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Organic Chemistry- Some Basic Principles and Techniques

1. Which of the following is strongest nucleophile? (a) Br– (b) OH– (c) CN– (d) C2H5O– Answer: (d) The order of nucleophilicity depends on (i) nature of alkyl group on which the nucleophile attacks. (ii) nature of solvent.

However, if the above factors are same, then nucleophilicity depends on the acidic strength. Weaker the acid stronger is the nucleophile.

HBr > HCN > H2O > C2H5OH ∴Nucleophilic strength = Br– < CN– < OH– < C2H5O–

Hence, (d) is the correct answer. 2. Which of the following carbocation is least stable?

(a)

CH2+

(b) +

= 2CH CH

(c) +

= −2 2CH CH CH (d) CH3 – C+

CH3

H

Answer: (b) Positive charge is on sp hybridized carbon. Hence, (b) is the correct answer.

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3. Rearrangement is a property of ______ (a) carbanion (b) carbocation (c) carbon free radical (d) none of the above Answer: (b) Only carbocation shows rearrangement. Hence, (b) is the correct answer. 4. A compound shows no optical rotation in a given solvent.

Which of the following statement is correct? (a) It may be a racemic mixture. (b) It may be meso compound. (c) It may not have chiral centre. (d) All of the above Answer: (d) In all cases, optical activity will be zero. Hence, (d) is the correct answer. 5. Which of the following statement is correct?

(a) Allyl carbonium ion (CH2=CH– +

2CH ) is more stable than propyl carbonium ion (b) Propyl carbonium ion is more stable than allyl carbonium ion

(c) Both are equally stable (d) None Answer: (a) Allyl carbonium ion undergoes resonance stabilization.

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Hence, (a) is the correct answer. 6. Numbers of possible isomers of glucose are (a) 10 (b) 14 (c) 16 (d) 20 Answer: (c)

Glucose has four dissimilar asymmetric carbon atoms; a = 24.

Hence, (c) is the correct answer. 7. The greater the s-character in an orbital the ------------ is its

energy. (a) greater (b) lower (c) both (d) None of the above Answer: (a) Bond energy order. sp–sp > sp2-sp2 > sp3-sp3 Hence, (a) is the correct answer. 8. The relative stability order of carbanions CH≡C–, –CH3 and

CH2 = –CH is_____ (a) CH ≡ C– > CH2 = –CH > CH3

– (b) CH3

– > CH2 = –CH > CH ≡ C– (c) CH ≡ C– < CH2 = –CH > CH3

– (d) CH2 = –CH < CH ≡ C– < CH3

–

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Answer: (c) Power to hold negative charge is in the following order sp > sp2 > sp3

Hence, (c) is the correct answer. 9. Aniline is weaker base than ethylamine. This is due to (a) –I effect of NH2 in aniline (b) –R effect of NH2 in aniline (c) +I effect of NH2 in aniline (d) +R effect of NH2 in aniline Answer: (d)

Lone pair on NH2 group is involved in resonance with the ring in aniline.

Hence, (d) is the correct answer. 10. Which among the following alcohol has greatest ease to breakheterolytic C–O bond? (a)CH3CH2OH (b)

CH3 – C – OH

CH3

CH3

(c) CH3 – CH – OH

CH3

(d) CH3 – C – CH2 – OH

CH3

CH3

Answer: (b) Ease of heterolytic C–O bond breaking is

Tertiary alcohol > Secondary alcohol > Primary alcohol

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Breaking of C–O bond in tertiary alcohol ( CH3 – C – OH

CH3

CH3

)

results in the formation of tertiary carbocation, which is more stable than primary and secondary carbocation.

11. The type of isomerism due to different types of alkyl group on either side of functional groups in the molecule of compounds CH3CH2OCH2CH3 and CH3OC3H7 is referred as

(a) metamerism (b) chain isomerism (c) functional isomerism (d) position isomerism Answer: (a) It is definition of metamerism. Hence, (a) is the correct answer. 12. Number of isomers possible for C4H8 are ______ (a) 4 (b) 3 (c) 2 (d) 5 Answer: (d)

C C

H

CH3

H

CH3

cis

C C

H

CH3

CH3

H

trans

H2C C

CH2CH2CH3

H

H2C C

C

CH3

HCH3

CH3

H2C

H2C CH2

CH2

Hence, (d) is the correct answer. 13. The most unlikely representation of resonance structures of

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p-nitrophenoxide ion is

(a) N+ OO-

O-

(b) N+ O-O-

O

(c) N

OO

O-

+

(d) N

OO-

O

+

Answer: (c)

In structure (c), N is pentavalent with a positive charge which is not possible.

Hence, (c) is the correct answer. 14. Arrange the following in order of increasing order of their

stability.

(I) (II) (III) (IV)

CH2

NO2

CH2

CH

+ +CH2

CH3

+ CH2

NH2

+

(a) I > III > II > IV (b) IV > III > II > I (c) III > II > IV > I (d) IV > I > III > II Answer: (b)

+M effect of NH2 group disperse the charge of carbocation, hence increases the stability and –CH3 group shows +I effect, so it will disperse the charge less than –NH2 group.

•

•CH is triplet carbene, which has least tendency to release electron, hence makes the carbocation less stable. Whereas –NO2 group shows –M effect due to which the positive

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charge on the carbocation increases, hence stability decreases.

Hence, (b) is the correct answer.

15.

Br

H OH

C2H5

CH3

H

HO

Br H

H

C2H5

CH3

The molecules represented by the above two structures are (a) identical (b) enantiomers (c) diastereomers (d)

epimers Answer: (c)

Since the two stereoisomers are not mirror image, they are diastereomers.

16. Which of the following compound is incapable of exhibiting tautomerism

(a)

O O (b)

O O

(c) O O (d) CH=CH—OH

Answer: (c) Benzoquinone is very stable as it is highly conjugated

system so it does not get enolized. 17. Compound (A) is formed on treating a benzilwith NaBH4. The number of possible meso isomers for (A) are (a) 1 (b) 2 (c) 4 (d) 0

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Answer: (a) On reduction of benzyl with NaBH4 compound Hydrobenzoin is formed.

Ph − C − C − Ph → 4NaBH Ph − CH − CH − Ph

O O OH OH

* *

Hydrobenzoin has 2 optically active and one optically inactive (meso) isomers.

18. Choose the most stable carbocation.

(a)

CH3

X

H ⊕ (b)

CH3

X

H

⊕ (c)

CH3

X H ⊕

(d)

CH3

X H

⊕

Answer: (d) Tertiary carbocation is most stable carbocation, thus option (d) is correct. 19. Choose the compound which shows Keto−enol isomerism

(a) C6H5 − C − CH3

O

(b) C6H5 − C − H

O

(c) C6H5 − C − C − C6H5

O CH3

CH3

(d) C6H5 − C − C6H5

O

Answer: (a) α−hydrogen containing compounds shows keto-enol isomerism. In acetophenoneα−hydrogen is present hence it will show keto-enol isomerism.

20. Which order for basic character of amine is correct for

following compounds?

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NH2 NHCH3 CH2NH2 NH2

NO2

NO2

NH2

(1) (2) (3) (4)(5)

(a) 3 > 1 > 2 > 5 > 4 (b) 3 > 2 > 1 > 5 > 4 (c) 3 > 1 > 2 > 4 > 5 (d) 3 > 2 > 1 > 4 > 5 Answer: (d)

Electron releasing group increasing basic character of amine and aromatic amine is less basic than aryl and aliphatic amine.

21. Which of the following is the most stable carbocation:

(a) CH2

+ (b) CH2

+

(c) CH2

CH2

CH + (d) CH2 CH+

Answer: (a) CH2

+ is the most stable carbocation. 22. Which of the following compounds is not chiral? (a) 3-chloro-2-methyl pentane (b) 2-chloropentane (c) 1-chloro-2-methyl pentane (d) 1-chloropentane Answer: (d)

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1-chloro Pentane has no asymmetric carbon atom. CH3CH2CH2CH2CH2Cl.

23. In which of the following compounds delocalisation is not possible

(a) 1, 4-pentadiene (b) 1, 3-butadiene (c) 1, 3, 5-hexatriene (d) benzene Answer: (a)

In 1, 4-pentadine π electron not participate in conjugation so delocatisation is not possible.

24. How many hyperconjugative structures are possible for t-butyl carbocation

(a) 3 (b) 6 (c) 9 (d) 5 Answer: (c)

No. of hyper conjugative structures = no. of alpha hydrogen atoms.

25. The arrangement of decreasing order of stability of 3 2 5 3 2C H ,C H , (CH ) C H

• • • and 3 3(CH ) C• free radical is:

(a) 3 2 5 3 2 3 3C H C H (CH ) C H (CH ) C• • • •

> > > (b) 3 3 3 2 2 5 3(CH ) C (CH ) C H C H C H• • • •

> > > (c) 2 5 3 3 2 3 3C H C H (CH ) C H (CH ) C

• • • •

> > > (d) 3 3 3 2 3 2 5(CH ) C (CH ) C H C H C H• • • •

> > > Answer: (b) Follow the concept of hyperconjugation.

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26. Which of these have only –R (negative resonance effect)? (a) -NO2, -OH, -CHO (b) –CHO, - COOH, -OCH3 (c) -NO2, -CN, -SO3H (d) –NHCOCH3, -COCH3, -SO3H Answer: (c)

–N

O

O–C N, S

O

O(-R)

O H(-R)

(-R)

27. The most stable carbonium ion is: (a) C6H5 C

C6H5

C6H5 (b) C6H5 C

CH3

C6H5

(c) CH3 CH

CH3

(d) CH3 C

CH3

CH3

Answer: (a) Due to resonance, extractability in 3° carbonium. 28. Resonance arises due to the (a) Migration of atoms (b) Migration of proton

(c) Delocalisation of σ-electron (d) Delocalisation of π-electron

Answer: (d) Resonance in a molecule is arised due to delocalisation of π-electrons.

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29. Which of the following intermediates is most stable?

(a) (b) +

(c)CH3

(d)C2H5

Answer: (a)

It is aromatic in nature.

30. Which one of the following is strongest acid (a) 2-chloropentanoic acid (b) 3-chloropentanoic acid (c) 5-chloropentanoic acid (d) 4-chloropentanoic acid Answer: (a) Inductive power ∝ 1

dis tan ce

31. Arrangement of (CH3)3 – C - , (CH3)2 – CH - , CH3 – CH2 – when attached to benzyl or an unsaturated group in increasing order of inductive effect is

(a) (CH3)3 – C - < (CH3)2 – CH - < CH3 – CH2

(b) CH3 – CH2 - < (CH3)2 – CH - < (CH3)3 – C –

(c) (CH3)2 – CH - < (CH3)3 – C - < CH3, - CH2

(d) (CH3)3 – C - < CH3 – CH2 – (CH3)2 – CH –

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Answer: (b)

–CH3 group has +I effect, as number of –CH3 group increases, the inductive effect increases. Therefore the correct order is

CH3 – CH2 < (CH3)2 – CH - < (CH3)3C –

32. The number of isomers for the compound with molecular formula C2 BrCl FI are

(a) 3 (b) 4 (c) 5 (d) 6 Answer: (d)

33. 29.5 mg of an organic compound containing nitrogen was

digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20mL of 0.1 M solution. The excess of the acid required 15mL of 0.1 M solution for complete neutralization. The percentage of nitrogen in the compound is

(a) 59.0 (b) 47.4 (c) 23.7 (d) 29.5 Answer: (c)

C=C

Cl

Br

F

I Z

C=C Cl

Br

I

F E

C=C

Cl

I

Br

F E

C=C Cl

I

F

Br Z

C=C

Cl

F

I

Br Z

C=C Cl

F

Br

I E

HCl

NaOH

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Weight of organic compound

Total millimoles of Millimoles used by Weight of Wight of nitrogen

34. Which one will have lowest value of pKb in aqueous phase? (a) Methanamine (b) N-Methylmethanamine (c) N, N-Dimethylmethanamine (d) Benzenamine Answer: (b)

N-Methylmethanamine is 2° amine and hence most basic in aqueous phase.

35. Among 1-pentene, cis-2-pentene and trans-2-pentene the

correct order of stability is (a) 1-pentene > cis-2-pentene > trans-2-pentene (b) 1-pentene > cis-2-pentene < trans-2-pentene (c) 1-pentene < cis-2-pentene > trans-2-pentene (d) 1-pentene < cis-2-pentene < trans-2-pentene. Answer: (d)

2-pentene (cis as well as trans) is more stable than 1-pentene due to more Hypercojugation structures.

29.5mg=

3 4NH HCl NH Cl+ →

2(remaining) 15 0.1M1.5 mmole

HCl NaOH NaCl H O×

=

+ → +

HCl 2=

3NH 2 1.5 0.5= − =

3NH 0.5 17mg 8.5 mg= × =

14 8.5mg 7mg17

= × =

7% Nitrogen 100 23.7%29.5

= × =

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In cis-isomers of 2-pentene, two alkyl groups on same side of the plane creates more steric hindrance, so trans-isomers is more stable than cis.

36. The most stable form of the following conformers are

(a) (b)

(c) (d)

Answer: (c)

Due to intramolecular H-bonding.

37. Out of the following the one containing only nucleophiles is

(a) (b) NH3,CN , (c) (d) Answer: (b)

and are electrophiles. Therefore, (b) is the only case, where all three i.e. are nucleophiles.

38. An Alkane with least boiling point is

(a) (b)

π

3 3 3AlCl ,BF ,NH 3CH OH

3 2 2AlCl ,NH ,H OΘ2 2RNH , : CX , H−

3 3AlCl , BF 2:CX

3 3NH ,CN ,CH OH−

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(c) (d) Answer: (d) Branching decreases the B.P.

39. The relative rates of solvolysis in 80% EtOH of the following bromides is in the order -

Br I

Br II

Br III

(a) I > II > III (b) III > II > I (c) II > III > I (d) II > I > III Answer: (b)

The relative rate of 3º bridgehead system towards solvolysis can be explained using. (i) strain energy & (ii) stability of carbocation intermediate.

40. Which of the following compound has the lowest barrier of

rotation about the indicated bond

(a)

(b)

(c)

(d)

Answer: (c)

σ bond with least crowding has minimum barrier of rotation.

41. Resonance energy of benzene is – 142 kJ/mol Given: Heat of atomization of graphite = 716 kJ/mol. Bond energy of

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C–H, C–C, C=C and H–H bonds are 490, 340, 620, 436 kJ/mol respectively. ∆Hºf of C6H6 is -

(a) – 358 kJ/mol (b) – 216 kJ/mol (c) – 74 kJ/mol (d)– 500 kJ/mol Answer: (a) 6 C(s) + 3H2 (g) → C6H6(g)

)lTheoretica(fºH∆ = 6(716) + 3 (436) – 3(340) – 3(620) – 6(490)

∆Hºf(Actual) = ∆Hºf(Theoretical) – 142 = – 358 kJ/mol. 42. Select the correct order of reactivity towards conjugate

addition of a soft nuclophile to the following compounds

R

O

(I)

R

(II) O

R

(III) O

(a) I > II > III (b) III > II > I (c) II > I > III (d) II > III > I Answer: (a)

Soft nucleophiles undergo conjugate addition easily where there is least steric hindrance.

43. If the reactivity factor for chlorine substitution through free radical by abstracting a primary H-atom is 1 then the ratio of amount of product A and B are -

Cl2 ∆

Cl +

Cl (A) (B) (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1 Answer: (c)

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]B[]A[ =

atomHof.NofactoractivityReatomHof.NofactoractivityRe

−×−× =

3161

×× =

12

44. The correct decreasing order of nucleo-philicity of the

following species are - (i) (CH3)3COΘ (ii) CH3OΘ (iii) (CH3)2CHOΘ (iv) CH3CH2OΘ (a) (iii) > (i) > (iv) > (ii) (b) (i) > (iii) > (iv) > (ii) (c) (ii) > (iv) > (iii) > (i) (d) (ii) > (iv) > (i) > (iii) Answer: (b)

Nucleophilicity increases as the basicity of nucleophile increases.

45. Trans-3-methyl-2-pentene is - (a) E (b) Z (c) both (d) none Answer: (b)

C C

Me

H

Et

Me Z