21 Organic Chemistry Basics

21 Organic Chemistry Basics Topic 1 Nomenclature and Isomerism Objective Questions I (Only one correct option) 5. What is the IUPAC name of the following compound? 1. The IUPAC name for the following compound is CH3 CH3 CH3 CH2 (a) 3-methyl-4-(3-methylprop-1-enyl)-1-heptyne (b) 3, 5-dimethyl-4-propylhept-6-en-1-yne (c) 3-methyl-4-(1-methylprop-2-ynyl)-1-heptene (d) 3, 5-dimethyl-4-propylhept-1-en-6-yne 2. The correct IUPAC name of the following compound is H H CH H3 C CH3 CH3 (a) 3-bromo-3-methyl-1,2-dimethylprop-1-ene (b) 3-bromo-1,2-dimethylbut-1-ene (c) 2-bromo-3-methylpent-3-ene (d) 4-bromo-3-methylpent-2-ene H H 6. The absolute configuration of (a) (2S, 3R) (c) (2R, 3R) CO2H OH Cl CH3 NO2 Cl Br is (b) (2S, 3S) (d) (2R, 3S) 7. The IUPAC name of the following compound is OH CH3 (a) 2-methyl-5-nitro-1-chlorobenzene (b) 3-chloro-1-methyl-1-nitrobenzene (c) 2-chloro-1-methyl 1-4-nitrobenzene CN (d) 5-chloro-4-methyl 1-1-nitrobenzene 3. Which of the following compounds will show the maximum Br ‘enol’ content? (a) CH3COCH3 (b) CH3COCH2COCH3 (c) CH3COCH2COOC2H5 (d) CH3COCH2CONH2 4. The IUPAC name of the following compound is CH3 OH ½ ½ H3C ¾ CH ¾ CH ¾ CH 2 ¾ COOH (a) (b) (c) (d) 4,4 - dimethyl -3-hydroxybutanoic acid 2-methyl-3-hydroxypentan-5-oic acid 3- hydroxy -4- methylpentanoic acid 4-methyl-3-hydroxypentanoic acid (a) 4-bromo-3-cyanophenol (b) 2-bromo-5-hydroxybenzonitrile (c) 2-cyano-4-hydroxybromobenzene (d) 6-bromo-3hydroxybenzonitrile 8. The number of stereoisomers obtained by bromination of trans-2-butene is (a) 1 (b) 2 (c) 3 9. The IUPAC name of C6 H5 COCl is (a) benzoyl chloride (b) benzene chloro ketone (c) benzene carbonyl chloride (d) chloro phenyl ketone (d) 4 Organic Chemistry Basics 1 21. The IUPAC name of the compound CH3 CH2 == CH ¾ CH(CH3 )2 is H 10. H (a) 1,1-dimethyl-2-butene (c) 2-vinyl propane 3 2 H H 22. An isomer of ethanol is CH3 C 2 is rotated anti-clockwise 120° about C 2 -C 3 bond. The resulting conformer is (a) partially eclipsed (c) gauche (a) methanol (c) acetone CH3 ½ H3 C ¾ C ¾ CH==CH2 ½ CH3 stereoisomerism? (b) 3-methylbutyne-1 (d) 2-methylbutanoic acid 12. The number of isomers for the compound with molecular formula C2 BrClFI is (a) 3 (b) 4 (c) 5 (d) 6 13. Which of the following compounds will exhibit geometrical isomerism ? (a) 1-phenyl-2-butene (c) 2-phenyl-1-butene (b) 3-phenyl-1-butene (d) 1,1-diphenyl-1-propene 14. The optically active tartaric acid is named as D-(+)-tartaric acid because it has a positive (b) diethyl ether (d) dimethyl ether 23. The IUPAC name of the compound having the formula is (b) eclipsed (d) staggered 11. Which of the following compounds exhibits, (a) 2-methylbutene-1 (c) 3-methylbutanoic acid (b) 3-methyl-1-butene (d) None of these (a) 3, 3, 3-trimethyl-1-propene (b) 1, 1, 1-trimethyl-2-propene (c) 3, 3-dimethyl-1-butene (d) 2, 2-dimethyl-3-butene 24. Which of the following compounds will exhibit cis-trans (geometrical) isomerism ? (a) 2-butene (c) 2-butanol (b) 2-butyne (d) butanal 25. The compound which is not isomeric with diethyl ether is (a) n-propyl methyl ether (c) 2-methyl propane-2-ol (b) butane-1-ol (d) butanone (a) optical rotation and is derived from D-glucose (b) pH in organic solvent (c) optical rotation and is derived from D-(+)-glyceraldehydes (d) optical rotation when substituted by deuterium Objective Questions II (One or more than one correct option) 26. The IUPAC name(s) of the following compound is (are) 15. How many optically active stereoisomers are possible for butane-2, 3-diol ? (a) 1 (b) 2 (c) 3 16. Isomers which can be interconverted through rotation around a single bond are (a) conformers (c) enantiomers (b) diastereomers (d) positional isomers 17. The enolic form of acetone contains (a) tert-butanol and 2-methylpropan-2-ol (c) 10 sigma bonds, 1 pi bond and 1 lone pair (b) tert-butanol and 1,1-dimethylethan-1-ol (d) 9 sigma bonds, 2 pi bonds and 1 lone pair (c) n-butanol and butan-1-ol 18. The number of isomers of C6 H14 is (c) 6 (a) 4-methylchlorobenzene (b) 4-chlorotoluene (c) 1-chloro-4-methylbenzene (d) 1-methyl-4-chlorobenzene with molecular formula C4 H10 O is/are (b) 8 sigma bonds, 2 pi bonds and 2 lone pairs (b) 5 Cl 27. The correct combination of names for isomeric alcohols (a) 9 sigma bonds, 1 pi bond and 2 lone pairs (a) 4 H3C (d) 4 (d) iso-butyl alcohol and 2-methylpropan-1-ol (d) 7 19. If two compounds have the same empirical formula but 28. Which of the given statement(s) about N, O, P and Q with respect to M is/are correct? different molecular formulae, they must have (a) different percentage composition (b) different molecular weight (c) same velocity (d) same vapour density 20. Which of the following will have least hindered rotation Cl HO H (b) Ethylene (d) Hexachloroethane M Cl HO H CH3 about carbon-carbon bond ? (a) Ethane (c) Acetylene HO H H CH3 OH H OH HO H CH3 CH3 N O Organic Chemistry Basics CH3 H HO 34. Which of the following have asymmetric carbon atom? CH3 OH HO H H HO H Cl P Cl Q (a) M and N are non-mirror image stereoisomers (b) M and O are identical (c) M and P are enantiomers (d) M and Q are identical 35. Keto-enol tautomerism is observed in 29. The correct statement(s) about the compound H3 C(HO) HC ¾ CH == CH ¾ CH(OH)CH3 ( X ) is/are (a) The total number of stereoisomers possible for X is 6 (b) The total number of diastereomers possible for X is 3 (c) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4 (d) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2 30. The correct statement(s) concerning the structures E , F and G is/are H3C O H3C H3C OH H3C CH3 H3C CH3 (E) CH3 H3C OH (F) ¾ ¾¾¾ ¾¾¾ H 36. Only two isomeric monochloro derivatives are possible for (a) n-butane (c) benzene (b) 2, 4-dimethyl pentane (d) 2-methyl propane Numerical Value Based Question 37. For the given compound X, the total number of optically active stereoisomers is ……… . HO HO This type of bond indicates that the configuration at the specific carbon and the geometry of the double bond is fixed This type of bond indicates that the configuration at the specific carbon and the geometry of the double bond is not fixed CH ¾ 3 H Assertion and Reason (a) the compound is optically active (b) the compound possesses centre of symmetry (c) the compound possesses plane of symmetry (d) the compound possesses axis of symmetry 32. Tautomerism is exhibited by ¾ ¾ ¾ CH== CH ¾ OH (b) O ¾ ¾O ¾ ¾ Read the following questions and answer as per the direction given below : (a) Statement I is correct; Statement II is correct; Statement II is the correct explanation of Statement I (b) Statement I is correct; Statement II is correct; Statement II is not the correct explanation of Statement I (c) Statement I is correct; Statement II is incorrect (d) Statement I is incorrect; Statement II is correct 38. Statement I Molecules that are non-superimposable on ¾ ¾O O ¾ ¾O (d) ¾ ¾ ¾ ¾ (c) O O ½½ ½½ (d) C6H5 ¾ C ¾ CH2 ¾ C ¾ CH3 ¾¾ ¾¾¾ ¾ Cl ¾ O ½½ (c) C6H5 ¾ C ¾ C6H5 HO is/are (a) O ½½ (b) C6H5 ¾ C ¾ CH3 HO 31. The correct statement(s) about the compound given below H 3C ¾ O ½½ (a) C6H5 ¾ C ¾ H (G) (a) E , F and G are resonance structures (b) E , F and E , G are tautomers (c) F and G are geometrical isomers (d) F and G are diastereomers Cl H Cl ½ ½ (b) H ¾ C ¾ C ¾ Cl ½ ½ H H H H ½ ½ (d) H ¾ C ¾ C ¾ CH3 ½ ½ Br OH Cl Br ½ ½ (a) H ¾ C ¾ C ¾ H ½ ½ H H H Cl ½ ½ (c) H ¾ C ¾ C ¾ D ½ ½ H H O 33. Which of the following compounds will show geometrical isomerism ? (a) 2-butene (c) 1-phenyl propene their mirror images are chiral. Statement II All chiral molecules have chiral centres. (b) propene (d) 2-methyl-2-butene Fill in the Blanks 39. Isomers which are ……… mirror images are known as ……… (superimposable, non-superimposable, enantiomers, diastereomers, epimers) Organic Chemistry Basics 46. True/False 40. 2, 3, 4-trichloropentane has three asymmetric carbon atoms. (a) Draw Newman’s projection for the less stable staggered form of butane. (b) Relatively less stability of the staggered form is due to (i) Torsional strain (ii) van der Waals’ strain 41. m-chlorobromobenzene is an isomer of m-bromochloro benzene. (iii) Combination of the above two Integer Answer Type Questions 47. Glycerine contain one ……… hydroxy group. 42. The total number of stereoisomers that can exist for M is 48. Identify the pairs of enantiomers and diastereomers from the H 3C following CH3 CH3 H HO H 3C M O 43. The total number(s) of stable conformers with non-zero dipole moment for the following compound is/are HO H OH H CH3 CH3 I II III 51. Give the IUPAC name of the following compound : Me stereoisomers possible for a compound with the molecular formula C5 H10 . Me Me Me Me (Me = methyl) 45. m obs = S m i xi 52. Write the IUPAC name of CH3 CH2 CH==CH ¾ COOH . 53. Write the structure of all the possible isomers of If m solution = 1.0 D and mole fraction of anti form = 0.82, find m gauche (b) Write most stable meso conformer of (CHDY )2. If H H CH3 44. Give the total number of cyclic structural as well as (a) Write stable conformer for Z ¾ CH2 ¾ CH2 ¾ Z in Newman’s projection. H HO CH3 ½ H3 C ¾ N ¾– C ¾ C H2 ¾ CH3 ½ ½ CH3 C2H5 CH3 where m i is the dipole moment of stable conformer and xi is the mole fraction of that conformer. HO 50. Write the IUPAC name of the following compound: CH3 Cl Subjective Questions OH 49. Write tautomeric forms of phenol. Cl Br Br CH3 CH3 dichloroethene. Which of them will have zero dipole moment ? 54. Write structural formulae for the isomeric alcohols having (i) Y = CH3 about C 2 -C 3 rotation and the molecular formula C4 H10 O. (ii) Y = OH about C1 -C 2 rotation. Topic 2 General Organic Chemistry Objective Questions I (Only one correct option) 1. 25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2 O. This unknown hydrocarbon contains (a) 20 g of carbon and 5 g of hydrogen (b) 22 g of carbon and 3 g of hydrogen (c) 24 g of carbon and 1 g of hydrogen (d) 18 g of carbon and 7 g of hydrogen 2. An organic compound A is oxidised with Na 2 O2 followed by boiling with HNO3 . The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is (a) (b) (c) (d) nitrogen phosphorus fluorine sulphur Organic Chemistry Basics 3. The increasing order of nucleophilicity of the following 10. Which amongst the following is the strongest acid? nucleophiles is (1) CH3 COÈ2 (2) H2O (a) (1) < (4) < (3) < (2) (c) (4) < (1) < (3) < (2) È (3) CH3 SO3È (4) O H (b) (2) < (3) < (1) < (4) (d) (2) < (3) < (4) < (1) 4. In chromatography, which of the following statements is incorrect for R f ? (a) (b) (c) (d) R f value depends on the type of chromatography Higher R f value means higher adsorption R f value is dependent on the mobile phase The value of R f can not be more than one differential absorption of the substances on the solid phase differential adsorption of the substances on the solid phase gravitational force capillary action 11. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation? NH2 (a) (b) N N+2Cl– NO2 (d) 12. Which of the following molecules is least resonance stabilised? (a) 6. In the following compound, NH2 dN e N H (b) CHI3 (d) CH(CN)3 (c) 5. The principle of column chromatography is (a) (b) (c) (d) (a) CHBr3 (c) CHCl 3 (b) O a (c) N b (d) N O N c 13. The distillation technique most suited for separating the favourable site/s for protonation is/are glycerol from spent lye in the soap industry is (a) (a) and (e) (c) (a) and (d) (a) fractional distillation (b) steam distillation (c) distillation under reduced pressure (d) simple distillation (b) (b), (c) and (d) (d) (a) 7. The correct match between items I and II is Item - I (Mixture) A. B. C. (a) (b) (c) (d) H2O : Sugar H2O : Aniline H2O : Toluene Item II (Separation method) P. Q. R. S. Sublimation Recrystallisation Steam distillation Differential extraction (A) ® (Q); (B) ® (R); (C) ® (S) (A) ® (Q); (B) ® (R); (C) ® (P) (A) ® (S); (B) ® (R); (C) ® (P) (A) ® (R); (B) ® (P); (C) ® (S) 14. The correct order of acidity for the following compounds is CO2H CO2H HO OH I II OH CO2H CO2H OH III IV 8. An organic compound is estimated through Dumas method and was found to evolved 6 moles of CO2 , 4 moles of H2 O and 1 mole of nitrogen gas. The formula of the compound is (a) C6H8 N (b) C12H8 N (c) C12H8 N2 (d) C6H8 N2 9. If dichloromethane (DCM) and water (H 2O) are used for differential extraction, which one of the following statements is correct? (a) DCM and H2O would stay as lower and upper layer respectively in the S.F. (b) DCM and H2O would stay as upper and lower layer respectively in the separating funnel (S.F.) (c) DCM and H2O will be miscible clearly (d) DCM and H2O will make turbid/colloidal mixture OH (a) I > II > III > IV (b) III > I > II > IV (c) III > IV > II > I (d) I > III > IV > II 15. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in 60 mL of M /10 sulphuric acid. The unreacted acid required 20 mL of M /10 sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is (a) 6% (c) 3% (b) 10% (d) 5% Organic Chemistry Basics 16. A gaseous hydrocarbon gives upon combustion 0.72 g of 24. The correct acidity order of the following is water and 3.08 g of CO 2. The empirical formula of the hydrocarbon is (a) C2H4 (b) C3H4 (c) C6H5 OH OH COOH COOH (d) C7 H8 17. The order of stability of the following carbocations CH2 CH2 CH CH2 ; CH3 CH2 CH2 ; (I) (I) (III) (b) II > III > I (d) III > I > II acidity OH ; OH ; (a) (III) > (IV) > (II) > (I) (b) (IV) > (III) > (I) > (II) (c) (III) > (II) > (I) > (IV) (d) (II) > (III) > (IV) > (I) 25. Hyperconjugation involves overlap of the following orbitals 18. Arrange the following compounds in the order of decreasing OH (b) s - p (a) s - s (c) p-p CH3 (I) (II) (a) II > IV > I > III (c) III > I > II > IV (d) p - p 26. The correct stability order for the following species is OH Ž ; ½ Å O (I) Cl CH3 (IV) (II) is (II) (a) III > II > I (c) I > II > III (III) Cl NO2 OCH3 (III) ½ Å (IV) (II) ½ O (b) I > II > III > IV (d) IV > III > I > II Å (III) (IV) 19. A solution of ( - l ) 1-chloro-1-phenylethane in toluene (a) (II) > (IV) > (I) > (III) (b) (I) > (II) > (III) > (IV) racemises slowly in the presence of a small amount of SbCl 5, due to the formation of (c) (II) > (I) > (IV) > (III) (d) (I) > (III) > (II) > (IV) (a) carbanion (c) carbocation 27. Among the following, the least stable resonance structure is (b) carbene (d) free radical Å 20. In allene (C3 H4 ), the type(s) of hybridisation of the carbon (a) sp and sp (b) sp and sp (c) only sp3 (2011) structure is + – (I) H2 C == N == N (II) H2 C ¾ N == N (III) H2 C ¾ N ºº N (IV) H2 C ¾ N == N + Å O – – + (a) (I) > (II) > (IV) > (III) (b) (I) > (III) > (II) > (IV) (c) (II) > (I) > (III) > (IV) (d) (III) > (I) > (IV) > (II) N Å O N ½ O O 28. When benzene sulphonic acid and p-nitrophenol are treated with NaHCO3, the gases released respectively are (a) SO2, NO2 (c) SO2, CO2 (b) SO2, NO (d) CO2, CO2 29. Which of the following is obtained when 4-methylbenzene sulphonic acid is hydrolysed with excess of sodium acetate? (a) H3C— —COONa migrate to the positively charged carbon is (b) H3C— + SO3 H H ½ ½ + 1 5 H3 C¾ C2 ¾ C3 ¾ C4 ¾ CH3 ½ ½ ½ OH H CH3 (c) H3C— —SO3Na + CH3COOH (d) H3C— —SO2OCOCH3 + NaOH 23. In the following carbocation; H/CH 3 that is most likely to (a) CH3 at C-4 (c) CH3 at C-2 ½ Å (d) ½ (b) p-hydroxybenzoic acid (d) p-toluic acid 22. The correct stability order of the following resonance – (c) O N O Å 21. Among the following compounds, the most acidic is + Å O 2 (d) sp2 and sp3 (a) p-nitrophenol (c) o-hydroxybenzoic acid Å (b) ½ atoms, is (are) 3 O Å N (a) (b) H at C-4 (d) H at C-2 Organic Chemistry Basics HOOC 30. For 1-methoxy-1, 3-butadiene, which of the following O– resonating structure is least stable? s Å (a) CH2 ¾ CH ¾ CH ==CH ¾ O ¾ CH3 s Å (b) CH2 ¾ CH== CH ¾ CH ==O ¾ CH3 (c) s Å (c) CH2 == CH ¾ CH— CH ¾ O ¾ CH3 s O2N CH Å (d) CH2 == CH ¾ CH ¾ CH== O ¾ CH3 31. H 3N + Z O– + NH3 Y HOOC OH COOH X Arrange in order of increasing acidic strength (a) X > Z > Y (c) X > Y > Z (b) Z < X > Y (d) Z > X > Y (d) O2N C– 32. Among the following, the molecule with the highest dipole O– moment is (a) CH3Cl (b) CH2Cl 2 (c) CH2Cl 2 (d) CCl 4 33. Which of the following represent the given mode of 2 2 hybridisation sp - sp - sp - sp from left to right? (c) H2C==C==C==CH2 (d) ½ ½½ (b) HC ºº C¾ C ºº CH ½½ (a) H2C==CH ¾ C ºº N 35. Which of the following acids has the smallest dissociation constant? (a) CH3CHFCOOH (b) FCH2CH2COOH (c) BrCH2CH2COOH d) CH3CHBrCOOH 36. Identify the correct order of boiling points of the following compounds: 34. HOOC CH3CH2CH2CH2OH (1) CH3CH2CH2COOH (3) (a) 1 > 2 > 3 (c) 1 > 3 > 2 OH 2 moles of NaNH2 O2N (b) 3 > 1 > 2 (d) 3 > 2 > 1 37. Which of the following hydrocarbons has the lowest dipole moment? CH OH The product A will be – OOC (a) cis-2-butene (b) 2-butyne (c) 1-butyne (d) H2C==CH—CººCH 38. The correct order of basicities of the following compounds is H3C¾ C OH (1) NH NH2 (CH3 )2 NH (a) CH3CH2CH2CHO (2) (3 ) O2N (a) 2 > 1 > 3 > 4 (c) 3 > 1 > 2 > 4 CH CH3CH2NH2 (2 ) CH3CONH2 (4 ) (b) 1 > 3 > 2 > 4 (d) 1 > 2 > 3 > 4 39. Among the following, the strongest base is O – – (a) C6H5NH2 OOC (b) p-NO2C6H4NH2 OH (c) m-NO2—C6H4NH2 (d) C6H5CH2NH2 (b) 40. Which of the following, has the most acidic hydrogen? O2N C– OH (a) 3-hexanone (b) 2, 4-hexanedione (c) 2, 5-hexanedione (d) 2, 3-hexanedione Organic Chemistry Basics 41. The most unlikely representation of resonance structures of 47. The number of sigma and pi-bonds in 1-butene 3-yne are p-nitrophenoxide ion is O– O– O N+ (a) 5 sigma and 5 pi (c) 8 sigma and 2 pi O N (b) 7 sigma and 3 pi (d) 6 sigma and 4 pi 48. The compound which gives the most stable carbonium ion on dehydration is (a) (a) CH3 ¾ CH ¾ CH2OH ½ CH3 CH3 ½ (b) CH3 ¾ C¾ OH ½ CH3 (c) CH3CH2 ¾ CH2 ¾ CH2OH OH ½ (d) CH3 ¾ CH ¾ CH2CH3 (b) O– O– – O O N – O– O + N + 49. Polarisation of electrons in acrolein may be written as (c) (d) d+ – O ½½ d(a) H2C== CH ¾ C ¾ H O O d+ 42. Among the following compounds, the strongest acid is (a) HCººCH (c) C2H6 (b) C6H6 (d) CH3OH OH d- O ½½ dd+ (c) H2C== CH ¾ C ¾ H O ½½ (d) H2C== CH ¾ C ¾ H d+ 50. The bond between carbon atom (1) and carbon atom (2) in 43. In the following compounds OH O ½½ (b) H2C== CH ¾ C ¾ H d- compound OH Nºº C ¾ CH== CH2 OH 1 2 involves the hybridisation as (a) sp2 and sp2 (c) sp and sp2 NO 2 CH 3 (II) (I) (III) NO 2 (IV ) (b) sp3 and sp (d) sp and sp Objective Questions II (One or more than one correct option) 51. Among P, Q, R and S, the aromatic compounds(s) is/are The order of acidity is Cl (a) III > IV > I > II (b) I > IV > III > II (c) II > I > III > IV AlCl3 NaH P Q (d) IV > III > I > II O 44. What is the decreasing order of strength of the bases? OH - , NH2- , H ¾ Cºº C - and CH3 ¾ CH2- (a) CH3 ¾ CH2- > NH2- > H ¾ Cºº C > OH (b) H ¾ Cºº C - > CH3 ¾ CH2- > NH2- > OH (c) OH - > NH2- > H ¾ Cºº C - > CH3 ¾ CH2(d) NH2- - - > H ¾ Cºº C > OH > CH3 ¾ CH2- 45. The hybridisation of carbon atoms in C ¾ C single bond H ¾ C ºº C ¾ CH==CH2 is (a) sp3 - sp3 (b) sp2 - sp3 (c) sp - sp2 (d) sp3 - sp 46. Amongst the following, the most basic compound is (a) benzylamine (c) acetanilide (b) aniline (d) p-nitroaniline (NH4)2CO3 O O (a) P 100-115°C (b) Q HCl R (c) R S (d) S 52. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to (a) s ® p (empty) and s ® p* electron delocalisations (b) s ® s* and s ® p electron delocalisations (c) s ® p (filled) and s ® p electron delocalisations (d) p (filled) ® s* and s ® p* electrons delocalisations 53. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is/are (a) H H2C C¾ C (c) H2C == C == O H CH2 (b) H ¾ C ºº C¾ C (d) H2C == C == CH2 H CH2 Organic Chemistry Basics 54. In the Newman’s projection for 2,2-dimethylbutane 57. Phenol is less acidic than X CH3 H 3C (b) p-methoxy phenol (c) p-nitrophenol (d) ethanol Assertion and Reason H H (a) acetic acid Y Read the following questions and answer as per the direction given below : (a) Statement I is correct; Statement II is correct; Statement II is the correct explanation of Statement I (b) Statement I is correct; Statement II is correct; Statement II is not the correct explanation of Statement I (c) Statement I is correct; Statement II is incorrect (d) Statement I is incorrect; Statement II is correct X and Y can respectively be (a) H and H (b) H and C2H5 (c) C2H5 and H (d) CH3 and CH3 55. The molecules that will have dipole moment are (a) 2, 2-dimethyl propane 58. Statement I p-hydroxybenzoic acid has a lower boiling (b) trans-2-pentene point than o-hydroxybenzoic acid. Statement II o-hydroxybenzoic acid has intramolecular hydrogen bonding. (c) cis-3-hexene (d) 2,2,3,3-tetramethyl butane 56. The compound in which C uses its sp 3 -hybrid orbitals for 59. Statement I p-nitrophenol is a stronger acid than bond formation is o-nitrophenol. (a) HCOOH (b) (H2N)2 CO (c) (CH3 )3 COH (d) CH3CHO Statement II Intramolecular hydrogen bonding make the o-isomer weaker acid than p-isomer. Match the Columns 60. Match the reactions in Column I with appropriate types of steps/reactive intermediate involved in these reactions as given in Column II. Column I H 3C O Column II O O A. p. aq NaOH Nucleophilic substitution 18 B. O O ½½ 18 CH2CH2CH2OH H2SO4 Electrophilic substitution r. Dehydration s. Nucleophilic addition t. Carbanion O O ½½ C. q. CH2CH2CH2Cl CH3MgI CH3 CH2CH2CH2C(CH3)2 D. H2SO4 OH H3C CH3 Organic Chemistry Basics 61. Match the reaction in Column I with appropriate options in Column II. Column II Column I 0°C q. Addition reaction r. Substitution reaction s. Coupling reaction t. Carbocation intermediate —OH OH OH O ½ ½ H 2SO 4 CH3 H3C — C — — C — CH3 ¾¾¾® C C—CH3 H3C ½ ½ CH3 CH3 CH3 O —–C CH3 C. Racemic mixture NaOH/H2O —N N— B. p. —OH —N2Cl + A. 1. LiAlH4 2. H3O + OH —–HC CH3 D. SH— —Cl Base S Fill in the Blanks 62. The kind of delocalisation involving sigma bond orbitals is called…… 63. The bond dissociation energy needed to form the benzyl Integer Answer Type Questions 73. The total number of contributing structures showing hyperconjugation (involving C — H bonds) for the following carbocation is radical from toluene is ……… than the formation of the methyl radical from methane. H 3C Å CH2CH3 ½ 64. The structure of the enol form of CH3 ¾ CO ¾ CH2 ¾ CO ¾ CH3 with intermolecular hydrogen bonding is ……… 65. The IUPAC name of succinic acid is ……… 66. The shape of (CH3 )+ is ……… 74. Amongst the following, the total number of compounds soluble in aqueous NaOH is H3C N CH3 COOH OH NO2 67. A…… diol has two hydroxyl groups on ……carbon atoms. 68. The terminal carbon atom in butane is …… hybridised. 69. ………ring is most strained. (cyclopropane, cyclobutane, cyclopentane) 70. The compound having both sp and sp 2 -hybridised carbon H 3C OCH2CH3 CH2OH OH N CH2CH3 CH2CH3 atoms is ……. (propane, propene, propadiene). 71. In acidic medium, ……….. behaves as the strongest base. (nitrobenzene, aniline, phenol) 72. Among the given cations, ……… is most stable. (sec-butyl carbonium ion, tert-butyl carbonium ion, n-butyl carbonium ion) COOH CH3 Organic Chemistry Basics p-methylbenzoic acid p-methoxybenzoic acid Subjective Questions 75. Which of the following is more acidic and why? + + NH3 NH3 36.2 ´ 10 -5 10.2 ´ 10 -5 79. Give reasons for the following: CH2 ==CH - is more basic than HCºº C - . 80. Explain, why o-hydroxybenzaldehyde is a liquid at room temperature while p-hydroxybenzaldehyde is a high melting solid? or F 81. Discuss the hybridisation of carbon atoms in allene (C3 H4 ) OH 76. Draw the resonating structures of and show the p-orbital overlaps. . 77. You have an ether solution containing 4-hydroxybenzoic acid and 4-aminobenzoic acid. Explain, how will you separate the two in not more than 3 steps? Give confirmatory tests with reagents and conditions for functional groups of each. 78. Match the following with their K a values Benzoic acid p-nitrobenzoic acid p-chlorobenzoic acid 82. Give reasons for the following in one or two sentences. The central carbon-carbon bond in 1, 3-butadiene is shorter than that of n-butane. 83. Although phenoxide ion has more number of resonating structures than benzoate ion, benzoic acid is a stronger acid than phenol. Why? 84. Arrange the following in the order of their increasing 4.2 ´ 10 -5 3.3 ´ 10 -5 6.4 ´ 10 -5 basicity. p-toluidine, N,N-dimethyl-p-toluidine, p-nitroaniline, aniline. Answers Topic 1 17. (d) 18. (c) 19. (c) 20. (b) 1. (d) 2. (c) 3. (b) 4. (c) 21. (c) 22. (b) 23. (d) 24. (a) 5. (d) 6. (a) 7. (b) 8. (a) 25. (b) 26. (d) 27. (a) 28. (d) 9. (c) 10. (c) 11. (d) 12. (d) 29. (c) 30. (c) 31. (a) 32. (a) 13. (a) 14. (c) 15. (b) 16. (a) 33. (a) 34. (a) 35. (c) 36. (b) 17. (a) 18. (b) 19. (b) 20. (a) 37. (b) 38. (b) 39. (d) 40. (b) 21. (b) 22. (d) 23. (c) 24. (a) 41. (b) 42. (d) 43. (d) 44. (a) 25. (d) 26. (b,c) 27. (a,c,d) 28. (a,b,c) 45. (d) 46. (a) 47. (b) 48. (b) 29. (a,d) 30. (b,c,d) 31. (a,d) 32. (a,c,d) 49. (d) 50. (c) 51. (a,b,c,d) 52. (a) 33. (a,c) 34. (c,d) 35. (b,d) 36. (d) 53. (b,c) 54. (b,d) 55. (b,c) 56. (c,d) 37. (7) 38. (c) 57. (a,c) 58. (d) 59. (a) 39. Non-superimposable, Enantiomers 40. False 60. A ® r, s, t; B ® p, s; C ® r, s; D ® q, r 41. False 44. (7) 61. A ® r, s; B ® t; C ® p, q; D ® r 42. (2) 43. (3) 62. hyperconjugation Topic 2 1. (c) 2. (b) 3. (c) 4. (b) 5. (b) 6. (b) 7. (a) 8. (d) 9. (a) 10. (d) 11. (b) 12. (d) 13. (c) 14. (a) 15. (b) 16. (d) 63. less 64. cyclic 66. triangular planar 68. sp3 65. butanedioic acid 67. geminal, same 69. cylopropane 70. propene 72. tert-butyl carbonium ion, 73. (6) 71. aniline 74. (4) Hints & Solutions Topic 1 Nomenclature and Isomerism While naming the compound, the longest chain that have principal functional group ¾ COOH is choosen and numbered in such a manner that the principal functional group gets the lowest possible number. ¾ OH act as substituent and used as prefix in nomenclature. 1. The IUPAC name for the given compound is 3, 5-dimethyl-4-propylhept-1-en-6-yne. CH3 7 CH H3C 4 2 3 1 Longest possible chain containing double and triple bond 5 6 CH3 5. While naming the compound, alkene gets priority over functional group ( ¾ Br ) and numbering starts from alkene side. Hence, IUPAC name: 4-bromo-3-methyl pent-2-ene CH2 1 CH3 If both double and triple bonds are present in the compound, the endings like-en-yne, a (numeral) dien-(numeral)-yne etc., are used. Numbers as low as possible are given to double and triple bonds as a set. 2. The IUPAC name of the given compound is 3 2 2- chloro-1-methy l- 4-nitrobenzene OH Cl 3 COOH O O H 1 OH 2S Cl 3R H H No additional stability with the ‘enol’ form 2 1 3 O CH3 OH 5 Acetylacetone This enol-form is highly stable because of intramolecular H-bonding (6-membered stability) and due to extended conjugation. O O (–R) 1 2 7. CN has the highest priority. 1 2 CN Br O O 1 2 (2-bromo-5-hydroxybenzonitrile) Br O Ethyl acetoacetate (EAA) NH2 H In both of the compounds, C -2 of C == O group is a part of the acid O O ½½ ½½ derivative (ester, ¾ C ¾ OEt and acid amide, ¾ C ¾ NH2). So, C2 2 3 OH Acetone O 1 For C-2, order of priority of substituents is OH > CH(Cl) (CH3 ) > COOH For C-3, order of priority of substituents is Cl > CH(OH)COOH > CH3 Hence, according to CIP rules, Here, the given compound contains two or more functional groups. So, the numbering is done in such a way that the sum of the locants is the lowest. O CH3 Br 4 CH3 CH3 3. H 4 COOH H H Cl 1 3 5 Correct way 4 Incorrect way H 6. NO2 CH3 2 CH3 H 8. C==C H3C H Br2 CCl4 H 3C given compound CH3 OH ½ 4 ½3 2 1 H3 C ¾ C H ¾ C H ¾ C H2 ¾ C OOH 5 Principal chain CH3 H Br H H resonance (–R) with the derivative group itself. IUPAC name of the 3-hydroxy-4-methylpentanoic acid. C CH3 O does not take part in enolisation, because it is already in 4. The C is Br Br CH3 (meso) O ½½ 9. C6H5 ¾ C ¾ Cl : Benzene carbonyl chloride Organic Chemistry Basics 1 CH3 H H H 120° 3 10. 19. Compounds with same empirical formula but different H 2 Anti-clockwise H molecular formula have same percentage composition of elements but different molecular weight. H 3 H 2 1 H H3C CH3 CH3 (gauche) H ½* 11. CH3 — CH2 — C ¾ COOH : Has a chiral carbon, optically active. ½ CH3 (2-methylbutanoic acid) F Cl 12. F C == C + I Br F Br F C == C + Cl I F Cl I 13. CH3 — O— CH3 and CH3 — CH2 — OH are functional isomers. Geometrical isomers CH 3 ½3 2 1 H3 C— C ¾ CH == CH2 : 3,3-dimethyl-l-butene ½ CH3 4 Geometrical isomers C == C Cl H C == C CH2 — Ph H3 C C == C H 23. Double bond has preference over alkyl group hence : Br CH2 — Ph H H3C cis 24. H trans HO HO H H H OH OH H OH CH3 CH3 I II III 18. IV III V H trans-2- butene disubstituted benzene is done in alphabetical order as Cl 1 2 6 5 3 4 CH3 Hence IUPAC name of this compound is 1-chloro-4-methyl benzene. Also, toluene is an acceptable name in IUPAC, hence this compound can also be named as 4-chloro toluene. Enol of acetone It has 9 s (6 with H, two C—C and one C ¾ O), one p-bond and two lone-pairs. II H3 C 26. Since, there is no principal functional groups, numbering of I and II are optically active while III is optically inactive (meso form ). 16. Conformers can be interconverted through rotation about C—C bond. H · · ·O · ½ 17. H— C == C ¾ CH3 : ½ H I C== C H saturated alcohols. Butanone (CH3CH2COCH3 ) is unsaturated, has two hydrogen less than the diethylether. H CH3 CH3 25. Diethyl ether (C2H5OC2H5 ) will be isomeric with all 4-carbon CH3 CH3 OH H cis-2- butene 14. The ‘D’ term in name is derived from D-glyceraldehyde. M 15. H CH3 C== C H 1-phenyl-2-butene CH3 4 carbons are always isomeric. Br I 3 22. Ethers and alcohols (saturated, acyclic) with same number of I F Br H3C Geometrical isomers C == C + 2 ½ CH3 Cl Cl C == C 1 21. CH2 == CH— CH — CH3 : 3-methyl-1-butene I C == C Br H H ½ ½ 20. H— C — C — H ; Ethane has the smallest sized group (H) ½ ½ H H bonded to carbons, hence there will be least hindered rotation about C—C bond. 27. PLAN This problem is based on structure and nomenclature of organic compound. Draw structure of each compound and write IUPAC name of the given compound. Match the molecular formula of given compound with molecular formula of compound given in choices. The combination of names for possible isomeric alcohols with molecular formula C4H10O is/are Organic Chemistry Basics Formula Names CH3CH2CH2CH2OH n-butyl alcohol / n-butanol / butan-1-ol CH3 ¾ CH ¾ CH2 ¾ OH ½ CH3 Iso-butyl alcohol / 2-methyl propan-1-ol M and P are non-superimposable mirror images, hence, enantiomers. CH3 HO H HO H H3C Cl HO Secondary butyl alcohol / butan-2-ol CH3 ¾ CH2 ¾ CH ¾ OH ½ CH3 CH3 ½ CH3 ¾ C ¾ OH ½ CH3 H 29. HO HO H HO H HO H CH3 HO CH3 Cl HO OH H Cl N N Since, M and N have ¾ OH on same side and opposite side respectively, they cannot be miror image, they are diastereomers. Cl H HO Cl H OH HO H Cl HO HO H HO H H CH3 O M and O are identical. CH3 O CH3 O NOTE Fischer projection represents eclipse form of Sawhorse projection. For comparision purpose, similar types of eclipse conformers must be drawn i.e. both vertically up or both vertically down. CH3 H OH H OH H 3C Cl HO H HO H H OH Cl OH CH3 HO Cl OH H OH H CH3 Q CH3 C C C==C H H OH H H CH3 CH3 C C C==C HO H C HO H OH C C H OH H OH CH3 P H C==C H H C CH3 H CH3 OH (VI) (meso) Total six isomers. In both cis and trans forms, there are two enantiomers each. 30. E, F and G are not resonance structures because movement of hydrogen between E and F are involved. E, F and E, G are tautomers in which E is keto form and both F and G are enol form of the same E. F and G are geometrical isomers. F and G are distereomers as they are stereo isomers but not related by mirror image relationship. 31. The compound is optically active as well as it possesses a two-fold axis of symmetry. O —CH==CH—OH r —CH2 ==C—H Cl H H CH3 C==C H H OH (II + III) (enantiomeric) (IV + V) (enantiomeric) 32. (a) H Cl P Cl CH3 H H OH HO (I) (meso) M CH3 H HO H H CH3 CH3 M H CH3 Cl Cl H M and Q are non-identical they are distereomers. 28. Converting all of them into Fischer projection. H HO HO Tertiary butyl alcohol / tert butanol / 2-methyl propan-2-ol Hence, choices (a), (c) and (d) are correct. HO H Cl Q ==O r (c) O —OH HO Organic Chemistry Basics ==O r (d) —OH O H3C 33. (a) C == C H3C + H (c) H Ph + CH3 cis-1-phenylpropene CH3 Ph H CH3 trans-1- phenylpropene H ½* (d) BrCH2 — C— CH3 ½ OH True / False 40. It has only two asymmetric carbon, carbon no.3 is not asymmetric. 41. They are identical. 42. Although the compound has two chiral carbons (indicated by stars), it does not has four optically active isomers as expected. It is due to its existence in cis-form only. 35. Carbonyl compounds containing a-H show keto-enol H3 C tautomerism. OH O ½ ½½ (b) C6H5 ¾ C ¾ CH3 r C6H5 — C == CH2 H 3C CH3 ½ + CH3 — C — CH3 ½ Cl 37. (7) As given in the question 3 stereocentres are visible, i.e. * * * HO OH Hence, the total number of stereoisomers = 23 = 8 But out of these the following one is optically inactive due to symmetry OH OH 43. PLAN This problem can be solved by using concept of conformational analysis of given organic compound. To solve the question draw the stable conformational structures of organic compound and determine the net resultant dipole moment. The conformations of the given compound are as follows Cl Br Cl CH3 Cl Br Br CH3 Cl Br = = Br Cl Br Cl CH3 CH3 H3C CH3 H H H ½ ½ ½ 1 5 C H3 ¾ C*2 ¾ C*3 ¾ C*4 ¾ CH3 ½ ½ ½ Cl Cl Cl Cl Br CH3 (Me-Me)gauche OH Hence, total number of optically active stereoisomers=7 O trans-form (only hypothetical) The above shown transformation does not exist due to restricted rotation about the bridge head carbons, hence only cis-form and its mirror image exist. CH3 HO CH3 O cis-form CH3 CH3 ½ ½ Cl 2 36. CH3 — C — CH3 ¾¾® CH3 — C — CH 2Cl hn ½ ½ H H CH3 * O OH ½½ ½ C6H5 ¾ C ¾ CH== C— CH3 OH CH3 CH3 * O O ½½ ½½ (d) C6H5 ¾ C ¾ CH2 ¾ C ¾ CH3 r OH H 39. Non-superimposable, Enantiomers. 34. A carbon bonded to four different atoms or groups is called asymmetric carbon : Cl ½ (c) CH3 — C*— D ½ H H Molecule is chiral but does not possesses any chiral carbon. C == C H CH3 C==C==C trans-2- butene H C == C H3C H C == C cis-2- butene H optically active and known as chiral molecule. However, for chirality of molecule, presence of chiral centre is not essential, e.g. O CH3 H 38. Molecules that are non-superimposable on its mirror image are CH3 Cl Br Br Br CH3 (Br-Me)gauche Br Cl Br Cl Cl Cl CH3 CH3 (Cl-Me)gauche Organic Chemistry Basics These three have non-zero dipole moment due to non-cancellation of all dipole moment created by C—Cl and C—Br bond. 44. CH3 CH3 OH 49. CH3 Mirror H non-superimposable while II is a meso form. Þ (I + III) = Enantiomers I + II and II + III = Diastereomers CH3 CH3 CH2CH3 CH3 48. I and III are mirror images of one another as well as they are CH3 O H enol H H CH3 H H CH3 meso pair of enantiomers Þ Total seven isomers 45. (a) 4 Z 3-(N, N-dimethyl)-3-methyl pentanamine or 3-methyl-3-(N, N-dimethyl) pentanamine Z H H H Z H H H 51. H Z H anti gauche CH3 O H (b) D H D CH3 (i) (ii) 1 Me 3 8 Me 10 Me 6 9 4 3 2 1 2-pentenoic acid or pent-2-enoic acid H 53. Cl Cl C== C Cl Cl C== C H C== C H Cl H H H I II III III is non-polar, has zero dipole moment. H 54. CH3 — CH2 — CH2 — CH2OH H Structure (ii) is more stable than its anti conformer because of intramolecular H-bonding. CH3 OH ½ CH3 — C ¾ CH3 ; ½ CH3 Cl CH3 — CH2 — CH — CH3 ½ OH enantiomeric ( ± ) CH3 — CH — CH2 OH ½ CH3 Total number of isomers (including stereoisomers) = 5 Topic 2 General Organic Chemistry 1. Hydrocarbon containing C and H upon burning produces CO2 H CH3 H H and water vapour respectively. The equation is represented as 46. (a) H less stable staggered form of butane (b) The less stability of above mentioned conformer is due to van der Waals’ repulsion between the adjacent methyl groups. 47. 2 52. CH3 CH2 — CH== CH— COOH H H Me 5,6-diethyl-3-methyl decane O D D Me 5 m anti = 0 Þ 1 = m gauche ´ 0.18 1 m gauche = = 5.55 D 018 . Þ 4 5 7 Mole fraction of anti form = 0.82 Mole fraction of gauche form = 0.18 m obs = 1 D 1 = m anti ´ 0.82 + m gauche ´ 0.18 Þ Q keto CH3 ½ 2 1 50. H3C— N — C— CH2 — CH3 3 ½ ½ 5 CH3 CH2CH3 HO OH OH glycerine It contains a secondary (2°) hydroxy group. CxH y + (x + y / 4 )O2 ¾® xCO2 + ( y / 2)H2O 12 Mass of carbon = ´ mass of CO2 44 12 = ´ 88 g = 24 g 44 2 Mass of hydrogen = ´ mass of H2O 18 2 = ´9 =1g 18 So, the unknown hydrocarbon contains 24 g of carbon and 1g of hydrogen. Organic Chemistry Basics 2. Organic compound ‘A’ contain phosphorus as it gives positive At ‘a’ and ‘e’, N-atoms are sp3 (s% = 25) hybridised, whereas at ‘ test with ammonium molybdate. Phosphorus present in organic compound ‘A’ get oxidised with Na 2O2 and form Na 3PO4. 2P + 5Na 2O2 ¾® 2Na 3PO4 + 2Na 2O b’, ‘c’ and ‘d’, N-atoms are sp2 (s% = 33) hybridised. So, ‘b’, ‘c’ and ‘d’ are the favourable sites for protonation (H+ is a Lewis acid, i.e. electrons acceptor). Compound Sodium phosphate Na 3PO4 in presence of HNO3 form H3PO4 and NaNO3. Na 3PO4 + 3HNO3 ¾¾® H3PO4 + 3NaNO3 Upon cooling, a few drops of ammonium molybdate solution are added. A yellow ppt. confirms the presence of phosphorus in the organic compound. H3PO4 +12 (NH4 )2 MoO4 + 21HNO3 ¾® (NH4 )3 PO4.12MoO3 ¯ +21NH4NO3+12H2O Yellow ppt. 3. Higher the basicity of a base, stronger will be its nucleophilic power. Again we know, a weaker acid produces a stronger base (conjugate), i.e. a stronger nucleophile. - HÅ CH3SOÈ3 CH3CO2H CH3COÈ2 H2O Acid strength È (A) ® (Q); (B) ® (R); (C) ® (S) (A) H2O and sugar mixture They do not react chemically. On heating, solubility of sugar in H2O increases and on rapid cooling of saturated solution, sugar recrystallises (Q). (B) H2O and aniline mixture Aniline is steam volatile but insoluble in H2O. So, steam distillation (R) is employed for their separation. (C) H2O and toluene mixture Toluene is steam non-volatile and also insoluble in H2O. So, differential extraction method (S) can be used to separate them. 8. In Dumas method, organic compound is heated with dry cupric Acid ¾¾® Conjugate base H3O+ H2O CH3SO3H 7. The correct option is : OH Base strength º Nucleophilicity Thus, the correct order of nucleophilicity of given nucleophiles are as follows : - H2O(2) < CH3SO-3 (3) < CH3CO-2 (1) < O H(4) 4. In chromatography, the expression of retention factor (R f ) is Distance travelled by the compound from origin Rf = <1 Distance travelled by the solvent from origin The value of R f signifies the relative ratio of migration of each component of the mixture with respect to the developing solvent used. R f value depends on the type of adsorption chromatography like TLC (Thin-Layer Chromatography), paper chromatography etc. The R f value is also the characteristic of a compound (sample) for a given developing solvent at a given temperature. When the compound in the sample (usually less polar) is weakly adsorbed the spot will travel a shorter distance from the origin and hence the R f value will be decreased. 5. In column chromatography, separation of mixture of compounds (adsorbate) takes place over a column of solid adsorbent (silica gel and Al 2O3 ) packed in a glass tube. When an appropriate eluant (liquid) is allowed to flow down the column, the compounds present in the mixture get adsorbed to different extent on the adsorbent column and thus complete separation takes place. Thus, column chromatography is based on the differential adsorption of the substance on the solid phase. 6. All sites (a, b, c, d, e) of the given molecule have lone pair on N-atoms. Higher the ease of donation of lp of electrons of N, more favourable will be the site for protonation. Ease of donation of lp of e s, i.e. Lewis basicity inversely depend on the percentage of s-character in hybridisation of ‘N’ which will decide the electronegativity of ‘N’. oxide in a combustion tube in the atmosphere of CO2. Upon heating, C and H present are oxidised to CO2 and water vapours while N 2 is set free. Let, the molecular formula of the organic compound (1 mol) be C xH yN z. In Dumas method, y y z y Cx H y N z + æç 2x + ö÷ CuO ® xCO2 + H 2 O + N 2 + æç 2x + ö÷ Cu 2 2 2 2 è ø è ø 1 mol x mol y mol 2 z mol 2 y z = 4 Þ y = 8 and = 1Þ z = 2 2 2 \ Molecular formula of the compound is C6H8 N2. Now, x = 6, 9. Dichloromethane, DCM (CH2Cl 2 ) is heavier (density = 13266 . g cm-3) than water (density = 1 g cm -3 ). So, DCM and H2O will stay as lower and upper layer respectively in the separating funnel (SF). 10. We know, a stronger acid produces its stable or weaker conjugate base. Here, CH(CN)3 produces the most stable conjugate base (NC)3 C- . Stronger -R and -I effects of the CN - group, make the carbanion (conjugate base) very stable. The resonance hybrid structure of [(NC)3 C]- is as follows: CN – d+ d– (NC)2C C N NC C y +H Å (NC) C H 3 (–I) C N (–R) y N C N C C C N Resonance hybrid structure of [NC)3 C]– However, halogen (X = Cl, Br, I) show -I effect but + R effect of halogens, destabilises the carbanion, X 3C – (conjugate base of the haloform, HCX 3 ). s X2C X (+R,–I ) Thus, CH(CN)3 is the strongest acid among the given options. Organic Chemistry Basics 11. Estimation of nitrogen through Kjeldahl’s method is not suitable for organic compounds containing nitrogen in ring or nitrogen in nitro or azo groups. It is because of the fact that nitrogen of these compounds does not show conversion to Ammonium sulphate ((NH4 )2 SO4 ) during the process. Hence, among the given compounds only aniline can be used suitably for estimation of nitrogen by Kjeldahl’s method. 12. Aromatic compounds are stable due to resonance while non-aromatics are not. According to Huckel’s rule (or 4 n + 2 rule), “For a planar, cyclic compound to be aromatic, its p cloud must contain (4 n + 2)p electrons, where, n is any whole number.” Thus, Mass of organic compound = 1.4 g Let it contain x m mole of N atom. Organic compound ¾® and , (6pe– system) ammonium sulphate and ammonium sulphate to sodium sulphate. (b) Calculate millimoles (m moles) of N present in organic compound followed by mass of N present in organic compound using the concept of stoichiometry. (c) At last, calculate % of N present in organic compound using formula Mass of N ´ 100 % of N = Mass of organic compound O N (4pe– + 2e– system) NH3 x m mole 2NH3 + (6pe– system) H2SO4 …(i) ¾® (NH4 )2 SO4 6 m mole initially taken are aromatic and stabilised by resonance. They follow Huckel’s rule. H2SO4 + 2 NaOH ¾® Na 2SO4 + 2 H2O ...(ii) 2 m mole NaOH reacted. is non-aromatic, hence, least stabilised by resonance. Hence, m moles of H2SO4 reacted in Eq. (ii) = 1 Þ m moles of H2SO4 reacted from Eq. (i) = 6 - 1 O 13. Glycerol with high boiling point (290°C) can be separated from spent lye by distillation under reduced pressure. This process is used to purify liquids having very high boiling points. By this process, liquid is made to boil at lower temperature than its boiling point by lowering the pressure on its surface. = 5 m moles Þ m moles of NH3 in Eq. (i) = 2 ´ 5 = 10 m moles Þ m moles of N atom in the organic compound = 10 m moles Þ Mass of N = 10 ´ 10-3 ´ 14 = 0.14 g % of N = Mass of N present in organic compound ´100 Mass of organic compound % of N = 0 .14 ´ 100 1.4 14. ¾ OH group displays both kinds of effect; an electron withdrawing acid-strengthening inductive effect from the meta-position and an electron-releasing acid weakening resonance effect from the para-position (at this position, resonance effect overweighs the inductive effect). Thus, III > IV. o-hydroxybenzoic acid (II) is far stronger than the corresponding meta and para isomers as the carboxylate ion is stabilised by intramolecular H-bonding. 2,6-dihydroxybenzoic acid (I) forms carboxylate ion which is further stabilised by intramolecular H-bonding, Thus, correct order is I > II > III > IV d- O C O d+ H O H O = 10% 16. 18 g H2 O contains 2g H \ 0.72 g H2O contains 0.08 g H 44 g CO2 contains 12 g C \ 3.08 g CO2 contains 0.84 g C \ C:H = d- d- d+ Þ C O d+ = 0.07 : 0.08 = 7 : 8 H O O 0.84 0.08 : 12 1 \ Empirical formula = C7 H8 17. The order of stability of carbocation will be II 15. I CH2+ (Most stable) THINKING PROCESS This problem is based on the estimation of percentage of N in organic compound using Kjeldahl’s method. Use the concept of stoichiometry and follow the steps given below to solve the problem. (a) Write the balanced chemical reaction for the conversion of N present in organic compound to ammonia, ammonia to > CH2 Benzyl ( resonance stabilised with more canonical forms) (III) CH CH2 > CH3 CH2 CH2 Allyl ( resonance stabilised) Propyl (stabilised by +I effect) (I) (II) Organic Chemistry Basics OH > 18. OH OH > NO2 Cl (–M, –I ) ( –I ) (III) (I) OH 22. I is most stable because it has more covalent bonds and negative charge on electronegative nitrogen. III is more stable than II and IV due to greater number of covalent bonds. Between II and IV, II is more stable since, it has negative charge on electronegative atom and positive charge on electropositive atom. Hence, overall stability order is > CH3 OCH3 (+I, (hyper conjugation) (+M) I > III > II > IV (IV) (II) 23. H at C 2 will migrate giving resonance stabilised carbocation. H H + ½ + ½ HH3C ¾ C ¾ C ¾ C ¾ CH3 ¾® H3C ¾ C ¾ CH2CH(CH3 )2 shift ½ ½ ½ ½ OH H CH 3 OH ¬® H3C ¾ C ¾ CH2 ¾ CH(CH3 )2 ½½ + OH Electron releasing group decreases while electron withdrawing group increases acidic strength by destabilising and stabilising the phenoxide ion formed respectively. Å SbCl 5 19. Cl ¾ CH ¾ CH3 ¾¾® Ph ¾ CH ¾ CH3 + SbCl6Toluene ½ Ph (Carbocation) (planar) (- ) ¾® Ph ¾ CH ¾ CH3 ½ Cl —OH < Cl— 24. ( d and l ) mixture I + SbCl 5 II 20. Allene is the name given to propdiene, H2C == C == CH2 . Hybridisation of an atom is determined by determining number of hybrid orbitals at that atom which is equal to number of sigma (s) bonds plus number of lone pairs at concerned atom. Pi(p ) bonds are not formed by hybrid orbitals, therefore, counted for hybridisation. the the the not + I effect decreases acid strength s C s H s H III IV p H – 25. H—C—CH—CH2 C s s H —COOH Cl has overall electron withdrawing effect, increases acid strength. H C —COOH < H3C— Allene H s —OH < H—C==CH—CH2 H H + 26. The s-electron of C ¾ H bond is delocalised with p-orbitals of Here, the terminal carbons have only three sigma bonds associated with them, therefore, hybridisation of terminal carbons is sp3. The central carbon has only two sigma bonds associated, hence hybridisation at central carbon is sp. p-bond. + + mono-substituted phenol as former being a carboxylic acid. Among the given substituted benzoic acid, ortho-hydroxy acid is strongest acid although —OH causes electron donation by resonance effect which tends to decrease acid strength. OH + O 2°, has resonnce stability O + III 1°, has no resonance stability IV – O has no resonance stability but 2° + O H+ + + II It is due to a very high stabilisation of conjugate base by intramolecular H-bond which outweigh the electron donating resonance effect of —OH. O 3°, has resonance stability O I 21. A mono-substituted benzoic acid is stronger than a COOH O H The overall order of acid-strength of given four acids is ortho-hydroxybenzoic acid (pKa = 2.98) > Toluic acid pKa = 4.37) > p-hydroxybenzoic acid (pKa = 4.58) > p-nitrophenol (pKa = 7.15) Therefore, overall stability order is : I > III > II > IV 27. The following structure has like charge on adjacent atoms, therefore, least stable + - + N O- O Organic Chemistry Basics 28. —SO3Na + H2O + CO2 —SO3H + NaHCO3 35. Weakest acid BrCH2CH2COOH has smallest dissociation constant. 36. Butanoic acid forms more exhaustive H-bonds than butanol and O2N— —OH + NaHCO butanal do not form intermolecular H-bonds. Hence, boiling point order will be 3 > 1 > 2. 3 O2N— —ONa + H2O + CO2 29. A spontaneous neutralisation will occur between strong acid and 37. CH3 ¾ C ºº C ¾ CH3 symmetric 38. I is most basic due to formation of resonance stabilised conjugate acid . NH strong base as H3C— strong base NH2 strong acid H3C— + H ¾® H3C ¾ C H3C ¾ C —SO3H + CH3COONa – + —SO3Na + NH2 N H2 ·· NH2 ¬® H3C ¾ C + CH3COOH NH2 weak acid + weak base + resonance stabilised conjugate acid - 30. H2C== CH ¾ CH ¾ CH ¾ O ¾ CH3 IV is amide, least basic. O O ½½ · · ½ + CH3 ¾ C ¾ N H2 ¬® CH3 ¾ C == NH2 Lone pair of oxygen is not the part of this mode of delocalisation. 31. Carboxylic acid is stronger acid than ammonium ion, hence + lone pair is part of delocalisation + —COOH(X) is most acidic. Z (NH3 ) is more acidic than Y (NH3 ) due to – I effect of —COOH on Z. Hence, overall acid strength order is Also, among alkyl amines, 2° is more basic than 1° amine. Hence, overall order of basic strength is 1>3>2>4 X > Z > Y 39. 32. CH3Cl has highest dipole moment. —CH2—NH2 33. H2C ==CH ¾ C ºº N ­ ­2 sp2 sp ­ ­ sp sp Lone pair is not taking part in resonance, most basic. In other cases, lone pair of nitrogen is part of delocalisation which decreases Lewis base strength. 34. In general, the order of acid strength is ¾ C ºº CH < ROH < H2O < PhOH < R ¾ COOH Therefore, during stepwise neutralisation of given acid, —COOH will be neutralised first. In the second step, the phenolic —OH, assisted by – I effect of ¾ NO2 at meta position will be neutralised. 40. A methylene ( ¾ CH2 ¾ ) with carbonyl on both side is highly acidic. O O ½½ ½½ CH3 ¾ C ¾ CH2 ¾ C ¾ CH2 ¾ CH3 * ­ HOOC has very acidic H. OH – 41. O— O2N ==N O 2 moles of NaNH2 CH O octet of nitrogen is violated. 42. Although alcohols are weaker acid than water, it is stronger than – OOC ammonia and terminal alkynes. OH OH + 2NH3 O2N CH O– 43. Nitro group from para position exert electron withdrawing resonance effect, increases acidity of phenol the most. This is followed by meta nitrophenol in which nitro group exert electron withdrawing effect on acidity. CH3— is an electron donating group, decreases acid strength. Hence, the overall order is IV > III > I > II Organic Chemistry Basics 44. CH3CH-2 > NH-2 > H ¾ C ºº C - > HOP Q R S It is because the order of acid-strength of their conjugate acid is CH3CH3 < NH3 < H ¾ C ºº C ¾ H < H2O : Acid strength. 45. H ¾ Cºº C ¾ CH == CH2 ­ sp ­ sp3 (4 n + 2) 2 6 6 6 —CH2—NH2 benzylamine Lone pair is not involved in resonance, most basic. In all other cases, lone pair of nitrogen is involved in resonance, less basic. H H ½ 47. C == C ¾ C ºº C ¾ H : It has 7 sigma and 3 pi bonds. H 1- butene-3-yne CH3 CH3 ½ ½ + H 48. CH3 ¾ C — OH ¾¾® CH3 ¾ C+ - H2 O ½ ½ CH3 CH3 (3°, most stable alkyl carbocation) PLAN Spreading out charge by the overlap of an empty p-orbital with an adjacent s-bond is called hyperconjugation. This overlap (the hyperconjugation) delocalises the positive charge on the carbocation, spreading it over a larger volume, and this stabilises the carbocation. Vacant p-orbital r r CH3—C—CH2—H – H CH2 ==CH—CH—CH3 In 2-butene, stabilisation is due to hyperconjugation between s - p* electron delocalisation. 53. In both (b) and (c), all the atoms are present in one single plane H H—C C—C straight C 50. N ºº C ¾ CH == CH2 sp2 sp2 H C H H C O straight H 2 In (a) 1, 3-butadiene, conformational change is possible between C2 — C3 bond in which atoms will be present in more than one single plane. In (d) allene, the terminals H—C—H planes are perpendicular to one another. CH3 sp sp2 — PLAN A species is said to have aromatic character if (a) ring is planar (b) their is complete delocalisation of p-electrons (c) Huckel rule i.e. ( 4 n + 2) rule is followed. where, n is the number of rings ( 4 n + 2) = p electron delocalised. 54. H3C—CH2—C—CH3 3 2 1 — 4 CH3 Cl + On C2 — C3 axis, X = CH3 , Y = CH3 On C1 — C2 axis, X = H, Y = C2H5 – + AlCl4 + AlCl3 CH3 ½ 55. H3C ¾ C — CH3 ½ CH3 H3C H C == C H C2H5 Aromatic P – + H2 + NaH Aromatic Q O O (NH4)2CO3 O D, 100–115°C H CH2 Å H—CH 2—CH==CH—CH 3 s p - 51. CH3—C CH3 CH3 tertiary butyl carbocation has one vacant p-orbital, hence, it is stabilised by s-p (empty) hyperconjugation. Od O O + d+ ½½ ½ ½½ 49. CH 2 == CH ¾ CH ¬® CH 2CH == CH ¬® CH 2 == CH ¾ C ¾ H 1 p electrons 2 6 (including lone pair) 6 (including lone pair on N) 6 In all cases there is complete delocalisation of p-electrons. 52. 46. n 0 1 1 1 – – – – OH HCl N H Aromatic R C2H5 C == C H Aromatic S m ¹0 unsymmetrical H5C2 r ; m =0 m ¹0 H CH3 CH3 ½ ½ CH3 ¾ C — C — CH3 ½ ½ CH3 CH3 symmetrical m =0 Organic Chemistry Basics O ½½ 56. H ¾ C ¾ OH ­ sp2 O ½½ H 2N ¾ C ¾ NH2 ­ sp2 CH3 ½ CH3 ¾ C — O ¾ H ½ sp 3 CH3 + O 18 OH C. Ph H OH + Ph :OH Nucleophilic addition –H+ O ½½ CH3 ¾ C ¾ H sp 3 D. Ph HO O Ph OH 57. Phenol is less acidic than a carboxylic acid (acetic acid). Nitro group from para position exert electron withdrawing resonance effect, increases acid strength. Therefore, phenol is less acidic than p-nitro phenol. On the other hand, methoxy group from para position, donate electrons by resonance effect, decreases acid strength of phenol. Also ethanol is weaker acid than phenol due to resonance stabilisation in phenoxide ion. –H2 O+ Ph O 18 Dehydration + H –H2O + FriedelCraft alkylation Electrophilic substitution 61. + —N2 + —OH Hence, ethanol < p-methoxyphenol < phenol < p-nitrophenol < acetic acid ¾¾increasing acid strength ¾¾® 58. Statement I is incorrect; Statement II is correct. Intramolecular H-bonding in ortho-hydroxy benzoic acid lowers the boiling point. 59. Statement I is correct; Statement II is correct; Statement II is the correct explanation of Statement I. —OH —N==N— this is an example of electrophilic substitution at para position of phenol, giving a coupling product. H O O– Ph—C—CH3 + H—Al–—H Intramolecular H-bonding discourage release of H+ to some extent, hence weaker acid than its para isomer. Ph—C—CH3 H H Nucleophilic addition H O OH O– Ph—C—CH3 + N H O Pianacol-pinacolone rearrangement, occur through carbocation intermediate. O Nucleophilic addition occur at sp2 (planar) carbon, generating a chiral centre, hence product will be a racemic mixture. 60. O A. : Ph—C—CH2—CH2—C—CH3 + OH – – O CH OH 2 O H2O O Ph Ph HS— —Cl Base —S– O Ph 62. Hyperconjugation. 63. Less, stable free radical is formed. Dehydration O– O B. Ph Cl CH3MgBr Ph SN 2 64. Cyclic H O CH Ph H 3C S + Cl Nucleophilic substitution carbanion –H2O —Cl Cl O O C H3C CH3 Organic Chemistry Basics 77. O O ½½ ½½ 65. HO ¾ C ¾ CH2 ¾ CH2 ¾ C ¾ OH OH OH OH NH2 Distillation butanedioic acid and 66. Triangular planar; carbon is sp2-hybridised 67. geminal, same COOH COOH 3 68. sp -hybridised COOH + ether solution 69. Cyclopropane: here the C—C—C bond angle is 60° while the requirement is 109°. COOH (ether layer) NH3Cl– HCl NH2 NaOH(aq) 70. Propene : CH2 == CH ¾ CH3 ­ sp2 ­ sp3 COOH COOH (aqueous layer) 71. Aniline 72. Tert-butyl carbonium ion because the three methyl group Tests of functional groups HOOC— stabilises carbocation by + I effect. 2 73. These are total 6a-H to sp carbon and they all can participate in hyperconjugation. + H H+ + H FeCl 3 ¾¾® violet colouration confirm phenolic group NaHCO3 + lime water ¾¾¾® CO2 ­ ¾¾® milky ¾® — COOH NH2 + —OH NC + Three structures + CHCl3 + KOH Two structures – NH2 group confirmed. 74. Aromatic alcohols and carboxylic acids forms salt with NaOH, COOH will dissolve in aqueous NaOH : COOH OH OH COOH characteristic foul smell of isocyanide COOH 78. p-methoxy benzoic acid is the weakest and p-nitrobenzoic acid is the strongest acid among these acids. Chloro group has overall electron withdrawing effect on ring, therefore, increases acid strength of benzoic acid. Methyl group decreases acid strength of benzoic acid by + I effect. Therefore, Ka p-methoxy benzoic acid 3.3 ´ 10- 5 p-methyl benzoic acid 4.2 ´ 10- 5 benzoic acid 6.4 ´ 10- 5 p-chlorobenzoic acid 10. 2 ´ 10- 5 p-nitrobenzoic acid 36.2 ´ 10- 5 N H3C CH3 + NH3 is more acidic due to – I effect of F. 75. 79. H ¾ C ºº C ¾ H is more acidic than CH2 == CH2. 80. Intramolecular H-bonding in ortho-hydroxybenzaldehyde F + OH 76. O¾H + - O¾H decreases its melting point as well as boiling point. Molecules of p-hydroxybenzaldehyde is symmetrical, associated together by intermolecular H-bonds, has higher boiling point and melting point. O - H O H Intramolecular H-bonding Organic Chemistry Basics H 81. H H H ºº C—C—C H C==C==C H H H Allene sp sp2 stable than phenoxide ion in which negative charge also moves on carbon atoms. - O O ½½ ½ CH3 ¾ C ¾ O- ¬® CH3 ¾ C == O – - 82. – + - —O ==O ==O – + The above shown resonance introduces some double bond character to central C—C bond. Therefore, the central C—C bond in 1, 3-butadiene is shorter (stronger) than C—C bond in butane. 83. In case of acetate ion, both the resonance structures are equivalent and negative charge always remains on electronegative oxygen. These factors makes acetate ion more ==O – 84. p-nitroaniline < aniline < p-toluidine < N,N-dimethyl-p-toluidine. Nitro group, by electron withdrawing resonance effect decreases the basic strength. Methyl group by electron donating inductive effect, increases basic strength.