JEE Chemistry Alcohols Phenols and Ethers MCQs Set 24

Answer: (a) 1

Explanation:
1. The four butanols are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol.
2. Only butan-2-ol has a carbon attached to four different groups (H, OH, CH3, C2H5).
3. So exactly one is chiral.

Teacher's Note:
1. A chiral centre needs four different substituents. Check each carbon bearing the OH group in the isomers.

Answer: (b) \(\ce{RO^-}\)

Explanation:
1. The stronger the base, the weaker its conjugate acid.
2. Conjugate acids: water, alcohol, phenol, m-nitrophenol. Alcohols are weaker acids than water, and phenols are far more acidic than both.
3. Hence alkoxide \(\ce{RO^-}\) is the strongest base.

Teacher's Note:
1. Alkyl groups push electrons (+I effect) and destabilise the alkoxide ion, making it a stronger base than hydroxide. Phenoxides are resonance stabilised, so they are weak bases.

Answer: (d) m-Chlorophenol

Explanation:
1. Phenols are more acidic than alcohols because the phenoxide ion is stabilised by resonance.
2. Chlorine at the meta position withdraws electrons by its -I effect (no +R effect operates at meta), stabilising the phenoxide further.
3. So m-chlorophenol is more acidic than phenol.

Teacher's Note:
1. Compare phenols by the substituent: electron-withdrawing groups raise acidity. Alcohols are weaker than phenols.

Answer: (d) b > d > c > a > e

Explanation:
1. p-Nitrophenol is the strongest (-I and -R), followed by m-nitrophenol (-I only).
2. At the meta position the methoxy group acts only by its -I effect, so m-methoxyphenol is slightly more acidic than phenol.
3. At the para position the +R effect of the methoxy group dominates, making p-methoxyphenol the weakest.
4. Order: b > d > c > a > e.

Teacher's Note:
1. A methoxy group is electron-withdrawing from meta (inductive only) but electron-donating from para (resonance). Remember this position dependence.

Answer: (c) b < c < a

Explanation:
1. The rate depends on the stability of the benzyl carbocation.
2. The nitro group strongly destabilises it (-I, -R), so (b) is the slowest.
3. Chlorine withdraws by -I but donates weakly by +R, so (c) is slightly less reactive than the unsubstituted benzyl alcohol (a).
4. Order: b < c < a.

Teacher's Note:
1. In \(\ce{S_N1}\) reactions the more stable the carbocation, the faster the reaction. Electron-withdrawing groups slow the reaction.

Answer: (a) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol

Explanation:
1. Boiling point rises with the number of carbon atoms (more surface area, stronger van der Waals forces).
2. Among isomers, branching lowers it, so butan-2-ol (about 99 C) boils lower than butan-1-ol (about 118 C).
3. Propan-1-ol is about 97 C and pentan-1-ol about 138 C.
4. Order: propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol.

Teacher's Note:
1. For alcohols, compare chain length first and then branching. A more compact molecule has a smaller surface area and boils lower.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Catalytic hydrogenation, \(\ce{LiAlH4}\) and \(\ce{NaBH4}\) all reduce an aldehyde to a primary alcohol.
2. \(\ce{RMgX}\) adds a new alkyl group to the carbonyl carbon and gives a secondary alcohol \(\ce{RCH(OH)R}\) from an aldehyde, so (iv) is wrong.

Teacher's Note:
1. Grignard reagents build new C-C bonds; reducing agents only add hydrogen. Methanal is the only aldehyde that gives a primary alcohol with a Grignard reagent.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. (i) is the Dow process and gives sodium phenoxide, hence phenol.
2. (ii) The diazonium salt is hydrolysed by warm water to phenol.
3. (iii) Sulphonation then alkali fusion gives sodium phenoxide, hence phenol.
4. (iv) Chlorobenzene is inert to aqueous NaOH under ordinary conditions because the C-Cl bond has partial double bond character.

Teacher's Note:
1. Aryl halides need drastic conditions or strongly electron-withdrawing groups for nucleophilic substitution.

Answer: (a) (i) and (iii) only

Explanation:
1. Phenol decolourises bromine water giving a white precipitate of 2,4,6-tribromophenol, and gives a violet colour with neutral \(\ce{FeCl3}\); ethanol does neither.
2. Both phenol and ethanol liberate hydrogen with sodium, so (ii) cannot distinguish them.

Teacher's Note:
1. Neutral ferric chloride is the classical test for phenols (violet, blue or green colour).

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Both statements are true.
2. The nitro group at the para position delocalises the negative charge of the phenoxide ion into the nitro group, which stabilises it and raises the acidity.
3. The Reason correctly explains the Assertion.

Teacher's Note:
1. Resonance stabilisation of the conjugate base is the main reason for the high acidity of nitrophenols.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. The Assertion is false: phenol is so activated that it reacts with bromine even without a Lewis acid (in water it gives 2,4,6-tribromophenol).
2. The Reason is a true statement about the role of a Lewis acid.
3. So the Assertion is false and the Reason is true.

Teacher's Note:
1. The -OH group strongly activates the ring, so no catalyst is needed for halogenation of phenol.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. o-Nitrophenol has intramolecular hydrogen bonding, so its -OH is not free to bond with water and it is less soluble (and steam volatile).
2. m- and p-isomers form intermolecular hydrogen bonds with water and with one another (association), so they are more soluble and less volatile.
3. Both are true and the Reason explains the Assertion.

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

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. In water (a polar solvent) phenol ionises partly and bromine attacks all three activated positions giving 2,4,6-tribromophenol; the Assertion is true.
2. In non-polar solvents such as \(\ce{CS2}\) bromine is not polarised, and only monobromophenols form; so the Reason is false.

Teacher's Note:
1. Solvent polarity decides the extent of bromination of phenol: water gives the tribromo product, \(\ce{CS2}\) at low temperature gives mainly the para monobromo product.

Answer: (a) 2-acetoxybenzoic acid (acetylsalicylic acid)

Explanation:
1. This is the Kolbe-Schmitt reaction: sodium phenoxide with \(\ce{CO2}\) gives sodium salicylate (ortho carboxylation).
2. Acidification gives salicylic acid.
3. Acetylation of the phenolic -OH with acetic anhydride gives 2-acetoxybenzoic acid, i.e. aspirin.

Teacher's Note:
1. Kolbe-Schmitt carboxylation takes place at the ortho position at about 125 C; at higher temperature with potassium phenoxide the para product forms.

Answer: (a) Acetyl salicylic acid

Explanation:
1. Aspirin is made by acetylating the phenolic -OH of salicylic acid, so it is acetylsalicylic acid (2-acetoxybenzoic acid).

Teacher's Note:
1. Phenyl salicylate is salol and methyl salicylate is oil of wintergreen. Do not confuse them with aspirin.

Answer: (b) o-nitrophenol shows intramolecular hydrogen bonding

Explanation:
1. In o-nitrophenol the -OH group hydrogen bonds with the adjacent nitro group within the same molecule.
2. The -OH is therefore not available to form hydrogen bonds with water, so the solubility is lower.
3. Steam volatility and melting point are consequences, not the cause.

Teacher's Note:
1. Intramolecular hydrogen bonding is the reason o-nitrophenol can be separated from its isomers by steam distillation.

Answer: (a) \(\ce{NH2NH2}\), \(\ce{OH^-}\)

Explanation:
1. The ketone C=O must be reduced completely to CH2 without touching the alkene or the alcohol.
2. The Wolff-Kishner reduction (hydrazine and base) does this under basic conditions.
3. Zn-Hg/HCl (acidic) may dehydrate or substitute the alcohol; Na/liquid ammonia reduces the alkene; \(\ce{NaBH4}\) only gives the alcohol.

Teacher's Note:
1. Choose Wolff-Kishner for acid-sensitive substrates and Clemmensen for base-sensitive ones.

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

Explanation:
1. KBr and \(\ce{KBrO3}\) in acid generate bromine in aqueous (polar) medium.
2. Phenol in polar medium is brominated at all activated positions, giving 2,4,6-tribromophenol.

Teacher's Note:
1. The bromide-bromate mixture is a convenient in situ source of bromine.

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

Explanation:
1. p-Nitrophenol is the most acidic (-I and -R).
2. m-Nitrophenol has only the -I effect, so it is less acidic than the para isomer but more acidic than phenol.
3. The methyl group of p-cresol donates electrons and lowers acidity.
4. Order: IV > III > I > II.

Teacher's Note:
1. A nitro group at para raises acidity more than at meta because of resonance (-R).

Answer: (b) dichlorocarbene \(\ce{:CCl2}\)

Explanation:
1. Hydroxide removes a proton from chloroform to give \(\ce{CCl3^-}\).
2. The anion loses chloride to give the electrophilic dichlorocarbene \(\ce{:CCl2}\).
3. The phenoxide ion attacks the carbene at the ortho position and hydrolysis gives the aldehyde.

Teacher's Note:
1. Carbene is electron-deficient (six electrons) and is attacked by the electron-rich phenoxide ring.

Answer: (d) I, III and IV

Explanation:
1. Primary alcohols react with HCl only in the presence of \(\ce{ZnCl2}\) (with heating), so I works and II does not.
2. Tertiary alcohols react readily with HCl without a catalyst, so III works.
3. Secondary alcohols react with HCl and \(\ce{ZnCl2}\), so IV works.

Teacher's Note:
1. Zinc chloride coordinates with the oxygen of the alcohol and makes the -OH a better leaving group.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. A compound dissolves in sodium hydrogen carbonate with effervescence only if it is a stronger acid than carbonic acid (pKa about 6.4).
2. Picric acid (pKa about 0.4), benzoic acid (4.2) and benzenesulphonic acid are stronger.
3. o-Nitrophenol (pKa about 7.2) is weaker than carbonic acid.

Teacher's Note:
1. NaHCO3 separates carboxylic acids and very acidic phenols from ordinary phenols.