CBSE Class 12 Chemistry Alcohols Phenols and Ethers MCQs Set 12

Answer: (a) H₂O

Explanation:
1. Due to +I-effect of the alkyl groups, water is a stronger acid than alcohols.

Teacher's Note:
1. Alkyl groups push electron density onto O, destabilising the alkoxide ion; so acidity falls as the alkyl group grows.

Answer: (d) CH₃OH

Explanation:
1. C₆H₆ and C₂H₆ being hydrocarbons are very weak acids. To decide whether CH≡CH or CH₃OH is a stronger acid, consider the reaction.
2. HC≡C⁻Na⁺ + CH₃OH → HC≡CH + CH₃O⁻Na⁺
3. (stronger acid)    (weaker acid)
4. Since a stronger acid displaces a weaker acid from its salts, therefore, CH₃OH is a stronger acid than HC≡CH.

Teacher's Note:
1. An O–H bond is more acidic than a C–H bond because O is more electronegative than C (pK_a about 16 for CH₃OH vs 25 for ethyne).

Answer: (d) CH₃CH₂C(OH)(CH₃)₂

Explanation:
1. Tertiary alcohols are the weakest acids and hence have highest pK_a values, i.e., option (d) is correct.

Teacher's Note:
1. Higher pK_a means weaker acid; acidity of alcohols falls in the order 1° > 2° > 3°.

Answer: (c) H₂O < CH₃OH < OH⁻ < CH₃O⁻

Explanation:
1. H₂O < CH₃OH < OH⁻ < CH₃O⁻, i.e., option (c) is correct.
2. Since CH₃OH is a weaker acid than H₂O, its conjugate base CH₃O⁻ is a stronger base than OH⁻; the neutral molecules are much weaker bases than the anions.

Teacher's Note:
1. Weaker acid ⇒ stronger conjugate base: use the acid order H₂O > CH₃OH to get the base order OH⁻ < CH₃O⁻.

Answer: (a) 1° > 2° > 3°

Explanation:
1. Among isomeric alcohols, b.p. decreases with branching, i.e., 1° > 2° > 3°.

Teacher's Note:
1. Branching makes the molecule more compact, reducing surface area and van der Waals forces – e.g. butan-1-ol 391 K, 2-methylpropan-2-ol 355 K.

Answer: (b) CH₃CH₂CH₂CH₂OH

Explanation:
1. n-Alcohols with larger number of carbon atoms has the highest boiling point, i.e., option (b) is correct.

Teacher's Note:
1. For b.p. questions: first look for H-bonding, then compare molecular mass, then branching.

Answer: (c) NO₂–

Explanation:
1. Electron-withdrawing –NO₂ increases the acid character of phenols.
2. CH₃O– and CH₃– are electron-releasing groups and decrease the acidity.

Teacher's Note:
1. Electron-withdrawing groups stabilise the phenoxide ion and increase acidity; electron-donating groups do the opposite.

Answer: (c) m-Chlorophenol

Explanation:
1. Due to –I-effect of Cl, m-chlorophenol is a stronger acid than phenol.

Teacher's Note:
1. Phenols are more acidic than alcohols (resonance-stabilised phenoxide); an electron-withdrawing Cl increases the acidity further.

Answer: (d) IV > III > I > II

Explanation:
1. –NO₂ is electron-withdrawing and increases the acidity of phenol; at the p-position it acts by both –R and –I effects, at the m-position only by –I effect. So IV > III > I.
2. –CH₃ is electron-releasing (+I, hyperconjugation) and decreases the acidity, so p-cresol is the weakest. Hence IV > III > I > II.

Teacher's Note:
1. The –R effect of NO₂ works only from o- and p-positions, where the negative charge of phenoxide can reach the nitro group.

Answer: (d) p-nitrophenol > o-nitrophenol > m-nitrophenol

Explanation:
1. NO₂ group does not show –R-effect at m-position, therefore, m-nitrophenol is the weakest acid. Due to intramolecular H-bonding (i.e., chelation), o-nitrophenol is less acidic than p-nitrophenol. Thus, option (d) is correct.

Teacher's Note:
1. Remember the pK_a values: p 7.1, o 7.2, m 8.3 – the m-isomer is clearly the weakest of the three.

Answer: (d) Reacting with Mg

Explanation:
1. Mg (activated with a little I₂) converts ethanol into magnesium ethoxide, which reacts with the traces of water: Mg(OC₂H₅)₂ + 2H₂O → Mg(OH)₂ + 2C₂H₅OH. Distillation then gives absolute alcohol.
2. Sodium would destroy ethanol itself, dry HCl reacts with ethanol, and simple distillation cannot remove water because ethanol and water form an azeotrope.

Teacher's Note:
1. Rectified spirit (95.6%) cannot be dried by simple distillation because it is an azeotrope; chemical drying with Mg (or quicklime) is used.

Answer: (d) Na

Explanation:
1. Sodium reacts with both phenol and ethyl alcohol, liberating H₂ gas, so it cannot distinguish them.
2. Phenol (not ethanol) dissolves in NaOH, gives a violet colour with FeCl₃ and a white precipitate of 2,4,6-tribromophenol with bromine water.

Teacher's Note:
1. A distinguishing test needs a visible result for one compound only; reactions shown by all O–H compounds (like with Na) are useless.

Answer: (b) ethylene

Explanation:
1. At 443 K, conc. H₂SO₄ brings about intramolecular dehydration of ethanol: CH₃CH₂OH \( \xrightarrow[443\ K]{\text{conc. } H_2SO_4} \) CH₂=CH₂ + H₂O.
2. At the lower temperature of 413 K, intermolecular dehydration gives diethyl ether.

Teacher's Note:
1. Temperature decides the product: 413 K (excess alcohol) gives ether, 443 K gives alkene.

Answer: (b) protonation of alcohol molecule

Explanation:
1. The initial step during dehydration of alcohols is protonation of the alcohol molecule.
2. Protonation converts –OH (poor leaving group) into –OH₂⁺, which then leaves as H₂O to form the carbocation.

Teacher's Note:
1. Mechanism order: protonation (fast) → loss of water to give carbocation (slow, rate-determining) → loss of H⁺ to give alkene.

Answer: (c) IV > III > II > I

Explanation:
1. The order of reactivity depends upon the stability of the carbocations formed, i.e., FCH₂C⁺HCH₃, FCH₂CH₂C⁺HCH₃, CH₃C⁺HCH₃ and PhC⁺H₂.
2. Since the relative stabilities of these carbocations follow the order : PhC⁺H₂ > CH₃C⁺HCH₃ > FCH₂CH₂C⁺HCH₃ > FCH₂C⁺HCH₃, therefore, the order of reactivity of the alcohols (I, II, III and IV) follows the sequence : IV > III > II > I.

Teacher's Note:
1. The –I effect of F destabilises a carbocation, and the effect weakens with distance – so F nearer the positive carbon (I) is worse than F further away (II).

Answer: (a) conc. H₃PO₄

Explanation:
1. To suppress substitution and to favour elimination, the acid used must have a non-nucleophilic counter ion. The counter ion of HCl or HCl/ZnCl₂ or HBr, i.e., Cl⁻ or Br⁻ ion is a strong nucleophile and hence forms either cyclohexyl chloride or cyclohexyl bromide by substitution reaction.
2. In contrast, the counter ion of H₃PO₄, i.e., PO₄³⁻ is non-nucleophilic and hence does not give substitution product but gives only the elimination product, i.e., cyclohexene.

Teacher's Note:
1. For dehydration choose an acid whose anion is a poor nucleophile (H₃PO₄, H₂SO₄); HX acids give alkyl halides instead.

Answer: (d) CH₃CH(OH)CH₂CH₃

Explanation:
1. Methyl carbinols, i.e., compounds having the CH₃CH(OH)– group linked to carbon or hydrogen, give yellow ppt. of CHI₃, i.e., option (d) is correct.

Teacher's Note:
1. The iodoform test needs CH₃CO– or CH₃CH(OH)– attached to C or H; acid derivatives such as esters, anhydrides and amides do not respond.

Answer: (a) (CH₃)₃COH

Explanation:
1. 3° Alcohols react fastest with Lucas reagent.
2. They form the most stable 3° carbocation, so turbidity appears immediately.

Teacher's Note:
1. Lucas test: 3° immediate turbidity, 2° in about 5 minutes, 1° no turbidity at room temperature.

Answer: (b) CH₃CHBr–C₆H₄–OH

Explanation:
1. Addition of a proton at β-carbon gives a carbocation (I) which is stabilized by resonance because of the electron-donating effect of the OH group. Subsequent addition of Br⁻ ion to α-carbon gives the addition product (II).
2. CH₂=CH–C₆H₄–OH \( \xrightarrow{H^+} \) CH₃–C⁺H–C₆H₄–OH (I) \( \xrightarrow{Br^-} \) CH₃–CHBr–C₆H₄–OH (II)

Teacher's Note:
1. HBr adds only to the side-chain C=C; the phenolic OH is not replaced by Br, since the C–O bond of phenol has partial double bond character.

Answer: (d) PCC

Explanation:
1. PCC oxidises 1° alcohols to aldehydes without any danger of their further oxidation to carboxylic acids.

Teacher's Note:
1. PCC (pyridinium chlorochromate, C₅H₅NH⁺CrO₃Cl⁻) is the standard mild reagent to stop oxidation at the aldehyde stage.

Answer: (a) (CH₃)₃C–OH

Explanation:
1. 3° Alcohols are resistant to oxidation.
2. They have no hydrogen on the carbon carrying the –OH group.

Teacher's Note:
1. Oxidation of an alcohol removes the H from the carbinol carbon; with no such H, 3° alcohols resist mild oxidation.

Answer: (c) CH₃–CH=CH–CHO

Explanation:
1. PCC oxidises only alcohols to aldehydes without affecting the double bond.
2. CH₃–CH=CH–CH₂OH \( \xrightarrow{PCC} \) CH₃–CH=CH–CHO

Teacher's Note:
1. PCC neither cleaves nor reduces a C=C bond – the double bond is carried through unchanged.

Answer: (d) X = 2-hydroxybenzoic acid, Y = 2-hydroxybenzaldehyde

Explanation:
1. With NaOH and CO₂ (Kolbe's reaction), phenol gives salicylic acid (2-hydroxybenzoic acid) = X.
2. With CHCl₃ and aq. NaOH (Reimer–Tiemann reaction), phenol gives salicylaldehyde (2-hydroxybenzaldehyde) = Y.

Teacher's Note:
1. Both reactions put the new group mainly at the ortho position: Kolbe gives –COOH, Reimer–Tiemann gives –CHO.

Answer: (c) o- and p-bromophenol

Explanation:
1. CS₂ cannot polarize Br₂, therefore, only monobromination occurs to give a mixture of o- and p-bromophenol.

Teacher's Note:
1. Solvent decides the product: in CS₂ (non-polar, low temperature) monobromination; in bromine water 2,4,6-tribromophenol.

Answer: (d) CH₃–C(CH₃)=CH₂

Explanation:
1. With strong bases, 3° alkyl halides exclusively undergo elimination to give alkenes.
2. (CH₃)₃C–Br + CH₃O⁻ → CH₂=C(CH₃)₂ + CH₃OH + Br⁻

Teacher's Note:
1. A 3° halide with an alkoxide gives an alkene, not an ether – a frequent trap in Williamson synthesis questions.

Answer: (d) C₆H₅–OH and CH₃Br

Explanation:
1. Protonation of anisole gives methyl phenyl oxonium ion (I). In this ion, O–CH₃ bond is weaker than O–C₆H₅ bond which has some double bond character between the O atom and the sp²-carbon of the phenyl ring due to resonance. As a result, weaker O–CH₃ bond breaks forming CH₃Br and phenol.
2. C₆H₅–O–CH₃ + H⁺ → C₆H₅–O⁺(H)–CH₃ (I) \( \xrightarrow[S_{N}2]{Br^-} \) C₆H₅OH + CH₃Br

Teacher's Note:
1. Alkyl aryl ethers always give phenol + alkyl halide with HX, never aryl halide.

Answer: (b) tert-butyl iodide and methanol

Explanation:
1. (CH₃)₃C–OCH₃ \( \xrightarrow{H^+} \) (CH₃)₃C–O⁺(H)–CH₃ \( \xrightarrow[-CH_3OH]{S_{N}1} \) (CH₃)₃C⁺ \( \xrightarrow{I^-} \) (CH₃)₃C–I

Teacher's Note:
1. With a 3° alkyl group, cleavage goes by S_N1 and the halide goes to the 3° carbon (stable carbocation); with two 1°/2° groups, S_N2 puts it on the smaller group.

Answer: (d) Di-tert-butyl ether.

Explanation:
1. Di-tert-butyl ether cannot be prepared by Williamson's synthesis as shown below :
2. (CH₃)₃C–Br + Na^{+ −}O–C(CH₃)₃ \( \xrightarrow[-NaBr]{} \) CH₃–C(CH₃)=CH₂ + (CH₃)₃C–OH
3. The 3° alkyl halide undergoes elimination instead of substitution.

Teacher's Note:
1. In Williamson synthesis the alkyl halide must be 1° (or methyl); a 3° group can be present only in the alkoxide.

Answer: (d) ethanal

Explanation:
1. CH₂=CH–O–CH₂CH₃ \( \xrightarrow{H^+} \) CH₂=CH–O⁺(H)–CH₂CH₃ \( \xrightarrow{I^-} \) [CH₂=CH–OH] + CH₃CH₂I (Iodoethane)
2. [CH₂=CH–OH] \( \xrightarrow{\text{Tautomerises}} \) CH₃CHO (Ethanal)

Teacher's Note:
1. Iodide attacks the sp³ ethyl carbon, never the vinyl (sp²) carbon; the vinyl alcohol (enol) formed tautomerises to the aldehyde.

Answer: (a) CH₃CH₂O–CH(OOH)–CH₃

Explanation:
1. Hydroperoxide formation occurs at the more reactive α-carbon.
2. CH₃CH₂–O–CH₂CH₃ + O₂ \( \xrightarrow{\text{light}} \) CH₃CH₂–O–CH(OOH)–CH₃

Teacher's Note:
1. Old ether samples contain these explosive peroxides; they are tested with FeSO₄ + KCNS (red colour) before distilling ether.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. p-Nitrophenol (pK_a 7.1) is a slightly stronger acid than o-nitrophenol (pK_a 7.2).
2. In o-nitrophenol the O–H hydrogen is held by intramolecular H-bonding with the NO₂ group, which makes its release more difficult. So R explains A.

Teacher's Note:
1. Intramolecular H-bonding (chelation) in o-nitrophenol also explains its lower b.p. and steam volatility.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Due to H-bonding, alcohols have higher b.p. than isomeric ethers.
2. Both statements are true, but isomerism does not explain the difference in boiling points.

Teacher's Note:
1. Ethers have no H on oxygen, so they cannot H-bond with each other; e.g. ethanol (351 K) vs dimethyl ether (248 K).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Benzenediazonium chloride decomposes to form C₆H₅⁺ which upon nucleophilic attack by water followed by loss of a proton gives phenol.
2. Both statements are true, but polarity of the C–N bond is not the reason.

Teacher's Note:
1. The driving force is the loss of N₂, an excellent leaving group; the reaction is C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion. Cyclopentylmethanol on heating with H₃PO₄ undergoes dehydration with ring expansion to give cyclohexene.
2. The 1° carbocation formed undergoes ring expansion to the more stable, strain-free six-membered 2° carbocation. So A is wrong but R is correct.

Teacher's Note:
1. Watch for a carbocation next to a small ring: a ring-expansion shift (5 → 6) often gives a more stable cation.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct reason. Since phenolic O–H group is more nucleophilic than the carboxyl group, therefore, acetylation occurs at the phenolic –OH group to give aspirin.
2. o-HO–C₆H₄–COOH (Salicylic acid) \( \xrightarrow[CH_3COONa]{(CH_3CO)_2O} \) o-CH₃COO–C₆H₄–COOH (Aspirin) + CH₃COOH

Teacher's Note:
1. Aspirin is acetylsalicylic acid: the acetyl group is on the phenolic oxygen and the –COOH stays free.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion. tert-Butyl alcohol undergoes dehydration to give 2-methylpropene.
2. (CH₃)₃C–OH (tert-Butyl alcohol) \( \xrightarrow[-H_2O]{Cu,\ 573\ K} \) CH₃–C(CH₃)=CH₂ (2-Methylpropene) + H₂O

Teacher's Note:
1. Heated Cu dehydrogenates 1° and 2° alcohols (to aldehydes and ketones) but dehydrates 3° alcohols to alkenes – a classic distinction test.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion. Neopentyl alcohol on treatment with HCl gives 2-chloro-2-methylbutane as shown below :
2. (CH₃)₃C–CH₂OH (Neopentyl alcohol) \( \xrightarrow[-H_2O]{H^+} \) (CH₃)₃C–C⁺H₂ (1° carbocation, less stable) \( \xrightarrow[\text{shift}]{1,2\text{-Methyl}} \) (CH₃)₂C⁺–CH₂CH₃ \( \xrightarrow{Cl^-} \) (CH₃)₂C(Cl)–CH₂CH₃
3. R is correct: neopentyl alcohol is a primary alcohol.

Teacher's Note:
1. Neopentyl systems almost always rearrange via a 1,2-methyl shift because the 1° carbocation is so unstable.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion. tert-Butyl methyl ether is prepared by the action of sodium tert-butoxide with methyl bromide.
2. With sodium methoxide, tert-butyl bromide undergoes elimination to give 2-methylpropene. So A is wrong but R is correct.

Teacher's Note:
1. For an unsymmetrical ether with a 3° group, take the 3° part as the alkoxide and the 1° part as the halide.

Answer: (d) A is false, but R is true.

Explanation:
1. Phenol (pK_a about 10) is a much stronger acid than ethanol (pK_a about 16), so C₂H₅ONa converts phenol into sodium phenoxide: C₆H₅OH + C₂H₅O⁻Na⁺ → C₆H₅O⁻Na⁺ + C₂H₅OH.
2. The phenoxide ion then attacks C₂H₅I to give ethyl phenyl ether (phenetole), C₆H₅–O–C₂H₅, not diethyl ether. So A is wrong.
3. R is correct: C₂H₅O⁻ is a better nucleophile than C₆H₅O⁻, whose negative charge is delocalised over the ring.

Teacher's Note:
1. In an acid-base step, the stronger acid (phenol) gives up its proton to the stronger base (ethoxide), so the nucleophile actually present is phenoxide.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct reason. Low polarity of ether favours S_N2 mechanism giving CH₃I and (CH₃)₃C–OH.
2. So A is correct but R is wrong.

Teacher's Note:
1. Solvent matters: in a polar medium (aq. HI) the 3° ether reacts by S_N1 to give (CH₃)₃C–I; in non-polar ether, S_N2 gives CH₃I.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion. Benzyl methyl ether on treatment with conc. HI gives benzyl iodide and methanol.
2. The protonated ether cleaves by S_N1 to give the resonance-stabilised benzyl carbocation, which combines with I⁻. So A is wrong but R is correct.

Teacher's Note:
1. When one group of an ether can form a stable (3° or benzylic) carbocation, the iodide ends up on that group.