JEE Chemistry Alcohols Phenols and Ethers MCQs Set 10

Chemistry Objective Questions and Answers: Alcohols Phenols and Ethers

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Multiple Choice Questions

Question 1: The compounds A, B and C in the sequence CH3CH2OH → (PBr3) A → (alc. KOH, heat) B → (Br2) C are, respectively:
(a) C2H5Br, CH3CH2OH, CH3CHBr2
(b) C2H5Br, HC≡CH, CH2=CHBr
(c) C2H5Br, CH2=CH2, BrCH2–CH2Br
(d) C2H5Br, CH3CH2OH, BrCH2–CH2Br
Show Answer & Explanation

Answer: (c) C2H5Br, CH2=CH2, BrCH2–CH2Br

Explanation:
1. PBr3 converts ethanol into bromoethane (A).
2. Alcoholic KOH eliminates HBr to give ethene (B).
3. Bromine adds across the double bond to give 1,2-dibromoethane (C).

Teacher's Note:
1. PBr3 replaces –OH by –Br, alc. KOH eliminates HBr and Br2 adds to alkenes.
2. Follow each step separately.

Question 2: Primary amines can be converted into alkanols by reaction with:
(a) aqueous HCl
(b) Tilden's reagent
(c) NaNO2 and dilute HCl
(d) hypochlorous acid
Show Answer & Explanation

Answer: (c) NaNO2 and dilute HCl

Explanation:
1. NaNO2 and dilute HCl produce nitrous acid in situ. A primary aliphatic amine gives an unstable diazonium salt, which is hydrolysed to the alcohol with release of N2: RNH2 + HNO2 → ROH + N2 + H2O.

Teacher's Note:
1. Nitrous acid converts RNH2 into ROH.
2. Aromatic amines give stable diazonium salts at 273 K.

Question 3: Which of the following alcohols gives a red colour in Victor Meyer's test?
(a) CH3CH2CH2OH
(b) (CH3)2CHOH
(c) (CH3)3COH
(d) CH3CH2CH(OH)CH3
Show Answer & Explanation

Answer: (a) CH3CH2CH2OH

Explanation:
1. Victor Meyer's test: the alcohol is converted to the iodide, then the nitro compound, and finally treated with HNO2 and NaOH. Primary alcohols give a red colour (nitrolic acid), secondary give blue and tertiary give no colour.

Teacher's Note:
1. Primary red, secondary blue, tertiary colourless.

Question 4:
Assertion (A): Chlorobenzene and benzyl chloride can be distinguished by treatment with AgNO3.
Reason (R): Chlorobenzene does not give a white precipitate with AgNO3 due to delocalisation of electrons.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Benzyl chloride gives white AgCl on warming with AgNO3. The C–Cl bond in chlorobenzene has partial double bond character because of delocalisation of the chlorine lone pair into the ring, so Cl− is not released.
2. Hence both statements are true and R explains A.

Teacher's Note:
1. Aryl halides do not ionise easily.
2. Benzylic halides give stable carbocations.

Question 5: Which of the following does NOT turn orange chromic acid to green?
(a) a primary alcohol
(b) a secondary alcohol
(c) a tertiary alcohol
(d) allyl alcohol
Show Answer & Explanation

Answer: (c) a tertiary alcohol

Explanation:
1. Primary and secondary alcohols are oxidised, so Cr(VI) (orange) changes to Cr(III) (green). Tertiary alcohols have no hydrogen on the carbon bearing –OH and are not oxidised under these conditions.

Teacher's Note:
1. This is a quick test to separate 3° alcohols from the others.

Question 6: Primary, secondary and tertiary alcohols can be distinguished by:
(a) Reimer–Tiemann reaction
(b) Tollen's reagent
(c) Lucas test
(d) Lassaigne's test
Show Answer & Explanation

Answer: (c) Lucas test

Explanation:
1. Lucas reagent (conc. HCl + anhydrous ZnCl2) gives turbidity immediately with 3° alcohols, in about five minutes with 2° alcohols, and only on heating with 1° alcohols.

Teacher's Note:
1. The reaction rate follows the stability of the carbocation.

Question 7: Ethanol is heated with concentrated H2SO4. Which of the following cannot be formed as a product under any conditions?
(a) C2H5OSO3H
(b) H2C=CH2
(c) HC≡CH
(d) CH3CH2OCH2CH3
Show Answer & Explanation

Answer: (c) HC≡CH

Explanation:
1. At low temperature ethyl hydrogen sulphate forms, at 413 K diethyl ether forms and at 443 K ethene forms. Ethyne cannot be formed by dehydration.

Teacher's Note:
1. Dehydration of an alcohol gives an alkene, not an alkyne.

Question 8:
Assertion (A): Phenol is a stronger acid than alcohols.
Reason (R): Phenoxide ion is stabilised by delocalisation of the negative charge into the ring, whereas alkoxide ions are not.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are true. The delocalisation of charge over the ring stabilises the phenoxide ion, which makes phenol more acidic than alcohols, so R explains A.

Teacher's Note:
1. Stability of the conjugate base decides acid strength.

Question 9: Deoxygenation of phenol (conversion to benzene) can be achieved by distillation with:
(a) Raney nickel
(b) lithium aluminium hydride
(c) sodium borohydride
(d) zinc dust
Show Answer & Explanation

Answer: (d) zinc dust

Explanation:
1. C6H5OH + Zn → C6H6 + ZnO. Zinc dust removes the oxygen as zinc oxide.

Teacher's Note:
1. Zinc dust distillation is a classical method for phenol to benzene.

Question 10: Which of the following compounds shows intramolecular hydrogen bonding?
(a) p-nitrophenol
(b) ethanol
(c) o-nitrophenol
(d) methanamine
Show Answer & Explanation

Answer: (c) o-nitrophenol

Explanation:
1. In o-nitrophenol the –OH is next to the –NO2 group, so a six-membered chelate ring forms through an internal O–H···O bond. p-Nitrophenol and ethanol show intermolecular bonding.

Teacher's Note:
1. Intramolecular H bonding makes o-nitrophenol steam volatile and less soluble in water.

Question 11: Sodium phenoxide reacts with acetyl chloride, CH3COCl, to give:
(a) methyl benzoate
(b) 2-hydroxyacetophenone
(c) phenyl acetate
(d) chlorobenzene
Show Answer & Explanation

Answer: (c) phenyl acetate

Explanation:
1. The phenoxide ion attacks the acyl carbon and chloride leaves: C6H5ONa + CH3COCl → C6H5OCOCH3 + NaCl. The product is phenyl acetate (an ester).

Teacher's Note:
1. This is O-acylation of a phenol.

Question 12:
Assertion (A): Addition of CH3OH to phenylmagnesium bromide gives benzene.
Reason (R): CH3OH is a stronger acid than benzene.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. C6H5MgBr is the conjugate base of benzene, a very weak acid. It takes a proton from the stronger acid CH3OH to give benzene. Both statements are true and R explains A.

Teacher's Note:
1. A strong base abstracts a proton from a stronger acid.

Question 13: Phenol can be distinguished from ethanol by all of the following reagents EXCEPT:
(a) iodine and alkali
(b) neutral ferric chloride
(c) acetyl chloride
(d) bromine water
Show Answer & Explanation

Answer: (c) acetyl chloride

Explanation:
1. Ethanol gives yellow iodoform with I2/NaOH while phenol does not; phenol gives a violet colour with neutral FeCl3 and a white precipitate of 2,4,6-tribromophenol with bromine water. Both react with acetyl chloride to give esters, so it cannot distinguish them.

Teacher's Note:
1. A reagent distinguishes only when the two compounds behave differently.

Question 14: Phenol on heating with NaNO2 and a few drops of conc. H2SO4 mainly gives:
(a) p-nitrophenol
(b) p-nitrosophenol
(c) o-nitrophenol
(d) m-nitrosophenol
Show Answer & Explanation

Answer: (b) p-nitrosophenol

Explanation:
1. NaNO2 and H2SO4 generate NO+, which attacks the para position of phenol (Liebermann nitroso reaction) to give p-nitrosophenol.

Teacher's Note:
1. Nitrosation, not nitration, takes place with nitrous acid.

Question 15: Phenol and benzoic acid can be distinguished by:
(a) Lucas reagent
(b) Victor Meyer test
(c) caustic soda
(d) sodium bicarbonate
Show Answer & Explanation

Answer: (d) sodium bicarbonate

Explanation:
1. Benzoic acid is a stronger acid than carbonic acid and liberates CO2 from NaHCO3 with brisk effervescence. Phenol is weaker than carbonic acid and does not react. Both dissolve in NaOH.

Teacher's Note:
1. Carboxylic acids dissolve in NaHCO3; phenols do not.

Question 16: Phenol can be distinguished from ethanol by the reaction with:
(a) Na
(b) Br2 in water
(c) neutral AlCl3
(d) both Na and Br2 water
Show Answer & Explanation

Answer: (b) Br2 in water

Explanation:
1. Phenol gives a white precipitate of 2,4,6-tribromophenol with bromine water; ethanol decolourises it only slowly or not at all. Both liberate H2 with sodium.

Teacher's Note:
1. Reaction with sodium is common to all compounds having an –OH group.

Question 17: In the sequence cyclohexanol → (CrO3) A → (CH3NH2) B → (H2/Pd) C, the final product C is:
(a) cyclohexanone
(b) 1-(methylamino)cyclohexanol
(c) N-methylcyclohex-1-enamine
(d) N-methylcyclohexanamine
Show Answer & Explanation

Answer: (d) N-methylcyclohexanamine

Explanation:
1. CrO3 oxidises the secondary alcohol to cyclohexanone.
2. CH3NH2 condenses with the ketone to form the imine, N-methylcyclohexanimine.
3. H2/Pd reduces the C=N bond to give N-methylcyclohexanamine.

Teacher's Note:
1. This is reductive amination of a ketone.

Question 18: In the sequence propan-2-ol → (PBr3) X → (Mg, ether) Y → (H2O/H+), the final product is:
(a) 2-methylpropan-2-ol
(b) propan-1-ol
(c) diisopropyl ether
(d) propane
Show Answer & Explanation

Answer: (d) propane

Explanation:
1. Propan-2-ol gives 2-bromopropane (X).
2. Magnesium gives the Grignard reagent (CH3)2CHMgBr (Y).
3. Water protonates the Grignard reagent to give propane.

Teacher's Note:
1. Grignard reagents are destroyed by water and alcohols to give alkanes.

Question 19: In the sequence A → (HBr) B → (C2H5ONa) ethoxyethane, A and B are:
(a) C2H6, C2H5Br
(b) CH4, CH3Br
(c) CH2=CH2, C2H5Br
(d) HC≡CH, CH2=CHBr
Show Answer & Explanation

Answer: (c) CH2=CH2, C2H5Br

Explanation:
1. Ethene adds HBr to give bromoethane (B). Williamson synthesis: C2H5Br + C2H5ONa → C2H5OC2H5 + NaBr.

Teacher's Note:
1. Williamson synthesis combines an alkyl halide with an alkoxide.

Question 20: What is formed when diethyl ether is treated with hot concentrated HI?
(a) ethyl alcohol
(b) ethyl iodide
(c) methyl iodide
(d) methanol
Show Answer & Explanation

Answer: (b) ethyl iodide

Explanation:
1. C2H5OC2H5 + 2HI → 2C2H5I + H2O (excess hot HI cleaves both C–O bonds).

Teacher's Note:
1. With one mole of HI the products are C2H5I and C2H5OH.

Question 21: Which of the following does NOT react with aqueous NaOH?
(a) phenol
(b) benzoic acid
(c) CH3COOH
(d) anisole (C6H5OCH3)
Show Answer & Explanation

Answer: (d) anisole (C6H5OCH3)

Explanation:
1. Phenol, benzoic acid and acetic acid are acids and form salts with NaOH. Anisole is a neutral ether and does not react.

Teacher's Note:
1. Only acidic compounds react with aqueous alkali.

Question 22: Methanol can be distinguished from ethanol by all of the following EXCEPT:
(a) reaction with iodine and alkali
(b) reaction with salicylic acid and conc. H2SO4
(c) reaction with Lucas reagent
(d) boiling point
Show Answer & Explanation

Answer: (c) reaction with Lucas reagent

Explanation:
1. Ethanol gives iodoform and methanol does not. Methanol gives oil of wintergreen with salicylic acid, and the boiling points differ (338 K and 351 K).
2. Both are primary alcohols and give no turbidity at room temperature with Lucas reagent.

Teacher's Note:
1. A test is useful only if the two compounds react differently.

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Alcohols Phenols and Ethers Objective Questions & Solutions for JEE Chemistry

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