IOE entrance Chemistry · Chapter 3
Organic Chemistry
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79 questions in 6 syllabus topics · 30 are 2-mark questions.
3.1 Fundamentals, nomenclature and isomerism
13 questions
1. A liquid organic compound decomposes at a temperature below its normal boiling point. The most suitable method to purify it is:
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How can a liquid be made to boil at a lower temperature?
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Answer: C. Distillation under reduced pressure
Lowering the external pressure lowers the boiling point, so the liquid boils below its decomposition temperature. This is vacuum (reduced pressure) distillation.
2. In Lassaigne's test for nitrogen, the Prussian blue colour is due to the formation of:
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Think of iron(III) combining with the ferrocyanide ion.
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Answer: B. Fe₄[Fe(CN)₆]₃
NaCN formed during fusion gives Na₄[Fe(CN)₆] with FeSO₄; with Fe³⁺ this forms ferric ferrocyanide, Fe₄[Fe(CN)₆]₃, which is Prussian blue. [Fe(SCN)]²⁺ is blood red (N and S together) and the nitroprusside complex is violet (S).
3. The IUPAC name of CH₃–CH(CH₃)–CH₂–CH₂–OH is:
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Give the carbon bearing the principal functional group the lowest number.
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Answer: D. 3-Methylbutan-1-ol
The longest chain containing the –OH carbon has 4 carbons; numbering from the –OH end gives OH at C-1 and the methyl at C-3: 3-methylbutan-1-ol.
4. The permanent electronic effect that is transmitted along a chain of σ-bonds and dies out quickly after about three carbon atoms is the:
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Which effect works through σ-bonds rather than π-electrons?
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Answer: D. Inductive effect
The inductive effect arises from electronegativity differences, is permanent, acts through σ-bonds and weakens rapidly with distance. The electromeric effect is temporary, and resonance involves π-electrons.
5. The hybridisation of the positively charged carbon in a carbocation such as (CH₃)₃C⁺ is:
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Count the σ-bonds on the charged carbon; the vacant orbital is unhybridised.
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Answer: B. sp²
The cationic carbon has three σ-bonds and an empty p-orbital, so it is sp² hybridised and trigonal planar.
6. Which of the following groups shows a −R (−M) effect when attached to a benzene ring?
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Look for a group with a multiple bond to a more electronegative atom and no lone pair to donate.
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Answer: D. –NO₂
–NO₂ withdraws π-electron density from the ring by resonance (−R). –OH, –NH₂ and –OCH₃ donate their lone pairs into the ring (+R).
7. Diethyl ether (C₂H₅–O–C₂H₅) and methyl propyl ether (CH₃–O–C₃H₇) are examples of:
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Same functional group, different alkyl groups on each side of it.
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Answer: D. Metamers
Both have the formula C₄H₁₀O and the same functional group, but the alkyl groups on either side of the oxygen differ. This is metamerism.
8. Which of the following is NOT an electrophile?
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Electrophiles accept electron pairs. Which one has a lone pair to give?
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Answer: B. NH₃
NH₃ has a lone pair that it donates, so it is a nucleophile. BF₃ and AlCl₃ are electron-deficient Lewis acids and NO₂⁺ is a cation, so all three are electrophiles.
9. The correct order of decreasing acid strength of (I) CH₃CH₂CH(Cl)COOH, (II) CH₃CH(Cl)CH₂COOH and (III) ClCH₂CH₂CH₂COOH is:
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The inductive effect weakens as the distance from –COOH increases.
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Answer: C. I > II > III
The −I effect of Cl stabilises the carboxylate ion and falls off rapidly with distance. Cl is closest to –COOH in I (α), then II (β), then III (γ), so the acidity order is I > II > III.
10. In a Kjeldahl estimation, the ammonia evolved from 0.5 g of an organic compound exactly neutralised 10 mL of 1 M H₂SO₄. The percentage of nitrogen in the compound is:
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Convert molarity of H₂SO₄ to normality first, then use %N = 1.4 NV/w.
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Answer: A. 56%
1 M H₂SO₄ is 2 N. %N = 1.4 × N × V / w = 1.4 × 2 × 10 / 0.5 = 56%.
11. In a Carius determination, 0.20 g of an organic compound gave 0.287 g of AgCl. The percentage of chlorine in the compound is:
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The fraction of Cl in AgCl is 35.5/143.5.
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Answer: B. 35.5%
Mass of Cl = 0.287 × 35.5/143.5 = 0.071 g. %Cl = 0.071/0.20 × 100 = 35.5%.
12. The number of α-hydrogen atoms available for hyperconjugation with the double bond in 2-methylbut-2-ene, (CH₃)₂C=CH–CH₃, is:
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Count only the H atoms on sp³ carbons directly bonded to the double-bond carbons.
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Answer: C. 9
The α-hydrogens are on carbons attached directly to the C=C. There are two CH₃ groups on C-2 (6 H) and one CH₃ on C-3 (3 H), giving 9. The vinylic H on C-3 is not counted.
13. The total number of structural isomers (alcohols and ethers) with the molecular formula C₄H₁₀O is:
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Count the alcohols and the ethers separately.
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Answer: D. 7
There are 4 alcohols (butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol) and 3 ethers (ethoxyethane, 1-methoxypropane, 2-methoxypropane), giving 7.
3.2 Hydrocarbons
13 questions
14. Propene reacts with HBr in the presence of benzoyl peroxide to give mainly:
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Peroxide changes the mechanism and therefore the orientation of HBr addition.
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Answer: B. 1-Bromopropane
In the presence of peroxide, HBr adds by a free-radical mechanism (anti-Markovnikov, the peroxide or Kharasch effect). Br attaches to the terminal carbon to give 1-bromopropane.
15. Ozonolysis of but-2-ene (CH₃CH=CHCH₃) followed by treatment with Zn/H₂O gives:
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Split the molecule at the double bond and put =O on each carbon.
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Answer: D. Two molecules of ethanal
Cleavage of the C=C in a symmetrical alkene gives two identical fragments: CH₃CHO + CH₃CHO (ethanal).
16. Which of the following hydrocarbons is the most acidic?
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Higher s-character in the carbon orbital stabilises the carbanion.
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Answer: D. Ethyne
In ethyne the C–H carbon is sp hybridised (50% s-character), so the conjugate base carbanion is the most stable. Ethyne therefore releases H⁺ most easily, for example forming sodium acetylide with Na.
17. Kolbe's electrolysis of an aqueous solution of sodium acetate gives which hydrocarbon at the anode?
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Two alkyl radicals from the carboxylate couple together.
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Answer: B. Ethane
2CH₃COO⁻ → 2CH₃• + 2CO₂ + 2e⁻ at the anode, and two methyl radicals combine to form CH₃–CH₃ (ethane).
18. Which of the following species is aromatic?
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Apply Hückel's rule: cyclic, planar, fully conjugated, (4n + 2) π-electrons.
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Answer: A. Cyclopentadienyl anion
The cyclopentadienyl anion is cyclic, planar and fully conjugated with 6 π-electrons (4n + 2, n = 1). Cyclopentadiene and cycloheptatriene each have an sp³ carbon breaking the conjugation, and cyclooctatetraene (8 π) is non-planar.
19. The electrophile that attacks benzene during nitration with a mixture of concentrated HNO₃ and H₂SO₄ is:
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The electrophile must be positively charged and is formed with the help of H₂SO₄.
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Answer: A. NO₂⁺
H₂SO₄ protonates HNO₃, which loses water to form the nitronium ion NO₂⁺. This is the attacking electrophile.
20. Hydrogenation of but-2-yne using Lindlar's catalyst (Pd/BaSO₄ poisoned with quinoline) gives:
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Catalytic surface addition delivers both H atoms from the same side.
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Answer: D. cis-But-2-ene
Lindlar's catalyst stops reduction at the alkene stage, and syn addition of H₂ gives the cis-alkene. Na in liquid NH₃ would give the trans-alkene.
21. Which of the following substituents on benzene directs an incoming electrophile to the meta position?
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Electron-withdrawing groups with a −R effect are meta directors.
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Answer: B. –CHO
–CHO withdraws electrons by −I and −R effects, deactivates the ring and makes the meta position relatively richer in electrons. –OH and –CH₃ are ortho/para directing, and –Cl is deactivating but still ortho/para directing.
22. An alkene with molecular formula C₆H₁₂ gives only propanone (acetone) on ozonolysis. The alkene is:
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Join the two carbonyl carbons of the products back together with a double bond.
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Answer: C. 2,3-Dimethylbut-2-ene
Two acetone units, (CH₃)₂C=O + O=C(CH₃)₂, join at the C=C to give (CH₃)₂C=C(CH₃)₂, which is 2,3-dimethylbut-2-ene (C₆H₁₂). Hex-3-ene would give propanal.
23. A hydrocarbon A (C₄H₆) gives a white precipitate with ammoniacal silver nitrate solution. A is:
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Which isomer has a hydrogen on a triply bonded carbon?
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Answer: C. But-1-yne
Only terminal alkynes (with an acidic ≡C–H) form insoluble silver acetylides. But-1-yne, CH≡C–CH₂CH₃, has a terminal H. But-2-yne does not, and the diene and cyclobutene are not alkynes.
24. What volume of oxygen at NTP is needed for complete combustion of 13 g of ethyne (C₂H₂)?
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Balance the combustion equation and use the 2 : 5 mole ratio.
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Answer: D. 28 L
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O. 13 g C₂H₂ = 13/26 = 0.5 mol, which needs 0.5 × 5/2 = 1.25 mol O₂. 1.25 × 22.4 = 28 L.
25. 3-Methylbut-1-ene, CH₂=CH–CH(CH₃)₂, reacts with HCl to give mainly 2-chloro-2-methylbutane rather than 2-chloro-3-methylbutane. This is because:
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Carbocations can rearrange to become more stable.
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Answer: D. The 2° carbocation first formed rearranges by a 1,2-hydride shift to a more stable 3° carbocation
Markovnikov protonation gives a 2° carbocation at C-2. A hydride shifts from C-3 to give the more stable 3° carbocation, which Cl⁻ then attacks to form 2-chloro-2-methylbutane.
26. The number of σ and π bonds in vinylacetylene (but-1-en-3-yne), CH₂=CH–C≡CH, is:
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Every bond has one σ; a double bond adds one π and a triple bond adds two.
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Answer: A. 7 σ and 3 π
The C–C framework has 3 σ-bonds and there are 4 C–H σ-bonds, so 7 σ. The C=C contributes 1 π and the C≡C contributes 2 π, so 3 π.
3.3 Haloalkanes and haloarenes
13 questions
27. Methyl bromide is heated with sodium metal in dry ether. The main product is:
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In the Wurtz reaction two alkyl groups join together.
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Answer: D. Ethane
Wurtz reaction: 2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr. Two methyl groups couple together.
28. Chlorobenzene and methyl chloride are treated with sodium in dry ether. The main product (Wurtz–Fittig reaction) is:
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The aryl group and the alkyl group couple together.
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Answer: A. Toluene
C₆H₅Cl + CH₃Cl + 2Na → C₆H₅–CH₃ + 2NaCl. The aryl and alkyl groups couple to give toluene. Biphenyl and ethane are only side products.
29. For an SN2 reaction R–X + OH⁻ → R–OH + X⁻, the rate of reaction:
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The '2' in SN2 means bimolecular.
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Answer: A. Depends on the concentration of both R–X and OH⁻
SN2 is a single concerted step in which the nucleophile attacks as the leaving group departs, so rate = k[R–X][OH⁻] (second order).
30. Which of the following is the most reactive towards an SN2 reaction?
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Think about steric crowding around the carbon being attacked.
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Answer: C. CH₃Br
SN2 needs backside attack by the nucleophile, which steric hindrance blocks. Reactivity is CH₃X > 1° > 2° > 3°, so CH₃Br is the most reactive.
31. When an optically active alkyl halide undergoes an SN1 reaction, the product is mainly:
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Consider the geometry of the carbocation intermediate.
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Answer: A. Racemic (a mixture of both enantiomers)
SN1 goes through a planar carbocation, which the nucleophile can attack from either face. This gives (largely) racemisation.
32. Ethyl bromide heated with alcoholic KOH gives mainly:
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Alcoholic KOH favours elimination; aqueous KOH favours substitution.
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Answer: A. Ethene
Alcoholic KOH supplies the strong base C₂H₅O⁻, which favours β-elimination (dehydrohalogenation): CH₃CH₂Br → CH₂=CH₂ + HBr. Aqueous KOH would give ethanol by substitution.
33. Chlorobenzene is much less reactive than chloroethane towards nucleophilic substitution mainly because:
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Think about resonance between the Cl lone pair and the benzene ring.
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Answer: C. The C–Cl bond in chlorobenzene has partial double-bond character due to resonance
A lone pair on Cl is delocalised into the ring, giving the C–Cl bond partial double-bond character. This makes it shorter and stronger and therefore harder to break. The sp² carbon also contributes.
34. The preparation of alkyl fluorides by heating alkyl bromides or chlorides with AgF or Hg₂F₂ is known as the:
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This halogen exchange specifically produces fluorides.
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Answer: A. Swarts reaction
In the Swarts reaction a heavier halogen is exchanged for fluorine using metallic fluorides such as AgF, Hg₂F₂ or CoF₂. The Finkelstein reaction uses NaI in acetone to make iodides.
35. 2-Bromobutane is heated with alcoholic KOH. The major product and the rule it follows are:
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The more highly substituted alkene is the more stable one.
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Answer: A. But-2-ene; Saytzeff (Zaitsev) rule
Dehydrohalogenation gives the more substituted (more stable) alkene as the major product, according to Saytzeff's rule. Here that is but-2-ene (CH₃CH=CHCH₃).
36. A mixture of methyl bromide and ethyl bromide is treated with sodium in dry ether. The hydrocarbons formed are:
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Consider every possible pairing of the two alkyl groups.
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Answer: C. Ethane, propane and butane
Random coupling gives CH₃–CH₃ (ethane), CH₃–C₂H₅ (propane) and C₂H₅–C₂H₅ (butane). This is why the Wurtz reaction is not useful for making odd-carbon alkanes from two different halides.
37. The correct order of reactivity towards an SN1 reaction is:
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Rank the carbocations that would form.
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Answer: A. C₆H₅CH₂Cl > (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl
SN1 rate follows carbocation stability. The benzyl cation is resonance-stabilised, then the 2° isopropyl cation, then 1° ethyl, and methyl is the least stable.
38. Which of the following is hydrolysed most easily by aqueous NaOH?
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More o/p electron-withdrawing groups help nucleophilic attack on the ring.
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Answer: B. 2,4,6-Trinitrochlorobenzene
Nitro groups at ortho and para positions withdraw electrons and stabilise the anionic intermediate in nucleophilic aromatic substitution. Three such groups make 2,4,6-trinitrochlorobenzene react most easily, even with warm water.
39. The number of structural isomers of C₄H₉Br is:
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Place Br at each distinct position on both the straight and branched C₄ chains.
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Answer: C. 4
The four isomers are 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. Stereoisomers are not counted here.
3.4 Alcohols, phenols and ethers
14 questions
40. In Williamson’s synthesis, a haloalkane is treated with
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It is a method of making ethers.
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Answer: C. sodium alkoxide
Williamson ether synthesis: R–X + R′–ONa → R–O–R′ + NaX, an SN2 reaction of a haloalkane with sodium alkoxide.
41. In the Lucas test (anhydrous ZnCl₂ + conc. HCl), turbidity appears immediately at room temperature with:
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Speed in the Lucas test follows carbocation stability.
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Answer: D. 2-Methylpropan-2-ol
Tertiary alcohols form a stable 3° carbocation quickly, so the insoluble alkyl chloride separates at once. 2° alcohols take about 5 minutes and 1° alcohols show no turbidity at room temperature.
42. Phenol is more acidic than ethanol because:
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Compare how stable the conjugate bases are.
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Answer: D. The phenoxide ion is stabilised by resonance
The negative charge on the phenoxide ion is delocalised over the benzene ring. The ethoxide ion has no such stabilisation (and is destabilised by the +I effect of the ethyl group).
43. Sodium ethoxide reacts with methyl iodide to give:
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The alkoxide displaces iodide by SN2.
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Answer: D. Methoxyethane
Williamson synthesis: C₂H₅O⁻Na⁺ + CH₃I → C₂H₅–O–CH₃ + NaI. The product is methoxyethane (ethyl methyl ether).
44. Which of the following is the most acidic?
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Electron-withdrawing groups stabilise the phenoxide ion.
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Answer: C. 4-Nitrophenol
The –NO₂ group at the para position stabilises the phenoxide ion by −I and −R effects. –CH₃ and –OCH₃ are electron donating and decrease acidity.
45. Phenol treated with bromine water gives a white precipitate of:
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In water, the strongly activating –OH group allows substitution at every o/p position.
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Answer: C. 2,4,6-Tribromophenol
–OH strongly activates the ring. In a polar solvent (water), bromination happens at all ortho and para positions, giving 2,4,6-tribromophenol.
46. Ethanol heated with excess concentrated H₂SO₄ at 443 K (170 °C) mainly gives:
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The higher temperature favours elimination.
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Answer: A. Ethene
At 443 K with excess acid, intramolecular dehydration gives CH₂=CH₂. At about 413 K with excess ethanol, intermolecular dehydration gives diethyl ether.
47. Phenol heated with chloroform and aqueous NaOH, followed by acidification, gives mainly:
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The electrophile here is dichlorocarbene.
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Answer: C. Salicylaldehyde (2-hydroxybenzaldehyde)
In the Reimer–Tiemann reaction the dichlorocarbene (:CCl₂) electrophile attacks the ortho position, and hydrolysis gives –CHO there. Using CCl₄ instead would give salicylic acid.
48. Oxidation of propan-2-ol with acidified K₂Cr₂O₇ gives:
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Secondary alcohols oxidise to ketones.
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Answer: C. Propanone
A secondary alcohol is oxidised to a ketone: CH₃CH(OH)CH₃ → CH₃COCH₃ (propanone). Further oxidation does not occur easily.
49. Methoxyethane (CH₃–O–C₂H₅) is heated with one equivalent of concentrated HI. The products are:
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With a mixed 1° ether, iodide attacks the smaller alkyl group.
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Answer: C. CH₃I and C₂H₅OH
After the ether is protonated, I⁻ attacks by SN2 at the less hindered methyl carbon. This gives CH₃I + C₂H₅OH.
50. The best pair of reagents for preparing tert-butyl methyl ether, (CH₃)₃C–O–CH₃, by Williamson synthesis is:
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Which halide can undergo SN2 without eliminating?
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Answer: D. (CH₃)₃C–ONa and CH₃Br
Williamson synthesis is SN2, so the halide must be methyl or primary. A strong base with a 3° halide such as (CH₃)₃CBr gives mainly elimination (isobutylene). The tert-butoxide must therefore be the nucleophile.
51. Propanone reacts with CH₃MgBr in dry ether, and the product is hydrolysed with dilute acid. The final product is:
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A Grignard reagent with a ketone gives a tertiary alcohol.
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Answer: C. 2-Methylpropan-2-ol
CH₃⁻ adds to the carbonyl carbon: (CH₃)₂C=O → (CH₃)₃C–OMgBr, and hydrolysis gives (CH₃)₃COH, a tertiary alcohol (2-methylpropan-2-ol).
52. 4.6 g of ethanol reacts completely with excess sodium metal. The volume of hydrogen evolved at NTP is:
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Two moles of alcohol give one mole of H₂.
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Answer: C. 1.12 L
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. 4.6 g = 0.1 mol ethanol, which gives 0.05 mol H₂. 0.05 × 22.4 = 1.12 L.
53. Sodium phenoxide is heated with CO₂ at about 400 K and 4–7 atm, and the product is then acidified. The final product is:
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CO₂ acts as a weak electrophile towards the phenoxide ring.
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Answer: A. Salicylic acid (2-hydroxybenzoic acid)
Kolbe's (Kolbe–Schmitt) reaction: the phenoxide ion attacks CO₂ at the ortho position, giving sodium salicylate. Acidification gives salicylic acid.
3.5 Aldehydes, ketones and carboxylic acids
13 questions
54. Which of the following undergoes the Cannizzaro reaction?
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Check for α-hydrogen atoms.
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Answer: B. HCHO
The Cannizzaro reaction is given by aldehydes that have no α-hydrogen. HCHO has none, while ethanal and propanal have α-H and undergo aldol condensation instead.
55. Which of the following does NOT give a silver mirror with Tollens' reagent?
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Tollens' reagent is a mild oxidising agent that ketones resist.
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Answer: B. Propanone
Aldehydes reduce Tollens' reagent to metallic silver. Methanoic acid also does, because it contains an –CHO-like H–C=O unit. Ketones such as propanone are not oxidised by this mild reagent.
56. Which of the following gives a positive iodoform test?
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Look for CH₃–CH(OH)– or CH₃–CO– in the molecule.
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Answer: A. Ethanol
The iodoform test requires a CH₃CO– group or a CH₃CH(OH)– group. Ethanol (CH₃CH₂OH) is oxidised to CH₃CHO by I₂/NaOH and then gives CHI₃. The other three lack these groups.
57. Which of the following is the strongest acid?
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Alkyl groups with a +I effect decrease acid strength.
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Answer: B. Methanoic acid (HCOOH)
HCOOH has no electron-donating alkyl group, so its anion is the most stable (pKa ≈ 3.75, compared with benzoic acid 4.2, acetic acid 4.76 and propanoic acid 4.87).
58. When an aldehyde is warmed with Fehling's solution, the red precipitate formed is:
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Copper is reduced from +2 to +1.
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Answer: C. Cu₂O
The aldehyde reduces Cu²⁺ (in the tartrate complex) to Cu⁺, which precipitates as red-brown cuprous oxide, Cu₂O.
59. Which of the following is the most reactive towards nucleophilic acyl substitution?
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The best leaving group gives the most reactive derivative.
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Answer: A. CH₃COCl
Reactivity order: acid chloride > anhydride > ester > amide. Cl⁻ is the best leaving group and Cl donates its lone pair least by resonance.
60. Benzoyl chloride is converted into benzaldehyde by H₂ over Pd/BaSO₄ (partially poisoned). This reaction is called:
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This reduction starts from an acid chloride.
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Answer: C. Rosenmund reduction
Rosenmund reduction: RCOCl + H₂ → RCHO + HCl over poisoned Pd/BaSO₄. The poisoning stops the reduction from going on to the alcohol.
61. Propanone is reduced to propane by:
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You need a reagent that removes the oxygen completely, not one that just gives an alcohol.
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Answer: B. Zn–Hg and concentrated HCl
Clemmensen reduction (zinc amalgam + conc. HCl) converts C=O to CH₂. LiAlH₄ and NaBH₄ give the alcohol, propan-2-ol.
62. Ethanal is treated with dilute NaOH and the product is then heated. The final product is:
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The aldol formed first dehydrates when heated.
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Answer: C. But-2-enal (crotonaldehyde)
The aldol reaction gives CH₃CH(OH)CH₂CHO (3-hydroxybutanal). On heating this loses water to give the α,β-unsaturated aldehyde CH₃CH=CHCHO (but-2-enal).
63. Benzaldehyde is heated with concentrated NaOH. The products are:
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One molecule is oxidised while another is reduced.
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Answer: D. Benzyl alcohol and sodium benzoate
Benzaldehyde has no α-H, so it undergoes the Cannizzaro reaction (disproportionation). One molecule is reduced to C₆H₅CH₂OH and the other is oxidised to C₆H₅COONa.
64. Among the following halogen-substituted ethanoic acids, the one with the lowest pKa is:
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Compare the total electron-withdrawing effect, not just the electronegativity of one atom.
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Answer: A. Cl₂CHCOOH
Two Cl atoms together exert a larger −I effect than one F atom. pKa values: Cl₂CHCOOH ≈ 1.3, FCH₂COOH ≈ 2.6, ClCH₂COOH ≈ 2.9 and CH₃COOH ≈ 4.8.
65. 0.1 mol of propanone is completely converted to iodoform (CHI₃) by I₂ and NaOH. The mass of iodoform formed is (I = 127):
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One mole of a methyl ketone gives one mole of iodoform.
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Answer: D. 39.4 g
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O. 1 mol of ketone gives 1 mol CHI₃ (M = 12 + 1 + 3 × 127 = 394). 0.1 × 394 = 39.4 g.
66. The mass of NaOH required to completely saponify 8.8 g of ethyl ethanoate (CH₃COOC₂H₅) is:
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The ester and NaOH react in a 1 : 1 mole ratio.
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Answer: A. 4 g
CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH. M(ester) = 88, so 8.8 g = 0.1 mol, which needs 0.1 mol NaOH = 4 g.
3.6 Nitro compounds and amines
13 questions
67. An amine warmed with chloroform and alcoholic KOH gives an offensive-smelling isocyanide. The amine must be:
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The reaction removes two hydrogens from nitrogen.
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Answer: C. A primary amine
The carbylamine reaction, RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O, needs two N–H hydrogens. Only primary amines (aliphatic or aromatic) give it.
68. Ethanamide (CH₃CONH₂) is heated with Br₂ and aqueous KOH. The product is:
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The amine formed has one carbon atom fewer than the amide.
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Answer: C. Methanamine
Hofmann bromamide degradation gives a primary amine with one carbon fewer: CH₃CONH₂ → CH₃NH₂ (methanamine).
69. Which of the following is the strongest base in aqueous solution?
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In water, solvation matters as well as the +I effect.
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Answer: B. (CH₃)₂NH
In water, the +I effect, solvation of the ammonium ion by hydrogen bonding and steric effects combine to make the 2° amine the strongest. The order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.
70. Aniline is a weaker base than methanamine because:
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Ask how available the nitrogen lone pair is.
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Answer: B. The lone pair on nitrogen in aniline is delocalised into the benzene ring
Resonance spreads the nitrogen lone pair of aniline over the ring, so it is less available to accept a proton. In methanamine, the +I effect of CH₃ increases electron density on N.
71. Aniline is treated with NaNO₂ and dilute HCl at 0–5 °C. The product is:
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The low temperature is needed to keep a reactive salt stable.
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Answer: B. Benzenediazonium chloride
Diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. The low temperature keeps the diazonium salt from decomposing.
72. Hinsberg's reagent, used to distinguish between primary, secondary and tertiary amines, is:
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It is a sulphonyl chloride.
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Answer: A. Benzenesulphonyl chloride
Hinsberg's reagent is C₆H₅SO₂Cl. Its products with 1° and 2° amines differ in their solubility in alkali, and 3° amines do not react.
73. Nitrobenzene is reduced with Sn and concentrated HCl, and the product is then treated with NaOH. The final product is:
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Strong reduction in acid takes –NO₂ all the way to –NH₂.
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Answer: D. Aniline
In acidic medium the –NO₂ group is reduced fully to –NH₂ (6 electrons), giving anilinium chloride. NaOH then liberates free aniline.
74. Benzenediazonium chloride reacts with phenol in mildly alkaline solution to give an orange dye. The product is:
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In coupling reactions the nitrogen atoms are retained.
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Answer: A. 4-Hydroxyazobenzene (p-hydroxyazobenzene)
In the coupling reaction the diazonium ion attacks the para position of the activated phenoxide ion, forming C₆H₅–N=N–C₆H₄–OH (para). The –N=N– azo link is kept.
75. An amine of formula C₃H₉N reacts with benzenesulphonyl chloride to give a product that is insoluble in aqueous KOH. The amine is:
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The product dissolves in alkali only if an acidic N–H remains.
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Answer: C. N-Methylethanamine
A 2° amine gives an N,N-disubstituted sulphonamide with no acidic N–H, so it is insoluble in alkali. N-methylethanamine is the only 2° amine listed. The 1° amines give alkali-soluble products and the 3° amine does not react.
76. The mass of methanamine obtained from 5.9 g of ethanamide by Hofmann bromamide reaction, assuming 100% yield, is:
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One mole of amide gives one mole of amine.
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Answer: D. 3.1 g
CH₃CONH₂ (M = 59) → CH₃NH₂ (M = 31) in a 1 : 1 mole ratio. 5.9 g = 0.1 mol, giving 0.1 × 31 = 3.1 g.
77. The correct order of basic strength of (I) aniline, (II) 4-methylaniline (p-toluidine) and (III) 4-nitroaniline is:
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Electron-donating groups increase basicity and electron-withdrawing groups decrease it.
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Answer: A. II > I > III
–CH₃ (+I, hyperconjugation) increases electron density on N, so p-toluidine is the most basic. –NO₂ (−I, −R) withdraws the lone pair, so p-nitroaniline is the least basic.
78. Benzenediazonium chloride is converted into benzene by treatment with:
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You need a reducing agent that supplies a hydrogen atom.
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Answer: D. H₃PO₂ (hypophosphorous acid)
H₃PO₂ (or ethanol) replaces –N₂⁺ by H, giving benzene. CuCN gives benzonitrile, warm water gives phenol and KI gives iodobenzene.
79. How many isomeric amines with molecular formula C₃H₉N give a positive carbylamine test?
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List all the isomers and pick out the primary amines.
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Answer: D. 2
C₃H₉N has four isomers: propan-1-amine (1°), propan-2-amine (1°), N-methylethanamine (2°) and trimethylamine (3°). Only the two primary amines give the carbylamine test.