Which of the following is an incorrect method of preparation of alcohols?
AHydroboration-oxidation of alkene (anti-Markovnikov addition of water)
BReaction of ketone with RMgBr (Grignard reagent) followed by hydrolysis
COzonolysis of an alkene followed by reduction
DReaction of alkyl halide with aqueous NaOH
✅ Correct: C
Ozonolysis of alkenes gives carbonyl compounds (aldehydes or ketones) — NOT directly alcohols. Ozonolysis cleaves the double bond. While ozonolysis followed by reduction (with NaBH₄) could give alcohols, simple ozonolysis (with H₂O₂ workup → oxidative) gives carboxylic acids. As a preparation of alcohols, ozonolysis is incorrect. Answer: C.
Q2
The water gas (CO + H₂) on reacting with cobalt catalyst forms:
AMethanoic acid (formic acid)
BMethanol
CMethanal (formaldehyde)
DEthanol
✅ Correct: B
CO + 2H₂ → CH₃OH (methanol) — industrial production using Cu/ZnO/Al₂O₃ catalyst at 250°C, 50-100 atm. This is not cobalt catalyst specifically for methanol; standard textbook: CO + H₂ → methanol with different catalysts. Standard answer: methanol.
Q3
Phenol (C₆H₅OH) reacts with concentrated HNO₃ and concentrated H₂SO₄ mixture at high temperature to give:
Ao-nitrophenol only
Bp-nitrophenol only
C2,4,6-trinitrophenol (picric acid)
Dm-nitrophenol
✅ Correct: C
Phenol with concentrated HNO₃/H₂SO₄ at high temperature undergoes trinitration to give 2,4,6-trinitrophenol (picric acid). OH group is a powerful ortho-para director. All three ortho/para positions get nitrated. With dilute HNO₃: mixture of o- and p-nitrophenol.
Q4
Phenolic group in phenol can be identified by which of the following tests?
ATollen's test
BPhthalein dye test (reaction with phthalic anhydride)
CCarbylamine test
DFehling's solution test
✅ Correct: B
The phthalein dye test (Liebermann-Burchard) involves heating phenol with phthalic anhydride in presence of conc. H₂SO₄ → forms phenolphthalein (turns pink/magenta in base). Alternatively, FeCl₃ gives violet/purple colour with phenols. Tollen's/Fehling's are for aldehydes; carbylamine is for primary amines.
Q5
Lucas test is used to distinguish between primary, secondary, and tertiary alcohols. Tertiary alcohols react with Lucas reagent (ZnCl₂/conc. HCl) at room temperature to give:
ANo reaction
BImmediate turbidity (within a few seconds)
CTurbidity after 5 minutes
DTurbidity only on heating
✅ Correct: B
Lucas test: ZnCl₂/conc. HCl at room temperature. Tertiary alcohols → immediate turbidity (within seconds) — most reactive because 3° carbocation forms easily. Secondary alcohols: turbidity in 5 min. Primary alcohols: no turbidity at room temperature (need heating).
Q6
The order of reactivity of alcohols toward dehydration (with conc. H₂SO₄) follows the order:
Phenol is more acidic than alcohol (ethanol) because:
APhenol has higher molecular weight
BThe phenoxide ion (C₆H₅O⁻) is stabilized by resonance with the benzene ring
CPhenol has an OH group directly bonded to benzene
DPhenol has a higher boiling point than ethanol
✅ Correct: B
Phenoxide ion (C₆H₅O⁻) is stabilized by resonance — the negative charge is delocalized over the benzene ring (ortho and para positions). Ethoxide ion (C₂H₅O⁻) has no such stabilization. More stable conjugate base → stronger acid. Hence phenol (pKa ≈ 10) is more acidic than ethanol (pKa ≈ 16).
Q8
Which of the following reactions is used to prepare ethers by the dehydration of alcohols?
AIntermolecular dehydration using conc. H₂SO₄ at 140°C
BIntramolecular dehydration using conc. H₂SO₄ at 170°C
CReaction with NaOH
DReaction with PCl₅
✅ Correct: A
Preparation of simple ethers: intermolecular dehydration at 140°C with conc. H₂SO₄: 2ROH → R−O−R + H₂O. At 170°C: intramolecular dehydration → alkene (E2). Temperature control determines the product: lower temp → ether, higher temp → alkene.
Q9
Kolbe's synthesis involves reaction of sodium phenoxide with CO₂ under high pressure at 125°C to give:
ABenzoic acid
BSodium salicylate (2-hydroxybenzoate)
CCarbonate ester of phenol
DSodium phenyl carbonate
✅ Correct: B
Kolbe's synthesis (Kolbe-Schmitt reaction): C₆H₅ONa + CO₂ → sodium salicylate (2-hydroxybenzoate, C₆H₄(OH)COONa). On acidification → salicylic acid. Used industrially: salicylic acid → aspirin (acetylsalicylic acid). The CO₂ adds to the ortho position of the phenoxide ion.
Q10
The reaction of ethanol with sodium metal produces:
ASodium oxide and ethylene
BSodium ethoxide and hydrogen gas
CSodium acetate and hydrogen gas
DEthyl sodium and water
✅ Correct: B
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑. Alcohols (and phenols) react with active metals like Na, K to give metal alkoxide and H₂ gas. This reaction shows the acidic nature of the O−H bond in alcohols (pKa ≈ 16-18). Phenols react more readily (pKa ≈ 10).
Q11
Reimer-Tiemann reaction involves conversion of phenol to:
ABenzoic acid
BSalicylaldehyde (2-hydroxybenzaldehyde)
CPicric acid
DSodium phenoxide
✅ Correct: B
Reimer-Tiemann reaction: Phenol + CHCl₃ + NaOH (aqueous) → 2-hydroxybenzaldehyde (salicylaldehyde). Mechanism: CHCl₃ + NaOH → :CCl₂ (dichlorocarbene) → electrophilic attack at ortho position of phenoxide → after hydrolysis → −CHO at ortho. Major product: o-hydroxybenzaldehyde.
Q12
The distinction between primary, secondary, and tertiary alcohols can be made using:
AFeCl₃ test
BVictor Meyer's test
CBaeyer's test
DFehling's test
✅ Correct: B
Victor Meyer's test: Primary alcohols → red colour (with nitrosoamine), secondary → blue colour, tertiary → no colour (colourless). This distinguishes all three classes. Process: alcohol → alkyl iodide (with I₂/P) → silver nitrite → nitroalkane → Victor Meyer's colour reactions.
Q13
In the reaction of phenol with bromine water, the product formed is:
ABromobenzene
Bo-bromophenol
Cp-bromophenol
D2,4,6-tribromophenol (white precipitate)
✅ Correct: D
Phenol reacts with bromine water (even without catalyst!) to give 2,4,6-tribromophenol as a white precipitate. The OH group strongly activates all three ortho/para positions. This reaction is so facile it can be used as a test for phenol (decolorisation of bromine water + white precipitate).
Q14
Williamson synthesis is used to prepare:
AAlcohols from alkyl halides
BEthers from alkyl halide and sodium alkoxide (or sodium phenoxide)
CEsters from alcohols and acids
DPhenols from aryl halides
✅ Correct: B
Williamson synthesis: R-ONa + R'X → R-O-R' + NaX. Sodium alkoxide (or sodium phenoxide) + alkyl halide → ether. Example: C₂H₅ONa + CH₃Br → C₂H₅OCH₃ + NaBr. Best for mixed (unsymmetrical) ethers. Uses SN2 mechanism → must use primary alkyl halide to avoid elimination.
Q15
Which of the following alcohols gives an iodoform test?
AMethanol
BEthanol
CPropan-1-ol
DButan-1-ol
✅ Correct: B
Iodoform test (I₂/NaOH): positive for CH₃CH(OH)− group (secondary alcohols with methyl next to OH) and ethanol (CH₃CH₂OH) — oxidized first to CH₃CHO → then iodoform. Also: propan-2-ol, butan-2-ol, methyl ketones. Methanol and primary alcohols with longer chain (propan-1-ol) are negative.
Q16
Catalytic dehydrogenation of a secondary alcohol (R₂CHOH) over Cu at 300°C gives:
Phenol is used as a disinfectant. This property is related to its:
AHigh boiling point
BAbility to denature proteins of bacteria
CAcidic nature
DAbility to form hydrogen bonds
✅ Correct: B
Phenol (carbolic acid) was the first antiseptic used by Lister (1865). Its disinfectant property is due to its ability to denature proteins of bacterial cell membranes — it disrupts protein structure and kills bacteria. Even dilute phenol solutions are effective antiseptics (Dettol contains chloroxylenol, a phenol derivative).
Q18
The product formed when diethyl ether (C₂H₅OC₂H₅) reacts with excess HI is:
AEthyl iodide only
BEthanol and ethyl iodide
CTwo moles of ethyl iodide
DDiethyl diiodide
✅ Correct: C
C₂H₅−O−C₂H₅ + HI → C₂H₅OH + C₂H₅I. With excess HI: C₂H₅OH + HI → C₂H₅I + H₂O. Net: C₂H₅OC₂H₅ + 2HI → 2C₂H₅I + H₂O. With excess HI, both molecules of ethanol formed are converted to ethyl iodide.
Q19
Which of the following gives a positive Tollen's test (silver mirror test)?
AAcetone (CH₃COCH₃)
BDiethyl ether
CAcetaldehyde (CH₃CHO)
DBenzophenone
✅ Correct: C
Tollen's test (Ag(NH₃)₂OH, silver-ammonia complex): positive for aldehydes only (not ketones). Aldehydes are oxidised to carboxylates → Ag⁺ reduced to Ag (silver mirror). CH₃CHO (acetaldehyde) gives positive test. Ketones (acetone, benzophenone) and ethers are negative.
Q20
The boiling point of ethanol (C₂H₅OH, 78°C) is much higher than dimethyl ether (CH₃OCH₃, −24°C) despite both having same molecular formula (C₂H₆O). This is because:
AEthanol has higher molecular weight
BEthanol molecules form intermolecular hydrogen bonds through O−H groups
CDimethyl ether has a different molecular formula
DEthanol has a larger dipole moment only
✅ Correct: B
Ethanol has an O−H group → forms strong intermolecular hydrogen bonds (O−H···O). More energy needed to break these bonds → much higher boiling point (78°C). Dimethyl ether (CH₃OCH₃) has no O−H group → cannot form H-bonds → only weak London dispersion forces → very low BP (−24°C).
R
Roshan
Expert · 5 Years Experience
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