JEE Chemistry Alcohols Phenols and Ethers MCQs Set 16

Answer: (a) Sodium salicylate

Explanation:
1. Kolbe-Schmidt reaction: carboxylation at the ortho position gives sodium salicylate.

Teacher's Note:
1. Kolbe-Schmidt reaction: sodium phenoxide with CO2 under pressure gives salicylic acid after acidification.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (c) Acetylene

Explanation:
1. Depending on temperature ethanol gives ethyl hydrogen sulphate (cold), ether (413 K) or ethene (443 K), never acetylene.

Teacher's Note:
1. Ethers are cleaved by HI, form explosive peroxides in air, and act as Lewis bases forming etherates.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (b) Tert-butyl > iso-propyl > ethyl > methyl

Explanation:
1. Alkyl groups stabilise a positive charge by inductive effect and hyperconjugation, so 3° > 2° > 1° > methyl.

Teacher's Note:
1. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (c) An alcohol with fewer carbon atoms is less soluble in water than one with more carbon atoms

Explanation:
1. Solubility in water decreases as the alkyl chain becomes longer, so the statement in option (c) is reversed.

Teacher's Note:
1. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (a) Brown or red then green then red then deep blue

Explanation:
1. Phenol with \( \ce{NaNO2} \) and conc.
2. \( \ce{H2SO4} \) gives brown/red, then green and, on dilution, red; with alkali it becomes deep blue.

Teacher's Note:
1. Nitro groups at ortho and para positions increase the acidity of phenol, and ortho-nitrophenol is steam volatile.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (d) \( \ce{C2H5OH} \) and \( \ce{C2H5Cl} \)

Explanation:
1. \( \ce{C2H5OH -> C2H5Cl ->[KCN] C2H5CN ->[H3O+] C2H5COOH} \)
2. A is ethanol and B is ethyl chloride.

Teacher's Note:
1. PCl5, PBr3 and SOCl2 replace the OH of alcohols by halogen; HX with ZnCl2 does the same via carbocations.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (c) p-Methylphenol, phenol, m-nitrophenol, p-nitrophenol

Explanation:
1. The electron-donating methyl group lowers acidity, nitro groups raise it; the para-nitro group (-M and -I) is more effective than meta (-I only). pKa values: about 10.2, 10.0, 8.4 and 7.15.

Teacher's Note:
1. Acidity order: carboxylic acids > carbonic acid > phenol > water > alcohols; nitro groups increase and alkyl groups decrease the acidity of phenols.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (c) Primary

Explanation:
1. Primary alcohols are the most acidic (least hindered, alkoxide best solvated), so they react fastest with sodium.

Teacher's Note:
1. Acidity order: carboxylic acids > carbonic acid > phenol > water > alcohols; nitro groups increase and alkyl groups decrease the acidity of phenols.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (c) 1/2 mole of \( \ce{H2} \)

Explanation:
1. 23 g is 1 mol Na; \( \ce{2CH3OH + 2Na -> 2CH3ONa + H2} \), so 1 mol Na gives 0.5 mol \( \ce{H2} \).

Teacher's Note:
1. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (d) \( \ce{PBr3} \)

Explanation:
1. \( \ce{3C4H9OH + PBr3 -> 3C4H9Br + H3PO3} \).

Teacher's Note:
1. PCl5, PBr3 and SOCl2 replace the OH of alcohols by halogen; HX with ZnCl2 does the same via carbocations.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (a) Chlorine

Explanation:
1. \( \ce{Cl2} \) would oxidise the alcohol to formaldehyde instead of substituting OH; HCl, \( \ce{PCl3} \) and \( \ce{PCl5} \) all replace OH by Cl.

Teacher's Note:
1. PCl5, PBr3 and SOCl2 replace the OH of alcohols by halogen; HX with ZnCl2 does the same via carbocations.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (d) \( \ce{CH3CH(OH)CH3} \)

Explanation:
1. Only a secondary alcohol gives a ketone: propan-2-ol gives acetone.

Teacher's Note:
1. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (d) Ethanol

Explanation:
1. Ethanol (351 K) is hydrogen bonded and larger than methanol (338 K); acetone and ether cannot hydrogen bond to themselves.

Teacher's Note:
1. Hydrogen bonding raises boiling point and solubility, while intramolecular bonding (o-nitrophenol) lowers boiling point.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (c) Ethylene

Explanation:
1. Conc.
2. \( \ce{H2SO4} \) at 443 K gives ethene.

Teacher's Note:
1. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (b) \( \ce{CH3CH(OH)CH3} \)

Explanation:
1. Propan-2-ol is oxidised by hypoiodite to acetone, which has the \( \ce{CH3CO-} \) group.

Teacher's Note:
1. The iodoform test is positive for ethanol, CH3CH(OH)R and CH3CO-; methanol and primary alcohols other than ethanol do not respond.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (c) They form a constant boiling mixture

Explanation:
1. Ethanol and water form an azeotrope (95.6 percent ethanol) that boils at a constant temperature.

Teacher's Note:
1. Ethanol and water form an azeotrope with 95.6 percent alcohol; absolute alcohol needs benzene (azeotropic distillation) or quicklime and further drying.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (d) \( \ce{CH3C#CH} \)

Explanation:
1. Propanol gives propene (X), then 1,2-dibromopropane (Y), and excess alcoholic KOH removes two HBr to give propyne (Z).

Teacher's Note:
1. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (a) Soluble in water

Explanation:
1. Lower alcohols are miscible with water through hydrogen bonding.

Teacher's Note:
1. Hydrogen bonding raises boiling point and solubility, while intramolecular bonding (o-nitrophenol) lowers boiling point.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (c) 2-Butanol

Explanation:
1. Butan-2-ol, \( \ce{CH3CH(OH)CH2CH3} \), is oxidised to butanone.

Teacher's Note:
1. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (b) \( \ce{C6H5OH} \)

Explanation:
1. Phenol is the only one acidic enough to turn litmus red slightly and dissolve in NaOH.

Teacher's Note:
1. Acidity order: carboxylic acids > carbonic acid > phenol > water > alcohols; nitro groups increase and alkyl groups decrease the acidity of phenols.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (b) o-Nitrophenol (together with p-nitrophenol)

Explanation:
1. Dilute nitric acid nitrates the ring at the ortho and para positions; picric acid needs conc.
2. \( \ce{HNO3} \)/\( \ce{H2SO4} \).

Teacher's Note:
1. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (a) Alkenes

Explanation:
1. Loss of water gives alkenes.

Teacher's Note:
1. Dehydration of alcohols follows carbocation stability, so tertiary alcohols dehydrate fastest; ethanol gives ether at 413 K and ethene at 443 K.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (a) Aldehydes and ketones respectively

Explanation:
1. Dehydrogenation at 573 K: primary alcohols give aldehydes, secondary alcohols give ketones.

Teacher's Note:
1. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (c) HCOOH

Explanation:
1. Methanol is oxidised to formaldehyde and then formic acid.

Teacher's Note:
1. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (a) Alcohols

Explanation:
1. Lucas reagent distinguishes primary, secondary and tertiary alcohols.

Teacher's Note:
1. In the Lucas test turbidity appears at once for tertiary, in about 5 minutes for secondary, and only on heating for primary alcohols.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (a) Primary amine

Explanation:
1. Phenol is converted to aniline (Bucherer-type amination).

Teacher's Note:
1. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (a) Phenol

Explanation:
1. The OH group strongly activates the ring.

Teacher's Note:
1. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (c) o-Nitrophenol

Explanation:
1. o-Nitrophenol has intramolecular hydrogen bonding, so its molecules are not associated and it is steam volatile.
2. The meta and para isomers are associated through intermolecular hydrogen bonds and boil much higher.

Teacher's Note:
1. Hydrogen bonding raises boiling point and solubility, while intramolecular bonding (o-nitrophenol) lowers boiling point.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.