QUICK ANSWER KEY
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| (b) | (c) | (b) | (b) | (b) | (b) | (b) | (b) | (b) | (c) | (b) | (a) | (b) | (d) | (b) |
| 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| (c) | (a) | (b) | (b) | (b) | (b) | (c) | (c) | (a) | (b) | (b) | (c) | (b) | (b) | (b) |
| 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
| (b) | (c) | (c) | (a) | (c) | (a) | (b) | (b) | (a) | (a) | (a) | (d) | (a) | (a) | (a) |
DETAILED SOLUTIONS
1. According to Huckel's rule, a planar cyclic conjugated system is aromatic if the number of delocalised π electrons is
Answer: (b) (4n + 2)
Huckel's rule: planar, cyclic, fully conjugated systems with (4n+2) π electrons (n = 0, 1, 2 ...) are aromatic. Benzene: n = 1 gives 6 π electrons.
2. Which of the following species is aromatic?
Answer: (c) Cycloheptatrienyl (tropylium) cation
Tropylium cation is planar, cyclic and has 6 π electrons (4n+2, n=1). Cyclobutadiene (4π) and cyclopentadienyl cation (4π) are antiaromatic; cyclooctatetraene is tub-shaped and non-aromatic.
3. All the six carbon–carbon bonds in benzene are of equal length (139 pm). This is best explained by
Answer: (b) Resonance / complete delocalisation of π electrons
The bond length lies between C–C (154 pm) and C=C (134 pm) because the π electrons are delocalised over all six carbons — every bond has partial double-bond character.
4. The hybridisation of each carbon atom and the C–C–C bond angle in benzene are respectively
Answer: (b) sp2, 120°
Benzene is a planar regular hexagon; each carbon is sp2 hybridised with bond angles of 120°. The unhybridised p-orbitals overlap sideways to form the delocalised π cloud.
5. The resonance energy of benzene is approximately
Answer: (b) 150 kJ mol–1
Benzene is about 150 kJ mol–1 more stable than the hypothetical cyclohexatriene. This extra stability is its resonance (delocalisation) energy.
6. Benzene prefers electrophilic substitution over addition reactions because
Answer: (b) Substitution preserves the delocalised aromatic sextet while addition destroys it
Addition would break the closed loop of six delocalised π electrons and destroy aromatic stabilisation; substitution regenerates the aromatic ring, so it is energetically favoured.
7. Cyclic polymerisation of ethyne when passed through a red hot iron tube at 873 K gives
Answer: (b) Benzene
Three molecules of ethyne (C2H2) cyclise to give one molecule of benzene (C6H6) — a standard laboratory preparation.
8. Benzene is obtained when sodium benzoate is heated with
Answer: (b) Soda lime (NaOH + CaO)
Decarboxylation: C6H5COONa + NaOH →(CaO, Δ) C6H6 + Na2CO3.
9. Phenol on distillation with zinc dust yields
Answer: (b) Benzene
Zinc dust reduces phenol by removing the oxygen: C6H5OH + Zn → C6H6 + ZnO.
10. The electrophile that attacks the benzene ring during nitration with a mixture of conc. HNO3 and conc. H2SO4 is
Answer: (c) NO2+
Sulphuric acid protonates nitric acid; loss of water gives the nitronium ion NO2+, the actual attacking electrophile. Here HNO3 behaves as a base.
11. The reaction of benzene with fuming sulphuric acid to give benzenesulphonic acid is
Answer: (b) Reversible and involves SO3 as the electrophile
Sulphonation is a reversible electrophilic substitution; the electron-deficient sulphur of SO3 attacks the ring. Heating with dilute acid/steam removes the –SO3H group.
12. Friedel–Crafts alkylation of benzene with methyl chloride is carried out in the presence of
Answer: (a) Anhydrous AlCl3
Anhydrous AlCl3 (a Lewis acid) abstracts Cl– to generate the electrophilic carbocation CH3+. Moisture destroys the catalyst.
13. Benzene reacts with acetyl chloride in presence of anhydrous AlCl3 to give
Answer: (b) Acetophenone
Friedel–Crafts acylation: the acylium ion CH3CO+ substitutes a ring hydrogen giving C6H5COCH3 (acetophenone).
14. Which of the following does not undergo the Friedel–Crafts reaction?
Answer: (d) Nitrobenzene
Nitrobenzene is strongly deactivated by the –NO2 group (–I and –R); the ring is too electron poor. Nitrobenzene is in fact used as a solvent for these reactions.
15. Benzene reacts with chlorine in the presence of bright sunlight (UV light) in absence of a halogen carrier to give
Answer: (b) Benzene hexachloride (gammexane)
In UV light a free-radical addition occurs: C6H6 + 3Cl2 → C6H6Cl6 (BHC / lindane / gammexane), an insecticide.
16. The number of moles of hydrogen required to convert one mole of benzene into cyclohexane in presence of Ni at 473–573 K is
Answer: (c) 3
Three moles of H2 add across the three formal double bonds: C6H6 + 3H2 → C6H12.
17. Ozonolysis of benzene followed by hydrolysis with Zn/H2O gives
Answer: (a) Three molecules of glyoxal
Benzene triozonide on reductive hydrolysis gives 3 molecules of glyoxal (OHC–CHO). Formation of only one product proves all three double bonds are equivalent.
18. Toluene on oxidation with hot alkaline KMnO4 followed by acidification gives
Answer: (b) Benzoic acid
The side chain is oxidised completely to –COOH while the ring is untouched, giving benzoic acid. Any alkylbenzene with at least one benzylic hydrogen gives benzoic acid.
19. Toluene reacts with Cl2 in presence of sunlight to give mainly
Answer: (b) Benzyl chloride
Sunlight promotes free-radical substitution in the side chain (benzylic position), giving C6H5CH2Cl. Remember: light → side chain, Lewis acid → ring.
20. Toluene reacts with Cl2 in presence of anhydrous FeCl3 to give
Answer: (b) o- and p-chlorotoluene
A halogen carrier promotes electrophilic substitution on the ring; the activating, o/p-directing –CH3 group sends chlorine to the ortho and para positions.
21. Which of the following groups is deactivating but ortho, para-directing?
Answer: (b) –Cl
Halogens deactivate the ring by a strong –I effect, but their lone pairs (+R effect) increase electron density at the o- and p-positions, so substitution occurs there.
22. Which of the following is the strongest activating group towards electrophilic substitution?
Answer: (c) –NH2
–NH2 donates its lone pair strongly into the ring (+R), greatly enriching the o- and p-positions. Order of activation: –NH2 > –OH > –OCH3 > –CH3.
23. Which one of the following is a meta-directing group?
Answer: (c) –COOH
–COOH withdraws electrons (–I, –R), leaving the meta position comparatively electron rich. All the other groups listed are activating and o/p-directing.
24. The correct order of reactivity towards electrophilic substitution is
Answer: (a) Anisole > toluene > benzene > chlorobenzene
–OCH3 is a strong activator (+R), –CH3 a weak activator (+I, hyperconjugation), –Cl a deactivator. Ring electron density decides the order.
25. Nitration of nitrobenzene with conc. HNO3 and conc. H2SO4 at 373 K gives mainly
Answer: (b) m-dinitrobenzene
–NO2 is deactivating and meta-directing, so the second nitro group enters the meta position; harsher conditions are needed than for benzene.
26. The number of possible isomeric dichlorobenzenes is
Answer: (b) 3
Ortho (1,2-), meta (1,3-) and para (1,4-) — three isomers only.
27. The number of possible aromatic isomers with molecular formula C8H10 is
Answer: (c) 4
Ethylbenzene plus o-, m- and p-xylene = 4 isomers.
28. The number of possible isomeric trichlorobenzenes is
Answer: (b) 3
1,2,3- ; 1,2,4- ; and 1,3,5-trichlorobenzene.
29. Benzene does not decolourise bromine water. This is because
Answer: (b) Benzene does not readily undergo addition reactions owing to the delocalised π sextet
Alkenes and alkynes decolourise bromine water by addition; benzene's aromatic stabilisation prevents easy addition, so the colour persists. This is a common test.
30. Chlorobenzene on treatment with methyl chloride and sodium in dry ether gives toluene. This reaction is called
Answer: (b) Wurtz–Fittig reaction
Wurtz–Fittig couples an aryl halide with an alkyl halide using sodium in dry ether. Coupling two aryl halides only is the Fittig reaction.
31. Two molecules of bromobenzene on treatment with sodium in dry ether give
Answer: (b) Biphenyl
Fittig reaction: 2C6H5Br + 2Na → C6H5–C6H5 + 2NaBr.
32. Which of the following polynuclear hydrocarbons is a well known carcinogen?
Answer: (c) Benzo(a)pyrene
Benzo(a)pyrene and 1,2-benzanthracene, formed on incomplete combustion of organic material (tobacco smoke, burnt fat), are carcinogenic.
33. The number of delocalised π electrons in naphthalene is
Answer: (c) 10
Naphthalene has 5 double bonds shared over the fused bicyclic system = 10 π electrons, satisfying (4n+2) with n = 2.
34. Sulphonation of naphthalene at 353 K gives mainly
Answer: (a) Naphthalene-1-sulphonic acid
At low temperature the kinetically favoured α (1-) product forms; above about 433 K the thermodynamically stable β (2-) isomer predominates.
35. The intermediate formed when an electrophile attacks the benzene ring is
Answer: (c) An arenium ion (sigma complex) that is resonance stabilised
The arenium ion is a resonance-stabilised carbocation in which aromaticity is temporarily lost; loss of a proton from the sp3 carbon restores the aromatic ring.
36. Toluene reacts with chromyl chloride (CrO2Cl2) in CS2 followed by hydrolysis to give benzaldehyde. This reaction is known as
Answer: (a) Etard reaction
The Etard reaction gives controlled (partial) oxidation of the methyl side chain to –CHO through a chromium complex intermediate.
37. Benzene treated with carbon monoxide and HCl in presence of anhydrous AlCl3/CuCl gives benzaldehyde. This is the
Answer: (b) Gattermann–Koch reaction
The Gattermann–Koch reaction is a formylation — effectively a Friedel–Crafts acylation using CO + HCl in place of an acyl chloride.
38. Aromatic hydrocarbons burn with a sooty (luminous) flame because
Answer: (b) They have a high carbon to hydrogen ratio
The high C : H ratio means incomplete combustion, so unburnt carbon particles glow and produce a smoky yellow flame — a quick identification test.
39. The essential conditions for a compound to be aromatic are
Answer: (a) Planarity, cyclic delocalisation and (4n+2) π electrons
All three conditions must hold together. Cyclooctatetraene fails because it is non-planar (tub shaped) and has 8 (= 4n) π electrons.
40. Assertion–Reason: Benzene undergoes electrophilic substitution reactions much more readily than addition reactions.
Answer: (a) Both A and R are true and R is the correct explanation of A
Both statements are correct and the reason is exactly why substitution is preferred — the aromatic stabilisation (about 150 kJ mol–1) is retained.
41. Assertion–Reason: Nitration of nitrobenzene requires more drastic conditions than nitration of benzene.
Answer: (a) Both A and R are true and R is the correct explanation of A
An electron-poor ring is less attractive to the NO2+ electrophile, so a higher temperature (about 373 K) is needed; the product is m-dinitrobenzene.
42. Assertion–Reason: Cyclooctatetraene is aromatic in nature.
Answer: (d) A is false but R is true
The assertion is false — the molecule is tub shaped (non-planar) and has 8 (= 4n) π electrons, so it is non-aromatic. The reason, as a statement of fact, is true.
43. Match the reagent/condition used with benzene in Column I with the product in Column II.
Answer: (a) A-iii, B-i, C-ii, D-iv
Acylation gives acetophenone; catalytic hydrogenation gives cyclohexane; sulphonation gives benzenesulphonic acid; UV light causes free-radical addition to give BHC.
44. Match the name reaction in Column I with the organic product in Column II.
Answer: (a) A-iii, B-i, C-iv, D-ii
Etard → benzaldehyde; Fittig couples two aryl halides → biphenyl; Wurtz–Fittig couples aryl + alkyl halide → toluene; Friedel–Crafts with C2H5Cl → ethylbenzene.
45. Match the substituent in Column I with its influence on electrophilic substitution in Column II.
Answer: (a) A-iv, B-ii, C-iii, D-i
–NH2 strong +R activator; –Cl deactivates by –I but directs o/p by +R; –NO2 strong deactivator, m-directing; –CH3 weak activator (+I, hyperconjugation).
HIGH-YIELD POINTS TO REMEMBER
1. Aromaticity checklist: cyclic + planar + complete conjugation + (4n+2) π electrons.
Benzene 6π, naphthalene 10π, anthracene 14π, tropylium cation 6π.
2. Structure: planar hexagon, sp2 carbon, 120°, all C–C = 139 pm,
resonance energy about 150 kJ mol–1.
3. Preparations: ethyne (red hot Fe tube, 873 K); sodium benzoate + soda lime; phenol + Zn dust.
4. Electrophiles: nitration NO2+; sulphonation SO3 (reversible);
halogenation X+ with Lewis acid; alkylation R+; acylation RCO+.
5. Directive influence: activating and o/p-directing — –NH2, –OH,
–OCH3, –CH3; deactivating but o/p-directing — halogens;
deactivating and m-directing — –NO2, –CN, –CHO, –COOH, –SO3H.
6. Toluene trap: light/heat → side-chain (benzyl chloride, benzoic acid);
Lewis acid catalyst → ring (o/p-chlorotoluene).
7. Name reactions: Wurtz–Fittig (aryl + alkyl halide/Na), Fittig (aryl + aryl),
Etard (CrO2Cl2 → benzaldehyde), Gattermann–Koch (CO + HCl → benzaldehyde).
8. Isomer counts: dichlorobenzene 3, trichlorobenzene 3, C8H10 aromatics 4.
9. Carcinogens: benzo(a)pyrene, 1,2-benzanthracene — from incomplete combustion.
10. Tests: sooty flame (high C:H ratio); no decolourisation of bromine water.