JEE Chemistry Alcohols Phenols and Ethers MCQs Set 05

Answer: (c) \( \ce{C6H5ONa} \) + RX, NaOH, heat

Explanation:
1. Sodium phenoxide with an alkyl halide gives an ether (Williamson synthesis), not a phenol.
2. The others give phenol by alkali fusion, diazonium hydrolysis and decarboxylation.

Teacher's Note:
1. Check the product of each reaction.

Answer: (b) sodium phenoxide, phenol, anisole

Explanation:
1. Chlorobenzene gives sodium phenoxide under Dow's conditions.
2. HCl converts it to phenol, and methyl iodide gives anisole (Williamson).

Teacher's Note:
1. Aryl halides need drastic conditions for nucleophilic substitution.

Answer: (c) sodium phenate, anisole, \( \ce{C6H5OH} \), \( \ce{CH3I} \)

Explanation:
1. Alkali fusion gives sodium phenoxide (A).
2. Methyl iodide gives anisole (B).
3. HI cleaves the aryl alkyl ether to phenol and methyl iodide.

Teacher's Note:
1. In aryl alkyl ethers, the alkyl-oxygen bond breaks.

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

Explanation:
1. KBr and KBrO3 in acid generate bromine in aqueous medium.
2. Phenol in water reacts with all three activated positions to give 2,4,6-tribromophenol.

Teacher's Note:
1. Bromine water substitutes all ortho and para positions.

Answer: (b) \( \ce{CH3COCH3} \)

Explanation:
1. Dehydration gives propene; Markovnikov hydration gives propan-2-ol.
2. Oxidation of the secondary alcohol gives acetone.

Teacher's Note:
1. Hydration follows Markovnikov's rule.

Answer: (c) \( \ce{CH3CH2OH + CH3COOH} \) (with acid catalyst)

Explanation:
1. In esterification, the O-H bond of the alcohol breaks and the alcohol acts as a nucleophile.
2. In the other three, the C-O bond breaks as the halogen replaces OH.

Teacher's Note:
1. Esterification: acid loses OH, alcohol loses H.

Answer: (d) tert-butyl alcohol

Explanation:
1. More branching reduces the surface area of the hydrocarbon part.
2. tert-Butyl alcohol is the most compact isomer and the most soluble.

Teacher's Note:
1. Branching increases solubility and lowers boiling point.

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

Explanation:
1. PCl5 gives 1-chloropropane (A), alcoholic KOH gives propene (B).
2. Acid-catalysed hydration follows Markovnikov's rule to give propan-2-ol (C).

Teacher's Note:
1. Hydration of an alkene goes to the more substituted carbon.

Answer: (b) 3

Explanation:
1. The three isomers are catechol (1,2), resorcinol (1,3) and hydroquinone (1,4).

Teacher's Note:
1. The two OH groups can be ortho, meta or para.

Answer: (c) dichlorocarbene

Explanation:
1. Chloroform and hydroxide give the trichloromethyl carbanion, which loses chloride to form dichlorocarbene.
2. The electrophilic carbene attacks the ring of the phenoxide ion.

Teacher's Note:
1. Hydrolysis finally gives the -CHO group.

Answer: (b) 4-hydroxyazobenzene, \( \ce{C6H5-N=N-C6H4-OH} \)

Explanation:
1. Under mildly alkaline conditions, diazonium coupling occurs at the para position of phenol.
2. The product is p-hydroxyazobenzene, an orange dye.

Teacher's Note:
1. Coupling is an electrophilic substitution.

Answer: (b) \( \ce{B2H6} \), then \( \ce{H2O2/OH-} \)

Explanation:
1. Hydroboration-oxidation adds H and OH in an anti-Markovnikov manner.
2. So styrene gives 2-phenylethanol.

Teacher's Note:
1. Acid-catalysed hydration would give 1-phenylethanol.

Answer: (d) \( \ce{C2H5OH, C2H5ONa, C2H5Cl} \)

Explanation:
1. Ethanol (A) with sodium gives sodium ethoxide (B).
2. Ethanol with PCl5 gives chloroethane (C).
3. B and C give diethyl ether by Williamson synthesis.

Teacher's Note:
1. Williamson: alkoxide plus alkyl halide.

Answer: (d) isopropylbenzene (cumene), phenol, acetone

Explanation:
1. The n-propyl carbocation rearranges to the isopropyl cation, giving cumene (P).
2. Cumene hydroperoxide on acid treatment gives phenol (Q) and acetone (R).

Teacher's Note:
1. This is the cumene process for making phenol.

Answer: (b) \( \ce{C6H5OH} \); NaOH; \( \ce{CH3I} \)

Explanation:
1. Phenol and NaOH give sodium phenoxide, which attacks methyl iodide in a Williamson reaction to give anisole.

Teacher's Note:
1. Anisole = methoxybenzene.

Answer: (c) o-nitrophenol

Explanation:
1. Only acids stronger than carbonic acid dissolve in NaHCO3 with CO2 evolution.
2. Picric acid, benzoic acid and benzenesulphonic acid are strong enough; o-nitrophenol is not.

Teacher's Note:
1. Picric acid is much more acidic than ordinary phenols.

Answer: (b) phenol

Explanation:
1. Cleavage occurs at the alkyl-oxygen bond because the aryl C-O bond is strong (partial double bond).
2. Products are phenol and methyl iodide.

Teacher's Note:
1. Aryl alkyl ethers always give phenol and alkyl halide.

Answer: (a) m-toluidine (3-methylaniline)

Explanation:
1. Diazotisation of m-toluidine and hydrolysis of the diazonium salt places -OH exactly at the same position.
2. Direct substitution on toluene would not give the meta product.

Teacher's Note:
1. Diazonium salts are used to introduce OH at a specific position.

Answer: (d) 2-iodopropane

Explanation:
1. HI first gives 1,2,3-triiodopropane, which loses I2 to form allyl iodide.
2. Allyl iodide adds HI (Markovnikov) to give 2-iodopropane after reduction with more HI.

Teacher's Note:
1. Excess HI reduces glycerol all the way to 2-iodopropane.

Answer: (a) two molecules of formaldehyde

Explanation:
1. Periodic acid cleaves the C-C bond between two vicinal hydroxyl groups.
2. Each CH2OH group is oxidised to HCHO.

Teacher's Note:
1. HIO4 cleaves vicinal diols.

Answer: (d) benzoic acid

Explanation:
1. Zinc dust reduces phenol to benzene (X).
2. Friedel-Crafts methylation gives toluene (Y), and KMnO4 oxidises the side chain to benzoic acid (Z).

Teacher's Note:
1. Any alkyl side chain with an alpha H is oxidised to -COOH.

Answer: (b) -CHO

Explanation:
1. The Reimer-Tiemann reaction forms a dichloromethyl intermediate, which is hydrolysed to the aldehyde.
2. Salicylaldehyde is the main product.

Teacher's Note:
1. With CCl4 the product would be salicylic acid.

Answer: (c) reaction with Lucas reagent

Explanation:
1. Ethanol gives iodoform and methanol does not.
2. Methanol gives the oil of wintergreen smell with salicylic acid.
3. Both are primary alcohols and give no turbidity with Lucas reagent.

Teacher's Note:
1. Choose the test that behaves the same for both.

Answer: (c) A is true, but R is false.

Explanation:
1. Alkali fusion of sodium benzenesulphonate gives sodium phenoxide, so A is true.
2. The sulphonate group is displaced by hydroxide, so it is a reasonable leaving group under these drastic conditions. R is false.

Teacher's Note:
1. Fusion with NaOH gives phenol after acidification.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In Williamson synthesis the phenoxide attacks the alkyl halide.
2. The reverse combination (aryl halide plus ethoxide) does not work because aryl halides resist nucleophilic substitution. R explains A.

Teacher's Note:
1. Choose the alkyl halide and aryl oxide.

Answer: (c) A is true, but R is false.

Explanation:
1. Propanol forms intermolecular hydrogen bonds, but methoxyethane cannot.
2. So its boiling point is lower and A is true.
3. Methoxyethane is less polar, so R is false.

Teacher's Note:
1. Hydrogen bonding raises boiling point.

Answer: (c) A is true, but R is false.

Explanation:
1. Cumene is oxidised to cumene hydroperoxide, which rearranges with acid to phenol and acetone.
2. The first step gives the hydroperoxide, not the alcohol, so R is false.

Teacher's Note:
1. Learn the cumene process steps.

Answer: (d) A is false, but R is true.

Explanation:
1. Propanal is more polar and boils higher (about 49 degrees C) than methoxyethane (about 8 degrees C).
2. Both statements are false.

Teacher's Note:
1. Carbonyl compounds have dipole-dipole attractions.