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. Answer (1) IIIrd is aromatic (most stable) and IInd is antiaromatic (least stable) 36. Which of the following organic molecule cannot form hydrogen bond in pure state but can form the same in water? O

(1) (CH3CH2)2NH (2) CH3CHO (3) CH3CH2COOH (4) CH3–– C NH2 Sol. Answer (2) HC 3   H C = O ----- H — O H

37. Correct stability order of the given free radicals is CH 3 CH3 CH >>CH— C > (CH ) CH > CH (1) 2 3 32 (2) 2 > CH33–– C > (CH )2 CH

CH3 CH3 CH CH3 3 > CH–– C > > (CH ) CH CH > CH–– C > (CH ) CH > (3) 3 CH2 32 (4) 2 3 32 CH CH3 3 Sol. Answer (2) Allyl C—H bond has lesser bond dissociation energy than benzylic C—H bond and hence corresponding allyl free radical is more stable.

38. In which of the following homolytic bond fission takes place? (1) Alkaline hydrolysis of ethylchloride (2) Addition of HBr to double bond (3) Photochlorination of methane (4) Nitration of benzene Sol. Answer (3) Halogenation of methane proceeds through free radical mechanism. In the initiation step of this reaction. Free radicals are generated through homolysis of a bond. Initiation

or h Cl — Cl Cl + Cl

39. The total number of structural monochloroderivatives possible in C5H12 is (1) 6 (2) 7 (3) 8 (4) 5 Sol. Answer (3) n-pentane, isopentane and neopentane form 3, 4 & 1 structural monohaloderivatives respectively. 10 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

40. In which of the following molecules positive charge is not delocalized because of resonance? NH  NH2  Me Me  (1) C (2) (3) C — N (4) C HN NH Me Me 3 2 NH2 NH2 Sol. Answer (1) Nitrogen atom cannot form five bonds due to absence of vacant d-orbitals.

41. Which of the following is correct order of dipole moment of o, m and p-methyl benzonitrile? CH CH3 CH3 CH3 CH3 3 CH3 CN CN (1) > > (2) >> CN CN CN CN CH CH CH 3 3 3 CH3 CH3 CH3 CN CN (3) >> (4) >> CN CN CN CN Sol. Answer (2)

22– 2 cosθ CH3– is electron donating group and –CN is electron withdrawing group  12   12

42. Which of the following reagents can be used to separate benzoic acid from a mixture of Benzoic acid, Phenol, Benzaldehyde and Toluene?

(1) aq. HCl (2) aq. NaHCO3 (3) Diethyl ether (4) NaOH Sol. Answer (2)

When benzoic acid reacts with NaHCO3 it forms water soluble sodium benzoate and hence through extraction. It can be separated.

43. Which of the following compounds will give negative Lassaigne’s test for Nitrogen? NO N 2 (1) NH2 (2) N N (3) (4)

Sol. Answer (2) It is azo compound. Azo compound gives negative Lassaigne’s test.

44. A mixture contains four solid organic compounds A, B, C and D. On heating, only C changes from solid to vapour state. The compound (C) can be separated from the mixture by (1) Distillation (2) Kinetic resolution (3) Crystallization (4) Sublimation Sol. Answer (4) Solid is directly converted to vapour.

45. The number of sp2 hybridised carbon in one benzyne is (1) 4 (2) 5 (3) 6 (4) 3 Sol. Answer (3)

 . All carbon atoms are sp2 hybridised. + Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 11

46. The number of structural isomers for C6H14 is [IIT-JEE-2007] (1) 3 (2) 4 (3) 5 (4) 6 Sol. Answer (3)

47. Hyperconjugation involves overlap of the following orbitals [IIT-JEE-2008] (1) – (2) –p (3) p – p (4) – Sol. Answer (2) Hyperconjugation involves -p conjugation.

48. Among the following, the least stable resonance structure is [IIT-JEE-2007]

 O O O O    (1) N (2) N (3)  N (4)  N |  | | | O O O O

Sol. Answer (1) 49. The correct stability order for the following species is [IIT-JEE-2008]

  O (I) (II)

 O  (III) (IV)

(1) (II) > (IV) > (I) > (III) (2) (I) > (II) > (III) > (IV) (3) (II) > (I) > (IV) > (III) (4) (I) > (III) > (II) > (IV) Sol. Answer (4) I & III are stabilized by resonance, hyperconjugation and + inductive effect.  order is (I) > (III) > (II) > (IV)

50. The correct acidity order of the following is [IIT-JEE-2009]

OH OH COOH COOH

Cl CH3 (I) (II) (III) (IV)

(1) (III) > (IV) > (II) > (I) (2) (IV) > (III) > (I) > (II) (3) (III) > (II) > (I) > (IV) (4) (II) > (III) > (IV) > (I) Sol. Answer (1) COOH COOH OH OH

>>>

CH3 Cl (+I effect) (–I effect) 12 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

51. The IUPAC name of the following compound is [IIT-JEE-2009] OH

CN Br

(1) 4-Bromo-3-cyanophenol (2) 2-Bromo-5-hydroxybenzonitrile (3) 2-Cyano-4-hydroxybromobenzene (4) 6-Bromo-3-hydroxybenzonitrile Sol. Answer (2)

OH

CN Br

2-Bromo-5 hydroxy benzonitrile.

52. In the following carbocation, H/CH3 that is most likely to migrate to the positively charged carbon is [IIT-JEE-2009] H H

1 + 5 H C—C24 —C—C —CH 333

HO H CH3

(1) CH3 at C-4 (2) H at C-4 (3) CH3 at C-2 (4) H at C-2 Sol. Answer (4)

H H H H 1 + 5 + H C—C24 —C—C —CH CH —C—C—C—CH 333 33

HO H CH3 :OH H CH3 More stable resonance stabilized carbocation

53. The correct stability order of the following resonance structures is [IIT-JEE-2009]

––––  HCNN2222 HC—NNHC—NNHC—NN   (I) (II) (III) (IV)

(1) (I) > (II) > (IV) > (III) (2) (I) > (III) > (II) > (IV) (3) (II) > (I) > (III) > (IV) (4) (III) > (I) > (IV) > (II) Sol. Answer (2) Resonating structures having maximum number of covalent bonds are more contributing. Among charge separated resonating structures, structures where opposite charge are close enough are more contributing. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 13

54. Among the following compounds, the most acidic is [IIT-JEE-2011] (1) p-nitrophenol (2) p-hydroxybenzoic acid (3) o-hydroxybenzoic acid (4) p-toluic acid Sol. Answer (3) O H O – C C O –H O H H O O Most acidic Stabilized by strong intramolecular hydrogen bonding

55. In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are) [IIT-JEE-2012] (1) sp and sp3 (2) sp and sp2 (3) Only sp2 (4) sp2 and sp3 Sol. Answer (2)

H ()sp H C = C = C H H ()sp2 ()sp2 (Allene)

SECTION - B Objective Type Questions (More than one options are correct) 1. Which of the following bicyclic compounds are isomers?

(1) (2) (3) (4)

Sol. Answer (1, 2, 3, 4) All of them have same molecular formula hence, they are isomers.

2. Which of the following will involve homolysis of a bond? O O h (1) Ph—C—O—O—C—Ph h (2) RO — Cl 

CH3 + BuLi H (3) CH —C—OH (4) 3 

CH3 Sol. Answer (1, 2)

O O O Ph—C—O—O—C—Ph h 2Ph—C—O

h R—O—Cl  R—O + Cl 14 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

3. Species which will exhibit geometrical isomerism among the following is/are

T D CH 3 H C C C N—OH (1) (2) Br (3) (4) N2H2 HC3 H Me

Sol. Answer (1, 3, 4)

4. Which of the following molecular formula will exhibit functional isomerism as well as metamerism?

(1) C4H10O (2) C4H11N (3) C4H8O (4) C4H9Cl Sol. Answer (1, 2, 3)

For molecular formula C4H11N only saturated amines are possible (1°, 2° & 3° amines are functional isomers)

Molecular formula C4H9Cl only represents saturated alkyl chloride. Therefore metamerism is not possible.

5. Out of the given isomeric hydrocarbons which will undergoes rearrangement reaction in acidic medium?

(1) (2) (3) (4)

Sol. Answer (2, 3)

(1) +H Stable tertiary carbocation

(2) +H Stable tertiary carbocation (No rearrangement possible)

6. Which of the following carbocation is stabilised by resonance?

CH CH 2   3 C (1) CH22 CH — CH (2) C (3) (4) CH3

Sol. Answer (1, 2, 3, 4)

7. Which of the following carbocations have potential to rearrange?

CH 3  CH   — CH 2 (1) CH3 — C — C — O (2) (3) (4)

CH3 CH3 Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 15

Sol. Answer (1, 3, 4)

CH3 CH 3 + CH —C—C = O CH —C+ + CO  3 3

CH3 CH 3

+ CH—CH3 + CH 3 

+ + CH 2 

8. Formic acid is more acidic than (1) Benzoic acid (2) Acetic acid (3) Phenol (4) Benzene sulphonic acid Sol. Answer (1, 2, 3)

In benzoic acid +R effect of benzene in acetic acid. +I effect of CH3–group and in phenol formation of less stable resonating structures makes formic acid more acidic.

9. Which of the following will have C  O bond length almost similar to C — O bond length? O O O

(1) (2) (3) (4) O  C  O

Sol. Answer (1, 2)

O– O– + +

Both are aromatic

10. The correct order of stability is/are

 (1) HCCHCCHCHCH (2) 23    2 HC  C > H2 C = CH > CH3 –CH2

CH2

  (3) HC C H23 C  CH  CH  CH2 (4)

Sol. Answer (2, 3, 4)  Due to percentage S-character (electronegativity) anion of sp hybridized carbon is most stable.  Due to lower bond energy of allylic H as compared to benzylic H, allyl radical is more stable than benzyl radical. 16 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

11. Which of the following is not stabilised by hyperconjugation? Me Me    CH  (1) CH3 (2) 2 (3) (4) Me Me Sol. Answer (1, 2, 4)

 +  In CH3 and there is no  hydrogen

+ CH  In 2 , stability is due to overlapping of p-orbital

12. Dichloro ethene shows (1) Geometrical isomerism (2) Position isomerism (3) Metamerism (4) Chain isomerism Sol. Answer (1, 2)

Cl

Cl – CH = CH – Cl can show geometrical isomerism and CH2 CH is its position isomer Cl

13. A compound having molecular formula C4H10O can show (1) Metamerism (2) Functional isomerism (3) Chain isomerism (4) Position isomerism Sol. Answer (1, 2, 3, 4) All four isomerism are possible. Optical isomer is possible by 2-methyl-1-butanol

14. Which of the following correctly represents the order of quality mentioned in bracket? (1) sp – sp > sp2 – sp2 > sp3 – sp3 (bond energy) (2) sp – sp3 > sp – sp2 > sp – sp (polarity in bond) (3) sp3 – s < sp2 – s < sp – s (% s character) (4) sp3 – sp3 > sp2 – sp2 > sp – sp (bond stability) Sol. Answer (2, 3, 4)

sp, sp2 & sp3 have 50%, 33% and 25% s-character

15. Consider the following compounds Which of the following statements are correct?

N N NH H 2 (I) (II) (III)

(1) I is more basic than II (2) II is more basic than I and III (3) III is more basic than II (4) I is weakly acidic Sol. Answer (2, 4)

Due to higher Kb for II, it is more basic than I and II Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 17

16. Which of the following correctly represent the acidic strength of given acids?

(1) Cl3CH > F3CH (2) CH3COOH > CH3CH2OH OH OH

> (3) H2O > CH3CH2OH (4)

NO2 Sol. Answer (1, 2, 3, 4) CCl – CCl 3 is more stable than CF3 due to presence of d-orbital in 3 hence CHCl3 is more acidic than CFCl3.

17. Which of the following correctly represents the stability of reactive intermediate? + (1) CH32– CH < CH3 OCH2 (2)

CH2 CH2 CH3 >  (3) (4) CH33–– CH C  < CH3 CH2

CH3 NO2 CH3 Sol. Answer (1, 2, 3)

 CH32 —O— CH is a resonance stabilized carbocation.

18. The correct order of basic strength in aqueous medium is/are

(1) (C2H5)2NH > (C2H5)3N > C2H5–NH2

(2) (CH3)2NH > CH3–NH2 > (CH3)3N

(3) CH – NH – CH – CH > CH33 CH CH > CH – CH – CH – NH 3 2 3 | 3 2 2 2

NH2

(4) C2H5 – NH2 > CH3 – NH2 > CH2 = CH – NH2 Sol. Answer (1, 2, 3, 4)

These orders are due to steric hindrance and different Kb value

19. Keto - enol Tautomerism is observed in

(1) C6H5–CHO (2) C6H5–CO–CH3 O

(3) (4) C6H5–CO–CH2–CO–CH3

NH2

Sol. Answer (2, 3, 4)

C6H5 CHO has no –H. Therefore it will not exhibit tautomerism O O H

H2N H NH2 18 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

20. Which of the following can exhibit geometrical isomerism? H H (1) C H CH = N–OH (2) (3) (CH ) C = C(CH ) (4) 6 5 3 2 3 2 Me Me Sol. Answer (1, 4)

Ph OH Ph C N and C N H H OH

Geometrical isomers

H H H Me and

Me Me Me H

Geometrical isomers

21. The compounds which cannot react with NaOH is/are OH

CH C CH (1) 3  (2) NH3 (3) C2H5OH (4)

Sol. Answer (1, 2, 3) – OH cannot neutrallise CH3—C  CH, NH3 and C2H5OH due to their very low acidic strength (weaker acids than H2O)

22. The hybridisation of N is correctly given in (1) sp3 in acetamide (2) sp2 in pyridine (3) sp2 in pyrrole (4) sp in methyl cyanide Sol. Answer (2, 3, 4)

O   CH—C—NH  CH — C N 3 2 N N 3  sp2  sp2 sp sp2 H

23. Which of the following Lewis structures are valid resonating structures for the azide ion? N

–  (1) N — N — N (2) N N N (3) N N N (4) N N

Sol. Answer (1, 3) NNN 2 Invalid resonating structure because of extension of octet around N. N

N N Invalid structure because position N atom is changing as well as no. of -electrons are also

changing. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 19

24. In the given compound the hybridisation states of C atom is/are not

(1) sp2 (2) sp3 (3) sp (4) dsp2 Sol. Answer (2, 3, 4) All carbon atoms are sp2 hybridised due to resonance

25. In Lassaigne’s test, the sodium extract of an organic compound containing both N and S on treatment with FeCl3 solution produces a blood-red colour. The appearance of this blood red colour is due to – 2+ (1) [Fe(SCN)4] (2) FeCl2(SCN) (3) [Fe(SCN)(H2O)5] (4) Na4[Fe(CN)5NOS] Sol. Answer (1, 2, 3) All of three are possible and (4) is Prussian blue coloured

26. 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) [IIT-JEE-2011] H H H C—C H—C C—C (1) (2)  (3) H2C = C = O (4) H2C = C = CH2 HC CH2 2 CH2 Sol. Answer (2, 3)

H C H H C sp2 C C C O C sp2 H H H

All atoms lie in one Planar plane in all conformation

H H C C C is non-planar H H

H H C C H H C C

H H

Non-planar in many conformations due to rotation along.

C2 — C3 bond.

27. Which of the following molecules, in pure form, is (are) unstable at room temperature? [IIT-JEE-2012] O O

(1) (2) (3) (4) 20 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

Sol. Answer (2, 3)

The compound is antiaromatic and hence unstable at room temperature. The other compound

O O– +

is also unstable at room temperature due to partial positive charge at carbonyl C-atom

28. Among P, Q, R and S, the aromatic compound(s) is/are [IIT-JEE-2013] Cl

AlCl 3 P

NaH Q

(NH ) CO 42 3 R 100-115 °C OO O HCl S

(1) P (2) Q (3) R (4) S Sol. Answer (1, 2, 3, 4) AlCl Cl 3 AlCl (P) (i) 4 (AROMATIC)

Na NaH (Q) + H (ii) 2 (AROMATIC)

(NH ) CO 42 3 R (iii) 100–115ºC O O

 2NH + CO + HO (NH42 ) CO 3 322

 IMPE + NH3 O O O O NH O OH 2 NH2 H

H H H H N N N  IMPE –2H O 2 HO OH O OH (R) AROMATIC

HCl O OH Cl (S) (iv) (AROMATIC) Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 21

29. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to [IIT-JEE-2013] (1)   p (empty) and   electron delocalisations (2)  * and   electron delocalisations (3)   p (filled) and   electron delocalisations (4) p (filled)  and   electron delocalisations Sol. Answer (1) In hyperconjugation   p (empty) electron delocalization for tert-butyl carbocation and   * electron delocalization for 2-butene will take place.

30. The correct combination of names for isomeric alcohols with molecular formula C4H10O is/are [JEE(Advanced)-2014] (1) Tert-butanol and 2-methylpropan-2-ol (2) Tert-butanol and 1, 1-dimethylethan-1-ol (3) n-butanol and butan-1-ol (4) Isobutyl alcohol and 2-methylpropan-1-ol Sol. Answer (1, 3, 4) Common name IUPAC name OH (1) tert-butanol 2-methylpropanol

OH (3) n-butanol butan-1-ol

(4) OH isobutyl alcohol 2-methyl propan-1-ol

SECTION - C Linked Comprehension Type Questions Comprehension-I Tautomerism, strictly defined could be used to describe the reversible interconversion of isomers. Interconversion of isomers is due to mobility of an atom or a group. R R

R—HC—C O R —CH C—O H H In the above examples the composition of the equilibrium mixture is of course, governed by the relative thermodynamic stability of the two forms under the particular conditions being studied.

1. In which of the following tautomeric equilibrium concentration of right hand product is more than left hand product? H O O O O CH3 O

(1) CH3 —N CH2 —N (2) O O—H HC OH HC3 O 2

O O OH O OH O (3) (4) C EtO OEt EtO OEt 22 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

Sol. Answer (3) In dienone phenol tautomerism phenol form is more stable due to its aromatic character.

O O H

H More stable

2. In which of the following solvent percentage enol content is maximum for 2,4-pentanedione?

(1) CH3CN (2) H2O (3) n-Hexane (4) Ethanol Sol. Answer (3) In non-polar solvent enol content would be maximum

Comprehension-II Names of organic compounds are under the latest guide line of IUPAC. IUPAC means international union of pure and applied chemistry. The main rules are longest chain rule, lowest number rule etc. We have to include the rules for naming the substituents, multiple bonds and even functional groups.

1. Write the IUPAC name of the following compound

CN

CN (1) 3,3-Diethenyl pentane1,5-dinitrile (2) 3,ethenyl,3-ethyl pentane 1,5-dinitrile (3) 3,3-diethenyl pentane 1,5-dicyanide (4) None of these Sol. Answer (1)

(4) (5) CN (3) (2)

CN (1)

OH

2. , IUPAC name is

OH

(1) 3,3-diethenyl pentane 1,5-diol (2) 2,2-diethenyl propane 1,3-diol (3) 2, 3-diethenyl propane 1,3-diol (4) 3,3-diethenyl propane 1,3-dialcohol Sol. Answer (2)

(3)

(2) OH (1)

OH Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 23

3. Write IUPAC name of 3 4 2 1 5 6 4 3 1 2

(1) 5-(3,3-dimethyl cyclobutyl) 1,2,2,3-tetra ethyl cyclohexane (2) 5-(3,3-dimethyl cyclobutyl) 1,3-diethyl 2,2-dimethyl cyclohexane (3) 1,2,2,3-tetra ethyl 5-(3,3-dimethyl cyclobutyl) cyclohexane (4) None of these Sol. Answer (1) Fact

Comprehension-III Weak Acid does not dissociate completely into its ions. It is in equilibrium with its conjugate base. Greater is the K stability of conjugate base, greater is value of k for that equilibrium making the equilibrium move in (HA a H+ + A–) forward direction. i.e., more is the degree of dissociation of that acid. Same is the case for weak bases. Factors affecting the stability of conjugate acid or base are electronic effect like resonance effect and inductive effect acting upon the species.

1. Which of the following is strongest acid? OH OH OH OH CH (1) (2) 3 (3) (4)

CH3

CH3 Sol. Answer (1) Alkyl group (electron donor) decreases acidic strength of phenol 2. Which of the following is having most acidic -Hydrogen? O

NO CH NO CH2 NO2 2 2 (1) CH3—NO2 (2) (3) (4) NO NO 2 2 O Sol. Answer (3) H

Conjugate base of ON2 C NO2 is

NO2

Stabilized by three strong electron withdrawing groups i.e., O O  NCN O O N OO

Stabilized by –I and –M effect of three –NO2 groups 24 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

3. Which of the following is most basic due to +R effect of any substituent? NH 2 NH2 .. N .. (1) (2) (3) (4) N OCH H 3 CH 3 OCH3 Sol. Answer (4)

+R effect of –OCH3 predominates over –I effect

Comprehension-IV Hyperconjugation is defined as No bond resonance. The concept of hyperconjugation arose from the discovery of electron releasing pattern for alkyl groups. It involves  electrons of C–H bond. Greater the number of C–H bond (  -hydrogen atom w.r.t. double bond) more will be hyperconjugative structures, more will be stability. Heat of hydrogenation of alkene are affected by hyperconjugative effects.

1. Which of the following is incorrect hyperconjugative structure?

H H H H H C H H C H H C H H C H

(1) (2) (3) (4)

Sol. Answer (3) H is released as H+ not as H–

2. Which of the following has highest magnitude of enthalpy of hydrogenation?

(1) (2) (3) (4)

Sol. Answer (1) Hydrocarbon (1) is least stable, as it is not stabilized by resonance, and has lesser number of hyper conjugative structures.

SECTION - D Assertion-Reason Type Questions 1. STATEMENT-1 : In naphthalene all C—C bonds are equal.

Naphthalene

and STATEMENT-2 : Like benzene naphthalene is also aromatic. Sol. Answer (4)

More contributing Less contributing

 All C—C bonds are not equal. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 25

2. STATEMENT-1 : p-Nitroaniline is more polar than nitrobenzene. and STATEMENT-2 : Nitro group has – M effect. Sol. Answer (2)

Both statements are correct but reason for this is combined effect of –NO2 and –NH2.

3. STATEMENT-1 : All C—C bonds are equal in [10]-Annulene. and STATEMENT-2 : [10]-Annulene is a non-aromatic compound. Sol. Answer (4) All C—C bonds are not equal in [10]-Annulene.

HO O OH

4. STATEMENT-1 : is less acidic than HO O

Squaric acid Phenol

and STATEMENT-2 : Conjugate base of phenol is resonance stabilized. Sol. Answer (4) Infact squaric acid is more acidic than phenol.

Br 5. STATEMENT-1 : When is treated with excess of Ag+ ion. One mole of the compound reacts Br Br Br (X)

with four moles of Ag+. and STATEMENT-2 : Ag+ is a Lewis acid hence it reacts with Br– ion on which is a Lewis base. Sol. Answer (4) Compound (X) will give only two moles of AgBr.

6. STATEMENT-1 : A compound with odd number of nitrogen always contains odd molecular weight. and STATEMENT-2 : Nitrogen has odd molecular mass. Sol. Answer (3) Molecular mass of nitrogen is 14.

7. STATEMENT-1 : Aldehydes and ketones having same molecular formulae are structural isomers. and STATEMENT-2 : Aldehydes and ketones are metamers. Sol. Answer (3) Aldehydes and ketones cannot be metamers. 26 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

8. STATEMENT-1 : Carbocationic rearrangement is known as electrophilic rearrangement. and STATEMENT-2 : Carbocations are stabilized by both hyperconjugation and +I effect. Sol. Answer (4) Carbocationic rearrangement is known as nucleophilic rearrangement.

9. STATEMENT-1 : Cyclopentanone exhibits keto-enol tautomerism. and STATEMENT-2 : Cyclopentanone has two hydrogen atoms attached to the carbon atom adjacent to carbonyl group. Sol. Answer (1) O OH

Tautomers

10. STATEMENT-1 : CH3 CH 2 CH2 is less stable than CH– N – CH 3 | 2 H and STATEMENT-2 : Carbocation with adjacent hetero-atom like N, O are less stable. Sol. Answer (3) If ‘N’ or ‘O’ are attached to carbocation. These groups participate in the delocalization of +ve charge.

11. STATEMENT-1 : Cyclopropane is more stable than cyclobutane. and STATEMENT-2 : Angle strain in cyclopropane is higher than cyclobutane. Sol. Answer (4)

1 Angle strain of cyclic compound.  stability

O || 12. STATEMENT-1 : Keto form is less stable than enol form. ||O

and STATEMENT-2 : Enol form is stabilized by aromaticity. Sol. Answer (1)

O OH

O OH Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 27

13. STATEMENT-1 : Aniline undergoes Friedel-Craft alkylation more readily than Toluene. and STATEMENT-2 : Aniline undergoes fast electrophilic substitution than Toluene. Sol. Answer (4) Aniline does not follow Friedel-Craft reaction.

14. STATEMENT-1 : Bridge head carbocation is less stable than Bridge head carbanion. and STATEMENT-2 : C atom in carbocation is sp2 generally hybridized. Sol. Answer (2) Carbocation and carbanion are generally sp2 and sp3 hybridised

SECTION - E Matrix-Match Type Questions 1. Match the following Column-I Column-II (Molecular formula) (Type of isomerism)

(A) C6H12O (p) Functional isomerism

(B) C4H11N (q) Geometrical isomerism

(C) C6H12 (r) Metamerism

(D) C5H12O (s) Tautomerism (t) Position isomerism Sol. Answer A(p, q, r, s, t), B(p, r, t), C(q, t), D(p, r, t)

C6H12O has one degree of unsaturation.  Carbonyl compounds, unsaturated alcohols and ethers are possible. This can also exhibit geometrical isomerism

C4H11N can form 1°, 2° & 3° amines which are functional isomer.

2. Match the following Column-I Column-II (Estimation) (Method)

(A) C (p) Liebig method

(B) H (q) Duma method

(C) N (r) Kjeldahl's method

(D) X (Chalcogen) (s) Carius method Sol. Answer A(p), B(p), C(q, r), D(s) Factual based 28 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

3. Match the following Column-I Column-II (Estimation) (Percentage) (A) 0.45 gm organic compound gives 1.1 gm (p) 66.6% C

CO2 and 0.3 gm H2O 7.4% H (B) C : H : N ratio in compound 18 : 2 : 7 (q) 26% N (C) 0.37 gm of a given compound gave 0.631 gm AgBr (r) Equivalent mass of acid is 122 (D) 0.122 gm of an organic acid required (s) 72.6% Br

N 3 10 cm 10 NaOH for neutralisation

Sol. Answer A(p), B(p, q), C(s), D(s)

12 Mass of CO 2 100 (A) Percentage of C = 44 Mass of organic compound

12 1.10 100 66.6% 44 0.45  

2 Mass of H O 2 100 Percentage of H = 18 Mass of compound

20.3 100 7.4% 18 0.45  

18 100 66.66% (B) Percentage of C = 27 

2 100 7.40% Percentage of H = 27 

7 100 26% Percentage of N = 27 

80 Mass of AgBr 100 (C) Percentage of Br = 188 Mass of compound

80 0.631 100 72.6 188 0.37

N 3 (D) 10 cm of 10 alkali required acid = 0.122 g

0.122 3 1000 10 1000 cm of 1 N alkali required acid = 10 

= 122 g But 1000 cm3 of 1 N alkali contain 1 gm equivalent of the alkali which must react with 1 gm equivalent of acid.  Equivalent weight of acid = 122 Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 29

4. Match the carbocation in column-I with the effect which is major, stabilizing factor for it in column-II Column-I Column-II

(A) Carbocation (p) Nucleophile

(B) Carbanion (q) Electrophile

(C) Free radical (r) Stabilized by +I effect

(D) Carbene (s) Stabilized by resonance Sol. Answer A(q, r, s), B(p, s), C(r, s), D(p, q) Carbocation is electrophile and stabilised by +I effect resonance and hyperconjugation. Carbanion is nucleophiles and stabilized by resonance and –I effect Free radical is electrophile and stabilised by resonance and hyperconjugation. Carbene can act as electrophile and nucleophile both.

5. Match the carbocation in column-I with the effect which is major, stabilizing factor for it in column-II Column-I Column-II

(A) (p) Aromatic character

(B) CH3 — CH (q) Resonance

CH3

(C) (r) Hyperconjugation

(D) CH3 — O— CH2 (s) Inductive effect Sol. Answer A(q), B(r, s), C(p, q), D(q, s)



is resonance stabilised but antiaromatic

+ CH3 CH has 6 –H for hyperconjugation and 2 methyl groups for + I effect

CH3

+

is aromatic resonance stabilised cation

  CH32 –O– CH is resonance stabilised because it can form CH32 – O CH 30 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

6. Match the following Column-I Column-II + (A) (p) Resonance stabilisation O

(B) (q) Aromatic molecule

CH3

(C) (r) Hyperconjugation effect is observed

2 (D) CH3 – CH = CH – CH = CH2 (s) At least one carbon is sp hybridised (t) Achiral compound Sol. Answer A(p, r, s, t), B(p, q, s, t), C(p, q, r, s, t), D(p, r, s, t) Resonance is possible in A and B. B and C are aromatic. (C) and (D) have hyperconjugation effect due to -hydrogen.

7. Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. [IIT-JEE-2008] Column I Column II  (A) HN—NHCl23 (p) Sodium fusion extract of the compound gives Prussian blue

colour with FeSO4  NH3 l (B) HO (q) Gives positive FeCI test COOH 3

 HO NH Cl (C) 3 (r) Gives white precipitate with AgNO3

 ON (D) 2 NH—NH3Br (s) Reacts with aldehydes to form the corresponding

NO2 hydrazone derivative Sol. Answer A(r, s); B(p, q); C(p, q, r), D(p)

SECTION - F Integer Answer Type Questions

1. One mole of a compound with molecular formula C30H43N absorbs 8 moles of H2 gas under catalytic hydrogenation. Then what is the ratio of number of  bonds to the number of rings in the compound? Sol. Answer (4)

43 1 Degrees of unsaturation in the compound = 31 = 31 – 21 = 10  2  Out of which 8  bonds are present

 bond 8 4 Hence ring 2 Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 31

2.

How many intermediates are possible (excluding stereoisomer) when above mentioned molecule is attacked by 1 equivalent H+? Sol. Answer (6)

3. On how many atoms positive charge is delocalized in the given ion?

Sol. Answer (4)

F FF F

Hence positive charge is delocalized over four atoms.

4. How many of the given species will behave as an electrophile?

CCl23343 , NH , CH , NH , CH , OH, BF 3 , AlCl 3 , Br Sol. Answer (5) – NH343 , NH , CH and OH are not electrophiles, infact they are nucleophiles

5. How many bicyclic isomers are possible for the molecular formula C6H12O? Sol. Answer (0) Index of hydrogen deficiency is 1.  Bicyclic system is not possible for this.

6. How many of the given species will evolve CO2 with NaHCO3? OH OH OH HO O CH CH ON O 2 2 NO2 O OH OH , , , , H , O HO HO OH NO2

O O OH OH

NO2 NO2 , , , , O O NO2 Sol. Answers (3) th Trinitrophenol, squaric acid and V product can give CO2 with NaHCO3.

7. How many total types of products are formed by dehydrohalogenation of 2-chlorobutane? Sol. Answer (3) 1-Butene, Cis-2-butene,trans-2-butene 32 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

8. The total number of structural dihaloderivatives possible in n-pentane are ____. Sol. Answer (9) (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (3, 3) Dihaloderivatives are nine different compounds.

9. Amongst the following, the total number of compounds soluble in aqueous NaOH is [IIT-JEE-2010]

H3C CH3 N COOH OCH23 CH OH

CH2 OH

NO2 OH CH23 CH COOH

CH23 CH

N HC3 CH3

Sol. Answer (4) Carboxylic acids and aromatic alcohols dissolves in aqueous NaOH.

COOH OH OH COOH

,,,

N HC3 CH3

will be soluble in aqueous NaOH.

10. The total number of contributing structures showing hyperconjugation (involving C-H bonds) for the following carbocation is [IIT-JEE-2011] HC CH CH 3 + 23

Sol. Answer (6) HC CHCH 3 + 23

It contains 6 -hydrogen atoms.  6 contributing structures are possible Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 33

11. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is [IIT-JEE-2010] Sol. Answer (5)

Index of hydrogen deficiency in the compound C4H6 is 2. Therefore either bicyclic compound or cycloalkenes are possible as cyclic isomers.

 Possible isomers are

,,,,

12. The number of resonance structures for N is [JEE(Advanced)-2015] OH NaOH N

Sol. Answer (9)

OH O NaOH

–H2 O I

O OO

II III IV

O

V

O O O

III VI VII

O O

VIII IX 34 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

SECTION - G Multiple True-False Type Questions 1. STATEMENT-1 : Phenol is more acidic than benzoic acid.

STATEMENT-2 : Fluorobenzene is less reactive than chlorobenzene towards electrophilic substitution.

STATEMENT-3 : Friedel Craft alkylation is not possible in tertiary butyl benzene.

(1) T F T (2) F F F (3) F T F (4) T T F

Sol. Answer (2)

– – Phenol is less acidic than benzoic acid due to less stability of C6H5O than C6H5COO . Fluorobenzene is more reactive than chlorobenzene.

O   2. STATEMENT-1 : CH32 O CH is more stable than ONCH 2 .

STATEMENT-2 : + is antiaromatic molecule.

STATEMENT-3 : Phenoxide is more stable than ethoxide.

(1) F F T (2) T T F (3) F T F (4) T F T

Sol. Answer (4)

..  CH O CH – – In 32.. , resonance is effective. + is aromatic (10 e ). Phenoxide is resonance

stabilised.

SECTION - H Aakash Challengers Questions 1. Rank the given species in the increasing order of water solubility?

CHO CHO OH CHO OH OH OHC OH HO

OH OHC HO OH CHO CHO OH IIIIII IV

Sol. Correct increasing order of water solubility would be

II < I < III < IV

Increasing solubility in water

Greater the number of inter molecular hydrogen bond with solvent molecule higher the solubility. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 35

2. The given six compounds are similarly sized, very similar molecular weight and number of electrons, but the boiling point of these compounds are quite different (30°C–141°C). Rank these compounds in the increasing order of boiling point (lowest boiling first)

OH O O NH2 N OH I II III IV V VI

VI < V < IV < II < I < III Sol. Increasing boiling point Strongest hydrogen bond is present in III. While weakest intermolecular force of attraction is present in VI.

3. For the given pair of compounds, identify the compound you expect to have the higher boiling point and explain your reasoning.

N — H and NH2

Sol. NH2

To predict relative boiling points, look for differences in (i) Hydrogen bonding (ii) Molecular weight and surface area and (iii) Dipole moment

4. Which of the following reactive intermediate is more stable and why?

III

Sol. II > III > I Bridge head carbocation are least stable. Because bridge head carbon cannot attain planarity.

 – 5. Salt of Ph3C —X can be stored for months. Explain why. Sol. Triphenyl methyl cation is very stable, due to resonance.

+ In triphenyl methyl carbocation positive charge can be delocalized in all the 3 ring systems C causing the lowering of energy and making the carbocation very stable.

6. For the dehydration reaction

CH OH CH CH 2 + 2 3 H + +

Explain the mechanism discussing formation of all the 3 products. 36 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

Sol. H CH OH CH O CH 2 + 2 2 H H

(H2 O being a rearrangement good leaving of carbocation group) elimination CH2 of H CH3

CH3

CH2 Ring expansion to bring release – H from ring strain

CH – OH CH 2 H+ 2

Ring expansion Hydride shift

CH3

-H elimination

7. Which of the following carbanion is less stable and why?

OCH3 OCH3

NH2 or

NH2 OCH3

NH2

Sol. is less stable, as a methoxyl group has negative (electron-withdrawing) inductive effect so

OCH3

negative charge will be less available on , making it more stable.

NH2 Hint : Electrons in both the carbanions are out of the plane of the  cloud, hence there is no resonance interaction, only the inductive effect works. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 37

NH2

8. Aromatic amines are weakly basic, whereas is strongly basic, explain this unusual behaviour.

Sol. Planarity between lone pair of nitrogen and  electron of ring is lost due to two bulky groups on ortho position.

9. Which of the following hydrocarbons can be readily deprotonated by NaOEt? Explain.

H ;; ;

I II III IV

Sol. Only III can be deprotonated by NaOEt as its conjugate base is aromatic.

10. Pick out the correct statements about the barrier of rotation about the indicated bond in the given compounds

O

IIIIII IV

(1) I and IV will have nearly same barrier of rotation (2) II and III will have nearly same barrier of rotation (3) At room temperature I will have frozen rotation (4) III will have relatively lower barrier of rotation as compared to IV Sol. Answer (1, 2, 3, 4) In II and III charge separated structures are more contributing.

11. Arrange the given species in the increasing acidic strength :

IIIIIIIV (1) II < I < IV < III (2) IV < III < I < II (3) II < I < III < IV (4) III < IV < II < I Sol. Answer (3) Stronger acids have more stable conjugate base.

–H

Least acidic Lesser delocalization of negative charge, hence least stable 38 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

12. Order of basicity of the following species is

N N N N N N H N N H IIIIIIIV

(1) I > II > IV > III (2) III > IV > II > I (3) IV > III > I > II (4) II > I > III > IV Sol. Answer (4) Conjugate acids of I and II are stabilized by resonance.

13. The correct stability order of the following resonance structures is

OMe O—Me Me Me

N N CH2 Me CH2 Me (I) (II)

O—Me OMe Me Me N CH N CH Me 2 Me 2

(IV) (III)

(1) I < III < IV < II (2) I < IV < III < II (3) I < IV < II < III (4) IV < I < II < III Sol. Answer (3) More number of covalent bonds more contributing structures.

14. What would be the major product of the given reaction? Justify mechanistically for the formation of your product?

HSO24 

CH2 —OH

Sol. Alcohols in the presence of H+ will result into formation of carbocation which will result into the formation of more stable carbocation.

15. Write mechanism for the following transformation

HNO3 NO2

Sol. HNO3 generates NO2 ion through self ionization which attacks on aromatic ring to give the desired product. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 39

16. Correct statements among the following is/are

D D D D

(1) and are identical compounds

D D D D

(2) and are structural isomers

D D D D (3) and are identical compounds

D D D D (4) and are structural isomers

Sol. Answer (2, 3)

is anti-aromatic while is aromatic.

17. Species in which all C—C bonds are not equal is/are

(1) (2) (3) (4)

Sol. Answer (1, 3, 4) (1) is anti-aromatic while (3) and (4) are non-aromatic. Therefore in these molecules less effective - electron delocalization will occur and hence all C—C bonds will not be equal.

18. Consider the following equilibrium between two conformers of methyl cyclohexane Me

Keq Me

Me at axial Me at equatorial position position

What is the percentage of axial conformer if the equilibrium constant for the given equilibrium is 18? 40 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

Sol. Answer (5) [Equatorial conformer] 18 K [Axial conformer] 1

[Equatorial conformer] Percentage equatorial conformer 100 [Equatorial] [Axial conformer]

18 100 18 1 = 95% Percentage axial conformer = 5%

19. Most stable carbocation among the following is

(1) (2) (3) (4)

I II III IV

Sol. Answer (2) Maximum delocalization of positive charge occur in II.

20. For which of the following compound tautomerization reaction is very slow?

O—H O O—H O

(1) C C (2) C C CH CH FC2 CH3 HF2 C CH3 HC2 3 HC3 3

O O

(3) (4) H O O O O H Sol. Answer (1) In case of highly fluorinated enols the enol form is less stable than keto form. Yet enol form can be kept at room temperature for long periods of time because the tautomerization reaction is very slow.

21. Aromatic species among the following is/are

(1) (2) (3) B—R (4)

Sol. Answer (3, 4) Both are planar and contain 6 & 14 e– respectively. Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques 41

22. A solution of the sodium salt of diphenyl methane, Ph2CHNa in diethyl ether is orange, due to absorption of light in the blue region by the delocalized diphenyl methide ion, when colorless triphenyl methane is added, the color changes to red. Explain this result. Would you expect the addition of diphenyl methane to the sodium salt of triphenyl methane, in diethyl ether to result in a change in color from red to orange? Give explanation.

 Sol. Ph23 CHNa Ph CH Ph 223 CH Ph CNa 23. Consider the following tautomeric equilibrium

Me H Me Et3 N 50%

DO2 50%

H H H

Correct statement about the above equilibrium is (1) This interconversion is an intermolecular process (2) It is an intramolecular process (3) Interconversion involves carbanion intermediate (4) Interconversion involves carbocation intermediate Sol. Answer (2) In the given example neither carbanion is formed nor carbocations is formed.

24. Most stable and least stable species respectively among the following are

O O O S S S O S S O O O O IIIIII IV

(1) II and I (2) IV and II (3) III and II (4) II and IV Sol. Answer (4) Maximum delocalization of negative charge is taking place in II. Electro negativity of S is less than oxygen.

25. Compare acidic strength of

O O O O

NH3 H—C CH3 —S—OH O—H H H O IIIIII IV

(1) II > I > III > IV (2) IV > II > I > III (3) III > IV > II > I (4) II > IV > I > III Sol. Answer (2) IV > II > I > III

Decreasing acidic strength

pKa these acids are

Acid pKa I5 II 3.5 III 10 IV 1 42 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

26. Which of the following species can be used as antioxidant? (Note : Antioxidants have capability to trap free radicals) OH O CH—CH2 OH O OH OH (1) (2) H (3) (4) HO OH Sol. Answer (2, 3, 4) Except naphthalene all three compounds can trap free radicals due to presence of oxygen.