28 Benzene and Alkyl Benzene

28 Benzene and Alkyl Benzene Objective Questions I (Only one correct option) 1. The major product obtained in the given reaction is CH3 O CH2 CH2 CH3 CH AlCl3 4. The major product of the following reaction is CH2CH3 Product (i) Alkaline KMnO4 Cl (ii) H3 O+ CH3 CH2CHO COOH (a) CH3 O (a) (b) H3C CH O CH2 (c) H3C CH CH3 (b) COCH3 CH2COOH (c) O (d) 5. The major product of the following reaction is CH3 (d) H3C CH2 O CH CH2 1. KOH (alc.) CH2 2. Free radical polymerisation Cl Cl CH3 2. The increasing order of the reactivity of the following compounds towards electrophilic aromatic substitution reaction is Cl CH3 (a) (b) n COCH3 n Cl (I) (II) (a) III < I < II (c) III < II < I Cl CH3 (III) (b) II < I < III (d) I < III < II (d) (c) n n 3. Increasing order of reactivity of the following compounds for SN 1substitution is CH3 CH3 CH2 Cl H 3C Cl H3CO (a) (b) (c) (d) 6. The increasing order of reactivity of the following (B) (A ) (C) (A) < (B) < (D) < (C) (B) < (C) < (D) < (A) (B) < (A) < (D) < (C) (B) < (C) < (A) < (D) Cl (D) OH OH Cl compounds towards aromatic electrophilic substitution reaction is Cl OMe Me CN A B C D Benzene and Alkyl Benzene (a) A < B < C < D (c) D < A < C < B (b) B < C < A < D (d) D < B < A < C 9. Among the following four aromatic compounds, which one will have the lowest melting point? 7. Polysubstitutiion is a major drawback in (a) (b) (c) (d) O Friedel-Craft’s alkylation Reimer-Tiemann reaction Friedel-Craft’s acylation Acetylation of aniline OH OH (a) (b) O CH3 8. The major product of the following reaction is O O (c) CH3 OH (d) CH3 (1) Cl2/hn (2) H2O,D 10. Which of the following compounds is not aromatic? Cl CHCl2 CO2H (a) (a) (b) N H (b) (c) Cl Cl CH2 OH CHO (d) N 11. In the following sequence of reactions KMnO (c) SOCl H / Pd 4 2 2 Toluene ¾¾® A ¾¾® B ¾¾® C, the product C is (d) BaSO 4 Cl (a) C6 H5 COOH (c) C6 H5 CH3 OH Cl (b) C6 H5 CH3 (d) C6 H5 CHO 12. Match the four starting materials given in Column I with the corresponding reaction schemes provided in Column II and select the correct answer using the code given below the lists. Column I A. H Column II H p. Scheme I (i) KMnO4 , HO- , heat (ii) H + , H 2O (iii) SOCl 2 (iv) NH 3 ¾® C7 H6N2O3 ? ¾¾¾¾¾¾¾¾¾¾¾¾ B. q. OH Scheme II (i) Sn / HCl (ii) CH 3COCl (iii) Conc. H 2SO4 (iv) HNO3 (v) Dil. H 2 SO4 , heat (vi) HO- ? ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾¾® C6H6N2O2 OH C. NO2 r. Scheme III (i) Red hot iron, 873 K (ii) fuming HNO3 , H 2SO4 , heat (iii) H 2S × NH 3 (iv) NaNO2 , H 2SO4 (v) hydrolysis ? ¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾ ¾® C6H5NO3 D. NO2 s. Scheme IV (i) Conc. H 2SO4 ,60°C (ii) Conc. HNO3 , conc. H 2SO4 (iii) Dil × H 2SO4 , heat ? ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾® C6H5NO4 CH3 Codes A B C D (a) p s q r (c) r s q p A B C D (b) r p s q (d) s p r q Benzene and Alkyl Benzene 13. For which of the following molecule significant m ¹ 0? Cl SH OH CN COOH (d) HO O C CN OH SH I II III IV (a) Only I (c) Only III (b) I and II (d) III and IV NO2 H3C NO2 Cl OCH3 NO2 O HBr —OCH3 15. In the reaction, the products are, 14. The compounds P, Q and S COOH (a) Br— OCH3 H 3C H3C P (c) O Q —OCH3 and H2 (b) C —Br and CH3Br (d) —Br and CH3OH —OH and CH3Br 16. In the following reaction, O O H3C N H S were separately subjected to nitration using HNO3 / H2 SO4 mixture. The major product formed in each case respectively, is COOH Conc. HNO3 Conc. H2SO4 The product X is O (a) OCH3 NO2 N H (a) HO H3C NO2 O NO2 O2N O (b) C O N H NO2 COOH O (c) OCH3 N H (b) NO2 HO H3C NO2 NO2 C N H O2N O NO2 COOH O (d) O AlCl 3 + Cl — CH2 CH2 — CH3 ¾¾¾® 17. OCH3 (i) O / D 2 P ¾¾¾¾® Q + Phenol (ii) H 3 O + (c) HO NO2 P and Q are respectively NO2 H 3C NO2 O C (a) + CH3CH2CHO (b) + CH3COCH3 O X Benzene and Alkyl Benzene 24. The compound that is most reactive towards electrophilic (c) + CH3COCH3 (d) + CH3CH2CHO substitution is (a) toluene (c) benzoic acid (b) benzene (d) nitrobenzene 25. Among the following, the compound that can be most readily 18. sulphonated is (a) benzene (c) toluene O HN CH3 H 3C Objective Questions II (One or more than one correct option) Fe/Br2 26. The reaction(s) leading to 1,3,5-trimethylbenzene is (are) product on monobromination of this compound is O HN CH3 H3C (b) H3 C (a) O O HN (b) nitrobenzene (d) chlorobenzene the formation of Conc. H2SO4 (a) D CH3 (b) Heated iron tube Me —— H 873 K O Br Br HN (c) CH3 H 3C (1) Br2, NaOH (2) H3O+ (c) O (3) Sodalime, D O O CHO Br HN (d) O CH3 H3C Zn/Hg, HCl (d) OHC CHO 27. Among the following reactions(s), which gives(give) Br tert-butyl benzene as the major product? 19. Identify the correct order of reactivity in electrophilic substitution reactions of the following compounds Benzene (1), Toluene (2), Chlorobenzene (3) and Nitrobenzene (4) (a) 1 > 2 > 3 > 4 (b) 4 > 3 > 2 > 1 (c) 2 > 1 > 3 > 4 (d) 2 > 3 > 1 > 4 NaOC2H5 H2SO4 (c) Cl 2 followed by treatment with aqueous NaOH gives (a) o-cresol (b) p-cresol (c) 2, 4-dihydroxy toluene (d) benzoic acid O O H 3C CH3 (a) (b) CH3 O O H H O 23. The reaction of toluene with chlorine in the presence of ferric chloride (FeCl 3 ) gives predominantly (a) benzoyl chloride (b) m-chlorotoluene (c) benzyl chloride (d) o- and p-chlorotoluene BF3,OEt2 H (d) NaOCl 22. Chlorination of toluene in the presence of light and heat (d) Radical initiator, O2 CH2==CH¾ CH 3,H + U T High pressure, Heat by chlorination with (b) SOCl 2 AlCl3 28. The major product U in the following reaction is 21. Benzyl chloride (C6 H5 CH2 Cl) can be prepared from toluene (a) SO2Cl 2 (b) OH (c) 20. A solution of (+) -2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amount of SbCl 5 , due to the formation of (a) carbanion (b) carbene (c) free-radical (d) carbocation Cl Br (a) O CH2 CH2 (c) O (d) O H Benzene and Alkyl Benzene 29. Which of the following molecules, in pure form, is/are unstable at room temperature? (a) F NaOH (aq) (b) (i) O2N (b) CH3 O O F (c) (d) (ii) H3C 30. An aromatic molecule will NaOH (aq) F – (not liberated) CH2NO2 (a) have 4n p-electrons (b) have ( 4 n + 2 )p-electrons (c) be planar (d) be cyclic NO NO (c) (i) NO NO2 Conc. HNO3 + Conc. H2SO4 31. Which compound(s) out of the following is/are not aromatic? (a) F– (liberated) NO2 (b) Å NO2 Å (c) (d) NO2 Conc. HNO3 (ii) Conc. H2SO4 NO2 s 32. Toluene, when treated with Br2 /Fe, gives p-bromotoluene as the major product because the CH3 group (a) is para-directing (b) is meta-directing (c) activates the ring by hyperconjugation (d) deactivates the ring Pd/C (d) 3.0 mol H2 33. A new carbon–carbon bond formation is possible in (a) Cannizzaro’s reaction (b) Friedel-Craft’s alkylation (c) Clemmensen reduction (d) Reimer-Tiemann reaction True or False 34. An electron donating substituent in benzene orients the not incoming electrophilic group to the meta-position. 35. In benzene, carbon uses all the three p-orbitals for hybridisation. Subjective Questions 36. Explain the followings : 37. 7-bromo-1, 3, 5-cycloheptatriene exists as ionic species in CH3 (a) (i) Br C2H5OH (aq) CH3 CH3 (ii) Br CH3 aqueous solution while 5-bromo-1,3-cyclopentadiene doesn't ionise even in presence of Ag + ( aq ). Explain. (2004) acidic solution C2H5OH (aq) 38. What would be the major product in the following reaction? O neutral solution N— Br2/Fe(1eq) Benzene and Alkyl Benzene 39. Give reasons for the following : 48. Identify the major product in the following reactions : (i) tert-butylbenzene does not give benzoic acid on treatment with acidic KMnO4. (ii) Normally, benzene gives electrophilic substitution reaction rather than electrophilic addition reaction although it has double bond. 40. Show the steps to carry out the following transformations. HNO3/H2SO4 —COO— (i) mononitration (ii) C6H5COOH + CH3MgI ? + ? 49. Write the structure of the major organic product expected from the following reaction (i) Ethylbenzene ® benzene (ii) Ethylbenzene ® 2-phenylpropionic acid + (CH3)2CHCH2Cl AlCl3 41. Write the structures of the products. 50. Give reasons in two or three sentences only for the following: Br 2 /Fe CH3 CONHC6 H5 ¾¾¾® 42. Give reasons for the following in one or two sentences “Nitrobenzene does not undergo Friedel-Craft’s alkylation.” 43. Complete the following, giving the structures of the principal organic products. Phenyl group is known to exert negative inductive effect, but each phenyl ring in biphenyl (C6 H5 ¾ C6 H5 ) is more reactive than benzene towards electrophilic substitution. 51. Arrange the following in increasing order of reactivity towards sulphonation with fuming sulphuric acid. Benzene, toluene, methoxy benzene, chlorobenzene. Me CH3 ½ 52. Answer the following with suitable equation wherever ½ + H3C—C—CH2Br necessary (i) How can you prepare benzene from lime ? Anhy. AlCl3 ½ (ii) How will you convert toluene to m-nitrobenzoic acid ? H 44. Toluene reacts with bromine in the presence of the light to give benzyl bromide, while in presence of FeBr3 it gives p-bromotoluene. Give explanation for the above observation. 45. Predict the structures of the intermediates/products in the following reaction sequences OMe 54. How many sigma and pi-bonds are present in a benzene molecule ? following using the reagents indicated against in parenthesis. O (i) AlCl3 (ii) H3O+ Zn(Hg)/HCl A MeO B B “Ethyl benzene from benzene.” [C2 H5 OH, PCl 5 , anhyd. AlCl 3 ] 56. Show with balanced equation what happens when the O O H3PO4 46. Predict the major product in the following reactions (i) CH CH COCl/AlCl 3 2 3 Benzene ¾¾¾¾¾¾¾® 55. Write down the reaction involved in the preparation of O + 53. Write down the main product of the following reaction H 2SO 4 + (CH3 )2 CHCH2 OH ¾¾® ‘p-xylene is reacted with concentrated sulphuric acid and the resultant product is fused with KOH’. 57. Give reasons for the following in one or two sentences : ‘Although benzene is highly unsaturated, normally it does not undergo addition reaction.’ Integer Type Questions 58. Among the following, the number of aromatic compound(s) is C2H5 (ii) r (i) Br2 , heat, light s ¾¾¾¾¾¾® (ii) NaCN 47. Give reason for the following ‘In acylium ion the structure R ¾ C ºº O+ is more stable than R ¾ C+ == O’ r r r Benzene and Alkyl Benzene Answers 1. (c) 2. (a) 3. (c) 4. (a) 21. (a) 22. (d) 23. (d) 24. (a) 5. (a) 6. (c) 7. (a) 8. (d) 25. (c) 26. (a,b,d) 27. (b,c,d) 28. (b) 9. (b) 10. (b) 11. (d) 12. (c) 29. (b,c) 30. (b,c,d) 31. (b,c,d) 32. (a,c) 13. (d) 14. (c) 15. (d) 16. (b) 33. (b,d) 34. False 35. False 54. (12s, 3p) 17. (c) 18. (b) 19. (c) 20. (d) 58. (5) Hints & Solutions 1. The given reaction takes place as follows : It will stabilise the carbocation by (+ve) hyperconjugation CH3 O Cl CH3 O AlCl3 – AlCl s CH3 4 CH3 O –HCl –AlCl3 r Intramolecular ArSE 2-reaction O r H CH3 – AlCl4 2. Key Idea In electrophilic aromatic substitution (Ar SE 2) reaction, the aromatic (benzene nucleus here) compound (substrate) acts as a nucleophile which gets attacked by an electrophile in the rate determining step (rds). Higher electron density on the nucleophile, i.e. benzene nucleus will fasten the rate of the reaction. Electron-donating groups (EDG) will increase electron density in benzene nucleus by any or both of their +R/+M and hyperconjugative (HPC) effects. Now, let us consider the given substrate. O Cl –I effect dominates the +R effect C—CH3 –R effect CH3 + HPC (I) (III) (II) As –R effect is more powerful than –I effect, the order of their reactivity, towards an electrophile will be III I (–R) < (–I>+R) II < (+ HPC) 3. Reactivity of substitution nucleophilic unimolecular (S N1) reaction depends on the formation of carbocation. Greater the stability of carbocation, greater will be its ease of formation of alkyl halide and faster will be the rate of reaction. So, the correct order of (S N1) reactivity is CH2Cl CH2—Cl > > > CH3 CH2Cl Cl OCH3 (+R) (C) (A ) (D) (B) In compound C, the carbocation formed is stabilised by activating group ( ¾ OCH3 ). Compound D forms benzyl carbocation (C6H5 ¾ CH2+ ) that is stabilised by resonance. Compound A produces a primary carbocation that further rearranges itself to secondary carbocation. Nus Cl r 1° carbocation Rearrangement r 2° carbocation Compound B produces primary carbocation which is least stable among all the given options. Benzene and Alkyl Benzene 4. The major product of the given reaction is benzoic acid CH2CH3 (C6H5COOH). On vigorous oxidation of alkyl benzene with acidic or alkaline KMnO4, aromatic acids are obtained. During oxidation of alkyl benzene, the aromatic nucleus remains intact and the entire chain is oxidised to ¾COOH group irrespective of the length of carbon chain. –+ CH2CH3 COOK C2H5 C2H5 C2H5Cl (1 eqv.) AlCl3 (anhyd.) C2H5 + C2H5Cl –HCl C2H5 AlCl3 C2H5Cl (excess) COOH C2H5 AlCl3 (anhyd.) (ii) H3O+ (i) Alk. KMnO4 Potassium benzoate Ethyl benzene C2H5 C2H5 H5C2 Benzoic acid H5 C2 C C2H5 2H5Cl + 5. In presence of alc. KOH, the given halide undergoes AlCl3 (Minor) C2H5 (Major) elimination reaction 8. Alc. KOH Cl Cl CH3 Cl The alkene produced undergoes polymerisation via free radical mechanism. This process involve three steps, i.e. initiation, propagation and termination. Cl hn/benzylic free radical substitution with Cl2 n Polymerisation Cl Cl2(1 eqv.) Cl2(2 eqv.) –HCl Cl Polymer Cl2(excess) –2HCl CH2Cl –3HCl CCl3 CHCl2 6. Aromatic electrophilic substitution reactions are fast in those compounds in which the attacking site possess high electron density. Electron withdrawing groups (EWGs ) reduces the electron density in the benzene ring due to its strong –I effect and makes further substitution difficult. Hence, called as deactivating groups. While electron releasing groups (ERGs) increases the electron density in the benzene due to resonance. Therefore, activates the benzene ring for the attack by an electrophile. Hence, called as activating groups. Cl s OH CH2OH So, the correct order is CN — CN > OH CH Cl Cl (Option-c) Gem-diol (unstable) ·· Strong + R effect –H2O — Cl Me OH C –H2O CH O O Strong – I effect (- I > - R) Cl OH C O—H OH Cl In given options, Activating groups are — CH3 < — OCH3 Deactivating groups are s –3H –3OH(aq) –2HCl 2H2O/D –Cl OH(Cl) ·· (+ I, + R) Cl Cl s s Cl – OCH3 COO as Cl (Option-d) < D < A Cl in basic medium < C H3Or B COOH 7. Through aromatic electrophilic substitution mechanism (ArS E2) when mono-alkylation (Friedel-Craft’s reaction) is performed, we get mono-alkylated benzene. As, the alkyl group is ring activating (towards ArS E 2) in nature, again o- and p-substituted alkyl benzene will be obtained and so on. e.g. Cl (Option-b) So, considering the second reaction condition, Cl- , H2O, D in the statement of the question, the correct answer is option (d). Benzene and Alkyl Benzene 9. Melting point of a compound depends on size and force of attraction between the molecules. Compound (c) has largest size and also possess weak intermolecular association due to dipole - dipole interactions. Compound (a) is a dicarboxylic acid and possess high melting point due to intermolecular hydrogen bonding. As a result, it exist as associated molecule. Compound (d) is an alcohol and possess intermolecular H-bonding. No interactions are present in hydrocarbon (naphthalene) compound (b). ~ 80°C), i.e. compound (b). Hence, melting point is lowest for naphthalene (- 10. Aromaticity of a compound can be decided by Huckel’s rule. In cyclopentadienyl cation (b), resonance takes place as follows: Þ Resonance hybrid H Hence, is anti-aromatic does not follow Huckel’s rule as it has conjugated 4p-electron (4 np, n = 1) system. Rest of the species are aromatic as each of them belongs to 6p-electron [(4 n + 2) p, n = 1] system. 11. CH3 COOH COCl CHO SOCl2 KMnO4 H2/Pd BaSO4 Toluene 12. Benzoic acid Benzoyl chloride Benzaldehyde PLAN This problem can be solved by using the various concepts synthesis of benzene, electrophilic substitution reaction and directive influence of various substituents, including oxidation and reduction. Þ —OH and —NH2 are o/p-directing groups. Þ N-acetylation is more favourable than C-acylation. Þ N-sulphonation is more favourable than C-sulphonation. Þ NO2 is a meta-directing group. Þ H2S× NH3 reduces only one NO2 group selectively in the presence of two NO2 groups. Using above concepts the correct sequence of reaction can be written as NO2 A. 3CH CH NH2 D N+2HSO4– NaNO2 H2S .NH3 Fuming HNO3 / H2SO4 Red hot iron 873K H2SO4 selective reduction NO2 OH NO2 Here, N-sulphonation is more favourable in comparisons to C-sulphonation. Hydrolysis NO2 NO2 OH + Conc. H2SO4 B. OH OH OH O2N O 2N NO2 60°C OH OH SO3H Dil.H2SO4 D Here, nitration takes place at o-position of less hindered side. OH SO3H OH Benzene and Alkyl Benzene NH2 NO2 NHCOCH3 C. NO2 CH3COCl Sn/HCl NHCOCH3 NHCOCH3 Conc.H2SO4 HNO3 Here, N-acetylation is more favourable than C-acetylation SO3H SO3H Dil.H2SO4 D NHCOCH3 NH2 NO2 NO2 NO2 NO2 KMnO4 D. NO2 NO2 H3O+ NO2 OH– SOCl2 NH3 – OH/D COO– CH3 COOH CONH2 COCl Above reaction proceeds through oxidation, chlorination and amide formation sequentially. \ A ® r, B ® s, C ® q, D ® p Hence, the correct choice is (c). 13. Draw the structure of organic compounds indicating net dipole moment which includes lone pair and bond angle also. OH In the quinol SH O H O H and thioquinol same as in thioquinol. SH OH — OH groups and —SH groups do not cancel their O dipole moment as they exist in different confirmation. 14. H H O COOH COOH HNO3 H2SO4 HO P HO NO2 — OH is activating while — COOH is deactivating group in SE Ar reaction. Therefore, electrophile attack to ortho of the activating — OH group. OCH3 OCH3 HNO3 H2SO4 H3C NO2 H 3C Q Both — OCH3 and — CH3 are activating ortho/para directing groups but — OCH3 is stronger activator, electrophile attack to ortho of — OCH3. NO2 O O C C O II I HNO3 H2SO4 O S Ring II is activated while ring I is deactivated in SE Ar reaction. Therefore, electrophile attack at para to ring-II, the less hindered position. Benzene and Alkyl Benzene ¾ OCH3 + HBr ¾® CH3Br + 15. 21. SO2Cl 2 brings about free-radical chlorination at —CH3 group. ¾ OH 22. Free radical chlorination occur at —CH3. —OH of phenol is not further substituted by bromide due to resonance with the ring. —CH3 + Cl2 hv Heat —CCl3 16. Ring attached to nitrogen is activated while ring attached to C==O is deactivated. Also, electrophilic substitution occur predominantly at para position of the activated ring due to immense steric hindrance at ortho position. NaOH —COOH benzoic acid O 23. The methyl group in toluene is ortho/para directing activating Conc. HNO3 N group : Conc. H2SO4 H CH3 O2N CH3 CH3 O N Cl FeCl3 + Cl2 + H Cl 17. AlCl3 + ClCH2CH2CH3 Friedel-Craft’s reaction O2/heat H3O+ P 24. Toluene is most reactive among these. Nitro and carboxylic groups are deactivating in aromatic electrophilic substitution reaction. 25. Toluene is most readily sulphonated among these because Q + Phenol, methyl group is electron donating (+ I effect), activate benzene ring for electrophilic aromatic substitution. O 26. Reaction shown in option (a) is aldol condensation in the , Q = CH3—C—CH3 P= Acetone Cumene 18. Ring attached to nitrogen is activated by electron donating presence of conc. H2SO 4 at high temperature. In summerised way the formation of mesitylene through this can be visualised as CH3 resonance effect while ring attached to carbonyl group is deactivated by electron withdrawing resonance effect : HN O H 3C HN CH3 Fe Br2 CH3 H3C Br aromatic substitution reaction. Also nitro group is stronger deactivating group. Methyl group is activator in electrophilic aromatic substitution. Hence, overall order of reactivity is : nitrobenzene < chlorobenzene < benzene < toluene H H ½ + Cl 20. C6H5 ¾ C ¾ Cl + SbCl 5 ¾® SbCl6- + C6H5 ¾ C ¾® ½ CH3 CH3 planar (+)-2-chloro2-phenylethane (±) -2-chloro-2-phenylethane carbocation (racemic mixture) HH O H—C CH3 19. Both chloro and nitro groups are deactivating in electrophilic C O H H C CH3 H C H2O O H C H H H2O O Conc. H2SO4 D H2O –3H2O C CH3 CH3 Mesitylene CH3 Reaction given in option (b) is simple polymerisation (trimerisation) reaction of alkyne i.e., Me H or CH3 C CH when passed through heated iron tube at 873K then mesitylene is formed as CH3 CH3 C CH HC C CH3 Heated iron tube 873 K C HC CH3 CH3 Mesitylene This reaction is also called aromatisation. CH3 Benzene and Alkyl Benzene (1) and (2) reactions of option (c) combined to give haloform reaction while (3) reaction given in this option is decarboxylation reaction i.e., 28. + CH2 — CHCH3 H COCH3 (1) Br2/NaOH – COONa+ O OH CH3 Cumene hydroperoxide 29. According to Huckel rule, the compounds which have + 3CHBr3¯ +– CH3 O2 Cumene hydroperoxide formed above is an intermediate in the synthesis of phenol. COCH3 – COONa+ Bromoform NaOOC + Cumene (2) H3O+ H3COC CH3 CH3 4 n (n = 0, 1, 2, 3¼) delocalised p-electrons in a close-loop are anti-aromatic and characteristically unstable. Compound B satisfy the criteria of anti-aromaticity as : The above product of haloform reaction on decarboxylation gives benzene as 4p – COONa+ Compound C is anti-aromatic in its resonance form : – O O (Sodalime) NaOH/CaO + – –3Na2CO3 – COONa+ NaOOC Benzene + The reaction given in option (d) is Clemmensen reduction i.e., Zn - Hg C == O ¾ ¾ ¾® HCl CH2 + H2O Hence, the final product of this reaction is also mesitylene which can be seen as anti-aromatic Compound A has 4p-electrons which are also delocalised but do not constitute close loop, hence non-aromatic. Compound D is aromatic, characteristically stable. CH3 CHO – O O Zn-Hg/HCl OHC CHO H3C Mesitylene 27. (a) Br Cl AlCl3 1, 2-hydride shift 1° carbocation SE (c) (d) H + OH + Order of stability Aromatic > Non-aromatic > Anti-aromatic 30. Aromatic molecule must. NaOC2H5 C2H5O- (a strong nucleophile) causes E1 reaction to form isobutene as the major product. (b) + Tropyllium ion, aromatic CH3 3° carbocation (Major) (b) have (4 n + 2)p electrons. (d) be cyclic. 31. A compound is considered to be aromatic, if it follows three rules: (a) Must be cyclic and planar. (b) Must have conjugated system in it. (c) It must follow Huckel rule which states that number of p-electrons = (4 n + 2) (A) Þ 2p-e- system [(4 n + 2)p , n = 0 ]Þ Aromatic (B) Þ 4p-e- system [ 4 np , n = 1]Þ Anti-aromatic (C) Þ 8p-e- system [ 4 np , n = 2 ]Þ Anti-aromatic (D) Þ The sp3-carbon does not support conjugation and Tertiary butyl benzene(X) (X) BF3,Et2O (c) be planar. sp3 also devoids planarity. Þ Non-aromatic Benzene and Alkyl Benzene CH3 32. CH3 N==O + + N==O N==O Fe + Br2 – (c) (i) Toluene – Br — CH3 in toluene is para directing group.It activates the ring by both inductive and hyperconjugation effect. Hence, —NO is ortho/para directing although deactivating group. Therefore, ortho/para nitro derivatives are formed on treatment with mixed acid. 33. In both Friedel-Craft’s reaction and Reimer-Tiemann reaction, NO2 new carbon-carbon bond is formed : + R—Cl R Friedel-Craft’s + HCl reaction AlCl3 OH (ii) NO2 + Conc. HNO3 Conc. H2SO4 NO2 OH + CHCl3 + OH CHO H+ – Nitro group is meta directing as well as strongly deactivating. H2O Reimer-Tiemann reaction 34. An electron donating substituent in benzene orient ortho/para substitution. Pd/C (d) H2 2 35. In benzene, carbons are sp -hybridised only, hence only two p-orbitals are involved in hybridisation. 36. (a) (i) CH3 CH3 C—Br C—OC2H5 + C2H5OH CH3 CH3 + HBr Acid CH3 (ii) Br CH3 OH —Br + Ag+ + F– O2N + O Nitro group from para position activate aromatic nucleophilic substitution. F Br2 N— 38. + NaOH (ii) Fe No reaction deactivated H 3C + AgBr anti-aromatic (unstable) 5-bromo-1,3-cyclopentadiene CH3 CH3 + Tropylium ion (aromatic, stable) No reaction + NaOH O2N AgBr¯ + —Br + Ag+(aq) 37. 7-bromo-1,3,5-cycloheptatriene + C2H5OH F (b) (i) In the above hydrogenation reaction, the anti-aromatic character of three cyclobutadiene rings are relieved that provide stability to the hydrogenation product. activated O CH2NO2 —NO2 is not in resonance with benzene ring, does not activate the aromatic nucleophilic substitution reaction, hence fluoride is not formed. N— —Br Benzene and Alkyl Benzene 39. (i) Oxidation of side-chain alkyl group of benzene occur through free radical mechanism and it initiates at a -C as. ½ ½ Ph — C— H ¾® Ph — C · + H · ½ ½ 42. Nitro group is a very strong deactivating group, the very slow Friedel-Craft’s reaction does not takes place in presence of nitro group. 43. CH3 CH3 CH3 In tertiary butyl benzene, there is no a - H, required to initiate free radical oxidation reaction, hence not oxidised with KMnO4. + CH3—C—CH2Br CH3 C—CH3 AlCl3 CH3 H CH3 CH3 + C—CH3 + KMnO4 CH3 No oxidation takes place CH3 tert butyl benzene (no a-H) CH3 (ii) Pi-electrons of double bonds are involved in aromatic delocalisation (aromaticity), hence electrophilic addition do not occur as it would destroy aromatic stability. However, electrophilic substitution do not destroy aromaticity. E H+ + CH3 C E+ + HBr 44. In presence of light, free radical reaction takes place at — CH3 group while in presence of FeBr3, electrophilic substitution occur in the ring. CH3 H Addition CH2Br Br2/heat or light Br aromaticity retained (occur preferably) electrophilic substitution Br2/FeBr3 loss of aromaticity (not occur) CH3 CH3 + CH2CH3 COOH Br K2Cr2O7 40. (i) OCH3 Br OCH3 O CaO/NaOH Heat CH2CH3 (ii) CH—CH3 Br2 + 45. O (i) AlCl3 Zn(Hg) (ii) H3O+ HCl O NaCN O Br heat CH3 CH—CH3 H /H2O CN NHCOCH3 41. OCH3 C—COOH + COOH Friedel-Craft’s reaction H3PO4 H Final product NHCOCH3 Br2/Fe COOH CH3 Br NHCOCH3 46. (i) + (CH3)2CHCH2OH C—CH3 H2SO4 CH3 + H Br NHCOCH3 is ortho/para directing although deactivating group. Therefore, Ortho/pare bromoderivatives are fomed or treatment. CH2CH3 (ii) (i) Br2/heat (ii) NaCN C—CH3 CN Benzene and Alkyl Benzene + + CaC2 + 2H2O ¾® Ca(OH)2 + 47. R — C ºº O ¬® R — C==O I Cu I is more stable resonance structure of acylium ion than II on the following grounds. It has more covalent bonds. It has complete octet of both carbon and oxygen. 3C2H2 ¾¾® D HNO3/H2SO4 mononitration —C—O— deactivated ring activated ring CH3 Conc. HNO3 Conc. H2SO4 NO2 O —NO2 + CH3CH2COCl 53. C—CH2CH3 AlCl3 CH3 CH3—C—CH2Cl AlCl3 Hydride + CH3—C—CH2 H shift H H H It has 12 sigma and 3 pi-bonds. C—CH3 C6H6 H 55. CH3CH2OH + PCl 5 ¾® POCl 3 + HCl + CH3CH2Cl CH3 + 3° carbocation (more stable) AlCl 3 + C6H6 + CH3CH2Cl ¾¾® C6H5— CH2CH3 – Friedel-Craft’s reaction — — — H 54. CH3 CH3 CH3—C—CH3 Friedel-Craft’s acylation H H 1° carbocation 50. COOH O (ii) C6H5COOH + CH3MgI ¾® CH4 + Mg(OOCC6H5 )I 49. COOH K2Cr2O7 —C—O— CH3 C6H6 Benzene (ii) —CH3 in toluene is ortho/para directing group, first oxidised to —COOH in order to make it meta directing and then nitrated. O 48. (i) C2H2 Acetylene II 56. H3C— + — — – —CH3 + Conc. H2SO4 p-xylene SO3H The above resonance activate an aromatic ring for electrophilic substitution reaction. 51. Both — CH3 and — OCH3 are activating groups but chloro is deactivating group in electrophilic aromatic substitution reaction. Also — OCH3 activate more than — CH3 due to resonance effect by the former group. Therefore, the overall order of reactivity is: Chlorobenzene < benzene < toluene < methoxy benzene H3C— CaO + Lime 3C Coke 2000 °C ¾¾¾® CaC2 + CO ­ KOH fusion H3C— —CH3 57. Pi-bonds of benzene are involved in aromaticity, not open for electrophilic addition reaction. Rather, it undergo electrophilic substitution reaction. 58. The aromatic systems are 52. (i) Lime is heated with coke at high temperature and then, hydrolysed to form acetylene. This acetylene on passing red hot tube polymerises to form benzene : —CH3 OH s r r , , , ,