JEE Chemistry Alcohols Phenols and Ethers MCQs Set 11

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Multiple Choice Questions

Question 1: Ethanol on heating with acetic acid in the presence of a few drops of H2SO4 gives the smell of:
(a) oil of wintergreen
(b) oil of mustard
(c) an ester (fruity smell)
(d) oil of bitter almonds
Show Answer & Explanation

Answer: (c) an ester (fruity smell)

Explanation:
1. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O (esterification). Ethyl acetate has a pleasant fruity smell.

Teacher's Note:
1. Esterification is reversible and acid catalysed.

Question 2: In the reaction sequence CH3CH2OH → (HBr) A → (Mg) B → (i CH3CHO, ii H3O+) C, the compounds A, B and C are:
(a) CH3CH2Br, CH3CH2MgBr, (CH3)3C–OH
(b) CH3CH2Br, (CH3CH2)2Mg, (CH3)2CHCH2OH
(c) CH3CH2Br, CH3CH2MgBr, CH3CH(OH)CH2CH3
(d) CH3CHBr2, CH3CH(MgBr)2, CH3CH(OH)CH3
Show Answer & Explanation

Answer: (c) CH3CH2Br, CH3CH2MgBr, CH3CH(OH)CH2CH3

Explanation:
1. Ethanol gives bromoethane (A) and the Grignard reagent ethylmagnesium bromide (B).
2. The Grignard reagent adds to the carbonyl carbon of acetaldehyde, and hydrolysis gives the secondary alcohol butan-2-ol (C).

Teacher's Note:
1. Aldehydes (other than HCHO) give secondary alcohols with Grignard reagents.

Question 3: Which of the following reactions will NOT lead to a phenol?
(a) C6H5SO3Na fused with NaOH, then H+
(b) C6H5N2Cl boiled with water
(c) C6H5ONa + RX, then NaOH, heat
(d) sodium salicylate heated with NaOH and CaO
Show Answer & Explanation

Answer: (c) C6H5ONa + RX, then NaOH, heat

Explanation:
1. Sulphonate fusion, hydrolysis of benzenediazonium chloride and decarboxylation of salicylate (soda lime) all give phenol. Sodium phenoxide with an alkyl halide gives an ether, not a phenol.

Teacher's Note:
1. The reaction of phenoxide with RX is Williamson synthesis.

Question 4: Chlorobenzene reacts with NaOH at 633 K to give A. A with HCl gives B and with CH3I gives C. A, B and C are, respectively:
(a) C6H5OH, C6H5Cl, C6H5CH3
(b) C6H5ONa, C6H5OH, C6H5OCH3
(c) C6H5ONa, C6H5Cl, C6H5I
(d) C6H5ONa, C6H5OH, C6H5I
Show Answer & Explanation

Answer: (b) C6H5ONa, C6H5OH, C6H5OCH3

Explanation:
1. Dow's process gives sodium phenoxide (A).
2. Acidification gives phenol (B).
3. Reaction with CH3I (Williamson) gives anisole (C).

Teacher's Note:
1. Phenoxide ion is a nucleophile that reacts with alkyl halides to form ethers.

Question 5: In the sequence sodium benzenesulphonate → (NaOH, 623 K) A → (CH3I) B → (HI) C + D, the compounds A, B, C and D are:
(a) sodium phenoxide, anisole, C6H5OH, CH3OH
(b) sodium phenoxide, phenetole, C2H5I, C6H5OH
(c) sodium phenoxide, anisole, C6H5OH, CH3I
(d) sodium phenoxide, phenetole, C6H5I, C2H5OH
Show Answer & Explanation

Answer: (c) sodium phenoxide, anisole, C6H5OH, CH3I

Explanation:
1. Fusion with NaOH gives sodium phenoxide (A).
2. CH3I forms anisole (B).
3. HI cleaves anisole into phenol (C) and methyl iodide (D), because the aryl C–O bond is too strong to break.

Teacher's Note:
1. Aryl alkyl ethers always give phenol and alkyl iodide with HI.

Question 6: Acetaldehyde is treated with CH3MgBr and then hydrolysed to give A. A on heating with H2SO4 gives B, and B on hydroboration–oxidation gives C. A and C are:
(a) identical
(b) functional isomers
(c) positional isomers
(d) optical isomers
Show Answer & Explanation

Answer: (c) positional isomers

Explanation:
1. A is propan-2-ol and B is propene.
2. Hydroboration–oxidation of propene is anti-Markovnikov, giving propan-1-ol (C). Propan-2-ol and propan-1-ol differ in the position of –OH, so they are positional isomers.

Teacher's Note:
1. Hydroboration–oxidation follows anti-Markovnikov addition.

Question 7: Propan-1-ol is dehydrated; the product is hydrated and then oxidised. The final compound is:
(a) prop-2-yn-1-ol
(b) propanone (acetone)
(c) propane-1,2-diol
(d) propane-1,2-diol monoether
Show Answer & Explanation

Answer: (b) propanone (acetone)

Explanation:
1. Dehydration gives propene, Markovnikov hydration gives propan-2-ol, and oxidation of this secondary alcohol gives propanone.

Teacher's Note:
1. Markovnikov hydration of propene gives the secondary alcohol.

Question 8: Phenol is heated with a mixture of KBr and KBrO3 in acid. The major product is:
(a) 4-bromophenol
(b) 2,4,6-tribromophenol
(c) 2-bromophenol
(d) 3-bromophenol
Show Answer & Explanation

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

Explanation:
1. KBr and KBrO3 in acid produce bromine in situ: 5Br− + BrO3− + 6H+ → 3Br2 + 3H2O. Bromine water brominates phenol at all activated positions to give 2,4,6-tribromophenol.

Teacher's Note:
1. Bromine water brominates phenol at all o- and p-positions.

Question 9: Ethyl isobutyl ether, (CH3)2CHCH2OC2H5, is treated with HI. Which compounds are mainly formed?
(a) (CH3)2CHI and CH3CH2OH
(b) (CH3)2CHCH2I and CH3CH2OH
(c) (CH3)2CHCH2OH and CH3CH2I
(d) (CH3)2CHCH2I and CH3CH2I
Show Answer & Explanation

Answer: (c) (CH3)2CHCH2OH and CH3CH2I

Explanation:
1. With HI, cleavage occurs by SN2 at the less hindered alkyl group. The ethyl group is less hindered than the isobutyl group, so ethyl iodide and isobutyl alcohol form first.

Teacher's Note:
1. For primary alkyl groups, the iodide ion attacks the smaller group.

Question 10: Ethane-1,2-diol on treatment with periodic acid gives:
(a) formaldehyde (2 moles)
(b) carbon dioxide
(c) formic acid
(d) glyoxal
Show Answer & Explanation

Answer: (a) formaldehyde (2 moles)

Explanation:
1. Periodic acid cleaves the C–C bond of 1,2-diols: HOCH2CH2OH + HIO4 → 2HCHO + HIO3 + H2O.

Teacher's Note:
1. Malaprade reaction: 1,2-diols are cleaved by HIO4.

Question 11:
Assertion (A): C6H5CH(OH)CH3 gives benzoic acid with NaOBr followed by hydrolysis.
Reason (R): First dehydration occurs, then oxidation.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. The assertion is true: the secondary alcohol is oxidised by hypobromite to acetophenone, which undergoes the haloform reaction to give benzoate.
2. The reason is false, since oxidation (not dehydration) occurs first.

Teacher's Note:
1. Haloform reaction needs a CH3CO– or CH3CH(OH)– group.

Question 12: Glycerol is treated with excess HI. The product is:
(a) allyl iodide
(b) propene
(c) glyceryl triiodide
(d) 2-iodopropane
Show Answer & Explanation

Answer: (d) 2-iodopropane

Explanation:
1. Glycerol → 1,2,3-triiodopropane (unstable) → allyl iodide → on excess HI gives 2-iodopropane (Markovnikov addition).

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

Question 13: Phenol on treatment with CHCl3 and aqueous NaOH, followed by acidification, gives mainly:
(a) salicylaldehyde
(b) p-hydroxybenzaldehyde
(c) m-hydroxybenzaldehyde
(d) salicylic acid
Show Answer & Explanation

Answer: (a) salicylaldehyde

Explanation:
1. This is the Reimer–Tiemann reaction. Dichlorocarbene attacks the ortho position of the phenoxide ion and hydrolysis gives salicylaldehyde (2-hydroxybenzaldehyde) as the chief product.

Teacher's Note:
1. Reimer–Tiemann: phenol + CHCl3 + NaOH gives o-hydroxybenzaldehyde mainly.

Question 14: Which of the following will NOT be soluble in sodium hydrogen carbonate?
(a) 2,4,6-trinitrophenol
(b) benzoic acid
(c) o-nitrophenol
(d) benzenesulphonic acid
Show Answer & Explanation

Answer: (c) o-nitrophenol

Explanation:
1. NaHCO3 dissolves only acids stronger than carbonic acid: picric acid (pKa about 0.4), benzoic acid and benzenesulphonic acid. o-Nitrophenol (pKa about 7.2) is weaker than carbonic acid.

Teacher's Note:
1. Phenols (except picric acid) do not dissolve in NaHCO3.

Question 15:
Assertion (A): Methoxyethane has a lower boiling point than propanol.
Reason (R): Methoxyethane is more polar than propanol.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Propan-1-ol has intermolecular hydrogen bonding, while methoxyethane has only weak dipole attraction, so its boiling point is lower. The assertion is true.
2. The reason is false: methoxyethane is less polar and cannot form hydrogen bonds.

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

Question 16: Which one of the following reagents produces anisole from a suitable reactant pair?
(a) CH3CHO; RMgX
(b) C6H5OH; NaOH; CH3I
(c) C6H5OH; neutral FeCl3
(d) C6H5CH3; CH3COCl; AlCl3
Show Answer & Explanation

Answer: (b) C6H5OH; NaOH; CH3I

Explanation:
1. Phenol with NaOH gives sodium phenoxide, which reacts with CH3I (Williamson synthesis) to give anisole.

Teacher's Note:
1. Williamson synthesis forms ethers from alkoxide and alkyl halide.

Question 17: Styrene (C6H5CH=CH2) is converted into 2-phenylethanol by:
(a) NaBH4 in C2H5OH after epoxidation
(b) B2H6 followed by H2O2/OH−
(c) HgSO4/dil. H2SO4
(d) dilute H2SO4
Show Answer & Explanation

Answer: (b) B2H6 followed by H2O2/OH−

Explanation:
1. Hydroboration–oxidation adds H and OH across the double bond in an anti-Markovnikov way, giving the primary alcohol 2-phenylethanol.

Teacher's Note:
1. Mercuric sulphate and dilute acid give the Markovnikov alcohol.

Question 18: A reagent that cannot be used to convert an alcohol into an alkyl halide is:
(a) red P and Br2
(b) KI in acetone
(c) thionyl chloride
(d) HCl/ZnCl2
Show Answer & Explanation

Answer: (b) KI in acetone

Explanation:
1. Red P/Br2 (PBr3), SOCl2 and HCl/ZnCl2 all replace –OH by halogen. KI in acetone is used for the Finkelstein exchange of an alkyl chloride or bromide, not for alcohols.

Teacher's Note:
1. The –OH group is a poor leaving group; it must first be activated.

Question 19:
Assertion (A): Isopropylbenzene (cumene) reacts with O2 and acid to give phenol.
Reason (R): Initially isopropylbenzene is converted into 2-phenylpropan-2-ol.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. The assertion is true (cumene process).
2. The first step is formation of cumene hydroperoxide, not 2-phenylpropan-2-ol, so the reason is false.

Teacher's Note:
1. Cumene hydroperoxide rearranges in acid to give phenol and acetone.

Question 20: In the sequence benzene + CH3CH2CH2Cl (anhydrous AlCl3) → P → (i O2, ii H3O+/heat) → Q + R, the products P, Q and R are:
(a) propylbenzene, benzaldehyde, propan-1-ol
(b) propylbenzene, benzaldehyde, benzoic acid
(c) cumene, phenol, propan-2-ol
(d) cumene, phenol, acetone
Show Answer & Explanation

Answer: (d) cumene, phenol, acetone

Explanation:
1. The primary carbocation rearranges to the secondary isopropyl cation, giving isopropylbenzene (cumene, P).
2. Air oxidation gives cumene hydroperoxide, which on acidic cleavage gives phenol (Q) and acetone (R). This is the cumene process.

Teacher's Note:
1. Friedel–Crafts alkylation with n-propyl chloride gives cumene because of rearrangement.

Question 21: The compound A reacts with Na to give B and with PCl5 to give C. B and C react together to give diethyl ether. A, B and C are, in order:
(a) C2H5OH, C2H6, C2H5Cl
(b) C2H5OH, C2H5Cl, C2H5ONa
(c) C2H5Cl, C2H6, C2H5OH
(d) C2H5OH, C2H5ONa, C2H5Cl
Show Answer & Explanation

Answer: (d) C2H5OH, C2H5ONa, C2H5Cl

Explanation:
1. A is ethanol. 2C2H5OH + 2Na → 2C2H5ONa + H2 (B).
2. C2H5OH + PCl5 → C2H5Cl + POCl3 + HCl (C).
3. C2H5ONa + C2H5Cl → C2H5OC2H5 + NaCl (Williamson synthesis).

Teacher's Note:
1. Sodium ethoxide and ethyl chloride give diethyl ether.

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