JEE Chemistry Alcohols Phenols and Ethers MCQs Set 27

Answer: (d)

Explanation:
1. H₂SO₄ protonates tert-butyl alcohol and loses water to give the tert-butyl cation.
2. Phenol is a very reactive arene; the electrophile attacks the ring at the para position (the ortho position is hindered), a Friedel-Crafts alkylation.
3. The product is 4-tert-butylphenol. An aryl ether is not formed because the phenolic oxygen is a poor nucleophile in acid, and a tertiary carbocation does not rearrange to an isobutyl group.

Teacher's Note:
1. Carbocation from a tertiary alcohol is a good electrophile for ring alkylation of phenol.
2. A bulky group goes para.

Answer: (c)

Explanation:
1. The hydroxyl group is strongly activating and bromine water is a polar medium, so all free ortho and para positions are attacked.
2. The carboxyl group on carbon 2 is displaced as carbon dioxide during substitution at that position (ipso attack).
3. The product is 2,4,6-tribromophenol.

Teacher's Note:
1. Bromine water with a phenol gives a white precipitate of the tribromo derivative.
2. Decarboxylative bromination is a known feature of salicylic acid.

Answer: (c) m-cresol

Explanation:
1. Solubility in NaOH but not in NaHCO₃ shows a weakly acidic phenol; anisole has no acidic hydrogen.
2. Three bromine atoms enter the ring only if three activated positions (two ortho and one para to OH) are free.
3. In m-cresol positions 2, 4 and 6 are free, giving 2,4,6-tribromo-3-methylphenol; o- and p-cresol allow only two.

Teacher's Note:
1. Count the free ortho and para positions of the hydroxyl group.
2. Phenols dissolve in NaOH but not in sodium bicarbonate.

Answer: (b) sodium phenoxide, anisole, C₆H₅OH, CH₃I

Explanation:
1. Alkali fusion of a sulphonate gives sodium phenoxide (A).
2. Methyl iodide alkylates the phenoxide ion (Williamson synthesis) to give anisole (B).
3. HI cleaves the ether; the aryl-oxygen bond is too strong to break, so phenol (C) and methyl iodide (D) are formed.

Teacher's Note:
1. Aryl alkyl ethers always cleave to a phenol and an alkyl halide with HI.
2. The C(aryl)-O bond has partial double-bond character.

Answer: (c) benzene and phenyl acetate

Explanation:
1. Zinc removes the oxygen of phenol, so A is benzene.
2. Ethanoyl chloride acylates the phenoxide oxygen (Schotten-Baumann conditions) to give an ester.
3. B is phenyl acetate (phenyl ethanoate).

Teacher's Note:
1. Zinc dust distillation is the classical reduction of phenol to benzene.
2. Acylation of phenol under basic conditions occurs on oxygen.

Answer: (b) 2-nitrophenol

Explanation:
1. Dilute nitric acid gives a mixture of 2-nitrophenol and 4-nitrophenol.
2. In 2-nitrophenol there is intramolecular hydrogen bonding between OH and NO2, so it is steam volatile and has the lower boiling point.
3. 4-Nitrophenol forms intermolecular hydrogen bonds and stays in the flask.

Teacher's Note:
1. Intramolecular hydrogen bonding lowers boiling point and raises volatility.
2. Steam distillation separates the ortho isomer.

Answer: (b) quinol (hydroquinone) and catechol

Explanation:
1. At low temperature (kinetic control) the ortho sulphonic acid forms; at 100 degrees C (thermodynamic control) the para isomer is the product.
2. Fusion with NaOH replaces the sulphonic acid group by OH, giving benzene-1,4-diol (x) from the para acid and benzene-1,2-diol (y) from the ortho acid.

Teacher's Note:
1. Sulphonation is reversible, so the more stable para product predominates at high temperature.
2. OH in the new position is para or ortho to the existing OH, never meta.

Answer: (c) 99

Explanation:
1. Y is 4-hydroxyazobenzene, C₁₂H₁₀N₂O.
2. Molecular weight = 144 + 10 + 28 + 16 = 198.
3. 198 / 2 = 99.

Teacher's Note:
1. Follow the transformations one at a time and count atoms or add atomic masses.
2. Integer-type answers are best checked by writing the formula of the final product.

Answer: (b) A, B, C and D

Explanation:
1. Phenol is brominated at the ortho and para positions (white precipitate of 2,4,6-tribromophenol).
2. Cyclohex-2-en-1-ol has a C=C bond that adds bromine.
3. In 2-hydroxybenzenesulphonic acid and salicylic acid, bromine substitutes at positions bearing the SO3H and COOH groups (ipso substitution) and also at free ortho/para positions, so these too decolourise bromine water.
4. All four decolourise it.

Teacher's Note:
1. Phenols with free or displaceable ortho/para positions react with bromine water.
2. An alkene double bond also decolourises bromine water.

Answer: (b) A and C only

Explanation:
1. Dichlorocarbene attacks the phenoxide ion mainly at the ortho position.
2. The para isomer is a minor product.
3. No meta product is formed because OH directs ortho/para.

Teacher's Note:
1. Dichlorocarbene is an electrophile.
2. Ortho attack is favoured by the interaction between the phenoxide and the sodium ion.

Answer: (d)

Explanation:
1. The primary carbocation rearranges, so X is cumene; the cumene process gives phenol (Y) and acetone (Z).
2. Acetone with Cl₂/OH- is a haloform reaction giving chloroform (W) and acetate.
3. Phenol with chloroform and alkali is the Reimer-Tiemann reaction, giving salicylaldehyde (P).

Teacher's Note:
1. The carbocation from 1-chloropropane rearranges to the more stable isopropyl cation.
2. W is chloroform.

Answer: (b) :CCl₂

Explanation:
1. Chloroform loses a proton to alkali to give CCl₃-, which loses chloride to form dichlorocarbene, :CCl₂.
2. Dichlorocarbene is a neutral electrophile that attacks the ortho carbon of the phenoxide ion.

Teacher's Note:
1. The electrophile is dichlorocarbene.
2. The dichloromethyl group is later hydrolysed to CHO.

Answer: (a) cumene and propanone (acetone)

Explanation:
1. The n-propyl carbocation rearranges (1,2-hydride shift) to the more stable isopropyl cation, so P is cumene.
2. Air oxidation gives cumene hydroperoxide, and acid-catalysed rearrangement gives phenol and acetone.

Teacher's Note:
1. Friedel-Crafts alkylation with primary halides often gives rearranged products.
2. Acetone is the by-product in the cumene process.

Answer: (d) :CCl₂

Explanation:
1. CHCl₃ is deprotonated to CCl₃-, which expels Cl- to give dichlorocarbene.
2. Dichlorocarbene, :CCl₂, is the electrophile.

Teacher's Note:
1. Carbenes are electron-deficient neutral species.
2. The carbene attacks the ortho position of the phenoxide.

Answer: (a)

Explanation:
1. Attack of :CCl₂ on the ring gives, after proton transfer, the dichloromethyl phenoxide.
2. Hydrolysis of the -CHCl₂ group by aqueous alkali gives the aldehyde group.

Teacher's Note:
1. -CHCl₂ hydrolyses to -CH(OH)₂, which loses water to give CHO.
2. The CCl3 compound would give a carboxylic acid.

Answer: (c)

Explanation:
1. Bromine water brominates the two ortho positions, and the third equivalent replaces the sulphonic acid group at the para position (ipso substitution, loss of SO₃).
2. The product is 2,4,6-tribromophenol.

Teacher's Note:
1. The sulphonic acid group is a good leaving group in electrophilic substitution.
2. Three equivalents of bromine give three bromine atoms.

Answer: (c)

Explanation:
1. T is cumene (isopropylbenzene).
2. Autoxidation by a radical mechanism abstracts the tertiary benzylic hydrogen, because the benzylic tertiary radical is the most stable.
3. U is cumene hydroperoxide.

Teacher's Note:
1. Benzylic radicals are stabilised by the ring.
2. Cumene hydroperoxide is the intermediate in the industrial production of phenol.

Answer: (c)

Explanation:
1. Picric acid is the common name of 2,4,6-trinitrophenol.
2. It has an OH group with a nitro group at each of the two ortho and the para position.

Teacher's Note:
1. It is a strong acid because of three nitro groups.
2. Picric acid is made from phenol (via sulphonation then nitration).

Answer: (d)

Explanation:
1. Three bromine atoms enter only when all three positions ortho and para to OH are free.
2. In m-cresol (3-methylphenol) positions 2, 4 and 6 are free.
3. o-Cresol and p-cresol have one of these positions blocked, and benzyl alcohol is not a phenol.

Teacher's Note:
1. Count free ortho/para positions to OH.
2. Benzyl alcohol does not react with bromine water.

Answer: (c) salicylic acid

Explanation:
1. Sodium phenoxide attacks carbon dioxide (Kolbe's reaction), mainly at the ortho position.
2. Acidification gives 2-hydroxybenzoic acid, salicylic acid.

Teacher's Note:
1. Kolbe reaction gives hydroxy acids.
2. Ortho product is major at about 400 K.

Answer: (a) 2,4,6-tribromophenol

Explanation:
1. KBr and KBrO₃ produce bromine in situ in the presence of acid.
2. Phenol is very reactive and, in this aqueous medium, all three ortho and para positions are brominated to give 2,4,6-tribromophenol.

Teacher's Note:
1. The bromide-bromate mixture is a source of bromine.
2. Polar aqueous medium gives polysubstitution.

Answer: (c)

Explanation:
1. Kolbe reaction at 125 degrees C and 5 atm gives sodium salicylate; acid gives salicylic acid (B).
2. Acetylation of the phenolic OH gives acetylsalicylic acid, aspirin (C).

Teacher's Note:
1. Aspirin is the O-acetyl derivative of salicylic acid.
2. Ac2O acylates oxygen, not the ring.

Answer: (a) and

Explanation:
1. Phenoxide attacks the carbonyl carbon of methyl chloroformate and gives methyl phenyl carbonate (A), an O-acylation.
2. The oxygen of the carbonate group still donates electrons to the ring, so it directs ortho/para; bromination gives the para product, methyl 4-bromophenyl carbonate (B).

Teacher's Note:
1. Acylation of a phenoxide occurs on oxygen.
2. The O-acyl group is a weaker activator than OH but remains ortho/para-directing.

Answer: (b)

Explanation:
1. Kolbe reaction gives salicylic acid (X) as the major product.
2. Acetylation of the OH group with acetic anhydride gives acetylsalicylic acid (aspirin).

Teacher's Note:
1. The acid group is not acetylated because it would give an unstable mixed anhydride.
2. Aspirin: 2-acetoxybenzoic acid.

Answer: (a) Gattermann reaction

Explanation:
1. HCN and HCl with ZnCl₂ introduce the CHO group into an activated ring: the Gattermann reaction.
2. Gattermann-Koch uses CO and HCl with AlCl₃/CuCl and does not work for phenols.

Teacher's Note:
1. Gattermann uses HCN; Gattermann-Koch uses CO.
2. Both introduce a formyl group.

Answer: (d) phenol and acetone

Explanation:
1. Air oxidation gives cumene hydroperoxide.
2. Acid causes rearrangement (phenyl migrates to oxygen) and hydrolysis, giving phenol and acetone.

Teacher's Note:
1. Cumene process: cumene to phenol + acetone.
2. Acetone is not reduced under these conditions.