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CBSE Class 12 Chapter Test: ALCOHOLS, PHENOLS & ETHERS
General Instructions:
- This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 6 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
Directions for Q7 & Q8: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): Phenoxide ion is resonance stabilized, whereas the ethoxide ion is not.
Reason (R): There is steric repulsion between the two bulky alkyl groups attached to the oxygen atom.
(i) Phenol and Ethanol
(ii) Propan-2-ol and 2-Methylpropan-2-ol
(i) Kolbe's Reaction
(ii) Williamson Synthesis
(iii) Friedel-Crafts alkylation of anisole
(i) The C-O bond length in phenol is shorter than that in methanol.
(ii) Ortho-nitrophenol is steam volatile, whereas para-nitrophenol is not.
(iii) Phenol is a weaker acid than p-nitrophenol.
(i) Cumene
(ii) Chlorobenzene (Dow's Process)
Ethers are the least reactive of the functional groups. The cleavage of the C-O bond in ethers takes place under drastic conditions with an excess of hydrogen halides (HX). The reaction of dialkyl ethers yields two molecules of alkyl halides. However, in the case of mixed (unsymmetrical) ethers, the alkyl halide formed depends on the nature of the alkyl groups. If one group is primary or secondary, the reaction follows an SN2 mechanism, and the halide attacks the smaller, less hindered alkyl group. But if one of the alkyl groups is tertiary, the reaction follows an SN1 mechanism, and the halide attacks the tertiary carbon due to the formation of a highly stable carbocation.
(a) What happens when diethyl ether reacts with an excess of HI at high temperatures? Write the products. [1 Mark]
(b) In alkyl aryl ethers (like anisole), the cleavage occurs at the alkyl-oxygen bond and not the aryl-oxygen bond. Why? [1 Mark]
(c) Predict the major products formed when tert-butyl methyl ether reacts with HI. Give the mechanism-based reason for your answer. [2 Marks]
(i) Propene to Propan-2-ol
(ii) Aniline to Phenol
(iii) Phenol to Picric acid
(iv) Phenol to p-Benzoquinone
(v) Benzyl chloride to Benzyl alcohol
(b) Arrange the following sets of compounds in order of their increasing boiling points:
Pentan-1-ol, Butan-1-ol, Butan-2-ol, Ethanol, Propan-1-ol, Methanol. [2 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (c) Butan-1-ol
Reasoning: Boiling points increase with an increase in molecular mass (number of carbon atoms). Among the 4-carbon isomers, Butan-1-ol (straight chain) has a larger surface area than Butan-2-ol (branched), leading to stronger van der Waals forces and a higher boiling point.
2. Answer: (b) Salicylaldehyde
Reasoning: The reaction described is the Reimer-Tiemann reaction. Phenol reacting with chloroform and aqueous NaOH introduces a formyl group (-CHO) at the ortho position, yielding Salicylaldehyde (2-Hydroxybenzaldehyde).
3. Answer: (b) 3° > 2° > 1°
Reasoning: The Lucas test relies on the SN1 mechanism, where the reactivity depends on the stability of the intermediate carbocation. Since tertiary (3°) carbocations are the most stable, 3° alcohols react the fastest (immediate turbidity).
4. Answer: (d) 2,4,6-Trinitrophenol
Reasoning: Acid strength increases with the presence of Electron Withdrawing Groups (EWG). The nitro group (-NO2) is strongly withdrawing (-I, -R effects). 2,4,6-Trinitrophenol (Picric acid) has three such groups, exceptionally stabilizing the phenoxide ion and making it the most acidic.
5. Answer: (b) Phenol and methyl iodide
Reasoning: In anisole (C6H5-O-CH3), the O-Phenyl bond has partial double bond character due to resonance, making it extremely difficult to break. Therefore, the cleavage occurs exclusively at the O-CH3 bond, yielding Phenol and Methyl iodide.
6. Answer: (b) ethyl chloride with sodium t-butoxide
Reasoning: In the Williamson ether synthesis, the alkyl halide must be primary (1°) to favor SN2 substitution. If a tertiary alkyl halide is used, the strong alkoxide base will cause elimination, yielding an alkene instead of an ether.
7. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: The enhanced acidity of phenol compared to aliphatic alcohols (like ethanol) is entirely due to the resonance stabilization of the resulting phenoxide ion, which disperses the negative charge over the benzene ring.
8. Answer: (d) A is false but R is true.
Reasoning: The assertion is false. The C-O-C bond angle in ethers (e.g., ~111.7° in methoxymethane) is slightly greater than the regular tetrahedral angle (109.5°). The reason given is true and precisely explains why the angle opens up: steric repulsion between the bulky alkyl groups.
9. Solution:
(i) Phenol and Ethanol: Neutral Ferric Chloride (FeCl3) test. Phenol reacts with neutral FeCl3 to give a characteristic purple/violet colour. Ethanol does not give this test.
(ii) Propan-2-ol and 2-Methylpropan-2-ol: Lucas Test (conc. HCl + anhydrous ZnCl2). 2-Methylpropan-2-ol is a 3° alcohol and produces turbidity immediately. Propan-2-ol is a 2° alcohol and produces turbidity after about 5 minutes.
10. Solution:
Hydroboration-oxidation involves the addition of borane (B2H6) to an alkene, forming a trialkylborane, followed by oxidation with H2O2 in an alkaline medium to yield an alcohol. The overall addition of water is Anti-Markovnikov.
Equation:
CH3-CH=CH2 + (BH3)2 → (CH3-CH2-CH2)3B
(CH3-CH2-CH2)3B + 3H2O2 →(OH-) 3 CH3-CH2-CH2-OH (Propan-1-ol) + B(OH)3
11. Solution:
(i) Kolbe's Reaction: Phenol →(1. NaOH) Sodium phenoxide →(2. CO2, 400K, then H+) Salicylic acid (2-Hydroxybenzoic acid).
(ii) Williamson Synthesis: R-X + R'-O-Na+ → R-O-R' + NaX. (e.g., C2H5Cl + CH3ONa → C2H5-O-CH3 + NaCl).
(iii) Friedel-Crafts alkylation of anisole: C6H5OCH3 + CH3Cl →(Anhydrous AlCl3) 2-Methoxytoluene (minor) + 4-Methoxytoluene (major) + HCl.
12. Solution:
(i) In phenol, the lone pair of electrons on the oxygen atom is involved in resonance with the benzene ring. This imparts partial double bond character to the C-O bond, making it shorter and stronger than the pure single C-O bond in methanol.
(ii) Ortho-nitrophenol forms intramolecular hydrogen bonds (within the same molecule), preventing association with other molecules, making it volatile. Para-nitrophenol forms intermolecular hydrogen bonds (between different molecules), causing association and lowering volatility.
(iii) The -NO2 group is an electron-withdrawing group (-I and -R effect). It withdraws electron density and stabilizes the resulting phenoxide ion, making p-nitrophenol a much stronger acid than phenol.
13. Solution:
(i) From Cumene:
C6H5CH(CH3)2 + O2 (air) → Cumene hydroperoxide [C6H5-C(CH3)2OOH]
Cumene hydroperoxide →(H+/H2O) Phenol (C6H5OH) + Acetone (CH3COCH3).
(ii) From Chlorobenzene (Dow's Process):
C6H5Cl + NaOH →(623 K, 300 atm) C6H5ONa (Sodium phenoxide)
C6H5ONa + HCl → C6H5OH (Phenol) + NaCl.
14. Solution (Case-Based):
(a) Reaction with excess HI yields two moles of alkyl halide (ethyl iodide) and water.
CH3CH2-O-CH2CH3 + 2HI →(Δ) 2 CH3CH2I + H2O.
(b) The aryl-oxygen bond acquires partial double bond character due to the resonance of the oxygen lone pair with the benzene ring. This makes the aryl-oxygen bond much stronger and more difficult to break than the single alkyl-oxygen bond.
(c) Products: tert-butyl iodide [(CH3)3C-I] and methanol [CH3OH].
Reason: Because one of the alkyl groups is tertiary, the cleavage proceeds via an SN1 mechanism. The ether is first protonated. Then, the C-O bond breaks to form a highly stable tertiary carbocation [(CH3)3C+], which is then attacked by the iodide ion (I-).
15. Solution:
(i) Propene to Propan-2-ol: Acid-catalyzed hydration (Markovnikov addition).
CH3-CH=CH2 + H2O →(H+) CH3-CH(OH)-CH3.
(ii) Aniline to Phenol: Diazotization followed by warming with water.
C6H5NH2 + NaNO2 + HCl (0-5°C) → C6H5N2+Cl-
C6H5N2+Cl- + H2O →(Warm) C6H5OH + N2 + HCl.
(iii) Phenol to Picric acid: Nitration with concentrated acids.
Phenol + Conc. HNO3 (in presence of Conc. H2SO4) → 2,4,6-Trinitrophenol (Picric acid).
(iv) Phenol to p-Benzoquinone: Oxidation.
Phenol + Na2Cr2O7 / H2SO4 → p-Benzoquinone (O=C6H4=O).
(v) Benzyl chloride to Benzyl alcohol: Nucleophilic substitution.
C6H5CH2Cl + aq. NaOH (or aq. KOH) →(Δ) C6H5CH2OH + NaCl.
16. Solution:
(a) Mechanism of Hydration of Ethene:
Step 1: Protonation of alkene to form a carbocation by electrophilic attack of hydronium ion (H3O+).
H2O + H+ → H3O+
CH2=CH2 + H3O+ ⇔ CH3-CH2+ + H2O
Step 2: Nucleophilic attack of water on the carbocation.
CH3-CH2+ + H2O ⇔ CH3-CH2-O+H2
Step 3: Deprotonation to form an alcohol.
CH3-CH2-O+H2 + H2O ⇔ CH3-CH2-OH (Ethanol) + H3O+.
(b) Increasing order of boiling points:
Boiling points increase with an increase in carbon chain length (molecular mass) and decrease with branching.
Methanol (1C) < Ethanol (2C) < Propan-1-ol (3C) < Butan-2-ol (4C, branched) < Butan-1-ol (4C, straight) < Pentan-1-ol (5C).
Final Order: Methanol < Ethanol < Propan-1-ol < Butan-2-ol < Butan-1-ol < Pentan-1-ol.
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