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Organic Chemistry : Some Basic Principles and Techniques

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Organic Chemistry : Some Basic Principles and Techniques Chapter 21 Organic Chemistry : Some Basic Principles and Techniques Solutions SECTION - A Objective Type Questions (One option is correct) 1. Compound having molecular formula C5H12O cannot show (1) Tautomerism (2) Position isomerism (3) Metamerism (4) Functional isomerism Sol. Answer (1) Factual 2. In which of the following reactions equilibrium will shift towards right? + NaHCO3 (1) (2) C C—H + NaOH H H O OH C OH (3) + NaHCO3 (4) + NaHCO3 Sol. Answer (4) NaHCO3 a weak base can deprotonate a very strong benzoic acid. Other acids are very less acidic, hence cannot be deprotonated by NaHCO3 O O O C H C O C ONa + NaO OH + . O CO + H O C 2 2 HO OH 2 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment 3. The IUPAC name of the compound CH 3 CH3 CH CH3 CH3 —CH—C—CH CH3 CH3 CH CH CH3 3 (1) 3-diisopropyl-2,4-dimethylpentane (2) 2, 4-dimethyl-3-diisopropylbutane (3) 2, 4-dimethyl-3-3-bis(1-methylethyl) pentane (4) None of these Sol. Answer (3) 5 carbon in longest chain 4. At conjugated position – NO imparts (1) + M and + I effect (2) – M and – I effect (3) + M and – I effect (4) – M and + I effect Sol. Answer (2) –N O (delocalization of p of double bonded N is difficult) Me 5. The IUPAC name of the compound Et (1) 3-methyl-6-ethylcyclohexene (2) 6-ethyl-3-methyl cyclohexene (3) 3-ethyl-6-methyl cyclohexene (4) 6-methyl-3-ethyl cyclohexene Sol. Answer (3) Priority of Et is alphabetically higher. 6. Strongest and weakest acid among the following is CH323 — NO CH — CHO CH3 — F CH 3 — CN I II III IV (1) I and II (2) III and IV (3) III and II (4) I and III Sol. Answer (4) Conjugate base of nitromethane (CH3—NO2) is stabilized by very strong – I and – M effect of – NO2 group O O O H—CH2 —N CH2 —N CH2 —N O O O B Very stable conjugate base Conjugate base of CH3—F is only stabilized by – I effect of F atom hence it is less stable and CH3F is least acidic. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 3 7. Correct order of acidic strength for the given species CH —CH —CH CH3 —C N—H 323 CH3 —C C—H CH2 —H IIIIII IV (1) I > II > IV > III (2) II > IV > III > I (3) III > I > IV > II (4) I > III > IV > II Sol. Answer (4) — B — CH3 —C—N—H CH3 —C—N highly unstable More stable due to presence of positive charge on highly en. sp—N 8. Which of the following compounds will not dissolve in aqueous NaOH? R O O H O S H C H (1) (2) OH (3) (4) OH HO OH Sol. Answer (3) H OH is least acidic. 9. Which of the following will have weakest indicated C—H bond? CH3 HC3 C HC ON CH —H 3 (1) 2 2 (2) CH —H HC3 2 HC3 C HC3 CH3 CH C 3 CH3 CH —H (3) 2 (4) CH3—H CH C 3 CH3 CH3 Sol. Answer (3) CH3 CH C 3 CH3 CH2 —H CH C 3 CH3 CH3 Therefore indicated C—H bond in (3) is weakest. 4 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment 10. In which of the following all electronic effects namely inductive, mesomeric and hyperconjugative effects are present? O O CH3 (1) (2) (3) (4) HC 3 CH3 O Sol. Answer (4) In (1), (2) and (3) hyperconjugate effect is absent. 11. Most acidic species among the following is OO SS O O (1) (2) (3) (4) O S O O Sol. Answer (2) Because its conjugate base is stabilized by p-d back bonding. 12. What is the index of hydrogen deficiency in the molecule C12H17NO? (1) 4 (2) 5 (3) 6 (4) 7 Sol. Answer (2) 17 1 (12 1) Index of hydrogen deficiency 2 = 13 – 8 = 5 13. Which of the following compound cannot have a bond? (1) C10H21N (2) C7H12 (3) C20H40Cl2O (4) C30H50Br2O2 Sol. Answer (3) Degrees of unsaturation in compound C20H40Cl2O is zero therefore the compound cannot have a bond 14. Which of the following experimental techniques can be used to detect carbon free radicals in a reaction mixture? (1) Magnetic susceptibility method (2) Polarimetry (3) NMR-spectroscopy (4) IR-spectroscopy Sol. Answer (1) This method indirectly gives some information about the presence of unpaired electrons. 15. Which of the following reactions will not generate a carbanion? O NH (1) NaOEt 2 (1) (2) Et-OH NO2 O SbF5 NaNH23 /NH (3) (4) CH3 — C C — H F Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 5 Sol. Answer (3) O O H SbF – 5 + [SbF6 ] H NaOEt + F O O NH C—H 2 CH —C C CH3 —C 3 16. Out of the given reactive intermediates which will be attracted towards the magnetic field? (1) Carbocation (2) Carbanion (3) Carbon free radial (4) Nitrene Sol. Answer (3) Carbon free radicals will have unpaired electrons and hence it will be attracted towards the magnetic field. 17. Which will have the highest melting point? (1) (2) (3) (4) Sol. Answer (3) Due to symmetrical structure neopentane will have most efficient stacking and hence highest melting point. 18. Which of the following species cannot behave as electrophile? + (1) BCl3 (2) AlCl3 (3) NH4 (4) SO3 Sol. Answer (3) NH4 does not have vacant orbital of appropriate energy. 19. Which of the following is the strongest base in water? (1) (2) NH2 (3) (4) N Sol. Answer (4) In this compound lone pair of N is nondelocalized. Above all, the N atom is sp3-hybridized and hence, this N is most basic. N sp3 H 20. Which of the following is most acidic? OH NH2 NH2 NH2 C (1) N (2) (3) CH3 – O – H (4) H H2N NH2 Sol. Answer (2) It forms most stable conjugate base 6 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment 21. Threedimensional arrangements which can be interconverted into one another due to rotation along a single bond are known as (1) Conformers (2) Diastereomers (3) Chain isomers (4) Positional isomers Sol. Answer (1) Conformers are formed due to rotation about single bond 22. In the given anion, –ve charge is delocalized on (1) One atom (2) Three atom (3) Four atom (4) Five atom Sol. Answer (3) (2) – (1) (3) (4) O 23. The compounds and are O (1) Chain isomers (2) Metamers (3) Position isomers (4) Both (1) & (2) Sol. Answer (4) Both have different carbon chain length. 24. Least stable carbocation among the following is CH3 H33 C – C – CH CH3 CH (1) (2) (3) CH33 – C – CH (4) H2C CH2 CH3 CH2 CH2 Sol. Answer (4) It has less number of resonating structures and less hyperconjugation effect. 25. Which of the following compound will give blood red colour while doing the Lassaigne’s test for N? NH2 (1) (NH2)2 C O (2) H2N (C6H4) SO3H (3) C6H5SO3H (4) Cl Sol. Answer (2) When both N and S is present in a compound. It gives blood red precipitate in Lassaigne’s test. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 7 26. The most stable free radical is _________ CH (1) (2) (3) (4) Sol. Answer (2) Due to resonance and hyperconjugation both. 27. Which of the following group will have the strongest electron donating mesomeric effect? H O Me N O (1) N (2) N (3) C = O (4) O – C O Me R HC3 Sol. Answer (2) Electron donating mesomeric effect of a group depends upon size, electronegativity and lone pair availability of key atom. 28. Among the following, the most stable carbocation is (1) CH33 CH CH (2) (3) (4) Sol. Answer (2) + is an aromatic cation. 29. Most acidic species among the following is O CH – C – CH H (1) 33(2) CH3 – O – H (3) CH3 – C C – H (4) Sol. Answer (2) —O—H bond is most polar due to maximum electronegativity difference 30. Which of the following double bond in the given molecule is most reactive towards a strong protic acid? Me A C Me B D Me (1) A (2) B (3) C (4) D Sol. Answer (1) Attack of H+ on double bond (A) gives most stable carbocation. CH 3 + CH3 CH 3 8 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment 31. Which of the following hydrocarbon is most acidic? (1) (2) (3) (4) Sol. Answer (1) is most acidic because it generates most stable aromatic conjugate base. 32. Resonance is not possible in (1) (2) (3) (4) O NH3 NO2 BH2 Sol. Answer (2) Nitrogen cannot form five bonds due to absence of vacant low energy d orbital 33. Which of the following compounds will have highest enolic content? O (1) (2) CH3COCH2CHO (3) CH3CHO (4) CH3COCH3 Sol. Answer (1) O OH H most stable enol form s tabilized by Aromatic character 34. Among the following, Cl Cl Cl I II III the correct order of reactivity of chloride is (1) I > II > III (2) III > II > I (3) II > I > III (4) II > III > I Sol. Answer (4) Stability order is + + + (Antiaromatic) Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 9 35. Which of the following represents the correct order of stability of the given carbocations? CH2 II I III (1) III > I > II (2) I > III > II (3) III > II > I (4) II > III > I Sol.
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