JEE Chemistry Alcohols Phenols and Ethers MCQs Set 18

Answer: (c) (i) o- and p-bromophenol and (ii) 2,4,6-tribromophenol

Explanation:
1. In the non-polar solvent ionisation of phenol is suppressed, giving monobromo products; in water the phenoxide ion forms and gives the tribromo product.

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: (a) Phenol to benzyl alcohol

Explanation:
1. Phenol does not give benzyl alcohol; the others are the usual copper-catalysed dehydrogenation and dehydration reactions.

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. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (d) Reacting with Mg

Explanation:
1. Mg reacts with ethanol to form magnesium ethoxide, and the ethoxide reacts with water to regenerate ethanol and give Mg(OH)2; this removes the last traces of water.

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) Allyl alcohol

Explanation:
1. At 260°C the intermediate loses CO2 and H2O to give allyl alcohol.

Teacher's Note:
1. Glycerol is propane-1,2,3-triol; it gives nitroglycerine, allyl iodide with HI, and acrolein on dehydration.
2. Learn the identification tests: Lucas, iodoform, FeCl3, Victor Meyer, Liebermann, bromine water.

Answer: (c) Aluminium oxide

Explanation:
1. Heated alumina dehydrates alcohols to 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. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.

Answer: (b) 1-Phenylethanol

Explanation:
1. 1-Phenylethanol, \( \ce{C6H5CH(OH)CH3} \), has the \( \ce{CH3CH(OH)-} \) 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. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (c) 10.22 g

Explanation:
1. Phenol: 2/94 = 0.0213 mol.
2. Three Br2 per phenol: 0.0638 mol x 160 = 10.2 g.

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: (b) Sodium metal

Explanation:
1. Ethanol releases H2 from sodium, forming sodium ethoxide, showing it can donate a proton.

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. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (a) Esterification

Explanation:
1. \( \ce{ROH + R'COOH -> R'COOR + H2O} \).

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) \( \ce{CH3OH} \)

Explanation:
1. Methanol is hydrogen bonded (338 K); the others are gases or low-boiling liquids.

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: (b) Formic acid

Explanation:
1. Formic acid (374 K) forms strong hydrogen-bonded dimers; ethanol boils at 351 K and is higher than propane or dimethyl ether.

Teacher's Note:
1. Ethers are cleaved by HI, form explosive peroxides in air, and act as Lewis bases forming etherates.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (c) Lower members are insoluble in water and organic solvents but solubility regularly increases with molecular weight

Explanation:
1. Lower alcohols are miscible with water and organic solvents, and solubility in water decreases with increasing molecular weight.

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: (d) None of these

Explanation:
1. Alcohols form crystalline addition compounds (alcoholates) with anhydrous \( \ce{CaCl2} \), so it cannot be used to dry them.

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) 1°, 2° and 3° alcohols

Explanation:
1. Rate of turbidity formation with \( \ce{ZnCl2/HCl} \) differs among 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: (d) Phenol

Explanation:
1. The -OH group is strongly activating, so phenol nitrates even with dilute \( \ce{HNO3} \).

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: (d) \( \ce{C6H6} \), \( \ce{C6H5NO2} \) and hydrazobenzene

Explanation:
1. Zinc dust removes OH to give benzene; nitration gives nitrobenzene; Zn in alkaline medium reduces it to hydrazobenzene.

Teacher's Note:
1. Nitro groups at ortho and para positions increase the acidity of phenol, and ortho-nitrophenol is steam volatile.
2. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (b) Salicylic acid

Explanation:
1. Reimer-Tiemann with \( \ce{CCl4} \) gives the ortho-\( \ce{CCl3} \) compound, hydrolysed to salicylic acid.

Teacher's Note:
1. Reimer-Tiemann reaction formylates phenol at the ortho position through dichlorocarbene; with CCl4 it gives salicylic acid.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (b) p > m > o

Explanation:
1. p-Nitrophenol (387 K) has strong intermolecular hydrogen bonding, m (370 K) less, and the chelated ortho isomer (318 K) the least.

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: (b) Methyl alcohol

Explanation:
1. Methanol has only one carbon and cannot form an alkene by dehydration.

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. Hydrogen bonding explains most physical trends: boiling point, solubility and volatility of ethers, ortho-nitrophenol.

Answer: (a) \( \ce{C2H5NH2} \)

Explanation:
1. Alcohol plus ammonia over alumina gives amines.

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) The OH of the acid is replaced by the OR group of the alcohol

Explanation:
1. The acid loses OH and the alcohol loses H (acyl-oxygen fission), so the alkoxy group takes the place of OH in the acid.

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) Primary alcohol

Explanation:
1. A primary alcohol (ethanol) is oxidised to aldehyde (Tollens-positive) and then to 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) Reimer-Tiemann reaction

Explanation:
1. Dichlorocarbene is the electrophile.

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) \( \ce{C2H5OH} \)

Explanation:
1. Primary alcohols give nitrolic acids, red with NaOH; secondary give blue pseudonitroles; tertiary give no colour.

Teacher's Note:
1. Nitro groups at ortho and para positions increase the acidity of phenol, and ortho-nitrophenol is steam volatile.
2. Dehydration and substitution with HX both go through carbocations, so 3 degree > 2 degree > 1 degree.

Answer: (b) Faster

Explanation:
1. The OH group activates the ring toward electrophiles.

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) o-Nitrophenol and p-nitrophenol

Explanation:
1. Dilute nitric acid gives a mixture of ortho and para isomers.

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: (b) 3 mol

Explanation:
1. Three bromine atoms are introduced, each from a molecule of \( \ce{Br2} \) (substitution releases HBr), so 3 mol \( \ce{Br2} \) are used.

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: (a) Reimer-Tiemann's reaction

Explanation:
1. Reimer-Tiemann reaction.

Teacher's Note:
1. Reimer-Tiemann reaction formylates phenol at the ortho position through dichlorocarbene; with CCl4 it gives salicylic acid.
2. Compare acidity through the stability of the conjugate base: electron-withdrawing groups strengthen the acid, electron-donating groups weaken it.