JEE Chemistry Alcohols Phenols and Ethers MCQs Set 28

Answer: (d) Phenol

Explanation:
1. Effervescence needs an acid stronger than carbonic acid (pKa about 6.4).
2. Phenol (pKa about 10) is weaker than carbonic acid, whereas nitrophenols with nitro groups at ortho or para positions are stronger acids.
3. Hence only phenol gives no carbon dioxide.

Teacher's Note:
1. Electron-withdrawing nitro groups at ortho and para positions stabilise the phenoxide ion.
2. Only carboxylic acids and strongly acidic phenols react with bicarbonate.

Answer: (b) aspirin

Explanation:
1. Sodium phenoxide with CO₂ under pressure (Kolbe reaction) gives sodium salicylate, and acidification gives salicylic acid (R).
2. Acetylation of the phenolic hydroxyl of salicylic acid by ethanoic anhydride gives 2-acetoxybenzoic acid.
3. That compound is aspirin.

Teacher's Note:
1. Kolbe reaction: phenoxide + CO₂ gives ortho hydroxy acid.
2. Aspirin is the acetyl ester of salicylic acid.

Answer: (d)

Explanation:
1. Dinitration gives 1-chloro-2,4-dinitrobenzene, and hydroxide replaces the activated chlorine to give 2,4-dinitrophenol.
2. Sn/HCl reduces both nitro groups to give 2,4-diaminophenol.
3. Diazotisation of both amino groups followed by warming with water replaces them by hydroxyl groups, giving benzene-1,2,4-triol.

Teacher's Note:
1. Nitro groups ortho and para to chlorine make nucleophilic substitution easy.
2. Warm water converts each diazonium group into a hydroxyl group.

Answer: (a)

Explanation:
1. The hydroxyl group is the stronger activator and directs to its ortho and para positions.
2. The position para to OH is also ortho to CHO and is less hindered than the position between the two groups.
3. The major product is 2-bromo-5-hydroxybenzaldehyde.

Teacher's Note:
1. Take the directing effect of the stronger activating group, OH.
2. Para to OH is generally preferred over ortho.

Answer: (b) (1), (2) and (3)

Explanation:
1. Chloroform and base give the electrophile :CCl₂, which attacks the phenoxide ring, so (1) and (2) are correct.
2. Aniline forms a complex with AlCl₃ (the amino group becomes positively charged) and nitrobenzene is strongly deactivated, so neither undergoes Friedel-Crafts reaction.
3. Hence all three statements are correct.

Teacher's Note:
1. Dichlorocarbene is an electrophile despite having no charge.
2. Lewis acid catalysts are bound by the lone pair on nitrogen.

Answer: (d)

Explanation:
1. AlCl₃ causes the Fries rearrangement; on heating the acyl group moves to the ortho position, giving 2-hydroxyacetophenone (X).
2. The methyl ketone gives the iodoform reaction with I₂/NaOH.
3. The COCH₃ group becomes COONa, giving sodium salicylate (Y), with iodoform as by-product.

Teacher's Note:
1. High temperature favours the ortho Fries product.
2. A methyl ketone is converted to a carboxylate and CHI₃.

Answer: (c) p-isopropyl-t-butylbenzene

Explanation:
1. An alkyl group is an ortho/para director.
2. The t-butyl group is very bulky and blocks the adjacent ortho positions.
3. The isopropyl cation therefore attacks the para position.

Teacher's Note:
1. Bulky substituents push the incoming group to para.
2. Alkyl groups activate the ring slightly.

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

Explanation:
1. The sulphonic acid group sits at the para position and can be replaced by an electrophile (ipso substitution).
2. With excess bromine water all three activated positions are brominated and SO₃H is lost as sulphur trioxide (as hydrogen sulphate).
3. The product is 2,4,6-tribromophenol.

Teacher's Note:
1. Sulphonation is reversible, so SO₃H is a good leaving group.
2. Phenols in bromine water give the tribromo product.

Answer: (d) Grignard reagent formation followed by CO₂

Explanation:
1. Bromobenzene with magnesium in dry ether gives phenylmagnesium bromide.
2. Reaction with dry ice (solid CO₂) followed by acidification gives benzoic acid.

Teacher's Note:
1. A Grignard reagent adds one carbon from carbon dioxide.
2. Reimer-Tiemann works on phenols.

Answer: (c) the ortho and para carbon atoms

Explanation:
1. Negative charge of the phenoxide ion is delocalised to the ortho and para carbons through resonance.
2. These carbons are the nucleophilic sites and attack the electrophile.

Teacher's Note:
1. Draw the resonance forms to see which carbons carry negative charge.
2. The meta position does not receive any charge.

Answer: (b) I > III > II

Explanation:
1. Aromaticity increases with the ability of the heteroatom to share its lone pair and decreases with electronegativity.
2. Sulphur is the least electronegative and its lone pair is the most delocalised, so thiophene is the most aromatic; oxygen is the most electronegative, so furan is the least aromatic.
3. The order is thiophene > pyrrole > furan.

Teacher's Note:
1. Lower electronegativity of the heteroatom gives greater aromaticity.
2. Resonance energies: thiophene 29, pyrrole 21, furan 16 kcal per mole.

Answer: (a) phenolic aldehyde

Explanation:
1. Phenol with chloroform and aqueous NaOH introduces a CHO group, mainly ortho to the OH.
2. The product is salicylaldehyde, a phenolic aldehyde.

Teacher's Note:
1. Dichlorocarbene is the electrophile; hydrolysis converts the CHCl₂ group to CHO.
2. Ortho product predominates.

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

Explanation:
1. Bromination at the ortho and para positions of the hydroxyl group is followed by loss of carbon dioxide from the carbon bearing the acid group.
2. The product is 2,4,6-tribromophenol.

Teacher's Note:
1. Replacement of a carboxyl group by bromine occurs because of the strongly activating OH.
2. A white precipitate forms.

Answer: (d)

Explanation:
1. NaOH converts the phenol to the phenoxide ion, a good oxygen nucleophile.
2. Benzyl chloride is attacked by oxygen in an SN2 (Williamson) reaction.
3. The product is 1-benzyloxy-4-methylbenzene, an ether, not a ring-alkylated phenol.

Teacher's Note:
1. Phenoxide with a reactive alkyl halide gives the ether.
2. Ring alkylation of this type needs a Lewis acid catalyst.

Answer: (d)

Explanation:
1. SOCl₂ converts the acid into its acid chloride.
2. Dry AlCl₃ causes intramolecular Friedel-Crafts acylation at the ring carbon ortho to the side chain, closing a six-membered ring.
3. The ketone tautomerises to the naphthol; the methoxy group, originally para to the side chain, ends up at carbon 7, giving 7-methoxynaphthalen-1-ol.

Teacher's Note:
1. Keep track of the positions relative to the side chain on the original ring.
2. Six-membered ring formation is favoured.

Answer: (b) S is benzaldehyde, W is benzene and T is benzoic acid

Explanation:
1. P is phenol; the Reimer-Tiemann reaction gives salicylaldehyde (Q) and 4-hydroxybenzaldehyde (R).
2. Zinc dust removes the hydroxyl group, giving benzaldehyde (S), which is oxidised to benzoic acid (T, C₇H₆O₂).
3. Soda lime decarboxylates T to benzene (W), which is also what phenol gives with zinc dust. So S is benzaldehyde, W is benzene and T is benzoic acid.

Teacher's Note:
1. Phenol with zinc dust gives benzene.
2. Q and R are the ortho and para hydroxy aldehydes.

Answer: (d) (1), (2), (3) and (4)

Explanation:
1. AlCl₃ coordinates with the phenolic oxygen, so acylation on the ring does not proceed readily.
2. In the nitrating mixture the amine is protonated and the NH₃+ group directs meta; in a non-polar solvent at low temperature bromination stops at the para product.
3. Carbon tetrachloride in alkali gives the Reimer-Tiemann-type carboxylation product, salicylic acid. All four are correct.

Teacher's Note:
1. Check each statement individually.
2. Carbon tetrachloride replaces chloroform when a carboxyl group is wanted.

Answer: (c) 14

Explanation:
1. X is 1-ethoxy-4-nitrobenzene and Y is 4-ethoxyaniline.
2. Diazotisation and coupling with phenol at the para position gives 4-ethoxy-4'-hydroxyazobenzene, C₁₄H₁₄N₂O₂.
3. It has 6 + 2 + 6 = 14 carbon atoms.

Teacher's Note:
1. The ethyl group adds two carbons to the two benzene rings.
2. Phenol couples at the para position.

Answer: (a) It gives a coloured solution with neutral FeCl₃

Explanation:
1. NH₄HS reduces only one nitro group, giving 3-nitroaniline (A).
2. Diazotisation and hydrolysis give 3-nitrophenol (T), which has one nitrogen atom and is weakly acidic.
3. Phenols give a coloured complex with neutral FeCl₃.

Teacher's Note:
1. Ammonium sulphide or hydrosulphide reduces one nitro group selectively (Zinin reduction).
2. A nitro-deactivated ring does not undergo Friedel-Crafts reactions.