C2-Allens Made Chemistry Exercise {PART-1}

It is possible to reduce the activity such groups (by side chain reaction) so that the reaction can be stopped after mono substitution then and again by a side chain reaction the original group is restored. Effective use can sometimes be made of removable blocking groups on the ring. 1 . Which of the following synthesis could be done in the ring step ? OH OH OH OH Br Br (A) (B) CH3 Br CH3 Br NH2 OH OH NH2 (C) Br (D) Br CH3 2 . Which of the following is the correct major product ? OCH3 OCH3CH3 CH3 O CH3 O CH3 (CH3)2CHBr CH3–C–Cl CH3 AlCl3,15–25°C CH3 (A) (B) AlCl3,CS2 5°C CH3 H3C CH3 H3C CH3 NH2 NH2 OH OH Br Br Br Br Br2–.H2O Br2 CS2,0°C (C) 40–50°C O (D) OH O Br OH Br 3 . Which of the following side chain reaction/s can be used to reduce the activity of strongly activating groups like —OH ? (A) benzoylation (B) acetylation (C) both of the above (D) none of the above Comprehension # 3 A third group is least likely to enter between two groups in the meta relationship. This is the result of steric hindrance and increases in importance with the size of the groups on the ring and with the size of the attacking species. When a Meat-directing group is meta to an ortho-para directing group, the incoming group primarily goes ortho th the meta directing group rather than para. 1 . Chlorination of m-chloro nitro benzene gives : NO2 NO2 NO2 NO2 Cl (A) (B) (C) Cl Cl Cl Cl (D) Cl Cl

CH3 OH 2 . E  ......... ? CH3 CH3 CH3 OH OH OH (A) (B) (C) (D) all of them E E E O HN CH3 3 . CH2 /CH3COOH ? OOO O HN CH3 HN CH3 HN CH3 HN Cl Cl Cl (A) Cl (B) (C) (D) CH3 CH3 Cl CH3 CH3 4. H NO 3 , unlikely to form is - H2SO4 NO2 CH3 CH3 (C) CH3 NO2 CH3 NO2 O2N NO2 (D) NO2 (B) NO2 (A) NO2 NO2 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3  True / False 3. T 4. F 5. F  1. F 2. F  – (v) CH3—C—CH3  Fill in the Blanks (iv) [CH 3 — Cl — AlCl3 ] ; CH3 1. (i) Br+ ;  (iii) SO ; 3 (ii) N O2 ; HH 3. (a) sp2 ; (b) sp3 ; (c) sp3 ; (d) sp2 2. (i) + SO3 (ii) + NO2  Match the Column 1. (A) q,r ; B q,r,s ; (C)  p ; (D) s 2. (A) q, r ; (B) p, r ; (C)  q, p ; (D) p  Assertion - Reason Questions 1. D 2. D 3. C 4. A 5. A 6. A 7. B 8. D 9. A 10. D  Comprehension Based Questions Comprehension #1 : 1. (A) 2. (B) 3. (A) 4. (C) Comprehension #2 : 1. (C) 2. (D) 3. (C) Comprehension #3 : 1. (A) 2. (D) 3. (B)

EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE H  Br 1. Br + FeBr4– H + FeBr3 H Br Why does the given step does not occur ................ 2 . Write resonance structures of arenium ions with : (i) N O  ,  2 (ii) C H3 3 . C6H4(NO2)CH3 can form different isomers. Write three isomers. (I) ........................... (II) ............................ (IIII) ............................. 4 . Write the structure of the products formed in the following SE reactions : CO2H CH3 (i) HNO3/ H2SO4  ............. (ii) CH3COCl/ AlCl3 .............. CO2H CH3 OH Cl (iii) Br2/ H2O ................. (iv) HNO3/ H2SO4  .................. CH3 (v) Br2Fe ................. 5 . Indicate the position of the electrophile in the following compounds : Cl (i) H3C (ii) CO2H (iii) CH3 HO2C 6 . Indicate the compound which can be nitrated most easily : CH3 CH3 CH3 CH3 (i) (ii) NO2 NO2 (iii) NO2

7 . Indicate wrong electron-migration : :O—CH3: :O+—CH3 :O—CH3 ::  HE  HE :II I EH III 8 . Indicate more stable carbocation and explain : OCH3 OCH3  (i) (ii)  EH HE 9 . In case of meta- attack on aniline (where —NH2 group is o-, p-directing) and in case of ortho-attack on nitrobenzene, resonating structures are possible. Which of them are more stable ? NH2 NO2  H E H E  (I) (meta-attack) (II) (ortho-attack) 1 0 . What are the major products of the following : CH3 OH NO2 (i) Br2/ Fe  A NO2 (ii) HNO3/ H2SO4  B OH (iii) Br2/ Fe  C NO2

CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) 1 . The given reaction does not occur because on addition the ring lost its aromaticity and on deprotonation of - complex it retains the aromaticity hence substitution reaction takes place. O– + O O– + O O– + O O– + O N N N N + + 2 . (i) + CH3 + CH3 CH3 CH3 + + + (ii) + CH3 CH3 CH3 NO2 3 . (i) (ii) (iii) NO2 NO2 CH3 OH Br Br COOH Cl 4 . (i) CO2H (ii) (iii) (iv) O2N Br CH3 Br NO2 COCH3 CH3 (v) + —Br E Cl 5 . (i) H3C E (ii) CO2H (iii) CH3 HO2C E CH3 6 . (i) ; because —CH3 is an activating group hence toluene is most reactive. 7 . (III) 8 . (I) is more stable, due to the + M effect of —OCH3. 9 . (I) CH3 OH Br OH Br NO2 (iii) NO2 1 0 . (i) (ii) NO2 NO2

EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . Unlike most hydrocarbons, azulene (C10H8) is highly coloured (deep blue), although its isomer, naphthalene, does not have significant zwitter-ionic character, azulene does. (i) Draw a resonance structure of azulene in which five-membered ring is anionic and the seven-membered ring cationic. (ii) Can azulene be considered aromatic ? 2 . Explain, which of these compounds is a stronger acid. : CN CN (i) or (ii) or (iii) or Ph Ph Ph Ph 3 . Arrange A, B, C in increasing potential energy : CH3 CH3 H NO2 NO2 + E+ H +H E + E + E+ +E + E+ , , C A B 4. C 6 H  h as electron withdrawing inductive effect yet each ring of biphenyl, C6H5–C6H5 is more reactive 5 than benzene towards SE reaction and chief products are ortho and para-isomer. Show how reactivity and orientation can be accounted for on the basis of resonance. 5 . What are the major products in the following reactions : CH3 (a) + CH3—C—CH2Br AlCl3  C (b) + CH2Cl2 AlCl3  D  CH3 (excess) 6 . Arrange the following in increasing order of electrophilic strength : ++ + I II III 7 . Which is more reactive towards SE and explain ? I II I II OH O OH OH (i) (ii) NO2

8. What are the major products when N O  (ni tronium) reacts with O . 2 —O—C— 9 . Which of the following compounds has the greater dipole moment ? O CO C 1 0 . Explain why pyridine does not undergo Friedel-Craft's reaction ?

BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(B) 1 . (i) (ii) Azulene is an aromatic specie. CN 2 . (i) (ii) (iii) 3. A < B < C Ph Ph +– + – 4. 5 . (a) CH3 (b) —CH2— —C—CH3 C2H5 6 . I > II > III 7. (i) (II) is more reactive, (ii) (II) is more reactive 8 . O2N— O —O—C— 9 . In the case of the compound on the left, the three membered ring becomes aromatic when the charges are separated. In the case of the compound on the rignt, the resonance contributor with separated charges does not have any additional aromaticity. Thus, the resonance contributor with separated charges is more stable and, therefore, makes a greater contribution to the hybrid for the compound on the left than for the compound on the rignt. The compound on the left, therefore, has the greater dipole moment. O O– C C+ O O– C C+ 1 0 . The Lewis acids AlCl3 or FeCl3, used as catalyst in Friedel - Craft's reaction coordinate with nitrogen of the pyridine through lone pair of electrons and form a complex. The +ve charge on N greatly diminishes the reactivity towards electrophilic substitution. + AlCl3 +N N –AlCl3 complex

EXERCISE - 05 [A] JEE-[MAINS] : PREVIOUS YEAR QUESTIONS 1. Which of these will not react with acetylene - [AIEEE-2002] (A) NaOH (B) ammonical AgNO (C) Na (D) HCl 3 2. What is the product formed when acetylene reacts with hypochlorous acid - [AIEEE-2002] (A) CH3COCl (B) ClCH2CHO (C) Cl2CHCHO (D) ClCH2COOH [AIEEE-2003] 3. 1–Butene may be converted to butane by reaction with - (A) Pd/H2 (B) Zn - HCl (C) Sn - HCl (D) Zn - Hg/HCl 4. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloroalkane. This alkane could be - [AIEEE-2003] (A) neopentane (B) propane (C) pentane (D) isopentane 5. Which one of the following is reduced with Zn-Hg/HCl to give the corresponding hydrocarbon (A) Butan-2-one (B) Acetic acid [AIEEE-2004] (C) Acetamide (D) Ethyl acetate 6. Which one of the following has the minimum boiling point : [AIEEE-2004] (A) isobutane (B) 1–butyne (C) 1–butene (D) n–butane 7. 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly[AIEEE-2005] (A) 2–bromo-2–methylbutane (B) 1–bromo-2–methylbutane (C) 1–bromo-3–methylbutane (D) 2–bromo-3–methylbutane 8. Alkyl halides react with dialkyl copper reagent to give [AIEEE-2005] (A) alkyl copper halides (B) alkenes (C) alkenyl halides (D) alkanes 9. Reaction of one molecule of HBr with one molecule of 1,3–butadiene at 40°C gives predominantly (A) 1–bromo–2–butene under thermodynamically controlled conditions [AIEEE-2005] (B) 3–bromobutene under kinetically controlled conditions (C) 1–bromo–2–butene under kinetically controlled conditions (D) 3–bromobutene under thermodynamically controlled conditions 10. Acid catalyzed hydration of alkenes except ethene leads to the formation of [AIEEE-2005] (A) secondary or tertiary alcohol (B) primary alcohol (C) mixture of secondary and tertiary alcohols (D) mixture of primary and secondary alcohols 11. Elimination of bromine from 2–bromobutane results in the formation of [AIEEE-2005] (A) predominantly 2–butene (B) equimolar mixture of 1 and 2–butene (C) predominantly 2–butyne (D) predominantly 1–butene Me 12. Me OH  N Et n Bu The alkene formed as a major product in the above elimination reaction is- [AIEEE-2006] Me Me (A) Me (B) CH2=CH2 (C) (D)

13. Reaction of trans-2-phenyl-1-bromocyclo pentane on reaction with alcoholic KOH produces- (A) 4-phenyl cyclopentene (B) 2-phenyl cyclopentene [AIEEE-2006] (C) 1-phenyl cyclopentene (D) 3-phenyl cyclopentene 14. Phenyl magnesium bromide reacts with methanol to give- [AIEEE-2006] (A) A mixture of anisole and Mg(OH)Br (B) A mixture of benzene and Mg(OMe)Br (C) A mixture of toluene and Mg(OH)Br (D) A mixture of phenol and Mg(Me)Br 15. Which of the following reactions will yield, 2, 2-dibromopropane [AIEEE-2007] (A) CH3—C CH + 2HBr  (B) CH3CH CHBr + HBr  (C) CH CH + 2HBr  (D) CH3 — CH CH2 + HBr  16. In the following sequence of reactions, the alkene affords the compound ‘B’ :- [AIEEE-2008] CH3CH=CHCH3 O3  A HZ2nO B. The compound B is (A) CH3CH2CHO (B) CH3COCH3 (C) CH3CH2COCH3 (D) CH3CHO 17. The hydrocarbon which can react with sodium in liquid ammonia is [AIEEE-2008] (A) CH3CH2CH2C  CCH2CH2CH3 (B) CH3CH2C  CH (C) CH3CH=CHCH3 (D) CH3CH2C  CCH2CH3 18. The treatment of CH3MgX with CH3C  C–H produces [AIEEE-2008] (A) CH3–CH=CH2 (B) CH3C  C–CH3 HH (D) CH4 || (C) CH3—C=C—CH3 19. The main product of the following reaction is [AIEEE-2010] C6H5CH2CH(OH)CH(CH3)2 Conc. H2SO4 ? H5C6CH2CH2 C = CH2 H5C6 C= C H CH(CH3)2 (A) (B) H3C H C6H5CH2 C= C CH3 C6H5 C= C CH(CH3)2 CH3 H (C) (D) H H 20. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is :- [AIEEE-2010] (A) ethene (B) propene (C) 1-butene (D) 2-butene 21. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :- [AIEEE-2011] (A) an isopropyl group (B) an acetylenic triple bond (C) two ethylenic double bonds (D) a vinyl group

22. Ozonolysis of an organic compound 'A' produces acetone and propionaldehyde in equimolar mixture. Identify 'A' from the following compounds :- [AIEEE-2011] (A) 2 - Methyl - 1- pentene (B) 1 - Pentene (C) 2 - Pentene (D) 2 - Methyl - 2 - pentene 23. 2–Hexyne gives trans –2–Hexene on treatment with :- [AIEEE-2012] (A) Li AlH4 (B) Pt/H2 (C) Li/NH3 (D) Pd/BaSO4 24. In the given transformation, which of the following is the most appropriate reagent ? [AIEEE-2012] CH=CHCOCH3 CH=CHCH2CH3 Re agent HO HO (A) NaBH4 (B) NH2 NH2,OH (C) Zn – Hg / HCl (D) Na, Liq.NH3 JEE-[MAIN] : PREVIOUS YEAR QUESTIONS ANSWER KEY EXERCISE -5[A] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 1 3 1 1 1 1 1 4 1 1 1 2 4 2 1 Que. 16 17 18 19 20 21 22 23 24 Ans 4 2 4 2 4 4 4 3 2

EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . The clorination of toluene in presence of ferric chloride gives predominantly : [IIT-86] (A) Benzyl chloride (B) m-chlorotoluene (C) Benzal chloride (D) o–and–p–chlorotoluene 2 . Aryl halides are less stable towards nucleophillic substitution reaction as compared to alkyl halides due to- (A) The formation of less stable carbonium ion (B) Resonance stabilization (C) Longer carbon-halogen bond (D) The inductive effect [IIT-90] 3 . The most basic compound among the following is - [IIT-90] (A)Benzylamine (B) Aniline (C) Acetanilide (D) p-nitro aniline 4 . Excess chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives : (A) o-cresol (B) p-cresol (C) 2, 4-dihydroxytoluene (D) Benzoic acid [IIT-90] 5 . Choose the correct statement from the ones given below for two anilium in - [IIT-93]  NH3 + NH3 (I) (II) (A) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions (B) II is not an acceptable canonical structure because it is aromatic (C) II is not an acceptable canonical structure because the nitrogen has 10 valence electrons (D) II is an acceptable canonical structure 6 . The major product of nitration of benzoic acid is : [IIT-93] (A) 3–nitrobenzoic acid (B) 4–nitrobenzoic acid (C) 2–nitrobenzoic acid (D) 2,4–dinitrobenzoic acid 7 . Most stable carbonium ion is : [IIT-95] (A) p–NO2–C6H4– +CH2 (B) C 6H + C H 2 5 (C) p–Cl–C6H4– +CH2 (D)p–CH3O–C6H4– +CH2 8 . Arrange in order of decreasing trend towards SE reactions, [IIT-95] Chlorobenzene, Benzene, Anilinium chloride, Toluene I II III IV (A) I > I > III > IV (B) III > I > II > IV (C) IV > II > I > III (D) I > II > III > IV 9 . Among the following statements on the nitration of arotmatic compounds, the false one is - [IIT-97] (A) The rate of nitration of benzene is almost the same as that of hexadeuterobenzene (B) The rate of nitration of toluene is greater than that of benzene (C) The rate of nitration of benzene is greater than that of hexadeuterobenzene (D) Nitration is an electrophilic substitution reaction 1 0 . Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and conc. H2SO4. In the nitrating mixture HNO3 acts as a - [IIT-98] (A) Base (B) Acid (C) Reducing agent (D) Catalyst 1 1 . Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with - [IIT-98] (A) SO2Cl2/h (B) SOCl2 (C) PCl5 (D) NaOCl

1 2 . The most unlikely representation of resonance structure of p-nitrophenoxide ions is - [IIT-98] O– + O O– + O– O+ O O– + O N N N N (A) (B) (C) :(D) O– O O– O 1 3 . The reaction of CH3–CH=CH– –OH with HBr gives - [IIT-98] (A) CH3–CHBrCH2 OH (B) CH3CH2CHBr OH (C) CH3CHBrCH2 Br (D) CH3CH2CHBr Br 1 4 . A new C–C bond is formed in - [IIT-98] (A) Cannizzaro' reaction (B) Friedel Crafts reaction (C) Both (A) & (B) (D) None of these 15. Amongst the following the strongest base is - [IIT-2000] (A) C6H5NH2 (B) p–O2NC6H4NH2 (C) m–O2NC6H4NH2 (D) C6H5CH2NH2 O NH Me ? Major product in this reaction is : [IIT-2004] Me Br2 16. Fe NH O NH O Me Me Me Me (B) (A) Br Br O O NH Me NH Me Me Me (C) (D) Br Br SO3H NO2 NO2 (I) NaHCO3 (II) NaHCO3 Gases released in reaction (I) and (II) are : [IIT-06] 17. (A) C O , CO2 OH (C) SO2, CO2 (D) SO2, NO 2 (B) SO2,NO2

1 8 . Order of boiling point for the following compounds is - [IIT-06] OH OH OH OH OH OH OH (III) (IV) (I) (II) (A) (I) < (II) < (III) < (IV) (B) (I) < (II) < (IV) < (III) (C) (IV) < (I) < (II) < (III) (D) (II) < (I) < (III) < (IV) 1 9 . Identify the correct order of reactivity in electrophilic substitution reaction of the following compounds [IIT-2002] CH3 Cl NO2 1 2 34 (A) 1 > 2 > 3 > 4 (B) 4 > 3 > 2 > 1 (C) 2 > 1 > 3 > 4 (D) 2 > 3 > 1 > 4 [IIT 1992] 20. C6H6 + (CH3)2CHCH2Cl AlCl3 ?  21. —COO— HNO3 / H2SO4 [IIT 1993] mononitration 2 2 . C6H6 + (CH3)2CH.CH2OH H2SO4  ? [IIT 1994] 2 3 . Fill in the blanks with appropriate structure of reaction products in the following transformation : O—HOOC—C6H4—CH2—C6H5 SOCl2 A anlydrousB ZnHgC [IIT 1995] AlCl3 HCl 2 4 . Predict the structures of the intermediates/products in the following reaction sequence. [IIT 1996] OMe O O + MeO O ((iii))HAl3COl3  A ZnHg/ HCl B H 3 PO 4  O 2 5 . Predict the product [IIT 1997] Me CH3 CCl3 (i) Br Anlydrous ? (ii) Cl2 + H3C AlCl3 /  Fe H 2 6 . Each of the following raction gives two products. Write the structure of the products : [IIT 1998] CH3CONHC6H5 Br2,Fe

2 7 . Predict the product [IIT 2002] O N— Br2/ Fe  ? (1eq.) 2 8 . Predict the product [IIT 2002] O Cl2,FeCl3 A NaHg/ HCl B HNO3/ H2SO4  C Cl CO2H CO2H 2 9 . Convert : [IIT 2003] F 3 0 . 7-bromo-1, 3, 5-cycloheptatriene exists as a ionic substance whereas 5-bromo-1, 3-cyclopentadiene does not ionize even in the presence of Ag+ ion. Explain ? [IIT 2004] CH3 31. (B) NaBr MnO2(A ) Conc.HNO3  (C) (D). (brown fumes & Pungent smell) (Explosive Product intermediate) Identify the missing compounds. Give the equation from A to B and A to C. [IIT 2005] 3 2 . Give reason : O O O N N N : : : (a) (i) Conc.HNO3  NO2 + C o nc.H 2 S O 4 NO2 NO2 NO2 (ii) Conc.HNO3  C o nc.H 2 S O 4 NO2 (b) Pd /C  is formed but not 3mole of H2

3 3 . CH3–CH=CH2 + NOCl  P CH3  CH  CH2 [IIT 2006] (B) | | Identify the adduct. [IIT 2007] NO Cl [IIT 2007] CH3  CH  CH2 [IIT 2007] (A) | | Cl NO NO | CH2  CH2  CH2 (D) | | (C) CH3  CH2  CH | NO Cl Cl 3 4 . The number of stereoisomers obtained by bromination of trans-2-butene is (A) 1 (B) 2 (C) 3 (D) 4 3 5 . The number of structural isomers for C6H14 is (A) 3 (B) 4 (C) 5 (D) 6 3 6 . The reagent(s) for the following conversion, ? H H is / are (A) alcoholic KOH (B) alcoholic KOH followed by NaNH2 (C) aqueous KOH followed by NaNH2 (D) Zn / CH3OH 3 7 . The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are [IIT-2010] (A) BrCH2CH2CH2CH2CH3 and CH3CH2CCH (B) BrCH2CH2CH3 and CH3CH2CH2CCH (C) BrCH2CH2CH2CH2CH3 and CH3CCH (D) BrCH2CH2CH2CH3 and CH3CH2CCH

JEE-[A DVANCE] : PREVIOUS YE AR QUESTIONS ANSWER KEY EXERCISE -5[B] 1. (D) 2. (B) 3. (A) 4.(D) 5. (C) 6. (A) 7.(D) 8. (C) 9. (C) 10. (A) 11. (A) 12. (C) 13. (B) 14. (B) 15. (D) 16. (D) 17. (A) 18. (C) 19. (A) 20. CMe3 21. —COO— —NO2 (Major) 22. C6H6 + (CH3)2CH.CH2OH H2SO4  CMe3 —CH2— , 23. , COCl O OMe OCH3 Me CCl3 24. (A) CO2H (B) CO2H 25. (i) (ii) O Cl CMe3 O O —Br 27. N— —Br 26. CH3—C—NH— O Cl Cl O NO2 Cl (C) 28. (A) (B) Cl Cl Cl 30. is aromatic whereas is antiaromatic.  CH3 CH3 O2N NO2  31. Br2 NaBr MnO2 H2SO 4 Conc.HNO3  ( NO 2 ) NO2 32. (a) (i) —NO is ortho and pera director. (ii) —NO is meta director. 2 (b) due to high resonence energy of the product. 33. A 34. A 35. C 36. B 37. D



EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Carbanion is a :- (A) Base (B) Nucleophile (C) Both (A) and (B) (D) None 2 . Electrophile is : (A) H2O (B) NH3 (C) AlCl3 (D) C2H5NH2 3 . Which of the following species has a pyramidal shape-  (A) CH3+ (B) BF3 (C) CH – (D) C H 3 3 4 . Increasing order (least basic first) of basic strength is shown by the set (A) ClNH2, NH3, CH3NH2 (B) ClNH2, CH3NH2, NH3 (C) NH3, ClNH2, CH3NH2 (D) CH3NH2, ClNH2, NH3 5 . The decreasing order of acid strength indicated by the following sequence of reaction is :- –C–H NaNH2 –C–Na C2H2 HC CNa+ Ph3CH H2O HC CH + NaOH (A) NH3 > Ph3CH > C2H2 > H2O (B) H2O > HC  CH > Ph3CH > NH3 (C) HC  CH > H2O > Ph3CH > NH3 (D) Ph3CH > HC  CH > H2O > NH3 6 . In which of the following compounds is hydroxylic proton the most acidic - F I O (C) O (D) O (A) O (B) H F H H H F 7 . Correct arrangement of the following acids in correct Ka order is :-  (I) H3N—CH2—COOH (II) NC—CH2—COOH (III) H3C—CH2—COOH (IV) OOC —CH2—COOH (A) I > II > III > IV (B) II > I > III > IV (C) I > III > II > IV (D) IV > III > II > I 8 . Which of the following orders of acid strength is correct :- (A) RCOOH > ROH > HOH > HC CH (B) RCOOH > HOH > ROH > HC CH (C) RCOOH > HOH > HC CH > ROH (D) RCOOH > HC CH > HOH > ROH 9 . Which of them is false for order of – I effect (A) –F > –Cl > –Br > I  (B) –NR3 > –NH3 > –NO2 (C) –OCH3 > –OH > –NH2 (D) > –C  CH > H 1 0 . The order of basicity of halides is : (B) F– < I – < Br– < Cl– (A) Cl– < Br– < I – < F– (D) Cl– < F– < I – < Br– (C) I – < Br– < Cl– < F–

1 1 . Which of the following molecule has longest C=C bond length - (A) CH =C=CH (B) CH –CH=CH 22 32 CH3 (D) CH3–C=CH2 (C) CH3–C–CH=CH2 CH3 CH3 1 2 . In the reaction CH3CN Hh3eOat  CH3COOH the hybridization state of the functional carbon changes from (A) sp3 to sp2 (B) sp2 to sp3 (C) sp to sp2 (D) sp2 to sp 1 3 . Most stable carbocation is :-     CH2 CH2 CH2 CH2 (A) (B) (C) (D) Cl NO2 OCH3 1 4 . Most stable carbanion is :- (A) HC C (B) C6H5 (C) ( C H 3 ) 3C — C H  (D) (CH3)2C CH 2 1 5 . Consider the following carbocations (a) CH3O–  (b)   –CH2 –CH2 (c) CH3– –CH2 (d) CH3— C H2 The relative stabilities of these carbocations are such that :- (A) d < b < c < a (B) b < d < c < a (C) d < b < a < c (D) b < d < a < c 1 6 . Among the following, the strongest base is :- (A) C6H5NH2 (B) p–NO2–C6H4NH2 (C) m–NO2–C6H4NH2 (D) C6H5CH2NH2 1 7 . The increasing order of base strength of Cl–, CH COO–, —OH and F— is : 3 (A) Cl– < F— < CH3COO– < —OH (B) Cl– > F— > CH3COO– > —OH (C) CH3COO– < Cl– < F— < —OH (D) None of these 1 8 . Formic acid is considered as a hybrid of the four structures O O O O  H–C–OH  H– C = O H  H–C–O–H  H–C–O–H  I II III IV Which of the following order is correct for the stability of the four contributing structures. (A) I > II > III > IV (B) I > II > IV > III (C) I > III > II > IV (D) I > IV > III > II 1 9 . Which of the following carbonium ion is most stable (A) Ph3C+ (B) (CH3)3 C+ (C) (CH3)2 CH+ (D) CH2 C H — C H + 2

2 0 . Among the following the aromatic compound is (A) (B) (C) (D)   2 1 . Match List- I with List- II and select the correct answer using the codes given below - List-I (Stability) List -II (Reason) H3C CH3 C (A) > CH3+ (1) Inductive effect  Br (2) Resonance (3) Hyperconjugation and resonance (B) H3C–C CH3 < CH3– (4) Hyperconjugation and inductive effect  CH3 > (C) H3C–C H C–CH + CH3 3 2 • H3C • CH2 CH2 (D) > CH3 Codes : (A) A - 2 ; B - 3 ; C - 4 ; D - 1 (B) A - 3 ; B - 1 ; C - 4 ; D -2 (C) A - 4 ; B - 3 ; C - 1 ; D - 2 (D) A - 3 ; B - 4 ; C - 2 ; D - 1 2 2 . Which of the following compounds is the strongest base - (A) NH2 (B) NH2 (C) CH3 NH2 (D) NHCOCH3 23. In which of the following molecule all the effect namely inductive, mesomeric & hyperconjugation operate: (A) —Cl (B) —CH3 (C) —COCH3 (D) CH3 CH3 (B) CH3–C–OH + HCOONa 2 4 . Which of the following reaction is possible : O OH ONa (D) H–CC–Na+H2O  (A) + NO2 SO3Na COOH (C) +

COOH 2 5 . , pKa value of the compound decreases if X is :- X (A) –NO2 (B) –NH2 (C) –OH (D) –OCH3 2 6 . Which one of the following results most stable carbon free radical : • • CH3 CH3 (A) (B) • • (C) (Ph) C = C(Ph) CH3 (D) CH = CH CH3 22 22 H  CH3–C–CH2 2 7 . 1,2 shift A (Major) ; Here A is :   CH3 CH–CH2 CH3–C–CH3 (B) (A) H (D) CH3–C= CH2 (C) CH3–C –H2C– Ph 2 8 . –CH2OH ; On dehydration with conc. H2SO4 forms predominantly : (A) =CH2 (B) –CH3 (C) –CH3 (D) –CH3 2 9 . Which of the following is incorrect : CH2CH3 •CHCH3 :O O O O (A) Cl• + HCl (B) HO–C–H+ C–H  HO–C–H + H–C–H HH H H–CH2 H (C) CH3–C Hshift  CH3  (D) CH 3CH 2CHO HO–  CH 3  HCHO  H2O –C–CH2 C CH3 H 3 0 . The K values of alcohols, water and phenol are of order 10–17, 10–14 and 10–10 respectively. Which of the a following reactions is possible : (I) C6H5O + H2O  C6H5OH + OH (II) C2H5O + H2O  C6H5OH + OH (A) Both I and II (B) only II (C) only I (D) neither of two

3 1 . Most stable carbanion is : (A) C H   CH2 CH2 3 (C) (D) (B) CH2  CH – CH2 NO2 3 2 . Stability of : (IV) CH3–C= CH–CH3 (I) CH3–CH=CH–CH3 CH3 (III) CH3–C= CH2 CH3 (II) CH3–C= C–CH3 CH3 (D) II < III < IV < I CH3 in the increasing order is : (C) I < IV < III < I (A) I < III < IV < II (B) I < II < III < IV 3 3 . –MgBr + CH2—CH2 H2O [A] ; the product (A) is : O (A) –CH2OH (B) –CH2CH2OH (C) —CH—CH2 (D) No reaction O 3 4 . Spin multiplicity of Triplet nitrene is : (A) 1 (B) 2 (C) 3 (D) 4 3 5 . The maximum probability of proton loss is in the case of :   H H H (A) CH3 – CH2 (B) CH3–C–CH3  CH3 (C) (D) (CH3)2C–CH2  3 6 . Select the least stable one :  H3C CH– CH2 H3C  (C) H3C (A) CH3–CH2 (B) CH3– CH2–CH2 (D) H3C C –CH2 H3C 3 7 . Which of the following will react fastest with conc. HCl : (A) –CH2CH2OH (B) –CH(OH)CH3 (C) –OH (D) –CH2OH

3 8 . Consider the following reaction : OCH3 + NaNH2 Product Br The product (P) and reaction (R) are : OCH3 OCH3 NH2 NH2 (A) ; addition-elimination (B) ; elimination addition OCH3 OCH3 (C) ; elimination addition (D) ; cine substitution NH2 NH2 3 9 . Energy of activation is lower for which reaction :   (A) RCH2 OH2  R C H2 (B) R2CH OH2  R2CH  (D) all have same E act. (C) R3C OH2  R3C CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. C C C A B D A B D C D C C A A Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. D A A A A B B C D A C A B C B Que. 31 32 33 34 35 36 37 38 39 Ans. D A B C C D B C C

EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . The correct order of decreasing basic strength is : NH2 NNN HH I II III IV (A) I > II > III > IV (B) II > III > I > IV (C) II > IV > I > III (D) II > III > IV > I 2 . The correct order of increasing dissociation constant of the following compound is :- OH OH OH OH NO2 NO2 NO2 NO2 NO2 NO2 NO2 I II III IV (A) II < IV < I < III (B) IV < III < I < II (C) IV < II < I < III (D) IV < I < II < III 3 . In the following arrange the H (numbered) for their ease of displacement during acid base reaction : (3) (2) HO OH H NaNH2(liquid NH3) (1) 2 equivalent (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 > 1 > 2 (D) 2 > 3 > 1 4 . Which one of the following is the strongest base in aqueous medium :- (A) (C2H5)3N (B) C2H5NH2 (C) NH3 (D) (C2H5)2NH 5 . The correct order of acid strength of the following compound is - (I) CH3–C–CH3 (II) CH3–C–C–CH3 (III) CH3–C–CH2–C–CH3 O OO OO (A) III > II > I (B) III > I > II (C) II > I > III (D) I > II > III 6. Which of the following shows the correct order of decreasing basicity in aqueous medium - (A) (CH3)3 N > (CH3)2NH > CH3NH2 > NH3 (B) (CH3)2 NH > (CH3)3N > CH3NH2 > NH3 (C) (CH3)2 NH > CH3NH2 > (CH3)3N > NH3 (D) (CH3)2 NH > CH3NH2 > NH3 > (CH3)3N 7 . For the compounds O NN N H H III I II (C) I > II > III the order of basicity is - (A) III > II > I (B) II > III > I (D) II > I > III

8 . Which of the following shows the correct order of decreasing acidity- (A) PhCO2H > PhSO3H > PhCH2OH > PhOH (B) PhSO3H > PhOH > PhCO2H > PhCH2OH (C) PhCO2H > PhOH > PhCH2OH > PhSO3H (D) PhSO3H > PhCO2H > PhOH > PhCH2OH 79 . Give the correct order of increasing acidity of the following compounds- OH OH COOH CCH I II III IV (C) I < II < IV < III (A) II < I < III < IV (B) IV < II < I < III (D) IV < I < II < III 1 0 . Which of the following is the most acidic (A) CH2 = CH2 (B) HC  CH (D) (C) CH2=CHCH2CH=CH2 1 1 . Which of the following substituents will increase the acidity of phenol - (A) –NO2 (B) –CN (C) –CHO (D) – CH3 1 2 . (I) (II) (III) Compare carbon-carbon bond rotation across I, II, III. (A) I > II > III (B) I > III > II (C) II > I > III (D) II > III > I (D) III and IV 1 3 . Which of the following s-bonds participate in hyperconjugation : (D) y > z > x I V H H CH3 II H IV III (A) I and II (B) I and V (C) I and V 1 4 . NH, N–H, N O (x) (y) (z) The correct order of decreasing basic strengths of x, y and z is : (A) x > y > z (B) x > z > y (C) y > x > z

1 5 . Set the following in increasing order of their pKa values: OO (x) CH3—S—O—H (y) CH3—C—O—H (z) CH3—OH (D) x < z < y (B) x < y < z (C) y > z > x O (A) y > x > z 1 6 . Rank the following alkenes in order of decreasing heats of hydrogenation (largest first): (I) (II) (III) (IV) (A) II > III > IV > I (B) II > IV > III > I (C) I > III > IV > II (D) I > IV > III > II 1 7 . Which of the following molecules has all the effects : inductive, mesomeric and Baker Nathan effect ? (A) C2H5Cl (B) CH3–CH=CH2 (C) CH2=CH–CH=CH2 (D) CH3–CH=CH –C–CH3 O 1 8 . Which nitorgen in LSD (Lysergic acid diethylamide) is more basic : I O H–N III –C–N(C2H5)2 II N CH3 (A) I (B) II (C) III (D) all are equally basic 1 9 . Which of the following substituted carboxylic acids has the highest Ka value : (A) CH3—CH2 —CH— COOH (B) CH3 —CH— CH2—COOH Cl Cl (C) CH2— CH2—CH2—COOH (D) CH3 —CH— CH2—COOH Cl Br SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 2 0 . Which of the following compounds yield most stable carbanion after rupture of (C1–C2) bond : O O (A) CH3—C—CCl3 (B) CH3—C—CBr3 12 12 O (D) None of these (C) CH3—CH2—C—CI3 12

N N NH2 21. pyridine (II) aniline (III) H pyrrole (I) Which is/are correct statements ? (A) I is more basic than II (B) II is more basic than I and III (C) III is more basic than II (D) all are aromatic bases 2 2 . Following is true for stability between the two structures : O O (A) CH3–CH–CH=CH2 > CH3–C=CH–CH3 CH2 (B) C H 2= C H – C H = C H– C H • > • 2 CH2–C–CH–CH3 CH3  (C) CH2=C–CH2 > CH3–CH=CH– C H2 O (D) —O < CH3–C–O 2 3 . Which have acidic hydrogen : (A) CH3COOH (B) —OH O (C) CH3—C—CH2—CN (D) NaNH2 2 4 . C—C and C=C bond lengths are unequal in : (A) Benzene (B) 1,3 buta-di-ene (C) 1, 3 cyclohexa-di-ene (D) None 2 5 . The acid strength of substituted carboxylic acids is known to be dependent on the nature and position of the substituent. In the following examples, an attempt has been made to arrange the acids in order of acid strength, the strongest first. One of the series is incorrect-which one : (A) CH3.CH2CH(Cl).COOH > CH3 .CH(Cl).CH2COOH > CH3.CH2.CH2.COOH (B) NO2.CH2.COOH > HOCH2. COOH > CH3.COOH (C) Cl3C.COOH > BrCH2.COOH > FCH2.COOH (D) CH3.COOH > CH3.CH2.COOH > (CH3)3C.COOH 2 6 . Which species is not aromatic : (A) (B) (C) NH (D) N N :

2 7 . Select the correct option : (A) carbonic acid is weaker acid than acetic acid (B) the boiling points of acids are higher than corresponding alcohols (C) chloroacetic acid is stronger acid than acetic acid (D) phenol is more acidic than ethanol 2 8 . Which reacts with AgNO3 to give ppt. : —Br (A) —Br (B) –CH=CH–CH2Br(C) (D) —Br 2 9 . Which statement is ture : (A) N–H2 is stronger base than O– H (B) NH2–OH is less basic than NH3 (C) CH3–NH2 is weaker base than N(CH3)3 in (aq) medium NH2 CH2–NH2 (D) is weaker base than 3 0 . Which is less acidic than phenol : OH OH (C) —NH2 (A) —CH3 (B) CH3–OH (D) H2O 3 1 . The precursor carbocation to the product in the following reaction is : C6H5 H OH H C6H5 H  (B) C6H5  (C) either of the two (D) none of the two (A) C6H5 3 2 . Which of the following ion is formed in the following reaction : H OH O (B) (A)   OH OH  (D) All of three CH2 (C) OH

OD 33. D NaOH [Intermediate] D P. Product is : CO2 OD OD HOOC D DOOC D (A) (B) OH DOOC D (D) Reaction not possible (C) 3 4 . Predict the nature of A in the following reaction : O + ClCH2COOC2H5 Base (A) O Cl (A) (B) CCOOC2H5 CH2COOC2H5 O O (C) CHCOOC2H5 (D) CHCOOC2H5 3 5 . Which of the following intermediate is most likely to be formed during addition of HBr on crotonic acid : CH3CH = CHCOOH + HBr(g)  CH3 CH–CH2COOH : Br   (A) CH3CH2 – C HCOOH (B) CH 3 CH – CH 2COOH OH (C) CH3CH=CH–C –OH (D) none of these 3 6 . Which will give most stable cation upon strong heating : CH2–Cl (B) (Ph)3C–Cl Cl CH2 (A) (C) (D) CH CH2–Cl 3 7 . Cl (A); Here (A) is : —Cl H (A) (B) (C)     (D)

F –F  38. [X] ; Here [X] is : Cl –C–C–F Cl F (A) Cl –C= C–F FF FF (D) C l – C = C – F Cl F (B) Cl –C= C–Cl (C) Cl –C= C–F Cl F 39. H3C C C CH3 CHCl3 H3C CH3 H H KOH the reaction proceeds via : HH (A) carbene (B) Carbon free radical (C) Carbocation (D) Carbanion :C Cl 4 0 . Cl (A) ; Here A is : •• N H (A) (B) N Cl Cl Cl N H (C) (D) N N 4 1 . Which of the following can give benzyne : Cl Cl Br NaNH2 (A) H NaNH2 (B)  (C) (D) All N2 4 2 . Dehydration of the following in increasing order is : (I) –OH (II) –OH (III) –OH (IV) –OH (A) I < II < III < IV (B) II < III < IV < I (C) I < III < IV < II (D) I < IV < II < III 4 3 . Increasing tendency for SN1 and SN2 reaction is : (I) CH3CH2CH2CH2Cl (II) CH3CH2CH – CH3 (III) CH3CHCH2 – Cl CH3 Cl CH3 (IV) CH3–C–Cl CH3 (A) SN1 ; I < III < II < I V (B) SN2 ; IV < II < III < I (C) (A) and (B) both are correct (D) Both incorrect 4 4 . Cyclopentyl ethyl ether can be obtained by : (A) –ONa+CH3CH2Br (B) –Br+CH3CH2ONa (C) –OH+CH3CH2Cl (D) none is correct

4 5 . Major product in the following reaction is : –Br + KOCH2CH3 Product (A) –OCH2CH3 (B) (C) —Br (D) OCH2CH3 4 6 . Mechanism of insertion reaction is of : (A) One step for triplet carbene (B) Two step for singlet carbene (C) One step for singlet carbene (D) Two step for both H3C 1 47. 6 2 NH2 (P). The most probable structure of intermediate P is : 5 3–Cl 4 6 OCH3 6 OCH3 6 OCH3 6 OCH3 (A) 1 (B) 1 (C) 1 (D) 1 2 2 2 2 5 5 5 5 3 3 3 3 4 4 4 4 CH3 1 4 8 . 6 2 NH2 (A). The most probable structure of intermediate A is : 5 3–Cl 4 6 CH3 6 CH3 6 CH3 6 CH3 (A) 1 (B) 1 (C) 1 (D) 1 2 2 2 2 5 5 5 5 3 3 3 3 4 4 4 4 4 9 . Hybridisation of arrow headed 'C' is (A) sp2 sp3 (B) sp3 sp3 (C) sp2 sp2 (D) sp3 sp3d 5 0 . Which of the following is most strongest electrophile : •• •• (C) singlet –C (D) triplet –C– (A) singlet H–N• (B) triplet H–N• • • 5 1 . In benzyne two external unshared electronic orbital : (A) do not overlap to form a  bond (B) can overlap to form a complete  bond (C) poorly overlap to form a very weak bond (D) contribute in benzene ring 5 2 . Which of the following is aromatic intermediate : (A) carbene • (C) Benzyne • (B) (D) •

5 3 . In which of the following case 'H' shift is more preferable than 'CH3' shift : CH3 CH3 H   (A) CH3–C–CH2 (B) CH3–C–CH–C–H H CH3 H CH3 (D) all  (C) CH3–C–CH–CH3 Ph 5 4 . Which of the following will not give 1, 2, shift :  Hg2 (C) OH (D) all (B) (CH3)3C CH—CH2 (A) (CH 3 )3C — C H2  5 5 . Which of the following reactions is wrong : (A) CH3CH2CHO OH  CH3 – CH – CHO CH2CH3 • (B) Cl• CHCH3+ HCl O O O O (C) HO–C–H+ H–C–H  HO–C–H + H–C–H HH CH3 (D) CH3 –C sHhyifdtirnidge CH3 –CH–CH2 CH3 CH3 5 6 . Methyl benzylic carbonium ion is the most stable one, hence it will react fastest consider the following reaction : Br CH3 + OH SN2 (P) ; product (P) will be : HH H CH3 HO CH3 H (B) H (A) HO H H CH3 OH CH3 (C) H H H H (D) H OH

5 7 . Consider the following reaction - (B) R–CH– CH2I R – CH = CH2+ IN3  A , A is : N3 (A) R–CH–CH2N3 I (D) R–CH–CH N (C) R–CH–CH2 N I 58. + CHBr3 1 BuOK product ; product will be : Br (A) Br (C) Br Br Br (B) Br Br Br (D) CH2 5 9 . HBr major product will be : Me CH2Br CH2Br CH3 (A) (B) (C) Br Br (D) CH3 60. O OH Product ; major product of the reaction can be : O CH3 O O O O (A) (B) (C) (D) 6 1 . CH3COCH3 + BrCH2COOEt (ii) H(i2)OZn/ H  (A) ; the product (A) is : (A) CH3 C OCH3 OH CH3 Br (B) CH3–C–CH2COOEt CH3 (C) CH3 C OH (D) CH3 C OCH2COOEt CH3 COOEt CH3 Br

6 2 . Which carbocation is more likely to be formed in the dehydration of OH H :    (A) (B) (C) (D)  6 3 . The most probable structure for (P) in the following reaction is : 2 conc H2SO4 (P) (A) (B) (C) (D) 6 4 . Which of the following alcohols is dehydrated most readily with conc. H2SO4 ? CHOHCH3 CHOHCH3 CHOHCH3 CHOHCH3 (A) (B) (C) (D) OCH3 NO2 Cl 6 5 . Dehydration of alcohols by conc. H2SO4 takes palce according to following steps : CH3 H CH3 –H2O CH3 step -I step -II ••   CH3–C–CH2O••H CH3–C–CH2OH2 CH3–C–CH2 H H H  CH3 –H CH3 1,2–H shift step -III CH3–C –CH3 step -IV CH2=C–CH3 the slowest and fastest steps in the above reaction are : (A) step I is slowest, while III is fastest (B) step II is slowest, while III is fastest (C) step II is slowest, while IV is fastest (D) all step proceed at equal rate

6 6 . Pyrrole is treated with alkaline chloroform to form two products A and B ? + CHCl3 KOH Cl N + H N CHO N H (A) (B) Which of the following intermediate is likely to be formed ? (I) H (II)  Cl  N N C  H (III) Cl (IV) N CCl2 (B) I and III •• (D) I, III and IV (A) I and IV N (C) III and IV 6 7 . conc. H2SO4 Major product is : OH (A) (B) (C) (D) BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A n s . D C B D A C D D B D A ,B,C C BBB Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A n s . D D B A A B,D A ,C A ,B,C C C B A ,B,C ,D B,D A ,D A ,B,D Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 Ans. A B C D C C C A A D D A A A B Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. C A B C B C C A B D A B C A B Que. 61 62 63 64 65 66 67 Ans. B C C B C B C

EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE OR FALSE : 1 . In benzene, carbon uses all the three p-orbitals for hybridization. 2 . All the carbon atoms in bicyclbutane are sp2-hybridized. 3 . Allene (CH2=C=CH2) is a planar molecule. 4 . The greater stability of trans-but-2-ene over cis-but-2-ene can be explained on the basis of hyperconjugation effect. 5 . In a two step reaction, the rate determining step has the lowest energy of activation. FILL IN THE BLANKS : 1 . Out of benzene, ethylene and acetylene, the carbon-carbon bond is longest in ...................... . 2 . The cyclopentadienyl cation is ...................... while cyclopentadienyl anion is ............... . 3 . The type of delocalization involving sigma bond orbitals with  bond or vacant orbital is called .............. 4 . Hydroperoxide ion is a stronger ...................... but weaker ................... than hydroxide ion. 5 . Catalytic hydrogenation of cis-2, 3-diphenyl-2-butene gives..................... . MATCH THE COLUMN Column-II ( p ) – m effect 1 . Match the column I with column II. ( q ) + m effect Column-I ( r ) + I effect ( s ) – I effect (A) —NO2 (B) —O– (C) —O—CH 3 ( D ) —C  N 2 . Match the column I with column II. Column-II Column-I ( A ) ( p ) Aromatic ( B ) ( q ) Non-aromatic (C) + ( r ) Anti-aromatic (D) ( s ) Aliphatic

3 . Observe the following compound P and match the column I with column II. HOOC SO3H HO CCH Column-I Column-II 1 ( A ) Strongest acidic group in P (p) 2 3 ( B ) Weakest acidic group in P (q) 4 ( C ) Number of Intra molecular (r) H-Bonding in P ( D ) Number of mole of hydrogen (s) gas that is liberated on reaction of 'P' with excess of sodium metal if one mole of P is used ASSERTION & REASON QUESTION : These questions contains, Statement-I (assertion) and Statement-II (reason). (A) Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False. (D) Statement-I is False, Statement-II is True. 1 . Statement-I : Greater the s-character of the hybrid orbital, the smaller is its size. Because Statement-II : Bond formed by the overlap of sp3-sp3 hybrid orbital is the longest while the bond formed by the overlap of sp-sp hybrid orbital is shortest.    2 . Statement-I : Me3 C is more stable than Me2 C H and Me2 C H is more stable than the Me C H2 . Because Statement-II : Greater the number of hyperconjugative structures, more is the stability of carbocation. 3 . Statement-I : Cyclopentadienyl anion is much more stable than allyl anion. Because Statement-II : Cyclopentadienyl anion is aromatic in nature. 4 . Statement-I : CH3OCH3 and C2H5OH have comparable molecular masses but boiling point of C2H5OH is higher than dimethyl ether. Because Statement-II : C2H5OH forms intermolecular H-bonding while CH3OCH3 forms intramolecular H-bonding. IV III II I 5 . Statement-I : In the compound, C H 3 — C H 2 — C  N , the most electroegative carbon is II. Because Statement-II : Carbon atom II has more s-character.

6 . Statement-I : pka1 of fumaric acid is more than maleic acid. Because Statement-II : Conjugate base of fumaric acid is stabilised by intramolecular H-bonding. 7 . Statement-I : Carbocation (II) is more stable than carbocation (I)   CH2 CH2 CH3 CD3 (I) (II) Because Statement-II : Carbocation (II) has more positive inductive effect of –CD3 group as compared to –CH3 group. 8 . Statement-I : Carbon-oxygen bonds are of equal length in acetate ion. Because Statement-II : Bond length decreases with the multiplicity of bond between two atoms. 9 . Statement-I : The potential energy barrier for rotation about C=C bond in 2-butene is much higher than that in ethylene. Because Statement-II : Hyperconjugation effect decreases the double bond character. 1 0 . Statement-I : Pyrrolidine (II) is more basic than pyrrole (I). •• •• NN HH (I) (II) Because Statement-II : Protonated pyrrole has resonance stabilization of positive charge in aromatic ring. COMPREHENSION BASED QUESTIONS : Comprehension # 1 The intramolecular delocalisation of  and /non-bonding electrons without any change in the position of atoms is called resonance. Delocalisation may occur in conjugated system involving carbon atom and atom other than the carbon. Delocalisation makes system stable. More is the number of resonating structures, more is the stability of the system. A conjugated structure is least stable when a higher electronegative atom has positive charge and when identical charges are present on adjacent atoms. 1 . The decreasing order of stability of the following resonating structures : ••  –+ + ••– is - CH2=CH—Cl CH2—CH=Cl CH2—CH= Cl (I) (II) (III) (A) I > II > III (B) II > III > I (C) III > II > I (D) I > III > II  2 . If A is Ph CH2 and B is CH2=CH— CH2 , the greater number of resonating structure is of - (A) A (B) B (C) both A and B (D) None of these

3 . Which of the following pairs represent resonance ? (A) CH2=CHOH ; CH3CHO – (B) C H2 –CHO ; H2C=CH–OH O OH + (C) CH3—C—CH3 ; CH3—C—CH2 OH OH (D) CH3—C—CH3 ; CH3—C—CH3 H 4 . Examine the structures I and II for nitromethane and choose the statement correctly : + O• • +2 O• • – CH3—N ••– CH3—N ••– (I) O (II) O —— — •• (A) II is less important because electrons have shifted to oxygen (B) II is less important because nitrogen has sextet of electrons (C) II is acceptable and important structure (D) none of these 5 . Examine the following two structures for pyrrole and choose the correct statement given below : + •• •• N N– OO (I) (II) (A) II is not an acceptable resonating structure because carbonium ions is less stable than nitride ion (B) II is not an acceptable resonating structure because there is charge separation (C) II is not an acceptable resonating structure because nitrogen has ten valance electrons (D) II is an acceptable resonating structure 6 . Delocalization of electrons increases molecular stability because : (A) Potential energy of the molecule decreases (B) Electron-nuclei attraction decreases (C) Both (A) and (B) (D) Electron-electron repulsion increases Comprehension # 2 Carbocation is a group of atoms with positively charged carbon atom having six electrons in the valence shell after sharing. Carbocations are formed in the heterolysis of a bond and are planar species. Stability of carbocation is determined by inductive effect, hyperconjugation and resonance effect. Greater the number of contributing structures, more is the stability of a Carbocation. Electron releasing groups (+I effect) increases the stability of a carbocation whereas the electron withdrawing groups (–I effect) have an opposite effect. 1 . Which of the following is most stable carbocation ?   + (A) C H 3 (B) CH 3 — C H — CH 3 (C) CH3 — C H2 (D) CH3—C—CH3 CH3

2 . The most stable carbocation among the following : (A)  (B) (C)  NO2  (D)  3 . In which of the following cases, the carbocation (I) is less stable than the carbocation (II) ? + + CH2 CH2  (B) (I), (II) (A) C6H5— C H2 (I), CH2= CH— C H2 (II)   (C) CH2= C H (I), CH3— C H2 (II) (D) H3C— C H 2 (I), CH2—CH2 (II) F 4 . The most stable and the least stable carbocation among  (I), CH2=CH— C H2 (II), C6H5— C H2 (III)  and CH3 — C H — CH3 (IV) are respectively : (A) II, I (B) III, IV (C) I, II (D) I, IV (D) C6H5CH2Br 5 . Most stable carbocation is formed by the heterolysis of : (A) (CH3)3CBr (B) (C6H5)3CBr (C) (C6H5)2CHBr Comprehension # 3 The most important condition for resonance to occur is that the involved atoms in resonating structure must be coplanar or nearly coplanar for maximum delocalisation. If this condition does not fulfil, involved orbitals cannot be parallel to each other and as a consequence delocalisation cannot occcur. Bulky groups present on adjacent atoms inhibit the of orbitats. This phenomenon is known as steric inhibition of resonance. Steric inhibition of resonance has profound effect on (1) Physical properties (2) Acidity and basicity (3) Reactivity of organic compounds 1 . Arrange the following in the increasing order of basicity : NH2 NH2 NH2 NH2 (I) (II) (III) (IV) NO2 NO2 (A) I > II > III > IV (B) IV > III > II > I (C) II > I > IV > III (D) I > IV > III > II 2 . Which of the following is most acidic : COOH COOH COOH COOH (B) (C) (D) (A) CH3 NO2 CH3

3 . Consider the following two structures and choose the correct statements - : : O+ O O+ O N N (I) (II) (A) carbon-nitrogen bond length structure I is greater than that in structure II (B) carbon-nitrogen bond length in structure I is less than in structure II (C) carbon-nitorogen bond length in both structure is same (D) It can not be compared 4 . Arrange the following carbocation in the increasing order of stability : CH2 CH2 CH2 CH3 NO2 CH3–CH CH3 (I) (II) CH3 CH–CH3 NO2 (III) (A) I < II < III (B) II < III < I (C) III < II < I (D) III < I < II Comprehension # 4 1-Butene (A) and 1,3-Butadiene (B) differ not only in the number of  bonds, but (B) also has  and  bonds at alternate positions. This type of the system is called conjugate system. Following are some of the conjugate system. OO CH2=CH–C–H CH2=CH–C–OH benzene acrolein acrylic acid in such systems,  electron shifting takes place consecutively giving permanent polarity on the chain. This type of -electron shift in the conjugate systems is called mesomeric effect or resonance. Rules for resonance forms : (i) Individual resonance forms are imaginary, not real (ii) Resonance forms differ only in the placement of their  electrons or nonbonding electrons. (iii) Different resonance forms of a substance don't have to be equivalent. (iv) Resonance forms must be valid Lewis structures and obey normal rules of valencey. (v) The resonace hybrid is more stable than any individual resonance. Rules for stability : (i) Structures with more covalent bonds are more stable than other structures. (ii) Structures in which all of the atoms have a complete valence shell of electrons (i.e. the noble gas structure) are especially stable and make large contribution to the hybrid. (iii) Stability is decreased by an increase in charge separation. (iv) Structure that carry negative charge on a more electronegative atom and positive charge on less electronegative atom are comparatively more stable.

1 . In which of the following compound, resonance is not possible :  (C) O N  OH2 (D)  (A) CH2  CH – NMe3 (B) : :: 2 . Which of the following statement is incorrect - (A) In a resonance hybrid all the molecules are the same. A resonance hybrid cannot be expressed by any single structure. (B) The value of the resonance energy of any resonance hybrid is not an absolute value (C) The energy of hybrid structures is always less than that of any resonating structure (D) Only one individual resonating structure explains all characteristics of the molecule 3 . Choose the correct option about stability of resonating structure :  O O (A) Cl: >  (B) < N Cl N  O O C O OO (C) < (D) H–C–O > CH3–C–O MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3  True / False 2. F 3. T 4. F 5. F 1. F  Fill in the Blanks 1. benzene 2. antiaromatic, aromatic 3. hyperconjucation 4. nucleophile, base 5. 2,3-diphenylbutane (optically inactive, meso)  Match the Column 1. (A) p,s ; B q,r ; (C)  q,s ; (D) p,s 2. (A) q ; (B) p ; (C)  p ; (D) r 3. (A) p ; (B) q ; (C)  p ; (D) q  Assertion - Reason Questions 1. B 2. A 3. A 4. C 5. A 6. C 7. D 8. B 9. D 10. C  Comprehension Based Questions Comprehension #1 : 1. (A) 2. (A) 3. (B) 4. (B) 5. (C) 6. (C) Comprehension #2 : 1. (D) 2. (C) 3. (C) 4. (D) 5. (B) Comprehension #3 : 1. (C) 2. (B) 3. (B) 4. (B) Comprehension #4 : 1. (A) 2. (D) 3. (C,D)

EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Rank the following amines in increasing basic nature : (a) N NH2 NH2 N (II) (III) (IV) H (I) NH2 NH2 NH2 CH2NH2 (b) CH3 CH3 CH3 (I) (II) (III) (IV) NH2 (c) NN (I) (II) H (III) 2 . Rank the following intermediates according to the stability (most stable first).   (a) CH3CH2CH2 C H2 , CH3 C HCH2CH3 , (CH3 )2 C CH2CH3 , (CH3 )3 C •• •• (b) CH3CH2CH2 C H2 , CH3 C HCH2CH3 , (CH3 )2 C CH2CH3 , (CH3 )3 C 3 . Rank the following intermediates according to the stability (most stable first). —— —— (a) CH3CH2CH2 C H2 , CH3 C HCH2CH3 , (CH3 )2 C CH2CH3 , (CH3 )3 C  (b) • C H 2 • , CH3CH •• , C 6H5CH •• , C 6 H 5 2 C •• 4 . Show why the phenolate ion, C H O–, has a greater resonance stabilization than phenol, C H OH. 65 65 5 . Is phenol a stronger acid than acetic acid. Explain your answer with proper reasoning. 6 . Arrange all these numbered H in order of their decreasing acidic strength. H 4 OO HH 12 H OH 3

•• •• 7 . Between –C=C–F and –C=C-Cl which lone pair- p conjugation will be more preffered. 8 . Which is more acidic and why ? OH OH 9 . Arrange the following compound in the correct order of acidity : COOH SO3H OH CH2OH (I) (II) (III) (IV) 1 0 . Why a cation like  is not possible. H O 11. OH (A) Product. O 1 2 . CH2=CHCHO + C2H5MgBr H3O  P ; Here P is : 1 3 . Within each set, select the compound which is more reactive in nucleophilic addition at carbonyl carbon : (I) CH3COCH2CH2Br or CH3COCH2Br (II) CH3CHO or CH3COCH3 (III) CH3COCH3 or CH3–CO–CF3 (IV) PhCHO or CH3CHO 1 4 . Rank of the following compounds in increasing order of electrophilicity : O CH2 NH (I) F3C H (II) CH3 H (III) CH3 OH 1 5 . Arrange the followings in increasing order of reactivity towards nucleophile : O OO O (A) H3C–C–NH2; CH3–C–Cl; CH3–C–OAc; H3C–C–OCH3 (I) (II) (III) (IV) OO OO (B) H3C–C–Cl; CH3CH2–C–Cl; (CH3)2CH–C–Cl; (CH3)3C–C–Cl (I) (II) (III) (IV) OO O (C) CH3–C–OCH3; CH3–C–OCH2CH3; H3C–C–OCH3 (I) (II) (III)

1 6 . Suggest the product of the reactions : Cl AlCl3 (B) CH3–C= O Cl (A) CH3–CH2 AlCl3/ CH3 KOH/Alc. (D) CH2=CH–CH=CH2 HBr A (C) CH3–C–Cl 40°C CH3 –80°C B O O–C–CH3 OH (E) AlCl3 E (F) + CO + HCl AlCl3 G O  C Cl O N (CH3)3 OH AlCl3 CH3–CH2 –CH–CH3 (G) H (H) 1 7 . Arrange the following in increasing order of basic character : (A) Cl, H2O, Br, H2S NH2 NH2 CH3–N–CH3 CH3–N–CH3 CH3 CH3, CH3 CH3, (B) , NH2 NH2 NH2 NH2 (C) , , , F Cl F Cl O O O–H O O O C–O C=O C OH C OH NO2 O (D) ; ; ; O–H 1 8 . Heat of hydrogenation of CH2=CH2 is greater than CH3–CH=CH2. Why ?  N(CH3)3 1 9 . In the case of CH3–CH2–CH–CH3 OH, less stable 1-Butene is obtained. Why ? 2 0 . When carbene attacks cis 2-butene singlet gives cis product while triplet gives trans product. Why ?

CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) 1 . (a) I > III > II > IV (b) IV > III > II > I (c) II > I > III    2 . (a) (CH 3 )3 C > (CH 3 )2 C CH 2CH 3 > CH3 C HCH2CH 3 > CH 3CH 2CH 2 C H2 •• • • (b) (CH 3 )3 C > (CH 3 )2 C CH 2CH 3 > CH 3 C HCH 2CH 3 > CH 3CH 2CH 2 C H 2 —— —— 3 . (a) CH 3CH2CH2 C H2 > CH 3 C HCH 2CH 3 > (CH 3 )2 C CH2CH 3 > (CH 3 )3 C  (b) • C 6H 5 2 C •• > C 6H5CH •• > CH3CH •• > C H 2 • 4 . Free charge has move reactivity than bound charge. 5. No Anion of acetic acid is more stabilised by resonance than phenoxide ion. 6 . 4 > 3 > 2 >1 •• 7 . In –C=C–F, lone pair - p conjugation will be more due to more overlaping of orbitals involved in conjugation. 8 . Phenol is more acidic as after loss of H+, phenoxide ion is Resonance stabilised. No such resonance is seen in conjugate base of cyclohexanol. 9 . II > I > III > IV 1 0 . Due to the non-planar structure. O OH 1 1 . 12. CH2=CH–C–C2H5 H

EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . Rank the following sets of intermediates in increasing order of their stability giving appropriate reasons for your choice. (a) ••CH2, ••C(C6H5)2 , CH3CH••,C6H5CH••, O2NCH2CH•• + + ++ O OO O (b) O + O O OO O (c) CH2—C—CH3, CH2—C—H, CH2—C—NH2 , CH2—C—CH2—OCH3 2 . For the following compounds, arrange the labelled proton in increasing order of their ease of deprotonation : C C—H3 H1 O H3 (i) H2 (ii) C CH2 CH2 H2 O O H1 3 . Benzene sulphonic acid is a stronger acid than benzoic acid, explain. 4 . Which is a stronger acid, A or B and why ? COOH COOH (I) NO2 (B) NO2 (A) H O Cl H H Cl C–OH H Cl O Cl C–OH (II) (B) (A) 5 . Descuss the following observations : (a) C—Cl bond in vinyl chloride is stronger than in chloroethane. (b) Carbon-carbon bond length in ethene is shorter than in CH2= CHOCH3. (c) CH SH is stronger acid than CH OH. 33 (d) CH CH NH is stronger base than CH =CHNH . 322 22

6 . Discuss the basic strength of two nitrogens in benzimidazole. N N H Benzimidazole 7 . In each of the following pair of compounds, which is more basic in aqueous solution ? Give an explaination for your choice : NH (i) CH3NH2 or CF3NH2 (ii) CH3CONH2 or H2N NH2 (iii) PhNH2 or CH3CN 8 . Answer the following questions : (i) Which of the indicated H is abstracted rapidly by bromine radical and why ? Ph Hc Hb Ha (ii) One of the indicated proton H or H , is approximately 1030 times more acidic than other, which is ab more acidic and why ? H H Ha Hb 9 . Explain the following (i) NC—CH2—CN is a stronger acid than CH3—CO—C2H5. (ii) CH3—CO—C2H5 is more acidic than CH3—CH=CH2. (iii) CH3—CH=CH—OH is more acidic than CH2=CH—CH2—OH. 1 0 . Compare the bond dissociation energies of C—H bonds in the following H (i) H—CH2—CH2—CH2—CH3 (ii) CH3 (iii) (CH3)3C—H H3C 1 1 . The most stable intermediatry carbocation and major product in following reactions : (i) H [C ] Product OH (ii) H [C ] Product OH OH (iii) HBr [C ] Product O

1 2 . Which one of the following is more basic and why ? N •• CH3 ••2 CH3 (a) –N CH3 (b) N (I) (II) N •• 1 1 3 . Find the product in following reaction : Ag A   (a) (b) + CH2Cl2 Bu O K B Br O H (d) H [E] (c) •• KOH C+ D Et Et + CHCl3 N H OD D (e) NaOH [Intermediate] (i) CO2 F (ii) D 1 4 . Write the mechanism of following reaction : OH (a) O+ dil H [A] H HO C5H5 NO2 Me (b) –O–O–H H [B] (c) Me H–Cl [A] O–H 3-Hydroperoxycyclohexene C2H5 Me (d) O BF3 [B] MeO (e) OTs KI A + B