii) C₆H₅OH + 3Br₂(aq) → 2,4,6-tribromophenol + 3HBr   (white precipitate)
(iv) C₆H₅ONa + CO₂ → o-HO–C₆H₄–COONa   (sodium salicylate)

Teacher's Note:
1. The violet colour with neutral FeCl₃ is the standard laboratory test that tells phenols apart from alcohols.

Question 21. Write:
(i) Friedel-Crafts reaction
(ii) Coupling reaction
Answer:
(i) C₆H₆ + CH₃Cl —(Anhyd. AlCl₃)→ C₆H₅CH₃ + HCl
(ii) C₆H₅N₂Cl + C₆H₅OH —(pH = 9–10)→ C₆H₅–N=N–C₆H₄–OH (p-hydroxyazobenzene, orange dye) + HCl

Teacher's Note:
1. Coupling with phenol is done in mildly alkaline medium (pH 9–10) because the phenoxide ion is the reactive species; coupling occurs at the para position.

Question 22. Give one reaction of alcohol involving cleavage of:
(i) C–O bond
(ii) O–H bond
Answer:
(i) CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl
(ii) 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂

Teacher's Note:
1. Reactions with Na, acids (esterification) break O–H; reactions with HX, PX₃, PCl₅, SOCl₂ and dehydration break C–O.

Question 23. Ethereal solution of an organic compound ‘X’ when heated with Mg gave ‘Y’. ‘Y’ on treatment with CH₃CHO followed by acid hydrolysis gave 2-propanol. Identify the compound ‘X’. What is ‘Y’ known as?
Answer:
CH₃Br (X) + Mg —(Dry ether)→ CH₃MgBr (Y)
CH₃CHO + CH₃MgBr —(Dry ether)→ CH₃–CH(OMgBr)–CH₃ —(H₃O⁺)→ CH₃–CH(OH)–CH₃
X is bromomethane (CH₃Br) and Y is a Grignard reagent (methylmagnesium bromide).

Teacher's Note:
1. Work backwards: 2-propanol = CH₃CHO + one CH₃ group, so the Grignard reagent must carry a methyl group.

Question 24. While separating a mixture of o- and p-nitrophenols by steam distillation, name the isomer which is steam volatile. Give reason.
Answer:
o-Nitrophenol is steam volatile. It has intramolecular hydrogen bonding (between the –OH and the adjacent –NO₂ group), so its molecules are not associated.
p-Nitrophenol shows intermolecular H-bonding (O₂N–C₆H₄–O–H····O–N(O)–C₆H₄–OH): H-bonding and association of molecules make it less volatile.

Teacher's Note:
1. Only an ortho group can form a six-membered H-bonded ring with –OH; that is why the ortho isomer alone distils with steam.

Question 25. Account for the following:
(i) Phenol has a smaller dipole moment than CH₃OH.
(ii) Phenols do not give protonation reactions readily.
Answer:
(i) Because phenol has electron attracting benzene ring (the C–O bond is less polar due to resonance).
(ii) Due to resonance, the oxygen acquires a +ve charge (its lone pair is delocalised into the ring), so it does not have tendency to accept a proton.

Teacher's Note:
1. The same resonance that makes phenol more acidic also makes its oxygen a poorer base.

Question 26. Write the reactions and conditions involved in the conversion of:
(i) Propene to propan-2-ol.
(ii) Phenol to salicylic acid.
Answer:
(i) CH₃CH=CH₂ + H₂O —(H₂SO₄ (dil.))→ CH₃–CH(OH)–CH₃
(ii) C₆H₅OH + CO₂ —(NaOH, Δ, 4–7 atm)→ o-HO–C₆H₄–COONa —(H₃O⁺)→ o-HO–C₆H₄–COOH

Teacher's Note:
1. Acid-catalysed hydration follows Markovnikov’s rule, so propene gives the 2° alcohol, not propan-1-ol (which needs hydroboration-oxidation).

Short Answer Questions (3 Marks)

Question 27. (i) Write the mechanism of the following reaction:
2CH₃CH₂OH \( \xrightarrow{H^+} \) CH₃CH₂–O–CH₂CH₃
(ii) Write the equations involved in the acetylation of Salicylic acid. [CBSE Delhi 2015]
Answer:
(i) 2CH₃CH₂OH \( \xrightarrow{H^+} \) CH₃CH₂–O–CH₂CH₃
CH₃–CH₂–OH + H⁺ → CH₃–CH₂–O⁺H₂
CH₃–CH₂–O(H) + CH₃–CH₂–O⁺H₂ → CH₃–CH₂–O⁺(H)–CH₂CH₃ + H₂O
CH₃–CH₂–O⁺(H)–CH₂CH₃ → CH₃CH₂–O–CH₂CH₃ + H⁺
(ii) o-HO–C₆H₄–COOH (Salicylic acid) + CH₃COCl (Acetyl chloride) → o-CH₃COO–C₆H₄–COOH (Aspirin) + HCl

Teacher's Note:
1. In aspirin the phenolic OH becomes an ester group, –OCOCH₃ (acetoxy), while the COOH group is unchanged.

Question 28. Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used.
(i) CH₃CH₂CH₂CH₂OH
(ii) 2-butanol
(iii) 2-methyl-1-propanol [CBSE Delhi 2012]
Answer:
(i) CH₃CH₂CH₂COOH, Butanoic acid
(ii) CH₃–CO–CH₂–CH₃, Butan-2-one
(iii) CH₃–CH(CH₃)–COOH, 2-Methylpropanoic acid

Teacher's Note:
1. Primary alcohols go all the way to carboxylic acids with excess oxidant, but a secondary alcohol stops at the ketone because the carbinol carbon has only one H.

Question 29. (i) Give mechanism of preparation of ethoxy ethane from ethanol.
(ii) How is toluene obtained from phenol? [CBSE Comptt. Delhi 2013]
Answer:
(i) 2CH₃CH₂OH \( \xrightarrow{H^+} \) CH₃CH₂–O–CH₂CH₃
CH₃–CH₂–OH + H⁺ → CH₃–CH₂–O⁺H₂
CH₃–CH₂–O(H) + CH₃–CH₂–O⁺H₂ → CH₃–CH₂–O⁺(H)–CH₂CH₃ + H₂O
CH₃–CH₂–O⁺(H)–CH₂CH₃ → CH₃CH₂–O–CH₂CH₃ + H⁺
(ii) C₆H₅OH (Phenol) \( \xrightarrow[-ZnO]{Zn} \) C₆H₆ (Benzene) \( \xrightarrow[\text{Friedel-Crafts reaction}]{CH_3Cl/AlCl_3} \) C₆H₅CH₃ (Toluene)

Teacher's Note:
1. The second step of ether formation is an S_N2 attack of an ethanol molecule on the protonated ethanol, so this method works well only for primary alcohols.

Question 30. (i) Why phenol is more acidic than ethanol?
(ii) Write the mechanism of acid dehydration of ethanol to yield ether:
2CH₃CH₂OH \( \xrightarrow[413\,K]{H^+} \) CH₃CH₂OCH₂CH₃ [CBSE Comptt OD Set-I, III 2017]
Answer:
(i) Due to resonance, phenoxide ion is more stable than phenol whereas there is no resonance in alkoxide ion / explained with the help of resonating structures.
(ii) (a) CH₃–CH₂–OH + H⁺ → CH₃–CH₂–O⁺H₂
(b) CH₃–CH₂–O(H) + CH₃–CH₂–O⁺H₂ → CH₃–CH₂–O⁺(H)–CH₂CH₃ + H₂O
(c) CH₃–CH₂–O⁺(H)–CH₂CH₃ → CH₃CH₂–O–CH₂CH₃ + H⁺

Teacher's Note:
1. Show the electron transfer in each step with curved arrows; in step (b) the arrow starts from a lone pair on the oxygen of the unprotonated ethanol.

Question 31. (i) Write Reimer-Tiemann reaction.
(ii) Write the mechanism of acid dehydration of ethanol to yield ethene:
CH₃CH₂OH \( \xrightarrow[443\,K]{H^+} \) CH₂=CH₂ + H₂O [CBSE Comptt. OD Set-2 2017]
Answer:
(i) C₆H₅OH (Phenol) \( \xrightarrow{CHCl_3 + aq.\,NaOH} \) [o-NaO–C₆H₄–CHCl₂] (intermediate) \( \xrightarrow{NaOH} \) o-NaO–C₆H₄–CHO \( \xrightarrow{H_3O^+} \) o-HO–C₆H₄–CHO (Salicylaldehyde)
(ii)
Step 1: Formation of protonated alcohol.
CH₃–CH₂–O–H + H⁺ \( \xrightarrow{\text{Fast}} \) CH₃–CH₂–O⁺H₂ (Protonated ethanol / 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 \( \xrightarrow{\text{Slow}} \) CH₃–CH₂⁺ + H₂O
Step 3: Formation of ethene by elimination of a proton.
H–CH₂–CH₂⁺ ⇄ CH₂=CH₂ (Ethene) + H⁺

Teacher's Note:
1. In Reimer-Tiemann reaction the –CHO group enters the ortho position of phenol; the reactive species is dichlorocarbene formed from CHCl₃ and NaOH.

Question 32. (i) Complete the following reaction and suggest a suitable mechanism for the reaction:
CH₃CH₂OH \( \xrightarrow{H^+,\ 443\,K} \)
(ii) Why ortho-Nitrophenol is steam volatile while para-Nitrophenol is less volatile? [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) C₂H₅OH \( \xrightarrow[443\,K]{H_2SO_4} \) CH₂=CH₂ + H₂O
Step 1: Formation of protonated alcohol.
CH₃–CH₂–O–H + H⁺ \( \xrightarrow{\text{Fast}} \) CH₃–CH₂–O⁺H₂ (Protonated ethanol / Ethyl oxonium ion)
Step 2: Formation of carbocation.
CH₃–CH₂–O⁺H₂ \( \xrightarrow{\text{Slow}} \) CH₃–CH₂⁺ + H₂O
Step 3: Formation of ethene by elimination of a proton.
H–CH₂–CH₂⁺ ⇄ CH₂=CH₂ (Ethene) + H⁺
(ii) o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding.

Teacher's Note:
1. Steam distillation is used in the lab to separate the o- and p-nitrophenols obtained by nitration of phenol.

Question 33. Explain the following behaviours:
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol.
(iii) Cumene is a better starting material for the preparation of phenol. [CBSE SQP 2017]
Answer:
(i) Because of H-bond formation between alcohol and water molecule.
(ii) Nitro being the electron withdrawing group stabilises the phenoxide ion.
(iii) Side product formed in this reaction is acetone which is another important organic compound.
Detailed Answer:
(i) Alcohols are able to form H-bonds with water due to presence of (–OH) group. Hence, they are soluble in water.
···H^δ+–O^δ−(R)···H^δ+–O^δ−(H)···H^δ+–O^δ−(R)···H^δ+–O^δ−(H)···
Whereas hydrocarbons are not able to form H-bonds with water hence are insoluble in water.
(ii) Nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O–H bond resulting in easier loss of proton. The o-nitrophenoxide ion formed after the loss of protons is stabilised by resonance. Hence, ortho-nitrophenol is a stronger acid. Whereas, methoxy is an electron-releasing group which increases the electron density in the O–H bond, releasing proton difficult. Thus ortho-nitrophenol is more acidic than ortho-methoxyphenol.
(iii) As oxidation of cumene results in the formation of phenol and acetone as by-product which is a commercial product used as chemical.

Teacher's Note:
1. Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, which on treatment with dilute acid gives phenol and acetone.

Question 34. Give reasons for the following:
(a) Protonation of Phenols is difficult whereas ethanol easily undergoes protonation.
(b) Boiling point of ethanol is higher than that of dimethyl ether.
(c) Anisole on reaction with HI gives phenol and CH₃–I as main products and not iodobenzene and CH₃OH. [CBSE OD Set-2 2016]
Answer:
(a) In phenol, the lone pair of electrons on oxygen involves in delocalization which results in their non-availability for the protonation. Whereas in ethanol, the electrons on oxygen atom are not delocalised which results in their availability for protonation.
(b) The variation can be established by intermolecular hydrogen bonding, where alcoholic hydrogen is bound to a strongly electronegative oxygen atom to give rise to dipole of the form, Et–O^δ−–H^δ+ which arises the boiling point. Whereas, C–O bonds in dimethyl ether are polar. Carbon bonded with hydrogen is moderately electronegative compared to oxygen. In the absence of intermolecular bonding, ether becomes more volatile.
(c) Attack of I⁻ ion on anisole is through S_N2 mechanism, which attacks the lesser hindered CH₃ group forming CH₃I. Also, C–O bond in ether is difficult to break due to resonance.

Teacher's Note:
1. In (c), the O–C₆H₅ bond has partial double bond character and the ring carbon is sp², so I⁻ cannot displace phenol; it attacks the methyl carbon instead.

Question 35. Explain the following with an example in each:
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis [CBSE Comptt. Delhi 2016]
Answer:
(i) Kolbe’s reaction: Sodium phenoxide is heated with CO₂ at 400 K under a pressure of 4-7 atm which on acidification results in salicylic acid.
C₆H₅OH \( \xrightarrow{NaOH} \) C₆H₅ONa (Sodium phenoxide) \( \xrightarrow[400\,K,\ 4\text{-}7\,atm]{+CO_2} \) o-HO–C₆H₄–COONa (Sodium salicylate) \( \xrightarrow{H^+} \) o-HO–C₆H₄–COOH (Salicylic acid)
(ii) Reimer-Tiemann reaction:
C₆H₅OH \( \xrightarrow[340\,K]{CHCl_3 + aq.\,NaOH} \) [o-NaO–C₆H₄–CHCl₂] \( \xrightarrow{NaOH} \) o-NaO–C₆H₄–CHO \( \xrightarrow{H_3O^+} \) o-HO–C₆H₄–CHO (Salicylaldehyde)
(iii) Williamson ether synthesis: An alkyl halide reacts with sodium alkoxide or sodium phenoxide to form ether.
R–X (Alkyl halide) + R′–O⁻Na⁺ (Sodium alkoxide) → R–O–R′ (Ether) + NaX

Teacher's Note:
1. Some students write Kolbe’s electrolytic reaction here; for phenols, Kolbe’s reaction means carboxylation of sodium phenoxide with CO₂.

Question 36. State the products of the following reactions:
(i) CH₃CH₂CH₂OCH₃ + HBr →
(ii) C₆H₅–OC₂H₅ + HBr →
(iii) (CH₃)₃C–O–C₂H₅ \( \xrightarrow{HI} \) [CBSE Foreign 2012]
Answer:
(i) CH₃CH₂CH₂OCH₃ + HBr → CH₃CH₂CH₂OH + CH₃Br
(ii) C₆H₅–OC₂H₅ + HBr → C₆H₅OH + C₂H₅Br
(iii) (CH₃)₃C–O–C₂H₅ + HI → (CH₃)₃C–I + C₂H₅OH

Teacher's Note:
1. Mention all the products formed; decide which C–O bond breaks by the mechanism – S_N2 at the smaller alkyl group, S_N1 at a tertiary group, and never at the aryl carbon.

Question 37. Write the structures of the main products in the following reactions:
(i)(ii) C₆H₅–CH=CH₂ + H₂O \( \xrightarrow{H^+} \)
(iii) C₆H₅–OC₂H₅ + HI → [CBSE OD 2018]
Answer:
(i) NaBH₄ reduces only the ring C=O group to CH–OH; the ester group is not affected:
(ii) C₆H₅CH(OH)CH₃
(iii) C₂H₅I + C₆H₅OH

Teacher's Note:
1. NaBH₄ is a mild reducing agent: it reduces aldehydes and ketones but leaves esters (and C=C bonds) untouched, unlike LiAlH₄.

Question 38. Name the reagents which are used in the following conversions:
(i) 1° alcohol to an aldehyde
(ii) Butan-2-one to butan-2-ol
(iii) Phenol to 2,4,6-tribromophenol
Answer:
(i) PCC (pyridinium chlorochromate), a complex of chromium trioxide with pyridine and HCl.
(ii) NaBH₄, sodium borohydride.
(iii) Br₂ (water), i.e. bromine water.

Teacher's Note:
1. PCC stops the oxidation at the aldehyde stage; acidified K₂Cr₂O₇ or KMnO₄ would go on to the carboxylic acid.

Long Answer Questions (5 Marks)

Question 39. (i) What happens when
(a) phenol reacts with bromine water?
(b) ethanol reacts with CH₃COCl/pyridine?
(c) anisole reacts with HI?
Write the chemical equations involved in the above reactions.
(ii) Distinguish between:
(a) Ethanol and phenol
(b) Propan-2-ol and 2-methylpropan-2-ol [CBSE (F) 2016]
Answer:
(i) (a) C₆H₅OH (Phenol) + 3Br₂(aq) → 2,4,6-Tribromophenol + 3HBr
(b) CH₃–CH₂–OH (Ethanol) + CH₃–CO–Cl (Ethanoyl chloride) →^Pyridine CH₃–CO–O–CH₂–CH₃ (Ethyl ethanoate) + HCl
(c) C₆H₅OCH₃ (Anisole) + HI → C₆H₅OH (Phenol) + CH₃–I (Methyl iodide)
(ii) (a) Phenol gives violet colouration with FeCl₃ solution but ethanol does not.
6C₆H₅OH (Phenol) + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ (Violet colouration complex) + 3H⁺ + 3HCl
(b) Propan-2-ol when warmed with I₂ in NaOH gives yellow precipitate of iodoform while 2-methylpropan-2-ol does not respond to this test.
CH₃–CH(OH)–CH₃ (Propan-2-ol) + 4I₂ + 6NaOH →^Δ CHI₃↓ (Iodoform, yellow ppt.) + CH₃–COO⁻Na⁺ + 5NaI + 5H₂O

Teacher's Note:
1. In anisole + HI the products are always phenol and methyl iodide: the O–aryl bond is too strong to break and iodide attacks the methyl carbon.

Question 40. (i) Write equations of the following reactions:
(a) Bromine in CS₂ with phenol
(b) Treating phenol with chloroform in the presence of aq. NaOH
(c) Anisole reacts with HI
(ii) Distinguish between
(a) Ethanol and Diethyl ether
(b) Propanol and t-butyl alcohol [CBSE South 2016]
Answer:
(i) (a) C₆H₅OH (Phenol) →^{Br₂ in CS₂, 273 K} p-Bromophenol (Major) + o-Bromophenol (Minor)
(b) C₆H₅OH (Phenol) →^{CHCl₃ + aq. NaOH} [o-Cl₂HC–C₆H₄–O⁻Na⁺] →^+NaOH o-OHC–C₆H₄–O⁻Na⁺ →^H+ o-HO–C₆H₄–CHO (Salicylaldehyde)
(c) C₆H₅OCH₃ (Anisole) + HI → C₆H₅OH (Phenol) + CH₃I (Methyl iodide)
(ii) (a) Ethanol when warmed with I₂ in NaOH gives yellow precipitate of iodoform while diethyl ether does not.
CH₃–CH₂–OH + 4I₂ + 6NaOH →^Δ CHI₃ (Iodoform, yellow ppt.) + HCOONa + 5NaI + 5H₂O
(b) tert-butyl alcohol on treatment with the Lucas reagent immediately gives turbidity while propanol does not produce turbidity at room temperature.
(CH₃)₃C–OH (tert-butyl alcohol) →^{Anhyd. ZnCl₂ + HCl (conc.), Lucas reagent} (CH₃)₃C–Cl (tert-butyl chloride, turbidity appears immediately)

Teacher's Note:
1. Ethanol and diethyl ether can also be told apart with sodium metal: only ethanol gives H₂ gas.