CBSE Class 12 Chemistry Alcohols Phenols and Ethers MCQs Set 10

Answer: (c) B, C

Explanation:
1. In an aromatic alcohol the –OH group is attached to an sp³ carbon of a side chain (–CH₂OH) on the benzene ring.
2. B and C have –CH₂OH groups, so they are aromatic alcohols; A and D have –OH directly on the ring, so they are phenols.

Teacher's Note:
1. OH directly on the ring = phenol; OH on a side-chain carbon = aromatic (benzylic) alcohol.

Answer: (a) 1

Explanation:
1. The alcohols of C₄H₁₀O are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol.
2. Only butan-2-ol, CH₃CH₂CH(OH)CH₃, has a carbon with four different groups, so only one is chiral.

Teacher's Note:
1. Look for a carbon bearing four different groups (H, OH, CH₃, C₂H₅ in butan-2-ol).

Answer: (a) 3-methylphenol

Explanation:
1. Cresols are methylphenols; in the meta isomer the CH₃ group is at position 3 from the –OH group, so m-cresol is 3-methylphenol.

Teacher's Note:
1. Cresol = methylphenol; anisole = methoxybenzene; resorcinol = benzene-1,3-diol. These common names are frequently mixed up.

Answer: (c) 2-methoxypropane

Explanation:
1. The larger alkyl group (isopropyl, three carbons) is the parent chain, propane; the smaller group OCH₃ is the methoxy substituent on C-2.
2. Hence the name is 2-methoxypropane.

Teacher's Note:
1. In ethers the bigger alkyl group gives the parent alkane and the smaller one is named as the alkoxy prefix.

Answer: (d) both (a) and (b)

Explanation:
1. In phenol the lone pair of oxygen is in resonance with the ring, giving the C–O bond partial double bond character.
2. Also, oxygen is attached to an sp² hybridised carbon of the ring. Both effects shorten the C–O bond (136 pm in phenol vs 142 pm in methanol).

Teacher's Note:
1. Remember the two reasons together: resonance (partial double bond) and the sp² carbon.

Answer: (d) (a), (b) and (c)

Explanation:
1. Aldehydes are reduced to primary alcohols by catalytic hydrogenation (H₂/Pd) and by the hydride reagents LiAlH₄ and NaBH₄.
2. Reaction with RMgX followed by hydrolysis would add an R group and give a secondary alcohol, not RCH₂OH.

Teacher's Note:
1. Reduction of RCHO gives a 1° alcohol and of RCOR′ a 2° alcohol; Grignard addition changes the carbon skeleton instead.

Answer: (d) (a), (b) and (c)

Explanation:
1. (a) Chlorobenzene fused with NaOH at high pressure gives sodium phenoxide, which on acidification gives phenol.
2. (b) Aniline is diazotised to benzenediazonium chloride, which on warming with water gives phenol.
3. (c) Benzene with oleum gives benzenesulphonic acid; fusion with NaOH and acidification gives phenol.

Teacher's Note:
1. Chlorobenzene does not give phenol with aqueous NaOH at room temperature and 1 atm – drastic conditions (623 K, 300 atm) are needed.

Answer: (b) (i) > (ii) > (iii)

Explanation:
1. Acid catalysed hydration goes through a carbocation. (i) gives a 3° carbocation, (ii) a 2° and (iii) a 1° carbocation.
2. More stable the carbocation, faster the hydration: (i) > (ii) > (iii).

Teacher's Note:
1. For any reaction with a carbocation intermediate, rank by carbocation stability: 3° > 2° > 1°.

Answer: (b) ^⊖OR

Explanation:
1. The weaker the acid, the stronger its conjugate base. Alcohols (ROH) are weaker acids than water and phenols.
2. Hence alkoxide ion RO⁻ is the strongest base; phenoxide ions are weak bases because their charge is delocalised over the ring.

Teacher's Note:
1. Order of acidity ROH < H₂O < C₆H₅OH < nitrophenol gives the reverse order of basicity of the anions.

Answer: (b) o-nitrophenol

Explanation:
1. The electron-withdrawing –NO₂ group at the ortho position stabilises the phenoxide ion by delocalising the negative charge, so o-nitrophenol is more acidic than phenol.
2. Ethanol is less acidic, and –CH₃ and –OCH₃ (electron-donating at ortho) decrease acidity.

Teacher's Note:
1. Electron-withdrawing groups at o/p increase the acidity of phenol; electron-releasing groups decrease it.

Answer: (a) C₆H₅OH > H₂O > ROH > HC≡CH

Explanation:
1. Phenoxide ion is resonance stabilised, so phenol is the strongest acid. The +I effect of R makes ROH weaker than water.
2. Ethyne is the weakest acid of the four, so C₆H₅OH > H₂O > ROH > HC≡CH.

Teacher's Note:
1. Approximate pK_a values help: phenol 10, water 15.7, ethanol 15.9, ethyne 25.

Answer: (b) (ii) > (iv) > (i) > (iii) > (v)

Explanation:
1. –NO₂ is electron-withdrawing and increases acidity; its effect is greater at the para position (−R and −I) than at the meta position (−I only). So (ii) > (iv) > phenol.
2. –OCH₃ is electron-releasing and decreases acidity, more strongly from the para position (+R). So phenol > (iii) > (v).

Teacher's Note:
1. A nitro group at meta acts only through its −I effect, so m-nitrophenol is less acidic than p-nitrophenol.

Answer: (c) 2-methylbut-2-ene

Explanation:
1. Neopentyl alcohol, (CH₃)₃CCH₂OH, gives the 1° carbocation (CH₃)₃CCH₂⁺, which rearranges by a 1,2-methyl shift to the 3° carbocation (CH₃)₂C⁺CH₂CH₃.
2. Loss of a proton gives the more substituted alkene, (CH₃)₂C=CHCH₃, 2-methylbut-2-ene.

Teacher's Note:
1. Whenever a 1° carbocation sits next to a quaternary carbon, expect a methyl shift to give a 3° carbocation.

Answer: (c) mixture of carboxylic acids containing lesser number of carbon atoms

Explanation:
1. Tertiary alcohols resist oxidation under mild conditions. Under strong conditions (KMnO₄, elevated temperature) C–C bonds break,
2. giving a mixture of carboxylic acids, each containing fewer carbon atoms than the alcohol.

Teacher's Note:
1. 3° alcohols have no H on the carbinol carbon, so any oxidation must break a C–C bond – hence fewer carbons in the products.

Answer: (c) 3° > 2° > 1°

Explanation:
1. The reaction proceeds through a carbocation (S_N1). Stability of carbocations is 3° > 2° > 1°, so reactivity of alcohols is 3° > 2° > 1°.

Teacher's Note:
1. This is the basis of the Lucas test: 3° turbidity at once, 2° in about 5 minutes, 1° not at room temperature.

Answer: (d) 2-methylpropan-2-ol

Explanation:
1. 2-Methylpropan-2-ol, (CH₃)₃COH, is a tertiary alcohol and forms a stable 3° carbocation, so it gives turbidity immediately with Lucas reagent.

Teacher's Note:
1. Identify the degree of each alcohol first; only the 3° one reacts instantly at room temperature.

Answer: (c) treatment with pyridinium chlorochromate

Explanation:
1. PCC is a mild oxidising agent that oxidises a primary alcohol only up to the aldehyde stage.
2. KMnO₄ would oxidise it further to ethanoic acid, while H₂ and LiAlH₄ are reducing agents.

Teacher's Note:
1. To stop a 1° alcohol at the aldehyde, use PCC (or Cu at 573 K); strong oxidants go all the way to the acid.

Answer: (d) CH₃CHOHCH₂CH₂CH₃

Explanation:
1. Pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃, is first oxidised to pentan-2-one, CH₃COCH₂CH₂CH₃.
2. On further oxidation the C–C bond next to C=O breaks, giving CH₃COOH (2 C) and CH₃CH₂COOH (3 C).

Teacher's Note:
1. The carbon count of the acid fragments must add up to that of the alcohol: 2 + 3 = 5 carbons.

Answer: (a) C₆H₅OH

Explanation:
1. Phenol is acidic enough to react with aqueous NaOH to form sodium phenoxide: C₆H₅OH + NaOH → C₆H₅ONa + H₂O.
2. Alcohols are weaker acids than water and do not react with aqueous NaOH.

Teacher's Note:
1. Benzyl alcohol is an alcohol, not a phenol – its OH is on a CH₂ group, so it does not react with NaOH.

Answer: (d) :CCl₂

Explanation:
1. In the Reimer–Tiemann reaction, CHCl₃ + OH⁻ gives ⁻CCl₃, which loses Cl⁻ to form dichlorocarbene, :CCl₂.
2. Dichlorocarbene is the electron-deficient electrophile that attacks the ortho position of the phenoxide ion.

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

Answer: (d) Benzyl alcohol

Explanation:
1. In sunlight, chlorination takes place in the side chain (free radical): C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl.
2. Hydrolysis with aq. NaOH gives benzyl alcohol, C₆H₅CH₂OH.

Teacher's Note:
1. Sunlight/heat = side-chain substitution; Lewis acid (FeCl₃) in the dark = ring substitution.

Answer: (c) 2-Butanol

Explanation:
1. Methyl ethyl ketone is CH₃COCH₂CH₃ (butan-2-one). It is formed by oxidation of the secondary alcohol CH₃CH(OH)CH₂CH₃, 2-butanol.

Teacher's Note:
1. A ketone comes only from a 2° alcohol with the same carbon skeleton.

Answer: (b) CH₃–C(CH₃)=CH₂

Explanation:
1. tert-Butyl bromide is a 3° alkyl halide and NaOCH₃ is a strong base, so elimination predominates over substitution.
2. The product is 2-methylpropene, (CH₃)₂C=CH₂.

Teacher's Note:
1. This is why Williamson synthesis uses a 1° alkyl halide; with a 3° halide an alkene is formed instead of an ether.

Answer: (b) substitution reaction

Explanation:
1. Conversion of alkyl halides into alcohols involves substitution reaction.
2. RX \( \xrightarrow{NaOH} \) ROH + NaX

Teacher's Note:
1. Here OH⁻ replaces X⁻ (nucleophilic substitution); with alcoholic KOH the same halide would instead undergo dehydrohalogenation to an alkene.

Answer: (b) ⁻OR

Explanation:
1. Weakest acid has the strongest conjugate base. Since R–OH is the weakest acid, therefore, ⁻OR is the strongest base.

Teacher's Note:
1. The alkyl group pushes electron density onto oxygen (+I effect), whereas in phenoxide ions the negative charge is spread into the ring, making them weaker bases.

Answer: (d) iv > ii > iii > i

Explanation:
1. Boiling point increases with molecular mass (number of carbon atoms), so pentan-1-ol is highest and propan-1-ol is lowest.
2. Among the isomeric butanols, branching (butan-2-ol) lowers the surface area and the boiling point, so iv > ii > iii > i.

Teacher's Note:
1. First rank by number of carbon atoms, then compare isomers: the more branched isomer always boils lower.

Answer: (c) ii > i > iii

Explanation:
1. –NO₂ is electron withdrawing and stabilises the phenoxide ion, so o-nitrophenol is more acidic than phenol.
2. –OCH₃ is electron releasing (+R effect) and destabilises the phenoxide ion, so o-methoxyphenol is less acidic than phenol. Order: ii > i > iii.

Teacher's Note:
1. Electron-withdrawing groups (–NO₂, –CN, –X) raise the acidity of phenol; electron-donating groups (–CH₃, –OCH₃) lower it.

Answer: (c) (CH₃)₂CH–CH₂–OH + CH₃–CH₂–I

Explanation:
1. Both alkyl groups are primary, so the cleavage is S_N2: I⁻ attacks the less hindered alkyl group of the protonated ether.
2. The ethyl group is less hindered than the isobutyl group, so ethyl iodide and 2-methylpropan-1-ol are formed.

Teacher's Note:
1. With 1°/2° groups, the halide goes to the smaller (less hindered) group; with a 3° group, the halide goes to the 3° carbon (S_N1).

Answer: (b) Benzoic acid

Explanation:
1. Zn dust reduces phenol to benzene (X); Friedel-Crafts methylation gives toluene (Y).
2. Alkaline KMnO₄ oxidises the side-chain –CH₃ to –COOH, so Z is benzoic acid.

Teacher's Note:
1. Alkaline KMnO₄ oxidises any alkyl side chain having a benzylic hydrogen all the way to –COOH, not just to –CHO.

Answer: (d) Phenol, C₆H₅–OH

Explanation:
1. In phenol the –OH is attached directly to the benzene ring and the phenoxide ion is stabilised by resonance.
2. In the other compounds –OH is on an sp³ carbon, so their alkoxide ions are not resonance stabilised. Phenol is the most acidic.

Teacher's Note:
1. Benzyl alcohol is an alcohol, not a phenol – the CH₂ group cuts off resonance of the O⁻ with the ring.

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

Explanation:
1. The compound has an isopropyl group and an n-propyl group on oxygen; its correct IUPAC name is 2-propoxypropane (or 1-(1-methylethoxy)propane), so the Assertion is incorrect.
2. The Reason correctly states how ethers are named in the IUPAC system (as alkoxy derivatives of hydrocarbons).

Teacher's Note:
1. Count the carbons on each side of the oxygen before naming; here both groups have three carbons, so there is no "ethoxy" group at all.

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

Explanation:
1. Ethanol has a higher boiling point than its isomer methoxymethane because ethanol molecules are associated by intermolecular hydrogen bonding.
2. There is no intramolecular hydrogen bonding in ethanol, so the Reason is incorrect.

Teacher's Note:
1. Intermolecular H-bonding (between molecules) raises boiling point; intramolecular H-bonding (within one molecule) lowers it.

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

Explanation:
1. Phenoxide ion is resonance stabilised but ethoxide ion is not, so ethanol is a weaker acid than phenol.
2. Ethanol is a weaker acid than water, so it does not react with aqueous NaOH; sodium ethoxide is made by reacting ethanol with sodium metal. The Reason is incorrect.

Teacher's Note:
1. Alcohols react with Na metal but not with NaOH; phenols react with both.

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

Explanation:
1. o-Nitrophenol has intramolecular hydrogen bonding, so its –OH group is not free to form hydrogen bonds with water; hence it is less soluble.
2. m- and p-nitrophenols do exist as associated molecules (intermolecular H-bonding), but this is not the explanation of the lower solubility of the ortho isomer.

Teacher's Note:
1. Intramolecular H-bonding in o-nitrophenol also explains its lower boiling point and its steam volatility.

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

Explanation:
1. The electron-withdrawing –NO₂ group at the para position delocalises the negative charge of the phenoxide ion by resonance and stabilises it.
2. Hence p-nitrophenol is more acidic than phenol, and the Reason explains the Assertion.

Teacher's Note:
1. The more stable the conjugate base (phenoxide), the stronger the acid.

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

Explanation:
1. 3° alcohols form stable carbocations and give turbidity with Lucas reagent immediately.
2. The Reason correctly describes Lucas reagent, but it does not explain why 3° alcohols react immediately.

Teacher's Note:
1. The real explanation is carbocation stability: 3° > 2° > 1°.

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

Explanation:
1. Oxidation of phenol with chromic acid gives benzoquinone, a conjugated diketone.
2. Phenol also turns pink on standing in air due to slow oxidation, but this fact does not explain the chromic acid reaction.

Teacher's Note:
1. Benzoquinone has two C=O groups para to each other on a six-membered ring.

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

Explanation:
1. The –OH group donates electrons to the ring by resonance (+R), so the carbocation (arenium ion) formed by attack of an electrophile on phenol is more resonance stabilised.
2. Hence phenol reacts faster than benzene, and the Reason explains the Assertion.

Teacher's Note:
1. The extra resonance structure has the positive charge on oxygen with all atoms having complete octets – this is what makes it so stable.

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

Explanation:
1. With dilute HNO₃ at low temperature, phenol gives o- and p-nitrophenols; with concentrated HNO₃ it gives 2,4,6-trinitrophenol (picric acid). So the Assertion is incorrect.
2. The –OH group is indeed ortho, para-directing, so the Reason is correct.

Teacher's Note:
1. Match the reagent to the product: dil. HNO₃ → mononitrophenols; conc. HNO₃ → picric acid.

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

Explanation:
1. Phenol is so strongly activated by the –OH group that bromine is polarised even without a Lewis acid; phenol is brominated without any catalyst. So the Assertion is incorrect.
2. The Reason is correct: a Lewis acid such as FeBr₃ polarises the Br₂ molecule.

Teacher's Note:
1. Phenol + bromine water gives a white precipitate of 2,4,6-tribromophenol at once, with no catalyst.

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

Explanation:
1. Phenol reacts with an acid chloride to form an ester (acylation). The reaction is reversible, so it is carried out in the presence of pyridine.
2. Pyridine, a base, neutralises the HCl formed and shifts the equilibrium forward, which explains the Assertion.

Teacher's Note:
1. Phenol + RCOCl gives an ester, not an ether; ethers of phenol are made by Williamson synthesis.