Organic Chemistry : Some Basic Principles and Techniques

SECTION - A

Objective Type Questions (One option is correct)

1. Compound having molecular formula C5H

12O 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?

(1)

H H

+ NaHCO3

(2) C C—H + NaOH

(3)

OH

+ NaHCO3

(4)+ NaHCO

3

C

O

OH

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

C

O

O

H

+ NaO

C

O

OH

C

O

ONa

+

C

O

OHHOCO + H O

2 2

.

Solutions

Chapter 21

Organic Chemistry : Some Basic

Principles and Techniques

2 Organic Chemistry : Some Basic Principles and Techniques Solution of Assignment

3. The IUPAC name of the compound

CH —CH—C—CH3

CH

CH3 CH

3

CH

CH3

CH3

CH3

CH3

CH3

(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)

5. The IUPAC name of the compound

Et

Me

(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

3 2 3 3 3

I III IVII

CH —NO CH —CHO CH —F CH —CN

(1) I and II (2) III and IV (3) III and II (4) I and III

Sol. Answer (4)

Conjugate base of nitromethane (CH3—NO

2) is stabilized by very strong – I and – M effect of – NO

2 group

H—CH —N2

O

OB

CH —N2

O

O

CH —N2

O

O

Very stable conjugate base

Conjugate base of CH3—F is only stabilized by – I effect of F atom hence it is less stable and CH

3F is

least acidic.

3Solution of Assignment Organic Chemistry : Some Basic Principles and Techniques

7. Correct order of acidic strength for the given species

CH —C N—H3

CH —CH —CH3 2 3

I II

CH —C C—H3 CH —H

2

III IV

(1) I > II > IV > III (2) II > IV > III > I (3) III > I > IV > II (4) I > III > IV > II

Sol. Answer (4)

CH —C3

———N—HB

CH —C3

———N

More stablehighly unstabledue to presence

of positive chargeon highly en. sp—N

8. Which of the following compounds will not dissolve in aqueous NaOH?

(1)

OR

HO

HO OH

(2)

C

OH

O

(3)

H

OH

(4)

SH

Sol. Answer (3)

H

OH is least acidic.

9. Which of the following will have weakest indicated C—H bond?

(1) O N2

CH —H2 (2) CH —H

2

C

CH3

H C3

H C3

C

H C3

H C3

H C3

(3) CH —H2

C

CH3

CH3

CH3

C

CH3

CH3

CH3

(4) CH3—H

Sol. Answer (3)

CH —H2

C

CH3

CH3

CH3

C

CH3

CH3

CH3

Therefore indicated C—H bond in (3) is weakest.


10. In which of the following all electronic effects namely inductive, mesomeric and hyperconjugative effects are

present?

(1) (2)

H C3 CH

3

O

(3)

O

O

(4)

CH3

Sol. Answer (4)

In (1), (2) and (3) hyperconjugate effect is absent.

11. Most acidic species among the following is

(1)

O O

O(2)

S S

S(3)

O O

OO

(4)

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 C12

H17

NO?

(1) 4 (2) 5 (3) 6 (4) 7

Sol. Answer (2)

Index of hydrogen deficiency 17 1

(12 1)2

= 13 – 8 = 5

13. Which of the following compound cannot have a bond?

(1) C10

H21

N (2) C7H

12(3) C

20H

40Cl

2O (4) C

30H

50Br

2O

2

Sol. Answer (3)

Degrees of unsaturation in compound C20

H40

Cl2O 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?

(1)

O

O

(1) NaOEt

Et-OH(2)

NO2

NH2

(3)F

SbF5

(4) 2 3NaNH /NH

3CH —C C—H


Sol. Answer (3)

O

H

O

O

O

NaOEtH

F

SbF5

++ [SbF ]

6

–

CH —C3

C—HNH

2

CH —C C3

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)

4NH

does not have vacant orbital of appropriate energy.

19. Which of the following is the strongest base in water?

(1) (2) NH2 (3)

N

(4)

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

Hsp

3

20. Which of the following is most acidic?

(1)

H

NH2

N (2)

OH

NH2 (3) CH

3 – O – H (4)

NH2

NH2

C

H2N

Sol. Answer (2)

It forms most stable conjugate base


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)

– (1)

(2)

(3)

(4)

23. The compounds

O

and

O

are

(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

(1)

H C – C – CH3 3

CH3

CH2

(2)

CH3

CH2

(3) CH – C – CH3 3

CH3

(4)

CH2

CH

H2C

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?

(1) (NH2)2 C O (2) H

2N (C

6H

4) SO

3H (3) C

6H

5SO

3H (4)

NH2

Cl

Sol. Answer (2)

When both N and S is present in a compound. It gives blood red precipitate in Lassaigne’s test.


26. The most stable free radical is _________

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

CH

Sol. Answer (2)

Due to resonance and hyperconjugation both.

27. Which of the following group will have the strongest electron donating mesomeric effect?

(1) N

O

O

(2) N

Me

Me(3)

N

H

C = O

H C3

(4) O – C

O

R

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)3 3

CH CH CH

(2) (3) (4)

Sol. Answer (2)

+

is an aromatic cation.

29. Most acidic species among the following is

(1) CH – C – CH3 3

O

(2) CH3 – O – H (3) CH

3 – C C – H (4) H

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

B

Me

Me

C

D

(1) A (2) B (3) C (4) D

Sol. Answer (1)

Attack of H+ on double bond (A) gives most stable carbocation.

CH3

+

CH3

CH3


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)O

(2)NH

3

(3)NO

2

(4)BH

2

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?

(1)

O

(2) CH3COCH

2CHO (3) CH

3CHO (4) CH

3COCH

3

Sol. Answer (1)

O

H

O H

most stable enolform s tabilized by

Aromatic character

34. Among the following,

I II III

Cl Cl Cl

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)


35. Which of the following represents the correct order of stability of the given carbocations?

I

CH2

II 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?

(1) (CH3CH

2)2NH (2) CH

3CHO (3) CH

3CH

2COOH (4) CH C NH

3– –

2

O

Sol. Answer (2)

C = O ----- H — O

H C3

H

H

37. Correct stability order of the given free radicals is

(1) CH2 CH C > (CH ) CH

3—

3 2

CH3

CH3

> > (2) CH2 CH C > (CH ) CH– –

3 3 2

CH3

CH3

> >

(3) CH2

CH C >– –

3

CH3

CH3

> (CH ) CH3 2> (4) CH

2CH C >

3– –

CH3

CH3

>(CH ) CH3 2>

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

Cl — Cl or h

Cl + Cl

39. The total number of structural monochloroderivatives possible in C5H

12 is

(1) 6 (2) 7 (3) 8 (4) 5

Sol. Answer (3)

n-pentane, isopentane and neopentane form 3, 4 & 1 structural monohaloderivatives respectively.


40. In which of the following molecules positive charge is not delocalized because of resonance?

(1) CH N

3NH

2

NH

(2)

(3) C — N

Me

Me

Me

Me

(4) CNH

2NH

2

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?

(1)

CH3

>

CN

CH3

CN

>

CH3

CN

(2)

CH3

CN

CH3

CN

CH3

CN

> >

(3) > >

CH3

CN

CH3

CN

CH3

CN

(4)

CH3

CN

CH3

CN

CH3

CN

> >

Sol. Answer (2)

CH3– is electron donating group and –CN is electron withdrawing group 2 2

1 2 1 2– 2 cosθ

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?

(1) NH2 (2) N N (3)

N(4)

NO2

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.


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]

(1)

N|O

O (2) N

|O

O

(3) N|O

O (4)

N|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

(I)

OH

(II)Cl

COOH

(III)

COOH

(IV)CH

3

(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

(+I effect)CH

3

> >

OH

>

Cl(–I effect)

OH


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

OH

Br

CN

(1) 4-Bromo-3-cyanophenol (2) 2-Bromo-5-hydroxybenzonitrile

(3) 2-Cyano-4-hydroxybromobenzene (4) 6-Bromo-3-hydroxybenzonitrile

Sol. Answer (2)

OH

Br

CN

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 C—C —C—C —CH3 3

2 41 + 5

H

HO H

H

CH3

3

(1) CH3 at C-4 (2) H at C-4 (3) CH

3 at C-2 (4) H at C-2

Sol. Answer (4)

H C—C —C—C —CH3 3

2 41

+5

H

HO H

H

CH3

3CH —C—C—C—CH

3 3

+

:OH H CH3

H H

More stable resonance

stabilized carbocation

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

– – – –

2 2 2 2H C N N H C—N N H C—N N H C—N N

(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.


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)

C

O

O

H

–H

C

O

O

H

O–

Stabilized by strongintramolecular hydrogen

bonding

Most acidic

O H

55. In allene (C3H

4), the type(s) of hybridisation of the carbon atoms is (are) [IIT-JEE-2012]

(1) sp and sp3 (2) sp and sp

2 (3) Only sp2 (4) sp

2 and sp

3

Sol. Answer (2)

C = C = CH

( )sp

( )sp2

(Allene)

H

H

H

( )sp2

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?

(1) Ph—C—O—O—C—Ph

O O

h (2)h

RO—Cl

(3) CH —C—OH3

CH3

CH3

H+

(4)

BuLi

Sol. Answer (1, 2)

Ph—C—O—O—C—Ph

O O

h2Ph—C—O

O

R—O—Clh

R—O + Cl


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

(1)

DT

MeH

(2)C C

CH3

Br (3) C N—OH

H

H C3

(4) N2H

2

Sol. Answer (1, 3, 4)

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

(1) C4H

10O (2) C

4H

11N (3) C

4H

8O (4) C

4H

9Cl


For molecular formula C4H

11N only saturated amines are possible (1°, 2° & 3° amines are functional isomers)

Molecular formula C4H

9Cl 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)

+H

Stable tertiary carbocation

+H

Stable tertiary carbocation

(No rearrangementpossible)

(1)

(2)

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

(1)2 2

CH CH—CH

(2) C

CH3

CH3

(3) C

(4)

CH

2

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

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

(1) CH — C — C O3

—

CH3

CH3

(2)

(3) — CH

CH

3

(4)

CH2



CH —C—C = O3

+

CH3

CH3

CH —C3

+

CH3

CH3

+ CO

CH—CH3

+

CH3

+

CH2

+ +

8. Formic acid is more acidic than

(1) Benzoic acid (2) Acetic acid

(3) Phenol (4) Benzene sulphonic acid


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?

(1)

O

(2)

O

(3)

O

(4) OCO

Sol. Answer (1, 2)

O–

+

O–

+

Both are aromatic

10. The correct order of stability is/are

(1)2 3 2

HC C H C CH CH CH

(2) HC C > H C = CH > CH –CH

3 22

(3)22 3

HC C H C CH CH CH

(4)

CH2


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.


11. Which of the following is not stabilised by hyperconjugation?

(1)3

CH

(2) CH

2

(3)

(4)

Me

Me

Me

Me


In3

CH

and + there is no hydrogen

In CH2

+

, 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 – CH = CH – Cl can show geometrical isomerism and CHCH2

Cl

Cl

is its position isomer

13. A compound having molecular formula C4H10

O 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 – sp

2 > sp3 – sp

3 (bond energy) (2) sp – sp3 > sp – sp

2 > sp – sp (polarity in bond)

(3) sp3 – s < sp

2 – s < sp – s (% s character) (4) sp3 – sp

3 > sp2 – sp

2 > sp – sp (bond stability)


sp, sp2 & sp

3 have 50%, 33% and 25% s-character

15. Consider the following compounds

Which of the following statements are correct?

NN

H NH2

(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


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

(1) Cl3CH > F

3CH (2) CH

3COOH > CH

3CH

2OH

(3) H2O > CH

3CH

2OH (4)

OH

NO2

>

OH

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

3CCl is more stable than CF

3

– due to presence of d-orbital in3CCl hence CHCl

3 is more acidic than CFCl

3.

17. Which of the following correctly represents the stability of reactive intermediate?

(1) CH CH3 2–

+

< CH OCH3 2 (2)

(3)

CH2

NO2

>

CH2

CH3

(4) CH CH– – <C CH CH

23 3 3

CH

3

CH

3


3 2CH —O—CH

is a resonance stabilized carbocation.

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

(1) (C2H

5)2NH > (C

2H

5)3N > C

2H

5–NH

2

(2) (CH3)2NH > CH

3–NH

2 > (CH

3)3N

(3) CH3

– NH – CH2

– CH3 >

3 3

2

CH CH CH|NH

> CH3

– CH2

– CH2

– NH2

(4) C2H

5 – NH

2 > CH

3 – NH

2 > CH

2 = CH – NH

2

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) C

6H5–CO–CH

3

(3)

O

NH2

(4) C6H5–CO–CH

2–CO–CH

3


C6H

5 CHO has no –H. Therefore it will not exhibit tautomerism

O O

NH2

H

H

H2N


20. Which of the following can exhibit geometrical isomerism?

(1) C6H5CH = N–OH (2) (3) (CH

3)2C = C(CH

3)2

(4)

Me

H H

Me

Sol. Answer (1, 4)

C N

Ph

H

OH

and C N

Ph

H OH

and

Geometrical isomers

H

Me

H

Me

H

Me

Me

H

Geometrical isomers

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

(1)3

CH C CH (2) NH3

(3) C2H

5OH (4)

OH


OH– cannot neutrallise CH3—C CH, NH

3 and C

2H

5OH 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) sp

2 in pyridine (3) sp2 in pyrrole (4) sp in methyl cyanide


CH—C—NH3 2

sp2

Nsp

2

N

H

sp2

O

3CH —C N

sp

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

(1) N — N — N–

(2) N N N (3) N N N

(4)N N

N

Sol. Answer (1, 3)

N N N2

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.


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

(1) sp2 (2) sp

3 (3) sp (4) dsp2


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

(1) [Fe(SCN)4]– (2) FeCl

2(SCN) (3) [Fe(SCN)(H

2O)

5]2+ (4) Na

4[Fe(CN)

5NOS]


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]

(1) C—C

H C2

H H

CH2

(2) H—C C—CH

CH2

(3) H2C = C = O (4) H

2C = C = CH

2

Sol. Answer (2, 3)

H

C

C

C

CH

Hsp

2

C C O

H

H

All atoms lie in oneplane in all conformation

Planar

H

sp2

C C C

H

H

H

H

is non-planar

C C

H

C C

H

H

H

H

H

Non-planar in many conformations due to rotation along.

C2 — C

3 bond.

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

(1) (2) (3)

O

(4)

O


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

P

Q

AlCl3

NaH

R(NH ) CO

4 2 3

100-115 °C

O O

SHCl

O

(1) P (2) Q (3) R (4) S

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

(i)Cl

AlCl3

AlCl4(P)

(AROMATIC)

(ii)

NaH(Q) + H

2

(AROMATIC)

Na

(iii)

O O

(NH ) CO4 2 3

100–115ºCR

(NH ) CO4 2 3

2NH + CO + H O

3 2 2

O O

+ NH3

O O NH2 O NH

2

OH

IMPE

N

H

–2H O

2

N

H

HO OH

IMPEN

H

O OH

H

(R) AROMATIC

H

(iv)

OHCl

OH Cl (S)

(AROMATIC)


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 C4H10

O 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


Common name IUPAC name

(1)

OH

tert-butanol 2-methylpropanol

(3)OH

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—HC—C O

R

H

R —CH C—O

R

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?

(1) CH —N3

O

O

CH —N2

O

O—H

(2)

O

H

O

OHH C2

OCH

3

H C3 O

(3)

O OH

(4)

O

C

OEt

O

EtO

OH

OEt

O

EtO


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) H

2O (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)

CN

CN

(1)

(2)(3)

(4)

(5)

2.

OH

OH

, IUPAC name is

(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)

OH

OH

(1)

(2)

(3)


3. Write IUPAC name of

1 2

3

4

5

6

1

2

3

4

(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

stability of conjugate base, greater is value of k for that equil ibrium making the equilibrium move in (HA K

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?

(1)

OH

(2)

OH

CH3

(3)

OH

CH3

(4)

OH

CH3

Sol. Answer (1)

Alkyl group (electron donor) decreases acidic strength of phenol

2. Which of the following is having most acidic -Hydrogen?

(1) CH3—NO

2(2) CH

2NO

2

NO2

(3) CH NO2

NO2

NO2

(4)

O

OSol. Answer (3)

Conjugate base of O N2 C NO2

NO2

H

is

Stabilized by three strong electron withdrawing groups

i.e.,

N C N

N

O O

O

O

O

O

Stabilized by –I and –M effect of three –NO2 groups


3. Which of the following is most basic due to +R effect of any substituent?

(1)N..

H

(2)N..

OCH3

(3)

NH2

CH3

(4)

NH2

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?

(1)

CH

H

H

(2)

CH

H

H

(3)

CH

H

H

(4)

CH

H

H

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.


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 –NH

2.

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.

4. STATEMENT-1 :

O

O

HO

HO

Squaric acid

is less acidic than

OH

Phenol

and

STATEMENT-2 : Conjugate base of phenol is resonance stabilized.

Sol. Answer (4)

Infact squaric acid is more acidic than phenol.

5. STATEMENT-1 : When

Br

Br

Br

Br

(X)

is treated with excess of Ag+ ion. One mole of the compound reacts

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.


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 : CH CH CH23 2 is less stable than CH N CH

23 – –

|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)

Angle strain 1

stability of cyclic compound.

12. STATEMENT-1 :

||

||

O

O

Keto form is less stable than enol form.

and

STATEMENT-2 : Enol form is stabilized by aromaticity.

Sol. Answer (1)

O

OH

OH

O


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 sp

3 hybridised

SECTION - E

Matrix-Match Type Questions

1. Match the following

Column-I Column-II

(Molecular formula) (Type of isomerism)

(A) C6H12

O (p) Functional isomerism

(B) C4H11N (q) Geometrical isomerism

(C) C6H12

(r) Metamerism

(D) C5H12

O (s) Tautomerism

(t) Position isomerism

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

C6H

12O has one degree of unsaturation.

Carbonyl compounds, unsaturated alcohols and ethers are possible. This can also exhibit geometrical

isomerism

C4H

11N can form 1°, 2° & 3° amines which are functional isomer.


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



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 H

2O 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

10 cm3 N

10 NaOH for neutralisation

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

(A) Percentage of C = 2

Mass of CO12100

44 Mass of organic compound

12 1.10100 66.6%

44 0.45

Percentage of H = 2

Mass of H O2100

18 Mass of compound

2 0.3100 7.4%

18 0.45

(B) Percentage of C = 18

100 66.66%27

Percentage of H = 2

100 7.40%27

Percentage of N = 7

100 26%27

(C) Percentage of Br =

80 Mass of AgBr100

188 Mass of compound

80 0.631100 72.6

188 0.37

(D) 10 cm3 of N

10 alkali required acid = 0.122 g

1000 cm3 of 1 N alkali required acid =

0.1221000 10

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


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) CH CH3 —

CH3

(q) Resonance

(C) (r) Hyperconjugation

(D) CH O CH3 2—

—

(s) Inductive effect

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

is resonance stabilised but antiaromatic

CH3 CH

CH3

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

+

is aromatic resonance stabilised cation

3 2CH –O–CH

is resonance stabilised because it can form

3 2CH –O CH



Column-I Column-II

(A)O

+

(p) Resonance stabilisation

(B) (q) Aromatic molecule

(C)

CH3

(r) Hyperconjugation effect is observed

(D) CH3

– CH = CH – CH = CH2

(s) At least one carbon is sp2 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) H N—NH Cl2 3

(p) Sodium fusion extract of the compound gives Prussian blue

colour with FeSO4

(B) HONH l

3

COOH

(q) Gives positive FeCI3 test

(C) HO NH Cl3

(r) Gives white precipitate with AgNO

3

(D) O N2 NH—NH

3Br

NO2

(s) Reacts with aldehydes to form the corresponding

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 C30

H43

N 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)

Degrees of unsaturation in the compound = 43 1

312

= 31 – 21 = 10

Out of which 8 bonds are present

Hence

bond 84

ring 2


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 F F F

Hence positive charge is delocalized over four atoms.

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

CCl , NH , CH , NH , CH , OH, BF , AlCl , Br2 3 3 4 3 3 3

Sol. Answer (5)

NH , NH , CH and OH are not electrophiles, infact they are nucleophiles3 4 3

–

5. How many bicyclic isomers are possible for the molecular formula C6H12

O?

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 NaHCO

3?

OH

,

OH

,

OH

,

O N2 NO

2

NO2

OHO

HO

O

,

O CH

H

CH2

OH OH

OHHO

O

,

NO2

NO2

OH

,

NO2

OH

,, ,

O

O

O

O

Sol. Answers (3)

Trinitrophenol, squaric acid and Vth 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


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]

NH C

3CH

3 COOH OCH CH2 3

OH

CH OH2

NO2

CH CH2 3

COOH

CH CH2 3

OH

NH C

3CH

3

Sol. Answer (4)

Carboxylic acids and aromatic alcohols dissolves in aqueous NaOH.

COOH OH COOHOH

NH C

3CH

3

, , ,

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]

H C3

CH CH2 3

+

Sol. Answer (6)

HC3

CHCH2 3

+

It contains 6 -hydrogen atoms.

6 contributing structures are possible


11. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H

6 is

[IIT-JEE-2010]

Sol. Answer (5)

Index of hydrogen deficiency in the compound C4H

6 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

NaOHN

Sol. Answer (9)

OH

NaOH

–H O2

O

I

O

II III IV

O O

O

V

O

III

O

VI

O

VII

O

VIII

O

IX


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) TFT (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 C6H

5O– than C

6H

5COO–. Fluorobenzene is

more reactive than chlorobenzene.

2. STATEMENT-1 : 3 2

CH O CH

is more stable than

2

O

O N CH

.

STATEMENT-2 : + is antiaromatic molecule.

STATEMENT-3 : Phenoxide is more stable than ethoxide.

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

Sol. Answer (4)

In 3 2CH O CH

..

..

, 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?

OHC

OH

CHO

OH

CHO

OH

HO

CHO

I II

OHC

OH

CHO

OH

CHO

OH

HO

III

OH

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.


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

I II

O

III

O

OH

NH2

IV V VI

N

Sol. VI < V < IV < II < I < III

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?

I II

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.

+

C

Triphenyl methyl cation is very stable, due to resonance.

In triphenyl methyl carbocation positive charge can be delocalized in all the 3 ring systems

causing the lowering of energy and making the carbocation very stable.

6. For the dehydration reaction

CH OH2

H+

+

CH2

+

CH3

Explain the mechanism discussing formation of all the 3 products.


Sol.

CH OH2

H+

CH2CH O

2

H

H

CH3

CH3

CH2

elimination

of H

rearrangementof carbocation

(H O being a

good leavinggroup)

2

CH2 Ring expansion

to bring releasefrom ring strain

– H

CH – OH2

H+

CH2

Ring expansion

CH3

Hydride shift

-H elimination

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

NH2

OCH3

NH2

OCH3

or

Sol.

NH2

OCH3

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

negative charge will be less available on

NH2

OCH3

, making it more stable.

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.


8. Aromatic amines are weakly basic, whereas

NH2

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

I II III 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 :

I II III IV

(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

Lesser delocalization of negative hence least stable

charge,

Least acidic


12. Order of basicity of the following species is

NN

NH N

N

N

H

NN

I II III IV

(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

N

Me

Me CH2

O—Me

N

Me

Me

CH2

OMe

N

Me

Me

CH2

O—Me

N

Me

Me

CH2

(I) (II)

(III)(IV)

(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?

CH —OH2

H SO

2 4

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 NO

2 ion through self ionization which attacks on aromatic ring to give the desired product.


16. Correct statements among the following is/are

(1)

DD

and

DD

are identical compounds

(2)

DD

and

DD

are structural isomers

(3)

D

D

and

D

D

are identical compounds

(4)

D

D

and

D

D

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)


(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 position

Me at equatorial position

What is the percentage of axial conformer if the equilibrium constant for the given equilibrium is 18?


Sol. Answer (5)

[Equatorial conformer] 18K

[Axial conformer] 1

Percentage equatorial conformer [Equatorial conformer]

100[Equatorial] [Axial conformer]

18100

18 1

= 95%

Percentage axial conformer = 5%

19. Most stable carbocation among the following is

(1)

I

(2)

II

(3)

III

(4)

IV

Sol. Answer (2)

Maximum delocalization of positive charge occur in II.

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

(1) C

O—H

CH3F C

2

C

O

CH3HF C

2

(2) C

O—H

CH3H C

2

C

O

CH3H C

3

(3)

O

OH

O

O

(4)

OH

O

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.


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. 2 3 2 2 3

Ph CHNa Ph CH Ph CH Ph CNa ��⇀↽��

23. Consider the following tautomeric equilibrium

MeH

H

Me

H

H

Et N 50%

D O 50%

3

2

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

OS

O

O

S

O

O

O

I II

S

O

O

III

SS

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

NH3 H—C

O

O—H

I II

H H

OO

CH —S—OH3

O

O

III 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

I 5

II 3.5

III 10

IV 1


26. Which of the following species can be used as antioxidant?

(Note : Antioxidants have capability to trap free radicals)

(1) (2)

O

O

CH—CH2

OH OHH

OHHO

(3)

OH

(4)

OH


Except naphthalene all three compounds can trap free radicals due to presence of oxygen.

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Organic Chemistry : Some Basic Principles and Techniques

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Transcript of Organic Chemistry : Some Basic Principles and Techniques