Download JEE MCQs for JEE Chemistry: Alcohols Phenols and Ethers
Review structured MCQ sets for JEE Chemistry Alcohols Phenols and Ethers. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Chapter-wise Objective Questions: Alcohols Phenols and Ethers
Navigate directly to the 50 objective questions for Alcohols Phenols and Ethers using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.
Multiple Choice Questions
(a) \( \ce{C2H5Br} \), \( \ce{CH3CH2OH} \), \( \ce{CH3CHBr2} \)
(b) \( \ce{C2H5Br} \), \( \ce{CH#CH} \), \( \ce{CH2=CHBr} \)
(c) \( \ce{C2H5Br} \), \( \ce{CH2=CH2} \), \( \ce{CH2Br-CH2Br} \)
(d) \( \ce{C2H5Br} \), \( \ce{CH3CH2OH} \), \( \ce{BrCH2-CH2Br} \)
Show Answer & Explanation
Answer: (c) \( \ce{C2H5Br} \), \( \ce{CH2=CH2} \), \( \ce{CH2Br-CH2Br} \)
Explanation:
1. PBr3 converts ethanol into bromoethane (A).
2. Alcoholic KOH removes HBr to give ethene (B).
3. Bromine adds across the double bond to give 1,2-dibromoethane (C).
Teacher's Note:
1. Alcoholic KOH gives elimination.
2. Aqueous KOH would give substitution.
(a) aqueous HCl
(b) Tilden's reagent
(c) \( \ce{NaNO2} \) and dilute \( \ce{HCl} \)
(d) hypochlorous acid
Show Answer & Explanation
Answer: (c) \( \ce{NaNO2} \) and dilute \( \ce{HCl} \)
Explanation:
1. Primary aliphatic amines react with nitrous acid (NaNO2 + HCl) to give unstable diazonium salts.
2. These lose nitrogen and react with water to give alcohols.
Teacher's Note:
1. Nitrous acid is generated in situ.
2. This is the standard route from -NH2 to -OH.
(a) \( \ce{CH3CH2CH2OH} \)
(b) \( \ce{(CH3)2CHOH} \)
(c) \( \ce{(CH3)3COH} \)
(d) \( \ce{CH3CH(OH)CH2CH3} \)
Show Answer & Explanation
Answer: (a) \( \ce{CH3CH2CH2OH} \)
Explanation:
1. The Victor Meyer test converts the alcohol to nitroalkane, then treats it with nitrous acid.
2. Primary nitroalkanes form nitrolic acid (red colour); secondary give blue; tertiary give no colour. Propan-1-ol is primary.
Teacher's Note:
1. Red = primary, blue = secondary, colourless = tertiary.
(a) bitter almonds
(b) oil of wintergreen
(c) rotten eggs
(d) mustard oil
Show Answer & Explanation
Answer: (b) oil of wintergreen
Explanation:
1. Esterification gives methyl salicylate.
2. Methyl salicylate has the smell of oil of wintergreen.
Teacher's Note:
1. Methyl salicylate is used in pain-relief balms.
(a) 1 degree alcohol
(b) 2 degree alcohol
(c) 3 degree alcohol
(d) allyl alcohol
Show Answer & Explanation
Answer: (c) 3 degree alcohol
Explanation:
1. Chromic acid oxidises primary and secondary alcohols (orange Cr(VI) to green Cr(III)).
2. A tertiary alcohol has no hydrogen on the carbon carrying OH and resists oxidation.
Teacher's Note:
1. This is a quick test for tertiary alcohols.
(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 (anhydrous ZnCl2 in conc. HCl) forms an alkyl chloride at a rate that depends on the alcohol class.
2. Tertiary alcohols turn turbid immediately, secondary in about 5 minutes and primary on heating.
Teacher's Note:
1. Rate follows stability of the carbocation.
(a) \( \ce{C2H5OSO3H} \)
(b) \( \ce{H2C=CH2} \)
(c) \( \ce{HC#CH} \)
(d) \( \ce{CH3CH2-O-CH2CH3} \)
Show Answer & Explanation
Answer: (c) \( \ce{HC#CH} \)
Explanation:
1. Cold conc. H2SO4 gives ethyl hydrogen sulphate, 140 degrees C gives diethyl ether and 170 degrees C gives ethene.
2. Ethyne cannot be formed, as a triple bond needs two eliminations.
Teacher's Note:
1. Temperature decides between ether and alkene.
(a) 2-nitrophenol
(b) 4-nitrophenol
(c) 3-nitrophenol
(d) 2,4,6-trinitrophenol (picric acid)
Show Answer & Explanation
Answer: (d) 2,4,6-trinitrophenol (picric acid)
Explanation:
1. The -OH group strongly activates the ring at the ortho and para positions.
2. Conc. HNO3 introduces three nitro groups to give picric acid.
Teacher's Note:
1. Dilute HNO3 gives a mixture of o- and p-nitrophenol.
(a) Raney nickel
(b) lithium aluminium hydride
(c) sodium borohydride
(d) zinc dust
Show Answer & Explanation
Answer: (d) zinc dust
Explanation:
1. Heating phenol with zinc dust removes the oxygen as zinc oxide and gives benzene.
Teacher's Note:
1. Zinc dust distillation is the classic reduction of phenol.
(a) p-nitrophenol
(b) ethanol
(c) o-nitrophenol
(d) methanamine
Show Answer & Explanation
Answer: (c) o-nitrophenol
Explanation:
1. In o-nitrophenol, the -OH and -NO2 groups are adjacent so they form a hydrogen bond within the same molecule (chelation).
2. In p-nitrophenol the groups are too far apart.
Teacher's Note:
1. Intramolecular H-bonding lowers the boiling point and makes it steam volatile.
(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 displaces chloride.
2. The product is the ester, phenyl acetate.
Teacher's Note:
1. This is O-acylation.
(a) Schotten-Baumann reaction
(b) Kolbe-Schmidt reaction
(c) Reimer-Tiemann reaction
(d) Lederer-Manasse reaction
Show Answer & Explanation
Answer: (b) Kolbe-Schmidt reaction
Explanation:
1. Heating sodium phenoxide with CO2 under pressure gives the ortho-hydroxybenzoate.
2. This is the Kolbe-Schmidt reaction.
Teacher's Note:
1. Acidification gives salicylic acid.
(a) iodine and alkali
(b) ferric chloride
(c) acetyl chloride
(d) bromine water
Show Answer & Explanation
Answer: (c) acetyl chloride
Explanation:
1. Ethanol gives iodoform with I2/NaOH but phenol does not.
2. Phenol gives a violet colour with FeCl3 and a white precipitate with bromine water; ethanol does not.
3. Both give esters with acetyl chloride, so it cannot distinguish them.
Teacher's Note:
1. Choose reagents that react with only one of them.
(a) o-cresol
(b) m-cresol
(c) anisole
(d) p-cresol
Show Answer & Explanation
Answer: (d) p-cresol
Explanation:
1. Friedel-Crafts alkylation occurs at the ortho and para positions, the para product being the major one because of less steric hindrance.
Teacher's Note:
1. The -OH group is ortho/para directing.
(a) p-nitrophenol
(b) p-nitrosophenol
(c) o-nitrophenol
(d) m-nitrophenol
Show Answer & Explanation
Answer: (b) p-nitrosophenol
Explanation:
1. Nitrous acid is formed in situ and nitrosates the ring at the para position.
2. This is the first step of the Liebermann nitroso reaction.
Teacher's Note:
1. The product shows green, then red, then blue colours.
(a) Na
(b) \( \ce{Br2} \)/water
(c) neutral \( \ce{AlCl3} \)
(d) 1 and 2 both
Show Answer & Explanation
Answer: (b) \( \ce{Br2} \)/water
Explanation:
1. Phenol decolourises bromine water, giving 2,4,6-tribromophenol.
2. Both react with sodium, so this does not distinguish them.
Teacher's Note:
1. Neutral FeCl3 (not AlCl3) gives the violet colour.
(a) 1-bromo-1-methylcyclopentane
(b) (bromomethyl)cyclopentane
(c) bromocyclohexane
(d) a dibromocyclohexane
Show Answer & Explanation
Answer: (c) bromocyclohexane
Explanation:
1. Protonation and loss of water give a primary carbocation, which rearranges by ring expansion to the more stable cyclohexyl cation.
2. This gives cyclohexene (A); HBr adds to give bromocyclohexane (B).
Teacher's Note:
1. Ring expansion relieves strain and gives a secondary cation.
(a) cyclohexanone
(b) 1-(methylamino)cyclohexan-1-ol
(c) an enamine, N-methylcyclohex-1-enamine
(d) N-methylcyclohexylamine
Show Answer & Explanation
Answer: (d) N-methylcyclohexylamine
Explanation:
1. \( \ce{CrO3} \) oxidises cyclohexanol to cyclohexanone.
2. Methylamine gives an imine, and \( \ce{H2/Pd} \) reduces it to N-methylcyclohexylamine.
Teacher's Note:
1. This is reductive amination.
(a) \( \ce{C2H6} \), \( \ce{C2H5Br} \)
(b) \( \ce{CH4} \), \( \ce{CH3Br} \)
(c) \( \ce{CH2=CH2} \), \( \ce{C2H5Br} \)
(d) \( \ce{CH#CH} \), \( \ce{CH2=CHBr} \)
Show Answer & Explanation
Answer: (c) \( \ce{CH2=CH2} \), \( \ce{C2H5Br} \)
Explanation:
1. Ethene adds HBr to give bromoethane (B).
2. Bromoethane and sodium ethoxide give ethoxyethane by Williamson synthesis.
Teacher's Note:
1. Ethoxyethane = diethyl ether.
(a) \( \ce{CH3CH2OCH2CH3} \)
(b) \( \ce{PhCH2OCH2Ph} \)
(c) \( \ce{PhCH2OCH2CH3} \)
(d) \( \ce{PhCH2-O-CH2-O-CH3} \)
Show Answer & Explanation
Answer: (d) \( \ce{PhCH2-O-CH2-O-CH3} \)
Explanation:
1. Bimolecular dehydration of the two alcohols gives three ethers: diethyl, dibenzyl and benzyl ethyl ether.
2. The acetal-type compound in option (d) cannot form.
Teacher's Note:
1. Mixed alcohols give a mixture of three ethers.
(a) \( \ce{C2H5-O-CH(CH3)-O^+H} \)
(b) \( \ce{CH3CH2-O^+(H)-CH2CH3} \)
(c) \( \ce{(C2H5)2O -> O} \)
(d) \( \ce{CH3CH2-O-CH2CH2-O^+(H)-O-H} \)
Show Answer & Explanation
Answer: (b) \( \ce{CH3CH2-O^+(H)-CH2CH3} \)
Explanation:
1. An ether oxygen has lone pairs and accepts a proton to form a trivalent oxonium ion, R-O+(H)-R.
Teacher's Note:
1. Ethers behave as Lewis bases in acids.
(a) phenol
(b) benzoic acid
(c) \( \ce{CH3COOH} \)
(d) \( \ce{CH3-O-C6H5} \)
Show Answer & Explanation
Answer: (d) \( \ce{CH3-O-C6H5} \)
Explanation:
1. Phenol, benzoic acid and acetic acid are acidic and form salts with NaOH.
2. Anisole is a neutral ether and does not react.
Teacher's Note:
1. Only acidic hydrogens react with NaOH.
(a) \( \ce{CH3CH2Br} \), \( \ce{CH3CH2MgBr} \), \( \ce{(CH3)3C-OH} \)
(b) \( \ce{CH3CH2Br} \), \( \ce{(CH3CH2)2Mg} \), \( \ce{(CH3)2CHCH2OH} \)
(c) \( \ce{CH3CH2Br} \), \( \ce{CH3CH2MgBr} \), \( \ce{CH3CH(OH)CH2CH3} \)
(d) \( \ce{CH3CHBr2} \), \( \ce{CH3CH(MgBr)2} \), \( \ce{CH3CH(OH)CH3} \)
Show Answer & Explanation
Answer: (c) \( \ce{CH3CH2Br} \), \( \ce{CH3CH2MgBr} \), \( \ce{CH3CH(OH)CH2CH3} \)
Explanation:
1. Ethanol gives bromoethane, which with Mg in ether gives ethylmagnesium bromide.
2. The Grignard reagent adds to acetaldehyde, and hydrolysis gives the secondary alcohol, butan-2-ol.
Teacher's Note:
1. Aldehyde (other than HCHO) + Grignard gives a secondary alcohol.
Assertion (A): Chlorobenzene and benzyl chloride can be distinguished by treatment with \( \ce{AgNO3} \).
Reason (R): Chlorobenzene does not give a white precipitate with \( \ce{AgNO3} \) due to resonance.
(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 a white precipitate of AgCl as it ionises easily.
2. In chlorobenzene the C-Cl bond has partial double bond character through resonance, so it does not ionise. R explains A.
Teacher's Note:
1. Aryl halides are inert to AgNO3.
Assertion (A): Phenol is a stronger acid than alcohols.
Reason (R): Phenol is stabilised by resonance whereas alcohols 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. The phenoxide ion is stabilised by delocalisation of the charge over the ring.
2. Alkoxide ions have no such stabilisation. R explains A.
Teacher's Note:
1. Resonance stabilisation of the conjugate base decides acidity.
Assertion (A): Addition of \( \ce{CH3OH} \) to phenyl magnesium bromide gives benzene.
Reason (R): \( \ce{CH3OH} \) 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. Grignard reagents are strong bases and take a proton from any acid stronger than the hydrocarbon.
2. Methanol is more acidic than benzene, so benzene forms. R explains A.
Teacher's Note:
1. Grignard reagents react with compounds having acidic H.
Assertion (A): Ethers behave as bases in the presence of mineral acids.
Reason (R): Ethers have lone pairs of electrons on oxygen.
(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. The lone pairs on oxygen can accept a proton to form oxonium salts.
2. So R explains A.
Teacher's Note:
1. Lone pairs make ethers Lewis bases.
Assertion (A): Benzyl phenyl ether forms phenol and benzyl iodide with HI.
Reason (R): Benzyl carbocation is more stable than phenyl carbocation.
(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. The ether is protonated and the C-O bond breaks on the side that forms a stable cation (benzyl).
2. The aryl C-O bond does not break. R explains A.
Teacher's Note:
1. Phenyl cations are very unstable.
Assertion (A): m-Nitrophenol is less acidic than p-nitrophenol.
Reason (R): p-Nitrophenol has intermolecular hydrogen bonding.
(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: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The p-nitro group stabilises the phenoxide by resonance (-M), which the meta group cannot.
2. Intermolecular H-bonding in p-nitrophenol is true but does not explain the acidity. Both are true, R is not the explanation.
Teacher's Note:
1. Resonance is the cause of the greater acidity.
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FAQs
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