CBSE Class 12 Chemistry Alcohols Phenols and Ethers MCQs Set 11

Answer: (a) :CCl₂

Explanation:
1. This is the Reimer-Tiemann reaction. CHCl₃ + OH⁻ → ⁻CCl₃ + H₂O; ⁻CCl₃ → :CCl₂ + Cl⁻.
2. Dichlorocarbene (:CCl₂) is the electron-deficient electrophile that attacks the phenoxide ring.

Teacher's Note:
1. :CCl₂ is neutral but has only six valence electrons on carbon, which is why it behaves as an electrophile.

Answer: (c) Salicylic acid

Explanation:
1. This is Kolbe’s reaction: sodium phenoxide reacts with CO₂ (electrophile) at the ortho position.
2. Acidification gives 2-hydroxybenzoic acid (salicylic acid).

Teacher's Note:
1. Do not mix up the two: Kolbe’s reaction (CO₂) gives salicylic acid; Reimer-Tiemann (CHCl₃) gives salicylaldehyde.

Answer: (b) n-propyl alcohol

Explanation:
1. A = CH₃CH₂I, B = CH₃CH₂MgI; addition to HCHO gives C = CH₃CH₂CH₂OMgI.
2. Hydrolysis gives D = CH₃CH₂CH₂OH (n-propyl alcohol).

Teacher's Note:
1. A Grignard reagent with HCHO always gives a 1° alcohol with one carbon more than the alkyl group.

Answer: (b)

Explanation:
1. Kolbe’s reaction gives sodium salicylate (B), which on acidification gives salicylic acid.
2. Acetylation of the –OH group by acetic anhydride gives acetylsalicylic acid (aspirin), with –OCOCH₃ and –COOH ortho to each other.

Teacher's Note:
1. Acetic anhydride acetylates the phenolic –OH (O-acylation), not the ring, so the –COOH stays ortho to the new –OCOCH₃ group.

Answer: (d) Acetone

Explanation:
1. A is cumene (isopropylbenzene) and B is cumene hydroperoxide.
2. Acid treatment of cumene hydroperoxide gives phenol and acetone (C).

Teacher's Note:
1. The cumene process is an important industrial method because acetone is obtained as a useful by-product along with phenol.

Answer: (d) propan-1-ol

Explanation:
1. The iodoform test is given by compounds having CH₃CO– or CH₃CH(OH)– group.
2. Propan-1-ol (CH₃CH₂CH₂OH) has neither group, so it does not give iodoform.

Teacher's Note:
1. Look for a methyl group on the carbon that carries C=O or CH–OH; a –CH₂OH end does not count.

Answer: (a) C₆H₅CH₂–I

Explanation:
1. The O–C₆H₅ bond has partial double bond character, so it is not cleaved.
2. The O–CH₂ bond breaks, giving benzyl iodide and phenol.

Teacher's Note:
1. Aryl–O bonds are never broken by HI; the halide always ends up on the alkyl (here benzyl) carbon.

Answer: (d) C₆H₅–OH + CH₃Br

Explanation:
1. Anisole is protonated and Br⁻ attacks the methyl carbon (S_N2), because the O–C₆H₅ bond has partial double bond character.
2. Hence the products are phenol and methyl bromide.

Teacher's Note:
1. Alkyl aryl ethers always give phenol and an alkyl halide with HX – never an aryl halide.

Answer: (d) Butan-2-ol

Explanation:
1. Butan-2-ol loses water to give a mixture of but-1-ene and but-2-ene; by Saytzeff rule the more substituted but-2-ene is the major product.
2. 2-Methylpropan-2-ol and 2-methylpropan-1-ol give 2-methylpropene; propan-2-ol gives propene.

Teacher's Note:
1. Saytzeff rule: in elimination, the H is removed from the carbon with fewer hydrogens, giving the more substituted alkene.

Answer: (c) Propan-2-ol

Explanation:
1. Propan-2-ol, CH₃CH(OH)CH₃, contains the CH₃CH(OH)– group and gives a yellow precipitate of CHI₃.
2. Butan-1-ol, propan-1-ol and methanol do not have this group.

Teacher's Note:
1. Ethanol is the only primary alcohol that gives the iodoform test; methanol does not.

Answer: (a) 4-Methoxyphenol

Explanation:
1. –OCH₃ at the para position releases electrons (+R effect) and destabilises the phenoxide ion, so 4-methoxyphenol is a weaker acid than phenol.
2. –NO₂ and –Cl are electron withdrawing and make phenol more acidic.

Teacher's Note:
1. Electron-releasing groups (–CH₃, –OCH₃) decrease the acidity of phenol; electron-withdrawing groups increase it.

Answer: (a) C₆H₅CH₂–I and C₆H₅–OH

Explanation:
1. The aryl C–O bond has partial double bond character due to resonance and is not cleaved.
2. The benzyl–O bond breaks; the products are benzyl iodide and phenol.

Teacher's Note:
1. Phenol is always one of the products when an alkyl aryl ether is cleaved by HI.

Answer: (b) t-butanol is more soluble in water than n-butanol

Explanation:
1. t-Butanol is more compact (branched), so its hydrophobic part is smaller and it dissolves better in water than n-butanol.
2. The boiling point of t-butanol is lower than that of n-butanol, because branching reduces surface area.

Teacher's Note:
1. Branching has opposite effects: it lowers the boiling point but raises the water solubility of isomeric alcohols.

Answer: (d) A–R, B–S, C–P, D–Q

Explanation:
1. Ethanol gives the iodoform test, methanol does not (A–R). Phenol gives a violet colour with FeCl₃, cyclohexanol does not (B–S).
2. tert-Butyl alcohol gives instant turbidity with Lucas reagent (C–P). Methanol liberates H₂ with Na, diethyl ether does not (D–Q).

Teacher's Note:
1. For each pair, pick a test that only one of the two compounds gives – a test both pass (or both fail) is useless.

Answer: (a) A–S, B–R, C–Q, D–P

Explanation:
1. Williamson synthesis: R–X + R′ONa → R–O–R′ (A–S). A 2° alcohol is dehydrogenated to a ketone by Cu at 573 K (B–R).
2. Reimer-Tiemann reaction gives salicylaldehyde (C–Q); Kolbe’s reaction gives salicylic acid (D–P).

Teacher's Note:
1. Cu at 573 K dehydrogenates 1° alcohols to aldehydes and 2° alcohols to ketones, but dehydrates 3° alcohols to alkenes.

Answer: (a) (i) and (iv)

Explanation:
1. (i) and (iv) are 2° alcohols.
2. (ii) is a 3° alcohol and (iii) is phenol (OH on an sp² ring carbon), not an alcohol.

Teacher's Note:
1. Classify an alcohol by counting the carbons attached to the carbon carrying –OH: one = 1°, two = 2°, three = 3°.

Answer: (c) C₆H₅–CH(CH₃)–OH

Explanation:
1. In benzylic alcohols, the –OH group is attached to an sp³ carbon which is directly bonded to a benzene ring, as in C₆H₅–CH(CH₃)–OH.
2. (a) is allylic, (b) is a phenol (p-cresol) and (d) has no benzene ring.

Teacher's Note:
1. Benzylic means the OH carbon is next to the ring; if OH is on the ring itself, the compound is a phenol.

Answer: (c) (CH₃)₂C=CH₂

Explanation:
1. (CH₃)₂C=CH₂ (Isobutylene) \( \xrightarrow[(ii)\ H_3O^+]{(i)\ \text{conc. } H_2SO_4} \) (CH₃)₃C–OH (tert-Butyl alcohol)

Teacher's Note:
1. Acid hydration follows Markovnikov's rule: the OH goes to the carbon that gives the more stable carbocation.

Answer: (c) 2-phenyl-2-propanol

Explanation:
1. C₆H₅(CH₃)C=CH₂ (2-Phenylpropene) \( \xrightarrow[\text{Mark. addn.}]{H^+/H_2O} \) C₆H₅C(CH₃)₂–OH (2-Phenyl-2-propanol)

Teacher's Note:
1. The intermediate here is a 3° benzylic carbocation, the most stable one possible, so OH ends up on C-2.

Answer: (a) CH₃–CH(OH)–CH₃

Explanation:
1. CH₃–CH=CH₂ \( \xrightarrow[(\text{Mark. addition})]{HBr} \) CH₃–CH(Br)–CH₃ \( \xrightarrow[(\text{Hydrolysis})]{\text{Aq. KOH}} \) CH₃–CH(OH)–CH₃

Teacher's Note:
1. Without peroxide, HBr adds by Markovnikov's rule; aqueous KOH then substitutes Br by OH (alcoholic KOH would eliminate instead).

Answer: (b) CH₃CHOHCH₃

Explanation:
1. Since alkenes give Markovnikov's addition products, therefore, 1° alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, Z must be 2-propanol, i.e.,
2. CH₃CHOHCH₃ (Z) \( \xrightarrow{PCl_5} \) CH₃CHClCH₃ (X) \( \xrightarrow[-HCl]{\text{Alc. KOH}} \) CH₃CH=CH₂ (Y) \( \xrightarrow[(ii)\ H_3O^+]{(i)\ \text{conc. } H_2SO_4} \) CH₃CHOHCH₃ (Z)

Teacher's Note:
1. When a sequence returns to its starting compound, check which options survive the last step – here acid hydration can only give a 2° or 3° alcohol from these alkenes.

Answer: (d) 2-Methylpropanal and isopropyl magnesium iodide

Explanation:
1. (CH₃)₂CH–CH=O (2-Methylpropanal) \( \xrightarrow[(\text{Grignard reaction})]{(i)\ (CH_3)_2CHMgI/(ii)\ H_3O^+} \) (CH₃)₂CH–CH(OH)–CH(CH₃)₂ (2,4-Dimethylpentan-3-ol)

Teacher's Note:
1. To plan a Grignard synthesis, break the bond between the carbinol carbon and one of its alkyl groups: that group comes from RMgX, the rest from the carbonyl compound.

Answer: (a) CH₃CH₂CH₂OH

Explanation:
1. CH₃CH=CH₂ (Propene) \( \xrightarrow[(\text{Hydroboration})]{B_2H_6/THF} \) (CH₃CH₂CH₂)₃B (Tri-n-propylborane) \( \xrightarrow[(\text{Oxidation})]{H_2O_2/OH^-} \) 3CH₃CH₂CH₂OH (1-Propanol)

Teacher's Note:
1. Hydroboration-oxidation gives the anti-Markovnikov alcohol: OH goes to the less substituted carbon.

Answer: (a) C₆H₅–CH₂CHOHCH₃

Explanation:
1. Mercuration-demercuration gives alcohols (without any rearrangement) according to Markovnikov's addition of H₂O to alkenes. Thus,
2. C₆H₅CH₂–CH=CH₂ (3-Phenylpropene) \( \xrightarrow[(ii)\ NaBH_4/OH^-]{(i)\ (CH_3COO)_2Hg/THF{-}H_2O} \) C₆H₅CH₂CHOHCH₃ (1-Phenyl-2-propanol)

Teacher's Note:
1. Oxymercuration-demercuration is Markovnikov hydration without carbocation rearrangement – a useful contrast with hydroboration-oxidation.

Answer: (c) C₆H₅COOCH₂C₆H₅

Explanation:
1. Esters in which acyl and alkyl/aralkyl groups contain the same number of carbon atoms would give two molecules of the same alcohol, i.e.,
2. C₆H₅COOCH₂C₆H₅ (Benzyl benzoate) \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4/\text{ether}} \) 2 C₆H₅CH₂OH (Benzyl alcohol)

Teacher's Note:
1. LiAlH₄ reduces RCOOR′ to RCH₂OH + R′OH; the two alcohols are the same only when R–CH₂ is identical to R′.

Answer: (d) ethyl alcohol

Explanation:
1. CH₃–Br (Methyl bromide) \( \xrightarrow[-KBr]{KCN} \) CH₃–CN (Acetonitrile) (A) \( \xrightarrow[(\text{Hydrolysis})]{H_3O^+} \) CH₃COOH (Acetic acid) (B) \( \xrightarrow[(\text{Reduction})]{LiAlH_4/\text{ether}} \) CH₃CH₂OH (Ethyl alcohol) (C)

Teacher's Note:
1. The cyanide step adds one carbon; LiAlH₄ reduces a carboxylic acid all the way to a 1° alcohol.

Answer: (a) CH₃–CH(CH₃)–CH₂CH₂OH

Explanation:
1. CH₃–CH(OH)–CH₃ \( \xrightarrow{PBr_3} \) CH₃–CH(Br)–CH₃ \( \xrightarrow{Mg/\text{ether}} \) CH₃–CH(MgBr)–CH₃
2. \( \xrightarrow{\text{ethylene oxide}} \) (CH₃)₂CH–CH₂CH₂OMgBr \( \xrightarrow[-Mg(OH)Br]{H_2O,\ \Delta} \) (CH₃)₂CH–CH₂CH₂OH

Teacher's Note:
1. A Grignard reagent with ethylene oxide gives a 1° alcohol with two extra carbons (R–CH₂CH₂OH).

Answer: (c) Cyclohexane (C₆H₁₂)

Explanation:
1. C₆H₁₁–Br \( \xrightarrow[\text{Dry ether}]{Mg} \) C₆H₁₁–MgBr (X) \( \xrightarrow{H_2O} \) C₆H₁₂ (cyclohexane)

Teacher's Note:
1. Water (or any compound with acidic H) protonates a Grignard reagent to the alkane – it does not give an alcohol.

Answer: (b) slightly less than 109°–28′

Explanation:
1. C–O–H bond angle in alcohols is slightly less than 109°–28′ due to greater lone pair-lone pair repulsion than lone pair-bond pair repulsions.

Teacher's Note:
1. The C–O–H angle in methanol is about 108.9°; compare C–O–C in dimethyl ether (111.7°), which is larger because of the bulky alkyl groups.

Answer: (a) isopropylbenzene and acetone

Explanation:
1. C₆H₆ + H₃C–CH=CH₂ \( \xrightarrow[\text{F.C. alkylation}]{AlCl_3/\text{ether},\ \Delta} \) C₆H₅CH(CH₃)₂ (Isopropylbenzene) (X) \( \xrightarrow{O_2/130^\circ C} \) C₆H₅C(CH₃)₂–O–OH (Y)
2. \( \xrightarrow[\text{heat}]{H^+/H_2O} \) C₆H₅OH (Phenol) + (CH₃)₂C=O (Acetone) (Z)

Teacher's Note:
1. This is the cumene process – the industrial route to phenol, with acetone as the valuable by-product.

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

Explanation:
1. The C–O–C bond angle in ethers (about 111.7°) is slightly greater than the tetrahedral angle, so the Assertion is incorrect.
2. This increase is due to the repulsive interaction between the two bulky alkyl groups, so the Reason is correct.

Teacher's Note:
1. Contrast with alcohols: the C–O–H angle (108.9°) is slightly less than tetrahedral because of lone pair repulsion.

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

Explanation:
1. Assertion is correct: alcohols boil higher than hydrocarbons of similar mass.
2. Reason is incorrect: the high boiling point is due to intermolecular hydrogen bonding between alcohol molecules, not intramolecular.

Teacher's Note:
1. Read the prefix carefully – "intra" (within one molecule) and "inter" (between molecules) is a favourite trap in A-R questions.

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

Explanation:
1. o-Nitrophenol has intramolecular H-bonding (chelation), so it does not associate and is steam volatile.
2. p-Nitrophenol molecules are associated by intermolecular H-bonding and are not steam volatile. The reason explains the assertion.

Teacher's Note:
1. Intramolecular H-bonding lowers boiling point and water solubility; intermolecular H-bonding raises both.

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

Explanation:
1. The C–O–H bond angle in alcohols (about 108.9°) is slightly less than the tetrahedral angle (109°28′).
2. This is due to greater repulsion between the two lone pairs on the oxygen atom. So both A and R are correct and R explains A.

Teacher's Note:
1. The order of repulsion lp–lp > lp–bp > bp–bp (VSEPR) explains most small deviations from ideal bond angles.

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

Explanation:
1. tert-Butyl alcohol is miscible with water while butan-1-ol is only partly soluble.
2. Branching makes the hydrocarbon part compact, so its non-polar surface area decreases and the H-bonding OH group dominates. So R correctly explains A.

Teacher's Note:
1. Branching lowers boiling point but raises water solubility among isomeric alcohols – do not mix up the two trends.

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

Explanation:
1. Correct assertion. tert-Butoxide is a strong base but a weak nucleophile.
2. Its bulky size hinders attack on carbon, so nucleophilicity depends upon the size of the alkoxide ion. Thus A is wrong but R is correct.

Teacher's Note:
1. Bulky bases like (CH₃)₃CO⁻ favour elimination over substitution – they can pull off an exposed H but cannot reach a crowded carbon.

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

Explanation:
1. Correct assertion. C₂H₅OH is a stronger base than phenol.
2. In phenol the lone pair on oxygen is delocalised into the ring by resonance and is less available for donation. So A is wrong but R is correct.

Teacher's Note:
1. Delocalisation of the O lone pair makes phenol both a weaker base and a weaker nucleophile than ethanol.

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

Explanation:
1. Dehydration of alcohols takes place through carbocations; since the stability of carbocations is 3° > 2° > 1°, the ease of dehydration follows the same order. So A is correct.
2. R is also correct (the more substituted alkene is more stable and is the major product), but it explains which alkene is formed, not the order of ease of dehydration.

Teacher's Note:
1. Ease of dehydration is decided by carbocation stability; Saytzeff's rule only tells you which alkene predominates.

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

Explanation:
1. The reactivity order HI > HBr > HCl is correct.
2. Correct reason. The nucleophilicity of halide ions decreases in the order : I⁻ > Br⁻ > Cl⁻. So A is correct but R is wrong.

Teacher's Note:
1. In protic solvents the larger, more polarisable I⁻ is the best nucleophile, and HI is also the strongest acid.

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

Explanation:
1. Correct explanation. Hyperconjugation at p-position increases the electron density in the O–H bond and hence decreases the acidity of p-cresol as compared to m-cresol since hyperconjugation through m-position cannot increase electron density in the O–H bond.

Teacher's Note:
1. Hyperconjugation, like resonance, operates only from o- and p-positions; this is why m-cresol is the strongest acid among the cresols.

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

Explanation:
1. Correct assertion. Traces of moisture from phenol cannot be easily removed by distilling with sodium metal.
2. Sodium reacts with phenol itself to form sodium phenoxide. Phenol (pK_a 10) is more acidic than water (pK_a 15.7), so R is correct.

Teacher's Note:
1. A drying agent must not react with the substance being dried; sodium reacts with every compound having an acidic O–H.