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Dear readers, I feel deeply indebted to all of you for such a tremendous response to earlier editions of

this book. The revised edition of this meticulously crafted book has been written based on the latest

syllabus of both engineering and medical entrance examinations.

Some major highlights of this edition are :

• Introduction of a new chapter entitled ‘Physical Properties of Organic Compounds’.

• Addition of new challenging problems in each chapter for JEE Advanced aspirants.

• Chapter on ‘Isomerism’ has been thoroughly revised for better understanding of the topic.

• Concept strengthening Matrix questions have been added to each chapter keeping in consideration

the pattern of entrance examinations.

• Sincere efforts to remove all errors.

Almost all suggestions received from learned fellow teachers, as well as, many students have been paid

due attention and incorporated to the best possible extent. I am confident and sure this book will help

students reinforce their fundamentals of Organic Chemistry.

Suggestions from readers for further improvement of the book are welcome.

Vaibhav Trivedi

Msc-IIT-R, NET

organicvaibhav@gmail.com

PREFACE

Dedicated to

my Parents

Chapter 1 THE LIVING WORLD 1-8

Chapter 2 NOMENCLATURE OF ORGANIC COMPOUNDS 9-18

Chapter 3 ISOMERISM 19-50

Chapter 4 REACTION MECHANISM (General Organic Chemistry) 51-150

Chapter 5 ACID & BASE 151-168

Chapter 6 HYDROCARBON (Alkane, Alkene & Alkyne) 169-202

Chapter 7 ALKYL HALIDE & GRIGNARD’S REAGENT 203-230

Chapter 8 ALCOHOL, ETHER, EPOXIDE & GRIGNARD’S REAGENT 231-264

Chapter 9 ALDEHYDE & KETONE 265-294

Chapter 10 CARBOXYLIC ACID & ITS DERIVATIVES 295-318

Chapter 11 AMINES 319-340

Chapter 12 BIOMOLECULES 341-354

Chapter 13 AROMATIC CHEMISTRY 355-394

Chapter 14 PRACTICAL ORGANIC CHEMISTRY 395-400

Chapter 15 PHYSICAL PROPERTIES OF ORGANIC COMPOUNDS 401-404

CONTENTS

Main Features

1. Hybridization refers to the phenomenon of mixing of two or more than two atomic orbital of same energy (or nearly same energy) to produce two or more than two orbital of same energy, identical shape & size.

Lone pair of electron (lp) + sigma bond pair of electron (s bp) + (–ve) charge = Hybridization for eg. s bp = 3 lp = 1 (–ve) charge = 0 sum = 3 + 1 + 0 = i.e. sp3 (3+1= 4)

sigma bp = 3, lp = 0, –ve charge = 0 Sum = 3 + 0 + 0 = 3 (sp2)

sigma bp = 3, –ve charge = 1, lp = 0 sum = 3 + 1 + 0 = 4 (sp3)

2. ‘N’ can acquire four states.

• (When four bonds are present then ‘N’ has +ve charge)

• (When three bonds are present then lone pair of e– is present on Nitrogen)

• (When two bonds are present on ‘N’ then there is a –ve charge along with lp on ‘N’

For eg. H — C σπ

πN Here N has a lp hence hybridization of ‘N’ is sp (one lp + 1bp)

3. The lone pair of electron in conjugation with double bond does not consider in hybridization. and

4. A carbon attached to one (or no) carbon atom is called a primary (1º) similarly if a ‘C’ is attached to two, three & four ‘C’ atoms

it is referred as secondary (2º) ,tertiary (3º), & quaternary (4º) carbon respectively.5. The hydrogen attached to 1º, 2º & 3º ‘C’ atoms is called 1º, 2º & 3º hydrogen respectively.

2 Problems in Organic Chemistry

6. Starting from the simplest member of the molecular formula of the compounds of a particular class arranged in an increasing number of ‘C’ atom then a series is obtained in which the member differ from the next member by ‘CH 2’ group. Such a series is called homologues series.

7. Mol. Wt. of homolog ∝ boiling point ∝ 1solubility in water

Objective Questions

1. Which type of hybridization is absent in CH C CH CH CH2 = = − =−( )

(a) sp2 (b) sp (c) sp3 (d) all are present

2. How many 3° H are present in

(a) 1 (b) 2 (c) 3 (d) 4

3. Sequence of hybridization in CH3—CH = CH — CN (a) sp3, sp2, sp, sp2, sp2 (b) sp3, sp2, sp2, sp2, sp2

(c) sp3, sp2, sp2, sp, sp (d) sp3, sp2, sp2, sp, sp2

4. which statement is false?

(a) Four carbon atoms are sp2 hybridized (b) One nitrogen is sp2 & other is sp3 hybridized (c) It contains four secondary – H – atoms (d) One nitrogen is 2º & other is 3º

5. which statement is correct for this compound?

(a) It contains two 2ºC atoms (b) It contains two 2º ‘D’ atoms (c) It contains 81- ‘H’ atoms (d) The ‘C’ joined with Cl is sp2 hybridized

6. which is correct?

(a) It contains four sp2 hybridized C atoms (b) It contains nine sp3 hybridized C atoms (c) It contains two 3º – H – atoms & two sp2 hybridized ‘C’ (d) It contains ten- H-atoms & two sp2 hybridized ‘C’ atoms.

7.

In which species bold ‘C’ atom is not sp2 hybridized (a) I (b) II (c) III (d) I & III

8. In which case state of hybridization is changing.

(a) (b)

(c) (d) (a) & (c)

The Living World 3

9. In which case state of hybridization is changing.

(a) ( ) ( )H

3 3 3 2CH — CH CH — CH+ −−

(b) ( ) ( )H

3 3 3 2CH — CH CH — CH− +−

(c) ( ) ( )H

2 2 2CH CH CH CH+ −−= =

(d) ( ) (–)H

CH CH CH C+− ≡≡

10. In which case state of hybridization on each atom is retained.

(a) ( ) ( )

2 2 3 2CH CH H CH — CH+ +

+ →= (b)

(c) 3

( ) ( )liq.

2NH

1CH CH Na H CH C Na

2

− ++ → +≡ ≡ (d) (b) & (c)

11. 3 2 2 3 3 2CH CH CH OH CH CHO CH CH OH HCHO

I II III IV

Which statement is correct? (a) I, II, III & IV are homolog (b) I & II, III & IV are homolog (c) B.P. of I will be lesser than III (d) Solubility of (IV) in water will be greater than II

12. which statement is correct?

(a) Two sp hybridized ‘C’ atoms & six - 2º - H atoms are present in it (b) Two 3ºC - atoms & two sp2 hybridized ‘C’ atoms are present in it. (c) N is sp3 & two - 2º - C atoms are present in it. (d) Two 3º – H – atoms, Eight – 2º – H – atoms & four – sp3 – hybridized ‘C’ atoms are present in it.

13.

State of hybridization of bold element is changing as:-

(a) 3 2 2sp sp sp→ → (b) 3 3 3sp sp sp→ →

(c) 3 3 2sp sp sp→ → (d) 3 2 2sp sp sp→ →

14. correct order of bond length is : (a) x > y > z (b) y > x > z (c) z > x > y (d) y > x = z

15. is called nitrene. Hybridization on ‘N’ is :

(a) sp (b) sp2 (c) sp3 (d) None

16. Which statement is incorrect about :

(a) Oxygen gives one of its electro pair to the empty p-orbital of 2nd carbon atom (b) Each carbon is sp2 hybridized (c) Each carbon has one e– in unhybridized pz orbital (d) Except 2nd C - atom each carbon has one e– in unhybridized pz orbital

17. Which type of bond is absent in benzyne?

(a) sp2 – sp2 s bond (b) sp2 – s s bond (c) sp2 – sp2 p bond (d) sp2 – p s bond

4 Problems in Organic Chemistry

Passage - I

Enzymes present in bacteria can convert p - amino benzoic acid in to folic acid.

NN

N

OH

NH N2

CH – NH2

(Folic acid)

C – N – C

O HCH CH COOH

2 2

H

C

O OH

Answer the question from 18 to 26

18. Which statement is in favour of ‘N’? (a) All are sp2 hybridized (b) All are sp3 hybridized (c) Four are sp2 hybridized & three are sp3 hybridized (d) Four are sp3 hybridized and three are sp2 hybridized

19. How many ‘C’ atoms are sp2 hybridized in the folic acid? (a) 14 (b) 21 (c) 12 (d) 15

20. How many ‘C’ are sp3 hybridized? (a) 4 (b) 3 (c) 5 (d) 6

21. How many p bonds are present in it? (a) 11 (b) 10 (c) 6 (d) 12

22. Number of lone pair of electron possessed by nitrogen is / are : (a) 3 (b) 5 (c) 7 (d) None

23. How many H – atoms are present in folic acid? (a) 19 (b) 23 (c) 18 (d) 17

24. How many ‘C = O’ bonds are present in folic acid? (a) 5 (b) 6 (c) 4 (d) 3

25. How many ‘C’ atoms (which are not the part of ring) are sp2 hybridized? (a) 4 (b) 3 (c) 2 (d) None

26. Ratio of N — H bond to C — H bond is : (a) 1 : 3 (b) 4 : 7 (c) 2 : 5 (d) 2 : 9

Passage - II

H2 / Ni can reduce almost all types of multiple bonds

2H2 3 2 2 2Ni

CH CH — C N CH CH CH NH→= ≡; 2H

3 3 3 3Ni

O OH| | |

CH — C— CH CH — CH —CH→

While PCC can convert alcohol (RCH2OH) in to aldehyde (R – CHO)

Consider the following compound.

Answer the question from 27 to 36

The Living World 5

27. In compound A ‘N’ does not exhibit : (a) sp hybridization (b) sp2 hybridization (c) sp3 hybridization (d) ‘N’ exhibits sp, sp2 & sp3 hybridization

28. In (A) carbon exhibits : (a) sp hybridization (b) sp2 hybridization (c) sp3 hybridization (d) All of these

29. In compound (C) : (a) All ‘N’ are sp3 hybridized (b) Two nitrogen’s are sp3 hybridized and one is sp2 hybridized (c) All nitrogen’s are sp2 hybridized (d) One nitrogen is sp3 hybridized & two are sp2 hybridized

30. In compound (C) (a) All ‘C’ are sp3 hybridized (b) 9 ‘C’ are sp2 hybridized & two are sp3 hybridized (c) 9 ‘C’ are sp3 hybridized & two are sp2 hybridized (d) One ‘C’ is sp hybridized while rests are sp3 hybridized

31. In compound (C) total ‘N — H’ bonds is : (a) 2 (b) 3 (c) 4 (d) 1

32. In compound (C) ratio of N — H to C — H bonds is : (a) 4 : 17 (b) 4 : 15 (c) 1 : 4 (d) 3 : 17

33. How many carbon atoms of (B) are sp2 hybridized which are not the part of ring? (a) 1 (b) 2 (c) 3 (d) None

34. In compound (B) ratio of ‘C = O’ to C = C bonds are : (a) 2 : 3 (b) 1 : 1 (c) 1 : 3 (d) 3 : 3

35. Total double bonds present in compound (B) are : (a) 4 (b) 3 (c) 5 (d) 6

36. In compound (B) (a) One ‘C’ is sp hybridized 5 ‘C’ are sp2 hybridized and rest ‘C’ are sp3 hybridized (b) On ‘C’ is sp hybridized 7 ‘C’ are sp2 hybridized and rest all are sp3 hybridized (c) One ‘C’ is sp hybridized 8 ‘C’ are sp2 hybridized and rest are sp3 hybridized (d) One ‘C’ is sp hybridized and rest all are sp3 hybridized

Passage - III

Succinoyl sulphathiazole was used as a chemotherapeutic agent beginning in 1942. Its structure is given below.

Answer the question from 37 to 40

37. Select the correct statement for the above compound:- (a) Both sulphur atoms are sp2 hybridized (b) Both sulphur atoms are sp3 hybridized (c) One sulphur atom is sp2 hybridized & other is sp3 hybridized. (d) One sulphur is sp3 hybridized and other is sp hybridized

38. Which statement is correct about nitrogen? (a) All N atoms are sp2 hybridized (b) All N atoms are sp3 hybridized (c) All N atoms are sp2 hybridized. (d) One N is sp3 hybridized while rest two N atoms are sp2 hybridized

6 Problems in Organic Chemistry

39. Total p bonds present in this compound are:- (a) 6 (b) 7 (c) 8 (d) 9

40. Which statement is true about pentagonal ring? (a) Both sulphur and Nitrogen are sp hybridized. (b) Both sulphur and nitrogen are sp2 hybridized (c) Sulphur is sp3 hybridized however nitrogen is sp2 hybridized (d) N is sp2 hybridized but sulphur is sp hybridized.

Answer Key

1. (c) 2. (a) 3. (c) 4. (b) 5. (c) 6. (c) 7. (a) 8. (c) 9. (b) 10. (c)

11. (d) 12. (b) 13. (c) 14. (c) 15. (c) 16. (c) 17. (d) 18. (a) 19. (d) 20. (a)

21. (a) 22. (c) 23. (a) 24. (d) 25. (b) 26. (a) 27. (c) 28. (d) 29. (a) 30. (a)

31. (c) 32. (a) 33. (b) 34. (b) 35. (c) 36. (b) 37. (c) 38. (a) 39. (b) 40. (b)

SOLUTION1. (c) CH2 = C = CH – CH = CH(–) ↑ ↑ ↑ ↑ ↑ sp2 sp sp2 sp2 sp2

2. (a)

3. (c)

4. (b)

Each ‘C’ atom present in ring is sp2 hybridized. Nitrogen containing positive charge is tertiary while other nitrogen is secondary.

5. (c)

Each ‘C’ atom is sp3 hybridized6. (c) Carbon of CO group is sp2 hybridized rest all carbon atoms are sp3 hybridized

Problems in Organic Chemistry for JEEMain & Advanced 3rd edition

Publisher : Disha Publication ISBN : 9789384089948 Author : Disha Publication

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