E0289.pdf - Navneet eCatalogue

Based on the Board's New Textbook

E0289

Salient features :1. �A complete book for understanding the syllabus of Chemistry XII.�2. �Chapter�Outline,�Important Terms, Units, Laws, Formulae, Properties and Reactions given at

the beginning of each chapter.�3. �Model answers to all the textual questions as well as additional questions for understanding

basic and advanced concepts given in the Textbook.�4. �Answers to inquisitive and brain stimulating questions under the titles ‘Can you tell ?’,

‘Just think’, ‘Use your brain power’, etc. are given in each chapter.�5. �Includes important questions from NCERT book for reference.�6. Includes well-formulated ample number of Multiple Choice Questions (MCQs).�7. Simple and lucid language.8. Neat fully-labelled, authentic and easily-reproducible diagrams.9. Very useful book to understand the subject well and to prepare thoroughly for HSC Board

Examination as well as other competitive examinations like NEET, JEE MAIN, MHT-CET, etc.

By

NAVNEET

New Edition : 2020

2

Published by : Navneet Education Limited, Dantali, Gujarat. Printed by : Navneet Education Limited, Dantali, Gujarat.

2001CTP

(30-6-2020)

Visit us at : www.navneet.com

★ Navneet Bhavan, Bhavani Shankar Road, Dadar (West),

Mumbai– 400 028.

Phone : (022) 6662 6565

e-mail : [email protected]

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

★ Navneet Bhavan, 1302, Shukrawar Peth,

Bajirao Road, Near Sanas Plaza,

Pune – 411 002.

Phone : (020) 2443 1007


- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

★ 63, Opp. Shivaji Science College,

Congress Nagar,

Nagpur– 440 012.

Phone : (0712) 242 1522


- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

★ Nirman Inspire, 2nd Floor,

Kanhere Wadi, Opp. Old CBS,

Nashik– 422 001.

Phone : (0253) 259 6950 ● Fax : (02531) 259 6950


© All rights reserved. No part of this book may be copied, adapted, abridged or translated, stored in any retrievalsystem, computer system, photographic or other system or transmitted in any form or by any means without a priorwritten permission of the copyright holders, M/s. Navneet Education Limited. Any breach will entail legal action andprosecution without further notice.

3

PREFACEIt gives us great pleasure to present you this first edition of Navneet Chemistry Digest (Part II) for

Standard XII. This book is prepared according to the Maharashtra Board’s New Textbook and latest

question pattern.

We understand that the Higher Secondary Examination is very crucial in a student’s career. As in the

past, this Navneet Digest will help you to triumph.

Each chapter of the Digest begins with important Terms, Units, Laws, Formulae, Properties, Reactions,

etc. for instant revision. This book includes all the questions given in the Board’s Textbook as well as

important questions from NCERT book and also many additional questions with their model answers

so as to cover every concepts given in the Textbook.

Besides these, all the other questions or informations given in various boxes in the textbook such as

'Can you tell?', ‘Do you know?’, 'Use your brain power', 'Observe and discuss', etc. are also included.

Ample number of Multiple Choice Questions (MCQs) have been given to enable the students to prepare

for any type of competitive examination, such as NEET, JEE MAIN, MHT-CET, etc.

Neat, accurate and easily reproducible diagrams are given in each chapter.

In short, this Chemistry Digest contains a lucid and clear explanation of the subject matter in simple

language.

We hope this Digest will help the students to master the subject. Besides it will also help them to secure

excellent marks in the examination and pave the way for realising their dreams of a bright and fruitful

career.

We have taken utmost care to see that this Digest proves to be very useful to the students as well as

the teachers. Suggestions for improvement of the Digest are most welcome and will be gratefully

acknowledged and appreciated.

—The Publishers

4

CONTENTSPage No.

9. Coordination Compounds . . . 155

10. Halogen Derivatives . . . 150

11. Alcohols, Phenols and Ethers . . . 116

12. Aldehydes, Ketones and Carboxylic Acids . . . 189

13. Amines . . . 277

14. Biomolecules . . . 339

15. Introduction to Polymer Chemistry . . . 383

16. Green Chemistry and Nanochemistry . . . 418

Chapter Nos. 1 to 8 are included in Chemistry Digest: Part 1, published separately.

COORDINATION COMPOUNDS

9. COORDINATION COMPOUNDS 5

9 COORDINATION COMPOUNDS

....

....

....

CHAPTER OUTLINE

Page No.

Important Terms … 6

Exercises

9.1 Introduction … 7

9.2 Types of ligands … 7

9.2.1 Monodentate ligands … 7

9.2.2 Polydentate ligands … 7

9.2.3 Ambidentate ligand … 7

9.3 Terms used in coordination chemistry … 9

9.3.1 Coordination sphere … 9

9.3.2 Charge number of complex ion and oxidation state of metal ion … 9

9.3.3 Coordination number (C.N.) of central metal ion … 9

9.3.4 Double salt and coordination complex … 9

9.3.5 Werner theory of coordination complexes … 9

9.4 Classification of complexes … 13

9.4.1 Classification on the basis of types of ligands … 13

9.4.2 Classification on the basis of charge on the complex … 13

9.5 IUPAC nomenclature of coordination compounds … 15

9.6 Effective Atomic Number (EAN) Rule … 17

9.7 Isomerism in coordination compounds … 19

9.7.1 Stereoisomers … 19

9.7.2 Structural isomers (Constitutional isomers) … 19

9.8 Stability of the coordination compounds … 30

9.8.1 Factors which govern stability of the complex … 30

9.9 Theories of bonding in complexes … 31

9.9.1 Valence Bond Theory (VBT) … 31

9.9.2 Octahedral complexes … 33

9.9.3 Tetrahedral complexes … 33

9.9.4 Square planar complex … 33

9.9.5 Limitations of VBT … 33

9.9.6 Crystal Field Theory (CFT) … 38

9.9.7 Factors affecting Crystal Field splitting parameter (��) … 38

9.9.8 Colour of the octahedral complexes … 38

9.9.9 Splitting of d-orbitals in tetrahedral and octahedral complexes … 38

9.10 Applications of coordination compounds … 46

Multiple Choice Questions … 47

1. Coordination compound : It consists of a central metal atom or ion surrounded by atoms, molecules or

anions called ligands bonded by coordinate bonds, e.g. cisplatin, Pt (NH�)�Cl

�, [Cu(NH

�)�

]SO�.

2. According to Werner’s theory, metal atom or ion in the complex has primary valence (outer sphere) and

secondary valence (inner sphere).

3. Coordination number (C.N.) : The number of ligand donor atoms directly bonded to the central metal atom

or ion by coordination bonds or number of electron pairs involved in the coordinate bonds is coordination

number (C.N.).

4. Ligands : IIIII

� (i) Monodentate ligands (Cl�, OH�, NH�, H

�O, etc.)

� (ii) Bidentate (en, C�O��

�etc.)

� (iii) Tridentate (dien)

� (iv) Tetradentate (trien)

� (v) Hexadentate (EDTA)

5. Classification of complexes on the basis of types of ligands :

(1) Homoleptic complexes Ni(CO)�, [Co(NH

�)�]Cl

�(2) Heterleptic complexes [Co(NH

�)�Cl

�H

�O]Cl

6. Classification of complexes on the basis of charge :

(1) Cationic sphere complexes, [Zn(NH�)�]��, [Co(NH

�)�Cl]��

(2) Anionic sphere complexes, [Ni(CN)�]��, [Fe(CN)

�]��

(3) Neutral sphere complexes, [Pt(NH�)�Cl

�], [Ni(CO)

�]

7. Effective atomic number : EANZX�Y

8. Isomerism in complexes : (1) Stereoisomerism : (i) Geometrical isomerism

(ii) Optical isomerism

(2) Structural isomerism : (i) Ionisation isomerism

(ii) Linkage isomerism

(iii) Coordination isomerism

(iv) Solvate isomerism

9. Bonding in complexes : IIIIIII

� (i) Valence bond theory (VBT)

� (ii) Crystal field theory (CFT)

� (iii) Ligand field theory (LFT)

� (iv) Molecular orbital theory (MOT)

10. Valence bond theory (VBT) : A central metal atom or ion undergoes hybridisation like sp�, dsp�, d�sp�, etc.

forming hybridised orbitals to accommodate the lone pairs of electrons from the ligands.

11. (1) Inner complex : (n1) d orbitals of metal are used

(2) Outer complex : nd orbitals of metal are used.

12. Crystal field theory (CFT) :

Degenerate d-orbitals : dxy , dyz , dzx , d(x�y�) and dz�

NAVNEET CHEMISTRY DIGEST : STANDARD XII (PART II)6

13. Crystal field splitting :

(1) For octahedral complex :

(2) For tetrahedral complex :

14. Factors affecting stability of the complex : (1) Charge to size ratio of the metal ion

(2) Nature of ligands

15. Applications of the coordination compounds : ––

� (1) In biology

� (2) In medicines

� (3) To estimate hardness of water

� (4) In electroplating

(★ ) Indicates question from the textbook.

Units

9.1 Introduction

9.2 Types of ligands

9.2.1 Monodentate ligands

9.2.2 Polydentate ligands

9.2.3 Ambidentate ligand

Q. 1. What are double salts ? (1 mark)

Ans. Double salts are crystalline molecular or addition compounds containing more than one salt in simple molecular

proportions soluble in water and in solution they ionise and exhibit all the properties of the constituent ions.

For example, K�SO

�·Al

�(SO

�)�

·24H�O

K�SO

�·Al

�(SO

�)�

·24H�O(aq) IIIIIJ 2K�(aq)�2Al��(aq) �4SO2

4(aq)�24H�O(l)

■

Q. 2. Define coordination compound. (1 mark)

Ans. Coordination compound : It consists of a central metal ion or atom surrounded by atoms, molecules or anions

called ligands by coordinate bonds, e.g. cisplatin Pt(NH�)�Cl

�, [Cu(NH

�)�

]SO�. ■


Q. 3. Define Lewis bases and Lewis acids with respect to a coordination compound. (2 marks)

Ans. Lewis bases : In a coordination compound the ligands being electron pair donors they are Lewis bases.

Lewis acids : The central metal atom or ion being electron acceptor behaves as a Lewis acid.

For example, in the coordination compound, [Cu(NH�)�

]��, NH�

is a Lewis base and Cu�� is a Lewis acid. ■

Q. 4. Write the formula of anticancer drug cisplatin. (1 mark)

Ans. Cisplatin : or Pt(NH�)�Cl

�.

■

★Q. 5. Name the Lewis acids and bases in the complex [Pt Cl�

(NH�)�]. (1 mark)

Ans. Lewis acid : Pt2�

Lewis bases : Cl� and NH�

■

★Q. 6. What are ligands ? What are their types ? Give one example of each type. (4 marks)

Ans. Ligands : The neutral molecules or negatively charged anions (or rarely positive ions) which are bonded by

coordinate bonds to the central metal atom or metal ion in a coordination compound are called ligands or donor

groups. For example in [Cu(CN)�

]��, four CN� ions are ligands coordinated to central metal ion Cu��.

Ligands can be classified on the basis of number of electron donor atoms in the ligand i.e. denticity.

(1) Monodentate or unidentate ligand : A ligand molecule or an ion which has only one donor atom with a lone pair

of electrons forming only one coordinate bond with metal atom or ion in the complex is called monodentate or

unidentate ligand. For example NH�, Cl�, OH�, H

�O, etc.

(2) Polydentate or multidentate ligand : A ligand molecule or an ion which has two or more donor atoms with the

lone pairs of electrons forming two or more coordinate bonds with the central metal atom or ion in the complex is

called polydentate or multidentate ligand. For example, ethylene diamine, H�N–(CH

�)�

–NH�.

According to the number of donor atoms they are classified as follows :

(i) Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine,

H�N–(CH

�)�

–NH�.

(ii) Tridentate ligand : This ligand molecule has three donor atoms or three sites of attachment.

E.g. Diethelene triamine, H�

. .N–CH

�–CH

�–

. .NH–CH

�–CH

�–

. .NH

�. This has three N donor atoms.

(iii) Tetradentate (or quadridentate) ligand : This ligand molecule

has four donor atoms.

Eg. Triethylene tetraamine which has four N donor atoms.

(iv) Hexdentate ligand : This ligand molecule has six donor atoms. E.g. Ethylenediamine tetracetato.

(3) Ambidentate ligand : A ligand molecule or an ion which has two or more donor atoms, however in the formation

of a complex, only one donor atom is attached to the metal atom or an ion is called ambidentate ligand. For

example, NO��

which has two donor atoms N and O forming a coordinate bond, M � ONO (nitrito) or M � NO�

(nitro).


(4) Bridging ligand : A monodentate ligand having more than one lone pairs of electrons, hence can attach to two or

more metal atoms or ions and hence acts as a bridge between different metal atoms is called bridging ligand.

For example : OH�, F�, SO��

, etc. ■

★Q. 7. What are bidentate ligands ? Give one example. (1 mark)

Ans. For answer, refer to Q. 6 (2) (i). ■

Use your brain power !

(Textbook page 192)

Q. Draw Lewis structures of the following ligands and identify the donor atom in them :

NH�, H

�O.

Ans.Ligand Lewis dot structure Donor atom

NH�

N

H�O O

Units

9.3 Terms used in coordination chemistry

9.3.1 Coordination sphere

9.3.2 Charge number of complex ion and oxidation state of metal ion

9.3.3 Coordination number (C.N.) of central metal ion

9.3.4 Double salt and coordination complex

9.3.5 Werner theory of coordination complexes

Q. 8. Define coordination sphere. Give example. (2 marks)

Ans. Coordination sphere : A coordination entity consisting of a central metal atom or ion and the coordinating

groups like neutral molecules or anions (ligands) written inside a square bracket is together called coordination

sphere. This is a discrete structural unit. The ionisable groups (generally ions) called counter ions are written

outside the bracket.

For example, in the coordination compound K�[Fe(CN)

�], the coordination sphere is [Fe(CN)

�]�� while K�

represents counter ion. ■

Try this …

(Textbook page 193)

Q. Can you write ionisation of [Ni (NH�)�] Cl

�?

Ans. [Ni (NH�)�] Cl

�IIIIIJ [Ni (NH

�)�]2�

�2Cl

Q. Identify coordination sphere and counter ions.

Ans. Coordination sphere : [Ni (NH�)�]��

Counter ions : Cl�


Q. 9. Define and explain charge number of a complex ion. (2 marks)

Ans. Charge number of a complex ion : The net charge carried by a complex ion or a coordination entity is called

its charge number.

Explanation :

(i) Charge number is equal to the algebraic sum of the charges carried by central metal atom or ion and all the ligands

attached to it.

(ii) E.g. consider anionic complex, [Fe(CN)�

]��.

Charge number of [Fe(CN)�

]��Charge on Fe�� ions�6�charge on CN�(�2)�6(1)4

Hence charge number of [Fe(CN)�

]�� is 4. ■

Q. 10. Explain the oxidation state of a metal in a complex. (2 marks)

Ans.

(i) The oxidation state of a metal atom or ion in the complex is the apparent charge carried by it in the complex.

(ii) It depends upon the atomic number and electronic configuration of the metal atom or ion.

(iii) The coordination number, the formula and geometry of a complex depend upon the oxidation state of the metal

atom or ion. ■

Q. 11. What is the charge on a monodentate ligand X in the complex, [NiX�

]�� ? (2 marks)

Ans. The charge number of the complex ion is 2. Nickel being divalent, its oxidation state is �2. If the charge on

monodentate ligand X is y, then

Charge numbercharge on Ni��charge on 6X

2�2�4�y

� y1

Hence the charge on ligand X is 1. ■

Q. 12. Calculate the oxidation state of a metal in the following complexes :

(a) [Fe(NH�)�](NO

�)�

(b) Ni(CO)�. (2 marks each)

Ans.

(a) [Fe(NH�)�](NO

�)�

� [Fe(NH�)�

]�� 3NO��

NH�

is a neutral ligand, and the charge number of complex ion is �3.

If the oxidation state of Fe is x then,

�3x�6(0)

� x�3

� The oxidation state of Fe is �3.

(b) Ni(CO)�

is a neutral complex and CO is a neutral ligand. If the oxidation state of Ni is x, then

zerox�5�(zero)

� xzero.

The oxidation state of Ni is zero.■

Can you tell ?

(Textbook page 193)

Q. A complex is made of Co (III) and consists of four NH�

molecules and two Cl� ions as ligands. What is the

charge number and formula of complex ion ?

Ans. The complex ion has formula, [Co(NH�)�Cl

�]�.

The charge number is �1.


Q. 13. Define and explain the term coordination number (C.N.) of a metal in the complex. (2 marks)

Ans. Coordination number or legancy (C.N.) : The number of (monodentate) ligands which are directly bonded by

coordinate bonds to central metal atom or ion in a coordination compound is called coordination number (C.N.) of

the metal atom or ion.

Explanation :

(i) The coordination number (C.N.) is a characteristic property of the metal and its electronic configuration.

(ii) C.N. takes the values from 2 to 10, of which 4 and 6 are very common.

(iii) The light transition metals show C.N. 4 and 6 while the heavier transition metals show C.N. 8.

(iv) The geometry and shape of a complex compound depends upon C.N. of the metal.■

Q. 14. Mention primary valence, secondary valence and coordination number in the following complexes :

(a) [Cu(NH�)�

]Cl�

(b) [Co(NH�)�Cl

�] (c) K

�[Fe(CN)

�] (d) [CoF

�]��

(e) [Pt(NH�)�Cl

�] (f) [Pt(NH

�)�(Py)�

Cl�] (g) Cr(CO)

�(h) [Ni(CN)

�]�� (2 marks each)

Ans.

ComplexPrimary valence /

Valency of metalSecondary valence Coordination number

(a) [Cu(NH�)�

]Cl�

2 4 4

(b) [Co(NH�)�Cl

�] 3 6 6

(c) K�[Fe(CN)

�] 2 6 6

(d) [CoF�]�� 3 6 6

(e) [Pt(NH�)�Cl

�] 2 4 4

(f) [Pt(NH�)�(Py)�

Cl�] 2 6 6

(g) Cr(CO)�

0 6 6

(h) [Ni(CN)�

]�� 2 4 4

■

Use your brain power !(Textbook page 193)

Q. Coordination number used in coordination of compounds is somewhat different than that used in solid

state. Explain.

Ans.

● In a coordination compound the coordination number is the number of donor atoms of ligands directly attached

to metal atom or ion.

● In a solid state, the number of closest constituent atoms or ions in contact with a particular atom in the crystal

lattice is called coordination number.

● In a coordination compound, coordination number depends upon nature of metal atom or ion, and its electronic

configuration.

● In a solid state, the coordination number depends upon the crystalline structure of the unit cell.


★ Q. 15. What is the coordination number and oxidation state of metal ion in the complex [Pt (NH�) Cl

�]� ?

Ans. Coordination number6

Oxidation state of Pt�4. ■

Can you tell ?

(Textbook page 194)

Q. What is the coordination number of (a) Co in [CoCl�(en)

�]�, (b) Ir in [Ir(C

�O

�)�Cl

�]�� and (c) Pt in

[Pt(NO�)�(NH

�)�] ?

Ans. (a) Coordination number of Co in [CoCl�(en)

�]�6

(b) Coordination number of Ir in [Ir (C�O

�)�Cl

�]��6

(c) Coordination number of Pt in [Pt (NO�)�(NH

�)�

]4

★ Q. 16. What is the difference between a double salt and a complex ? Give an example. (2 marks) OR

Illustrate with example, the difference between a double salt and a coordination compound (complex).

(2 marks)

Ans. Double salt Coordination compound (complex)

(1) Double salts exist only in the solid state and

dissociate into their constituent ions in the aqueous

solutions.

(1) Coordination compounds exist in the solid state as

well as in the aqueous or non-aqueous solutions.

(2) Double salts lose their identity in the solution. (2) They do not lose their identity completely.

(3) The properties of double salts are same as those of

their constituents.

(3) The properties of coordination compounds are

different from their constituents.

(4) Metal ions in the double salts show their normal

valence.

(4) Metal ions in the coordination compounds show

two valences namely primary valence and second-

ary valence satisfied by anions or neutral molecules

called ligands.

(5) For example in K�SO

�. K

�SO

�. Al

�(SO

�)�. 24H

�O.

The ions K�, Al�� and SO��

show their properties.

(5) In K�[Fe(CN)

�], ions K� and [Fe(CN)

�]�� ions

show their properties.

■

Remember …

(Textbook page 194)

When a complex is dissolved and a solution is formed, it does not dissociate into simple metal ions. When

[Co(NH�)�]Cl

�, is dissolved in water it does not give the test for Co�� or NH

�. However, on reacting with AgNO

�a curdy white precipitate of AgCl corresponding to 3 moles is observed.


Q. 17. What are the postulates of Werner theory ? (4 marks)

Ans. The following are the postulates of Werner theory :

Postulate (1) Unlike metal salts, the metal in a complex possesses two types of valencies : primary (ionizable)

valency and secondary (nonionizable) valency.

Postulate (2) The ionizable sphere consists of entities which satisfy the primary valency of the metal. Primary

valencies are generally satisfied by anions.

Postulate (3) The secondary coordination sphere consists of entities which satisfy the secondary valencies and are

nonionizable.

The secondary valencies for a metal ion are fixed and satisfied by either anions or neutral ligands. Number of

secondary valencies is equal to the coordination number.

Postulate (4) The secondary valencies have a fixed spatial arrangement around the metal ion. Two spheres of

attraction in the complex [Co(NH�)�]Cl

�are shown.

■

(Textbook page 194)Question 9.1 :

Q. One mole of a purple coloured complex of CoCl�

and NH�

on treatment with excess AgNO�

produces two

moles AgCl. Write the formula of the complex if the coordination number of Co is 6.

Ans. : One mole of the complex gives 2 moles of AgCl. It indicates that two Cl� ions react with Ag� ions. The

complex has two ionisable Cl� ions. The formula of the complex is then [Co(NH�)�

Cl]Cl�.

Can you tell ?(Textbook page 194)

Q. One mole of a green coloured complex of CoCl�

and NH�

on treatment with excess of AgNO�

produces 1

mole of AgCl. What is the formula of the complex? (Given : C.N. of Co is 6)

Ans. : Since 1 mol of complex gives 1 mol of AgCl, it indicates the complex on dissociation gives 1 mol Cl�.

Hence complex has one mol ionisable Cl� ions. Since C.N. of Co is 6, the formula of the complex is

[Co(NH�)�

Cl�

]Cl.

Units

9.4 Classification of complexes

9.4.1 Classification on the basis of types of ligands

9.4.2 Classification on the basis of charge on the complex

Q. 18. Define the following terms. (2 marks)

(1) Homoleptic complexes : Consider [Co (NH�)�]��. Here only one type of ligands surrounds the Co�� ion. The

complexes in which metal ion is bound to only one type of ligands are homoleptic.

(2) Heteroleptic complexes : Look at the complex [Co (NH�)�

Cl�]�. There are two types of ligands, NH

�and Cl

attached to Co�� ion. Such complexes in which metal ion is surrounded by more than one type of ligands are

heteroleptic. ■



(Textbook page 195)

Q. Classify the complexes as homoleptic and heteroleptic : (a) [Co (NH�)�Cl]SO

�, (b) [Co (ONO)(NH

�)�]Cl

�,

(c) [CoCl (NH�)(en)

�]�� and (d) [Cu (C

�O

�)�]��.

Ans. Homoleptic Complexes : (d) [Cu (C�O

�)�]��

Heteroleptic Complexes : (a) [Co (NH�)�Cl]SO

�(b) [Co (ONO)(NH

�)�]Cl

�, (c) [CoCl (NH

�)(en)

�]��

★Q. 19. What are cationic, anionic and neutral complexes ? Give one example of each. (3 marks)

Ans.

(1) Cationic sphere complexes : A positively charged coordination sphere or a coordination compound having a

positively charged coordination sphere is called cationic sphere complex.

For example : [Zn(NH�)�]�� and [Co(NH

�)�Cl]SO

�are cationic complexes. The latter has coordination sphere

[Co(NH�)�Cl]��, the anion SO

�� makes it electrically neutral.

(2) Anionic sphere complexes : A negatively charged coordination sphere or a coordination compound having

negatively charged coordination sphere is called anionic sphere complex. For example, [Ni(CN)�]�� and

K�[Fe(CN)

�] have anionic coordination sphere; [Fe(CN)

�]�� and three K� ions make the latter electrically

neutral.

(3) Neutral sphere complexes : A neutral coordination complex does not possess cationic or anionic sphere.

[Pt(NH�)�Cl

�] or [Ni(CO)

�] are neither cation nor anion but are neutral sphere complexes. ■


Q. Classify the complexes as cationic, anionic or neutral : Na�[Fe(CN)

�], Co(NH

�)�Cl

�,

Cr(H�O)

�(C

�O

�)��, PtCl

�(en)

�and Cr(CO)

�.

Ans. Cationic complexes : [Co(NH�)�]Cl

�Anionic complexes : Na

�[Fe(CN)

�], [Cr(H

�O)

�(C

�O

�)�]��

Neutral complexes : Cr(CO)�, Pt Cl

�(en)

�

★Q. 20. Is the complex [CoF�] cationic or anionic if the oxidation state of cobalt is �3 ? (1 mark)

Ans. In the complex, Co carries �3 charge while 6F� carry 6 charge. Hence the net charge on the complex is 3.

Therefore it is an anionic complex. ■

★Q. 21. Classify following complexes as homoleptic and heteroleptic complex :

(a) [Cu(NH�)�]SO

�; (b) [Cu(en)

�(H

�O)Cl]�� (c) [Fe(H

�O)

�(NCS)]�� (d) Tetraaminezinc(II) nitrate.

( 12 mark each)

Ans. Homoleptic complex :

(a) [Cu(NH�)�]SO

�(d) Tetraaminezinc(II) nitrate : [Zn(NH

�)�](NO

�)�

Heteroleptic Complex :

(b) [Cu(en)�(H

�O)Cl]�� (c) [Fe(H

�O)

�(NCS)]�� ■


Unit

9.5 IUPAC nomenclature of coordination compounds

Q. 22. Summarise the rules of IUPAC nomenclature of coordination compounds. (3 marks)

Ans. Following rules are followed for naming coordination compounds recommended by IUPAC :

(1) In case of a complex ion or a neutral molecule, name the ligand first and then the metal.

(2) The names of anionic ligands are obtained by changing the ending -ide to -o and -ate to -ato.

(3) The name of a complex is one single word. There must not be any space between different ligand names as well as

between ligand name and the name of the metal.

(4) After the name of the metal, write its oxidation state in Roman number which appears in parentheses without any

space between metal name and parentheses.

(5) If complex has more than one ligand of the same type, the number is indicated with prefixes, di-, tri-, tetra-, penta-,

hexa- and so on.

(6) For the complex having more than one type of ligands, they are written in an alphabetical order. Suppose two

ligands with prefixes are tetraaqua and dichloro. While naming in alphabetical order, tetraaqua is written first and

then dichloro.

(7) If the ligand itself contains numerical prefix in its name, then display number by prefixes bis for 2, tris for 3, tetrakis

for 4 and so forth. Put the ligand name in parentheses. For example, (ethylenediamine)�

or (en)�

would appear as

tris (ethylenediamine) or tris(ethane-1, 2-diamine).

(8) The metal in cationic or neutral complex is specified by its usual name while in the anionic complex the name of

metal ends with ‘ate’. ■

Try this ...(Textbook page 197)

Q. Write the representation of the following :

(i) Tricarbonatocobaltate(III) ion.

(ii) Sodium hexacyanoferrate(III).

(iii) Potassium hexacyanoferrate(II).

(iv) Aquachlorobis(ethylenediamine)cobalt(III).

(v) Tetraaquadichlorochromium(III) chloride.

(vi) Diamminedichloroplatinum(II).

Ans. (i) [Co(CO�)�]��

(ii) Na�[Fe(CN)

�]

(iii) K�[Fe(CN)

�]

(iv) Co(en)�(H

�O)(Cl)

(v) [Cr(H�O)

�Cl

�]Cl

(vi) Pt(NH�)�Cl

�


★Q. 23. Write the formula for tetraamineplatinum(II) chloride. (1 mark)

Ans. Formula of tetraamineplatinum(II) chloride : [Pt(NH�)�]Cl

�Table 9.1 : IUPAC names of anionic and neutral ligands

Anionic ligand IUPAC name Anionic ligand IUPAC name

Br�, Bromide Bromo CO��, Carbonate Carbonato

Cl�, Chloride Chloro OH�, Hydroxide Hydroxo

F�, Fluoride Fluoro C�O

��, Oxalate Oxalato

I� Iodide Iodo NO��, Nitrite Nitro (For N-bonded ligand)

CN�, Cyanide Cyano ONO�, Nitrite Nitrito (For O-bonded ligand)

SO��, Sulphate Sulphato SCN�, Thiocyanate Thiocyanato (For ligand

donor atom S)

NO��, Nitro Nitrato NCS�, Thiocyanate Isothiocyanato (For ligand

donor atom N)

Neutral ligand IUPAC name Neutral ligand IUPAC names

NH�, Ammonia Ammine (Note the spelling) H

�O, water Aqua

CO, Carbon monoxide Cabonyl en, Ethylene diamine Ethylenediamine

Table 9.2 : IUPAC names of anionic complexes

Metal Name

Al Aluminate

Cr Chromate

Cu Cuprate

Co Cobaltate

Au(Gold) Aurate

Metal Name

Fe Ferrate

Pb Plumbate

Mn Manganate

Mo Molybdate

Ni Nickelate

Metal Name

Zn Zincate

Ag Argentate

Sn Stannate

Table 9.3 : IUPAC names of some complexes

Complex IUPAC name

(i) Anionic complexes :

(a) [Ni(CN)�]��

(b) [Co(C�O

�)�]��

(c) [Fe(CN)�]��

Tetracyanonickelate(II) ion

Trioxalatocobaltate(III) ion

Hexacyanoferrate(II) ion

(ii) Compounds containing complex anions and metal cations :

(a) Na�[Co(NO

�)�]

(b) K�[Al(C

�O

�)�]

(c) Na�[AIF

�]

Sodium hexanitrocobaltate(III)

Potassium trioxalatoaluminate(III)

Sodium hexafluoroaluminate(III)


Complex IUPAC name

(iii) Cationic complexes :

(a) [Cu(NH�)�]��

(b) [Fe(H�O)

�(NCS)]��

(c) [Pt(en)�(SCN)

�]��

Tetraamminecopper(II) ion

Pentaaquaisothiocyanatoiron(III) ion

Bis(ethylenediamine)dithiocyanatoplatinum(IV)

(iv) Compounds containing complex cation and anion :

(a) [PtBr�(NH

�)�]Br

�(b) [Co(NH

�)�CO

�]Cl

(c) [Co(H�O)(NH

�)�]I

�

Tetraamminedibromoplatinum(IV) bromide,

Pentaamminecarbonatocobalt(III) chloride,

Pentaammineaquacobalt(III) iodide

(v) Neutral complexes :

(a) Co(NO�)�(NH

�)�

(b) Fe(CO)�

(c) Rh(NH�)�(SCN)

�

Triamminetrinitrocobalt(III)

Pentacarbonyliron(0)

Triamminetrithiocyanatorhodium(III)

★Q. 24. Write formula of the following complexes :

(a) Potassium amminetrichloroplatinate(II)

(b) Dicyanoaurate(I) ion. (2 marks)

Ans. (a) Potassium amminetrichloroplatinate(II) (b) Dicyanoaurate (I) ion

K[Pt(NH�)Cl

�] [Au(CN)

�]� ■

Unit

9.6 Effective Atomic Number (EAN) Rule

Q. 25. State effective atomic number (EAN). (1 mark) OR

State and explain effective atomic number (EAN). How is it calculated ? (2 marks)

Ans. Effective atomic number (EAN) : It is the total number of electrons present around the central metal atom or ion

and calculated as the sum of electrons of metal atom or ion and the number of electrons donated by ligands.

It is calculated by the formula : EANZX�Y

where, ZAtomic number of metal atom

XNumber of electrons lost by a metal atom forming a metal ion

YTotal number of electrons donated by all ligands in the complex.

Generally the value of EAN is equal to the atomic number of the nearest inert element.

Explanation : Consider a complex ion [Co(NH�)�]��

Oxidation state of cobalt is �3 hence X3.

There are six ligands, hence Y2�612

Atomic number of cobalt, Z27

� EANZX�Y273�1236.■


Try this …

(Textbook page 197)

Q. Find out the EAN of (a) [Zn(NH�)�

]��

(b) [Fe(CN)�]��

Ans. (a) For the complex ion, [Zn(NH�)�]�� :

Atomic number of ZnZ30

Charge on metal ion�2

� Number of electrons lost by Zn atomX2

Total number of electrons donated by 4NH�

ligandsY2�48

EANZX�Y

302�8

36

(Note : This is atomic number of the nearest

inert element��

Kr.)

(b) For the complex ion, [Fe(CN)�]�� :

For Fe, Z26 (Atomic number)

X2 (Due to �2 charge on Fe)

Y12 (Due to 6 CN� ligands)

� EANZX�Y

262�12

36


(Textbook page 197)

Q. Do the following complexes follow the EAN rule (a) Cr(CO)�, (b) Ni(CO)

�, (c) Mn(CO)

�,

(d) Fe(CO)�

?

Ans. (a) Cr(CO)�

: EANZX�Y

240�8

32

(b) Ni(CO)�

: EANZX�Y

280�8

36

(c) Mn(CO)�

: EANZX�Y

250�10

35

(d) Fe(CO)�

: EANZX�Y

260�10

36

Conclusion : (a) Cr(CO)�

and (c) Mn(CO)�

do not follow EAN Rule.

Q. 26. Find effective atomic number (EAN) in the following complexes :

(1) [Ni(CO)�] (2) [Fe(CN)

�]�� (3) [Co(NH

�)�]�� (4) [Zn(NH

�)�]�� (5) [Pt(NH

�)�]�� (2 marks each)

Atomic number Number of Number of

Ans. Complex Metal of metal electrons lost electrons from EANZX�Y

by metal ligands

Z X Y

(1) [Ni(CO)�] Ni 28 0 4�28 280�836 (Kr)

(2) [Fe(CN)�]�� Fe 26 2 6�212 262�1236 (Kr)

(3) [Co(NH�)�]�� Co 27 3 6�212 273�1236 (Kr)

(4) [Zn(NH�)�]�� Zn 30 2 4�28 302�836 (Kr)

(5) [Pt(NH�)�]�� Pt 78 4 6�212 784�1286 (Rn)

■


Q. 27. What is effective atomic number (EAN) in the following complexes ?

(1) [Fe(CN)�]�� (2) [Cu(NH

�)�]�� (3) [Pt(NH

�)�]��. (2 marks each)

Atomic number Number of Number of

Ans. Complex Metal of metal electrons lost electrons from EANZX�Y

by metal ligands

Z X Y

(1) [Fe(CN)�]�� Fe 26 3 6�212 263�1235

(2) [Cu(NH�)�]�� Cu 29 2 4�28 292�835

(3) [Pt(NH�)�]�� Pt 78 2 4�28 782�884

■

Q. 28. Calculate EAN in the following complexes :

(1) [Cr(H�O)

�(NH

�)�(en)]Cl

�; (2) [Ni(en)

�]SO

�; (3) Na

�[Cr(C

�O

�)�]. (2 marks each)

Atomic number Number of Number ofAns. Complex Metal of metal electrons lost electrons from EANZX�Y

by metal ligandsZ X Y

(1) [Cr(H�O)

�(NH

�)�(en)]Cl

�Cr 24 3

2H�O4

2NH�4

en4 � 12 243�1233

(2) [Ni(en)�]SO

�Ni 28 2 2en8 282�834

(3) Na�[Cr(C

�O

�)�] Cr 24 3 3C

�O��

�12 243�1233

■

Units

9.7 Isomerism in coordination compounds

9.7.1 Stereoisomers

9.7.2 Structural isomers

Q. 29. Define in coordination compounds : (1) Isomerism (2) Isomers. (2 marks)

Ans.

(1) Isomerism : It is the phenomenon in coordination compounds having same molecular formula but different

physical and chemical properties due to different arrangements of the ligands around the central metal atom or ion

in the space.

(2) Isomers : The isomers are the coordination compounds having same molecular formula but different physical and

chemical properties due to the difference in arrangements of the ligands in the space. ■

Q. 30. Mention the types of isomerisms in coordination compounds. (2 marks)

Ans. There are two principal types of isomerisms in coordination compounds as follows :

(A) Stereoisomerism (B) Structural isomerism (OR Constitutional isomerism)

(A) Stereoisomerism is further classified as :

(i) Geometrical isomerism

(ii) Optical isomerism


(B) Structural isomerism is further classified as :

(i) Ionisation isomerism

(ii) Linkage isomerism

(iii) Coordination isomerism

(iv) Solvate (or hydrate) isomerism ■

Q. 31. Why does stereoisomerism arise in the coordination compounds ? (2 marks)

Ans. In the coordination compounds (complexes) the ligands are linked to the central metal atom or ion by coordinate

bonds which are directional in nature and hence give rise to the phenomenon of stereoisomerism.

In this isomerism, the different stereoisomers have different arrangements of ligands (atoms, molecules or ions) in

space around the central metal atom or ion. Hence they have different physical and chemical properties and give

rise to the phenomenon of stereoisomerism. ■

Q. 32. Define, in coordination compounds : (1) Stereoisomerism (2) Stereoisomers. (1 mark each)

Ans.

(1) Stereoisomerism : The phenomenon of isomerism in the coordination compounds arising due to different spatial

positions of the ligands in the space around the central metal atom or ion is called stereoisomerism.

(2) Stereoisomers : The coordination compounds having same molecular formula but different stereoisomerism

due to different spatial arrangements of the ligand groups in the space around the central metal atom or ion are

called stereoisomers. ■

Q. 33. Define : (1) Geometrical isomerism and (2) Geometrical isomers. (1 mark each)

Ans.

(1) Geometrical isomerism : The phenomenon of isomerism in the heteroleptic coordination compounds with the

same molecular formula but different spatial arrangement of the ligands in the space around the central metal atom

or ion is called geometrical isomerism.

(2) Geometrical isomers : The heteroleptic coordination compounds having same molecular formula but different

geometrical isomerism due to different spatial arrangements of the ligands in the space around the central metal

atom or ion are called geometrical isomers. ■

Q. 34. Define cis and trans isomers in the coordination compounds. (2 marks)

Ans.

(1) Cis-isomer : A heteroleptic coordination compound in which two similar

ligands are arranged adjacent to each other is called cis-isomer. For

example,

Cis-Diamminedichloroplatinum(II)

(2) Trans-isomer : A heteroleptic coordination compound in which two

similar ligands are arranged diagonally opposite to each other is called

trans-isomer. For example,

Trans-Diamminedichloroplatinum(II)■


Q. 35. Write structures for geometrical isomers of Diamminebromochloroplatinum(II). (1 mark)

Ans.

Cis-Diamminebromochloroplatinum(II) Trans-Diamminebromochloroplatinum(II) ■

Q. 36. Explain the geometrical isomerism of the octahedral complex of the type [MA�B�] with suitable example.

(2 marks)Ans.

(1) Consider an octahedral complex of a metal M with coordination number six and monodentate ligands a and b

having formula [MA�B

�].

(2) Cis-isomer is obtained when both the B ligands occupy adjacent (1, 2) positions.

(3) Trans-isomer is obtained when the ligands B occupy the opposite (1, 6) positions.

(4) For example, consider a complex [Co(NH�)�Cl

�]�. The structures of cis and trans isomers are

(a)

(b)

Fig. 9.1 (a) and (b) : Cis and trans-isomers of [CoCl�(NH

�)�]� ■


● Draw structures of cis and trans isomers of [Fe(NH�)�(CN)

�]


Q. 37. Explain the geometrical isomerism of the octahedral complex of the type [M(AA)�B

�]n� with a suitable

example. (3 marks)

Ans.

(1) Consider an octahedral complex of metal M with coordination number six and a bidentate ligand AA and

monodentate ligand B having molecular formula [M(AA)�B

�]n� .

(2) Bidentate ligand AA has two identical coordinating atoms.

(3) Cis- isomer is obtained when two bidentate AA ligands as well as two ‘B’ ligands are at adjacent positions.

(4) Trans-isomer is obtained when two AA ligands and two B ligands are at opposite positions.

(5) For example, consider a complex [Co(en)�Cl

�]�.

(a)

(b)

Fig. 9.2 (a) and (b) : cis and trans-isomers of [CoCl�(en)

�]� ■

Q. 38. Explain the geometrical isomerism of the octahedral complex of the type [MA�BC] with suitable example.

(3 marks)

Ans.

(1) Consider an octahedral complex of metal M with coordination number six and monodentate ligands A, B

and C.

(2) Cis-isomer is obtained when both the ligands B and C occupy adjacent (1, 2) positions.

(3) Trans-isomer is obtained when the ligands B and C occupy opposite positions.

(4) For example, consider a complex [Pt(NH�)�BrCl] of the type [MA

�BC].


(a)

(b)

■

Q. 39. Define : (1) Optical isomerism (2) Optical isomers. (1 mark each)

Ans.

(1) Optical isomerism : The phenomenon of isomerism in which different coordination compounds having same

molecular formula have different optical activity is called optical isomerism.

(2) Optical isomers : Different coordination compounds having same molecular formula but different optical activity

are called optical isomers. ■

Q. 40. Explain : (1) Plane polarised light (2) Optical activity. (1 mark each)

Ans.

(1) Plane polarised light : A monochromatic light having vibrations only in one plane is called a plane polarised

light. This light is obtained by passing monochromatic light through NICOL prism.

(2) Optical activity : A phenomenon of rotating a plane of a plane polarised light by an optically active substance is

called optical activity. This substance is said to be optically active. ■

Q. 41. Explain : (1) Dextrorotatory substance (2) Laevorotatory substance. (1 mark each)

Ans.

(1) Dextrorotatory substance : An optically active substance which rotates the plane of a plane polarised light to

right hand side is called dextrorotatory or d isomer denoted by d.

(2) Laevorotatory substance : An optically active substance which rotates the plane of a plane polarised light to the

left hand side is called laevorotatory or l isomer and denoted by l .■

Q. 42. What are the conditions for the optical isomerism in coordination compounds ? (2 marks)

Ans.

(1) Optical isomerism is exhibited by those coordination compounds which possess chirality.

(2) There should not be the presence of element of symmetry which makes the complex optically inactive.

(3) The mirror images of the complex molecule or ion must be non-superimposable with the molecule or ion. ■


Q. 43. What are enantiomers ? (1 mark)

Ans. Enantiomers : The two forms of the optical active complex molecule which are mirror images of each other

are called enantiomers.

There are two forms of enantiomers, d form and l form. ■

Remember ...

(Textbook page 199)

Our hands are non superimposable mirror images. When you hold your left hand up to a mirror the image

looks like right hand.

Q. 44. Draw diagrams for the optical isomers of a complex, [Co(en)�

]��. (2 marks)

Ans. The complex [Co(en)�

]�� has two optical isomers.

Fig. 9.3 : Optically active forms of [Co(en)�

]�� ■

Q. 45. Explain the optical isomerism in the octahedral complex with two symmetrical bidentate chelating

ligands. (2 marks)

Ans. The octahedral complexes of the type [M(AA)�Q

�]n�, in which two symmetrical bidentate chelating ligands

like AA and two monodentate ligands like a are coordinated to the central metal atom or ion exhibit optical

isomerism and two optical isomers d and l can be resolved. For example, [Pt Cl�(en)

�]�.

The cis-form is unsymmetrical and optically active while trans-form is symmetrical and hence optically inactive.

The optical isomers of cis-form (d and l ) of this complex along with trans-form are shown below,

Fig 9.4 : Optically active (cis) and optically inactive (trans) forms of the complex [CoCl�(en)

�]� ■


Try this...(Textbook page 199)

(1) Draw enantiomers of [Cr(OX)�]�� where OXC

�O��

�:

(2) Draw (A) enantiomers and (B) cis and trans isomers of [Cr(H�O)

�(OX)

�]� :

(A) Enantiomers :

(B) Cis and trans isomers :


Q. 46. When are optical isomers called chiral ? (1 mark)

Ans. When the mirror images of optical isomers of the complex are nonsuperimposable they are said to be chiral.

For example, [Co(en)�(NH

�)�

]��. ■

★Q. 47. Predict whether the [Cr(en)�(H

�O)

�]�� complex is chiral. Write structure of its enantiomers.

(2 marks)

Ans. (i) Complex is chiral.

(ii) The following are its enantiomers

■

★Q. 48. Draw isomers in each of the following :

(a) [Pt(NH�)�ClNO

�]

(b) [Ru(NH�)�Cl

�]

(c) [Cr(en�)Br

�]� (1 mark each)

Ans.

(a) [Pt(NH�)�ClNO

�]

Geometric isomers :

(b) [Ru(NH�)�Cl

�]

Geometric isomers :


(c) [Cr(en�)Br

�]�

Optical isomers :

Geometric isomers :

■

★Q. 49. Draw geometric isomers and enantiomers of the following complexes.

(a) [Pt(en)�]�� (b) [Pt(en)

�ClBr]�� (2 marks)

Ans. The complex [Pt(en)�]�� has two optical isomers.


Optical isomers of [Pt(en)�ClBr]��

■

Q. 50. Define and explain ionisation isomerism. (2 marks)

Ans. Ionisation isomerism : The phenomenon of isomerism in the metal complexes in which there is an exchange of

ions between coordination (or inner) sphere and outer sphere is known as ionisation isomerism.

Explanation : (i) Ionisation isomers have same molecular formula but different arrangement of ions in the inner

sphere and outer sphere in the complex. (ii) Hence on ionisation, these ionisation isomers produce different ions

in the solution. This ionisation isomerism is also called ion-ion exchange isomerism.

Examples : (A) [Co(NH�)�Cl

�] Br and (B) [Co(NH

�)�ClBr] Cl

Ionisation :

(A) [Co(NH�)�Cl

�] Br � [Co(NH

�)�Cl

�]� � Br�

(B) [Co(NH�)�ClBr]Cl � [Co(NH

�)�ClBr]��Cl�

In the isomers (A) and (B), there is an exchange of ions namely Br� and Cl�.■

★Q. 51. What are ionisation isomers ? Give an example. (2 marks)

Ans. Ionisation isomers : The coordination compounds having same molecular composition but differ in the

compositions of coordination (or inner) sphere and outer sphere and produce different ions on ionisation in the

solution are called ionisation isomers. For example, Pentaamminesulphatocobalt (III) bromide [Co(NH�)�SO

�] Br,

Pentaamminebromocobalt(III) sulphate [Co(NH�)�Br] SO

�. ■

Q. 52. Define : (1) Linkage isomerism (2) Linkage isomers. OR

What is linkage isomerism ? Explain with an example. (2 marks)

Ans.

(1) Linkage isomerism : The phenomenon of isomerism in which the coordination compounds have same metal

atom or ion and same ligand but bonded through different donor atoms or linkages is known as linkage isomerism.

(2) Linkage isomers : The coordination compounds having same metal atom or ion and ligand but bonded through

different donor atoms or linkages are called linkage isomers.

For example : Nitro complex [Co(NH�)�NO

�]Cl

�and nitrito complex [Co(NH

�)�ONO] Cl

�(Yellow) (Red) ■

Q. 53. Explain linkage isomers with NO�2 group as a ligand. (2 marks)

Ans.

(1) Nitro group (–NO��

) is an ambidentate ligand. NO��

group may link to central metal atom, through N or O.

(2) The two linkage isomers are, [Cl : � Ag � : NO�]� and [Cl : � Ag � O–NO]�

Choloronitroargentate(I) ion Chloronitritoargentate(I) ion■


Q. 54. Write linkage isomers of a complex having constituents Co��, 5NH�

and NO�2 . (2 marks)

Ans.

(i) NO��

is an ambidentate ligand which can be linked through N or O.

(ii) The linkage isomers are as follows :

(a) [Co(NH�)�(NO

�)]�� Pentaamminenitrocobalt(III) ion

(b) [Co(NH�)�(ONO)]�� Pentaamminenitritocobalt(III) ion

■

Can you tell ?

(Textbook page 200)

Q. Can you write IUPAC names of isomers (I) [Co(NH�)�SO

�]Br and (II) [Co(NH

�)�Br]SO

�?

Ans.Coordination compound IUPAC name

(I) [Co(NH�)�SO

�]Br Pentaamminesulphatocobalt(III) bromide

(II) [Co(NH�)�Br]SO

�Pentaamminebromocobalt(III) sulphate

Can you tell ?

(Textbook page 199)

Q. Write linkage isomers of [Fe(H�O)

�SCN]�. Write their IUPAC names.

Ans.Linkage isomers IUPAC name

(a) [Fe(H�O)

�SCN]� Pentaaquathiocyanatoiron(II)

(b) [Fe(H�O)

�NCS]� Pentaaquaisothiocyanatoiron(II)

Q. 55. Define : (1) Coordination isomerism (2) Coordination isomers. (1 mark each)

Ans.

(1) Coordination isomerism : The phenomenon of isomerism in the ionic coordination compounds having the same

molecular formula but different complex ions involving the interchange of ligands between cationic and anionic

spheres of different metal ions is called coordination isomerism.

(2) Coordination isomers : The ionic coordination compounds having same molecular formula but different complex

ions due to interchange of ligands between cationic and anionic spheres of different metal ions are called

coordination isomers.

For example, (I) [Co(NH�)�]��[Cr(CN)

�]�� and (II) [Cr(NH

�)�]��[Co(CN)

�]��

(cationic) (anionic) (cationic) (anionic) ■

Q. 56. Give three examples of coordination isomers. (1 mark each)

Ans.

(1) [Cu(NH�)�][PtCl

�] and [Pt(NH

�)�][CuCl

�]

(2) [Cr(NH�)�][Cr(CN)

�] and [Cr(NH

�)�(CN)

�][Cr(NH

�)�(CN)

�]

(3) [Cr(NH�)�][Cr(SCN)

�] and [Cr(NH

�)�(SCN)

�][Cr(SCN)

�(NH

�)�] ■

Q. 57. Define Solvate or Hydrate isomerism. (1 mark)

Ans. Solvate or Hydrate isomerism : The phenomenon of isomerism in the coordination compounds arising due to

the exchange of solvent or H�O molecules inside the coordination sphere and outer sphere of the complex is known

as solvate or hydrate isomerism. ■


Q. 58. Define solvate or hydrate isomers. OR

What are hydrate isomers ? Explain with examples. (2 marks)

Ans. Solvate or Hydrate isomers : The coordination compounds having the same molecular formula but differ in the

number of solvent or H�O molecules inside the coordination sphere and outer sphere of the complexes are called

solvate or hydrate isomers.

For example : [Cr(H�O)

�] Cl

�; [Cr(H

�O)

�Cl]Cl

�·H

�O; and [Cr(H

�O)

�Cl

�] Cl·2H

�O. ■

Q. 59. A coordination compound has formula CoCl�

·6H�O. Write the hydrate isomers of the complex.

(2 marks)

Ans. The possible hydrate isomers of the coordination compounds having molecular formula CoCl�

·6H�O are

as follows :

(1) [Co(H�O)

�]Cl

�; (2) [Co(H

�O)

�Cl] Cl

�·H

�O

(3) [Co(H�O)

�Cl

�] Cl·2H

�O (4) [Co(H

�O)

�Cl

�] · 3H

�O. ■

★Q. 60. Consider the complexes [Cu(NH�)�] [PtCl

�] and [Pt(NH

�)�] [CuCl

�]. What type of isomerism these two

complexes exhibit ?

Ans. Since in these two given complexes, there is an exchange of ligands between cationic and anionic constituents, they

exhibit coordination isomerism.■

Units

9.8 Stability of the coordination compounds

9.8.1 Factors which govern stability of the complex

★Q. 61. How can stability of the coordination compounds be explained in terms of equilibrium constants ?

(2 marks)

Ans. Stability of the coordination compounds : The stability of coordination compounds can be explained on the

basis of their stability constants. The stability of coordination compounds depends on metal-ligand interactions. In

the complex, metal serves as electron-pair acceptor (Lewis acid) while the ligand as Lewis base (since it is electron

donor). The metal-ligand interaction can be realized as the Lewis acid-Lewis base interaction. Stronger the

interaction greater is stability of the complex.

Consider the equilibrium for the metal-ligand interaction :

Ma��nLx

� � [MLn]a�(nx)

where a, x, [a�(nx)] denote the charge on the metal, ligand and the complex, respectively. Now, the

equilibrium constant K is given by

K[MLn]

[a�(nx)]

[Ma�] [Lx�]n

Stability of the complex can be explained in terms of K. Higher the value of K larger is the thermodynamic

stability of the complex hence K is called stability constant, and denoted by Kstab.

The equilibria for the complex formation with the corresponding K values are given below.

Ag��2CN� � [Ag(CN)�]� K5.5�10 �

Cu��4CN� � [Cu(CN)�]�� K2.0�10��

Co��6NH�

� [Co(NH�)�]�� K5.0�10��

From the above data, the stability of the complexes is [Co(NH�)�]�� [Cu(CN)

�]�� [Ag(CN)

�]�. ■


★Q. 62. Name the factors governing the equilibrium constants of the coordination compounds. (2 marks)Ans. The equilibrium constant of the complex depends on the following factors :

(a) Charge to size ratio of the metal ion : Higher the ratio greater is the stability. For the divalent metal ion

complexes their stability shows the trend : Cu�� Ni�� Co�� Fe�� Mn�� Cd��. The above stability

order is called Irving-William order. In the above list both Cu and Cd have the charge �2, however, the ionic

radius of Cu�� is 69 pm and that of Cd�� is 97 pm. The charge to size ratio of Cu�� is greater than that of Cd��.

Therefore the Cu�� forms stable complexes than Cd��.

(b) Nature of the ligand : A second factor that governs stability of the complexes is related to how easily the ligand

can donate its lone pair of electrons to the central metal ion that is, the basicity of the ligand. The ligands those are

stronger bases tend to form more stable complexes. ■


● The stability constant K of the [Ag(CN)�]� is 5.5�10 � while that for the corresponding [Ag(NH

�)�]� is

1.6�10�. Explain why [Ag(CN)�]�� is more stable.

Ans. Stability constant of [Ag(CN)�]�� is larger than that of [Ag(NH

�)�]� and hence [Ag(CN)

�]�� is more stable.

Also, CN� is a stronger ligand than NH�.

Units

9.9 Theories of bonding in complexes9.9.1 Valence Bond Theory (VBT)

Q. 63. Explain the steps involved in describing the bonding in coordination compounds using valence bondtheory. (3 marks)

Ans.(1) Vacant d-orbitals of metal ion form coordination bonds with ligands.

(2) s, p orbitals along with vacant d-orbitals of metal ion take part in hybridisation.

(3) The number of vacant hybrid orbitals formed is equal to number of hybridising orbitals which is equal to the

number of ligand donor atoms or coordination number of the metal.

(4) The metal-ligand coordination bonds are formed by the overlap between the vacant hybrid orbitals of metal

and the filled orbitals of the ligands.

(5) The hybrid orbitals used by the metal ion point in the direction of the ligands.

(6) When inner (n1) d orbitals of metal ion are used in the hybridisation then the complex is called (a) inner orbital

complex while when outer nd orbitals are used, complexes are called (b) outer orbital complexes. ■

Q. 64. Explain the steps involved in the metal-ligand bonding. (3 marks)Ans.(1) Find the oxidation state of central metal ion in the complex.

(2) Write the valence shell electronic configuration of metal ion.

(3) From the formula of the complex determine the number of ligands and find the number of metal ion orbitals

required for bonding.

(4) Find the orbitals of metal ion available for hybridisation and the type of hybridisation involved.

(5) Represent the electronic configuration of metal ion after hybridisation.

(6) Exhibit filling of hybrid orbitals after complex formation.

(7) Determine the number of unpaired electrons and predict magnetic property of the complex.

(8) Find whether the complex is low spin or high spin (applicable for octahedral complexes with d� or d� electronic

configuration.)■


Q. 65. What are the salient features of valence bond theory (VBT) ? (3 marks)

Ans. The salient features of valence bond theory (VBT) are as follows :

(1) According to this theory, a central metal atom or ion present in a complex provides a definite number of vacant

orbitals (s, p, d and f ) to accommodate the electrons from the ligands for the formation coordinate bonds with the

metal ion / atom.

(2) The number of vacant orbitals provided by the central metal atom or ion is the same as the coordination number of

the metal. For example : Cu�� provides 4 vacant orbitals in the complex. [Cu(NH�)�

]��.

(3) The vacant orbitals of metal atom or ion undergo hybridisation forming the same number of hybridised orbitals,

since the bonding with the hybrid orbitals is stronger.

(4) Each ligand has one or more orbitals containing one or more lone pairs of the electrons.

(5) The shape or geometry of the complex depends upon the type of hybridisation of the metal ion / atom.

(6) When inner orbitals namely (n1) d orbitals in transition metal atom or ion hybridise, the complex is called inner

complex and when outer orbitals i.e., nd orbitals hybridise then the complex is called outer complex.

(7) When the central metal atom or ion in the complex contains one or more unpaired electrons the complex is

paramagnetic while if all the electrons are paired, the complex is diamagnetic. ■

★Q. 66. What is the shape of a complex in which the coordination number of central metal ion is 4. (1 mark)

Ans. A complex with the coordination number of central metal ion equal to 4 may be tetrahedral or square planar.

■

★Q. 67. What are strong field and weak field ligands ? Give one example of each.

Ans. The ligands are then classified as (a) strong field and (b) weak field ligands. Strong field ligands are those in which

donor atoms are C,N or P. Thus CN�, NC�, CO, HN�, EDTA, en (ethylenediammine) are considered to be strong

ligands. They cause larger splitting of d orbitals and pairing of electrons is favoured. These ligands tend to form

low spin complexes. Weak field ligands are those in which donor atoms are halogens, oxygen or sulphur. For

example, F�, Cl�, Br�, l�, SCN�, C�O

��. In case of these ligands the R

�parameter is smaller compared to the

energy required for the pairing of electrons, which is called as electron pairing energy. The ligands then can be

arranged in order of their increasing field strength as

I� � Br� � Cl� � S�� F� � OH� � C�O

�� H

�O � NCS� � EDTA � NH

�� en � CN� � CO.

■

★Q. 68. [CoCl�]�� is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in

hybridisation. (2 marks)

Ans.��

Co [Ar] 3d� 4s�

Oxidation state of Co�2

Co�� [Ar] 3d� 4s�

Since Cl� is a weak ligand, there is no pairing of electrons. Since C.N. is 4, there is sp� hybridisation.

■


★Q. 69. With the help of crystal field energy level diagram explain why the complex [Cr(en)�]�� is coloured ?

(2 marks)

Ans. Cr��[Ar]3d�4s�

Since (en) is a strong field ligand there is pairing of electrons. The electrons occupy the t�g orbitals of lower energy.

It has one unpaired electron. Due to d-d transition, it is coloured. ■

Q. 70. What is the spin pairing process in the coordination compound ? (3 marks)

Ans. When the ligands approach the metal atom or ion for the formation of a complex, they influence the valence

electrons of metal atom or ion. Accordingly the ligands are classified as (A) strong ligands and (B) weak ligands.

(A) Strong ligands :

(i) They cause the pairing of unpaired electrons present in the metal atom or ion.

(ii) Spin pairing process :

(a) The process of pairing of unpaired electrons in metal atom or ion due to the presence of ligands in the

complex is called spin pairing process.

(b) This spin pairing process decreases the number of unpaired electrons and hence decreases the

paramagnetic character of the complex.

(c) The strong ligands also promote the outer ns electrons to the vacant inner (n1)d orbitals.

(B) Weak ligands : The weak ligands have no effect on the electrons in the valence shell of a metal atom or ion.

Strong ligands : CO, CN� , ethylenediammine (en), NH�, EDTA, etc.

Weak ligands : Cl�, I�, OH�, etc.

[Note : If a complex has n number of unpaired electrons then the magnetic moment, � is given by ‘spin only’ formula �� n(n�2) B.M.

where B.M. (Bohr Magneton) is the unit of magnetic moment. Hence from the magnitude of �, the number of unpaired electrons in the complex

and its structure can be evaluated.] ■

Units

9.9.2 Octahedral complexes

9.9.3 Tetrahedral complexes

9.9.4 Square planar complex

9.9.5 Limitations of VBT

Q. 71. Explain the structure of octahedral complex, [CO(NH�)�

]�� on the basis of valence bond theory.

(3 marks)

Ans.

(1) Hexaamminecobalt(III) ion, [CO(NH�)�]�� is a cationic complex, the oxidation state of cobalt is �3 and the

coordination number is 6.

(2) Electronic configuration :��

Co [Ar] � 3d� 4s�

Electronic configuration : Co�� [Ar] � 3d� 4s� 4p�

Co�� (Ground state)

3d� 4s� 4p�

2/Navneet Chemistry Digest : Std. XII (Part II) E0289 33

(3) Since NH�

is a strong ligand, due to spin pairing effect, all the four unpaired electrons in 3d orbital are paired

giving two vacant 3d orbitals.

Co�� (Excited state)

3d� 4s� 4p�

ZIIIII d�sp� hybridisation IIIIIJ

(4) Since the coordination number is 6, Co�� ion gets six vacant orbitals by hybridisation of two 3d vacant orbitals,

one 4s and three 4p orbitals forming six d�sp� hybrid orbitals giving octahedral geometry. It is an inner complex.

Co��

3d� d�sp� hybrid orbitals

(5) 6 lone pairs of electrons from 6NH�

ligands are accommodated in the six vacant d�sp� hybrid orbitals. Thus six

hybrid orbitals of Co�� overlap with filled orbitals of NH�

forming 6 coordinate bonds giving octahedral geometry

to the complex.

[Co(NH�)�]��

Co��

3d�

NH�

NH�

NH�

NH�

NH�

NH�

��

d�sp� hybrid orbitals

Since the complex has all electrons paired, it is diamagnetic.

[Co(NH�)�

]��

Fig. 9.5 : Structure of [Co(NH�)�

]��

Complex Hybridisation Geometry Magnetic property

[Co(NH�)�

]�� d�sp� (inner) Octahedral Diamagnetic (�0)■

Q. 72. Explain the geometry of [CoF�]�� on the basis of valence bond theory. (3 marks)

Ans.

(1) Hexafluorocobaltate(III) ion, [CoF�]�� is an anionic complex, the oxidation state of cobalt is �3 and the

coordination number is 6.


Co [Ar] � 3d� 4s� 4p� 4d�

Electronic configuration : Co�� [Ar] � 3d� 4s� 4p� 4d�

Co�� (Ground state)

3d� 4s� 4p� 4d�

ZIIIIII sp�d� hybridisation IIIIIIJ

(3) Since F� is a weak ligand, there is no spin pairing effect and Co�� possesses 4 unpaired electrons.

(4) Since the coordination number is 6, the Co�� ion gets six vacant orbitals by hybridisation of one 4s orbital, three 4p

orbitals and two 4d orbitals forming six sp�d� hybrid orbitals giving octahedral geometry.

Co��

3d� sp�d� hybrid orbitals


(5) 6 lone pairs of electrons from 6F� ligands are accommodated in the six vacant sp�d� hybrid orbitals. Thus six

hybrid orbitals of Co�� overlap with filled orbitals of F� forming 6 coordinate bonds giving octahedral geometry

to the complex. It is an outer complex.

[CoF�]��

Co��

3d�

F� F� F� F� F� F�

��

sp�d� hybrid orbitals

As the complex has 4 unpaired electrons it is paramagnetic.

Magnetic movement � is, ��n(n�2)�4(4�2)4.9 B.M.

[CoF�

]��

Fig. 9.6 : Structure of [CoF�

]��


[CoF�

]�� sp�d� (outer) Octahedral Paramagnetic (�4.9 B.M.)■

Q. 73. Explain the structure of tetrachloronickelate(II) [NiCl�]�� on the basis of valence bond theory.

(3 marks)

Ans.

(1) Tetrachloronickelate(II) ion is an anionic complex, oxidation state of Ni is �2 and the coordination number is 4.


Ni [Ar] � 3d� 4s� 4p�

Electronic configuration : Ni�� [Ar] � 3d� 4s� 4p�

Ni�� (Ground State)

3d� 4s� 4p�

sp� hybridisation

(3) Since the coordination number is 4, it gets 4 vacant hybrid orbitals by sp�-hybridisation of one 4s and three

4p orbitals giving tetrahedral geometry to the complex.

Ni��

3d� sp� hybrid orbitals

(4) As Cl� is a weak ligand, 2 unpaired electrons in Ni�� remain undisturbed.

(5) 4 lone pairs of electrons from 4Cl� ligands are accommodated in the vacant four sp� hybrid orbitals. Thus four

sp� hybrid orbitals of Ni�� overlap with filled orbitals of Cl� forming 4 coordination bonds, giving tetrahedral

geometry to the complex.

[NiCl�]��

Ni��

3d�

Cl� Cl� Cl� Cl�

��

sp� hybrid orbitals

Since the complex has 2 unpaired electrons, it is paramagnetic.


Magnetic moment � is, ��n(n�2)�2(2�2)2.83 B.M.

Fig. 9.7 : Structure of [NiCl�

]��

Hybridisation Geometry Magnetic property

sp� Tetrahedral Paramagnetic (�2.83 B.M.)■

Q. 74. Explain the structure of [Ni(CN)�

]�� on the basis of valence bond theory. (3 marks)

Ans.

(1) Tetracyanonickelate (II) ion, [Ni(CN)�]�� is an anionic complex, oxidation state of Ni is �2 and the coordination

number is 4.


Ni [Ar] � 3d� 4s� 4p�

Electronic configuration : Ni�� [Ar] � 3d� 4s� 4p�

Ni�� (Ground state)

3d� 4s� 4p�

(3) Since CN� is a strong ligand, one of the unpaired electrons in 3d orbital is promoted giving two paired electrons

and one vacant 3d orbital.

Ni�� (Excited state)

3d� 4s� 4p�

� dsp� hybridisation �

(4) Since the coordination number is 4, Ni�� gets 4 vacant hybrid orbitals by hybridisation of one 3d, one 4s and two

4p orbitals forming four dsp� hybrid orbitals. This has square planar geometry.

Ni��

3d� dsp� hybrid orbitals

(5) 4 lone pairs from 4CN� ligands are accommodated in the vacant four dsp� hybrid orbitals. Thus four dps� hybrid

orbitals of Ni�� overlap with filled orbitals of CN� forming 4 coordinate bonds giving square planar geometry to

the complex. It is an inner complex.

[Ni(CN)�]��

Ni��

3d�

CN� CN� CN� CN�

��

dsp� hybrid orbitals

Since the complex ion has all electrons paired, it is diamagnetic.

Fig. 9.8 : Structure of [Ni(CN)�

]��



[Ni(CN)�

]�� dsp� (inner) Square planar Diamagnetic (�0)■

Remember ...(Textbook page 202)

Complete the missing entries.

Coordination number Geometry of complex Hybridisation

2 Linear sp

4 Tetrahedral sp�

4 Square planar dsp�

6 Octahedral d�sp�/sp�d�

(Note : The missing entries are underlined.)

Table 9.3 : Type of hybridisation and geometry of a complex

Type of hybridisation

Geometry

sp

Linear

sp�

Triangular plane

sp�

Tetrahedral

dsp�

Square planar

dsp�

Trigonal bipyramidal

Type of hybridisation

Geometry

d�sp�, sp�d�

Octahedral

d�sp�

Pentagonal bipyramidal

(Note : Number of hybridised orbitalsCoordination number)

Q. 75. What are the limitations of valence bond theory ? (3 marks)

Ans. In case of the coordination compounds, the valence bond theory has the following limitations :

(1) It cannot explain the spectral properties (colours) of the complex compounds.

(2) Even if the magnetic moments can be calculated from the number of unpaired electrons, it cannot explain the

magnetic moment arising due to orbital motion of electrons.

(3) It cannot explain why the metal ions with the same oxidation state give inner complexes and outer complexes with

different ligands.

(4) In every complex, it cannot explain magnetic properties based on geometry of the complex.

(5) Quantitative interpretations of thermodynamic and kinetic stabilities of the coordination compounds cannot be

accounted.

(6) The complexes with weak field ligands and strong field ligands cannot be distinguished.

(7) It cannot predict the tetrahedral and square planar geometry of complexes with coordination number 4.

(8) The order of reactivity of inner complexes of d�, d�, d� and d� metal ions cannot be explained.

(9) It cannot explain the rates and mechanisms of reactions of the coordination compounds. ■

Try this …(Textbook page 204)

Q. Based on the VBT predict structure and magnetic behaviour of the [Ni(NH�)�

]�� complex.

Ans.��

Ni [Ar] 3d� 4s�

Ni�� [Ar] 3d� 4s�

Hybridisation : sp�d�

Geometry : Octahedral

Magnetic property : Paramagnetic


Units

9.9.6 Crystal field theory (CFT)9.9.7 Factors affecting Crystal Field splitting parameter (�

�)

9.9.8 Colour of the octahedral complexes9.9.9 Splitting of d-orbitals in tetrahedral and octahedral complexes

Q. 76. What are the assumptions of Crystal Field Theory (CFT) ? (3 marks)Ans. Bethe and van Vleck developed Crystal Field Theory (CFT) to explain various properties of coordination

compounds. The salient features of CFT are as follows :(1) In a complex, the central metal atom or ion is surrounded by various ligands which are either negatively charged

ions (F�, Cl�, CN�, etc.) or neutral molecules (H�O, NH

�, en, etc.) and the most electronegative atom in them

points towards central metal ion.

(2) The ligands are treated as point charges involving purely electrostatic attraction between them and metal ion.

(3) (i) The central metal ion has five, (n1) d degenerate orbitals namely d��

, d��

, d��

, d(x�y�) and dz�.(ii) When the ligands approach the metal ion, due to repulsive forces, the degeneracy of d-orbitals is destroyed

and they split into two groups of different energy, t�g and eg orbitals. This effect is called crystal field splitting

which depends upon the geometry of the complex.(iii) The d-orbitals lying in the direction of ligands are affected to a greater extent while those lying in between

the ligands are affected to a less extent.(iv) Due to repulsion, the orbitals along the axes of ligands acquire higher energy while those lying in between the

ligands acquire less energy.(v) Hence repulsion by ligands give two sets of split up orbitals of metal ion with different energies.

(vi) The energy difference between two sets of d-orbitals after splitting by ligands is called crystal field splittingenergy (CFSE) and represented by �

�or by arbitrary term 10Dq. The value of � or 10Dq depends upon the

geometry of the complex.

(4) The electrons of metal ion occupy the split d-orbitals according to Hund’s rule, aufbau principle and those orbitalswith minimum repulsion and the farthest away from the ligands.

(5) CFT does not account for overlapping of orbitals of central metal ion and ligands, hence does not consider covalentnature of the complex.

(6) From the crystal field stability energy, the stability of the complexes can be known.

Fig. 9.9 : Shapes of d–orbitals ■


Q. 77. What is crystal field splitting ? (1 mark)

Ans. The splitting of five degenerate d-orbitals of the transition metal ion into different sets of orbitals (t2g and eg)

having different energies in the presence of ligands in the complex is called crystal field splitting. ■

Try this …

(Textbook page 202)

Q. Give VBT description of bonding in each of following complexes. Predict their magnetic behaviour.

(a) [ZnCl�]��

(b) [Co(H�O)

�]�� (high spin)

(c) [Pt(CN)�]�� (square planar)

(d) [CoCl�]�� (tetrahedral)

(e) [Cr(NH�)�]��

Ans.Complex ion Hybridisation Geometry Magnetic property

(a) [ZnCl�]�� sp� Tetrahedral Diamagnetic

(b) [Co(H�O)

�]�� sp�d� Octahedral Paramagnetic

(c) [Pt(CN)�]�� dsp� Square planar Diamagnetic

(d) [CoCl�]�� sp� Tetrahedral Paramagnetic

(e) [Cr(NH�)�]�� d�sp� Octahedral Paramagnetic

Q. 78. What is crystal field stabilisation energy ? (1 mark)

Ans. Crystal field stabilisation energy (CFSE) : It is the change in energy achieved by preferential filling up of the

orbitals by electrons in the complex of metal atom or ion.

CFSE is expressed as a negative quantity i.e., CFSE � 0. Higher the negative value more is the stability of the

complex. ■

Q. 79. Explain the factors affecting Crystal Field Splitting parameter (��). (2 marks)

Ans. Crystal Field Splitting parameter (��) depends on, (a) Strength of the ligands and (b) Oxidation state of

the metal.

(a) Strength of the ligands : Since strong field ligands like CN�, en, etc. approach closer to the central metal ion, it

results in a large crystal field splitting and hence ��

has higher values.

(b) Oxidation state of the metal : A metal ion with the higher positive charge draws the ligands closer to it which

results in large separation of t�g and eg set of orbitals. The complexes involving metal ions with low oxidation state

have low values of ��. For example [Fe(NH

�)�

]�� has higher ��

than [Fe(NH�)�]��. ■

Q. 80. Explain the octahedral geometry of complexes using crystal field theory. (3 marks)

Ans.

(1) In an octahedral complex [MX�

]n�, the metal atom or ion is placed at the centre of regular octahedron while six

ligands occupy the positions at six vertices of the octahedron.

(2) Among five degenerate d-orbitals, two orbitals namely dx�y� and dz� are axial and have maximum electron density

along the axes, while remaining three d-orbitals namely dxy, dyz and dzx are planar and have maximum electron

density in the planes and in-between the axes.


Fig. 9.10 : Octahedral geometryhaving central metal (M) at thecentre and six ligands (L) at the

vertices of the octahedron.

(3) Hence, when the ligands approach a metal ion, the orbitals dx�y� and dz�

experience greater repulsion and the orbitals dxy, dyz and dzx experience less

repulsion.

(4) Therefore the energy of dx�y� and dz� increases while the energy of dxy, dyz and

dzx decreases and five d-orbital lose degeneracy and split into two point groups.

The orbitals dxy, dyz and dzx form t�g group of lower energy while dx�y� and dz�

form eg group of higher energy.

Thus t�g has three degenerate orbitals while eg has two degenerate orbitals.

(5) Experimental calculations show that the energy of t�g orbitals is lowered by

0.4�o or 4Dq and energy of eg orbital is increased by 0.6�o or 6Dq. Thus energy

difference between t�g and eg orbitals is �

�or 10Dq which is crystal field

splitting energy.

Fig. 9.11 : Crystal field splitting in octahedral complex

(6) CFSE increases with the increasing strength of ligands and oxidation state of central metal ion. ■

Q. 81. Explain the tetrahedral geometry of complexes using crystal field theory. (3 marks)

Ans.

(1) In the tetrahedral complex, [MX�

]n�, the metal atom or ion is placed at the centre of the regular tetrahedron and the

four ligands, are placed at four corners of the tetrahedron.

(2) The ligands approach the central metal atom or ion in-between the three coordinates x, y and z. The orbitals dxy , dyz

and dzx are pointed towards ligands and experience greater repulsion while the axial orbitals dx�–y� and dz� lie

in-between metal–ligand bond axes and experience comparatively less experience.

(a) (b)

Fig. 9.12 (a) and (b) : Tetrahedral geometry having central metal atom (M) at the centre and four ligands (L) at the four corners

(3) Therefore energy of dxy , dyz and dzx orbitals is increased while that of dx�-y� and dz� is lowered. Hence 5d-orbitals lose

their degeneracy and split into two point groups, namely t�g of higher energy (dxy, dyzand dzx) and eg of lower energy

(dx�y� and dz�).


Fig. 9.13 : Crystal field splitting in tetrahedral complex

(4) The experimental calculations show that the energy of t�g orbitals is increased by 0.4 �

�or 4Dq and energy of eg is

lowered by 0.6 ��

or 6Dq. Thus energy difference between t�g and eg orbitals is �

�or 10Dq which is crystal field

splitting energy (CFSE).

(5) This explains that the entry of each electron in eg orbitals, stabilises the complex by 0.6��

or 6Dq. While the entry

of each electron in t�g orbitals destabilises tetrahedral complex by 0.4�

�or 4Dq.

(6) In case of strong field ligands, the electrons prefer to pair up in eg orbitals giving low spin (LS) complexes while in

case of weak field ligands, the electrons prefer to enter higher energy t�g orbitals giving more unpaired electrons

and hence form high spin (HS) complexes.

Table 9.4 : Properties of complexes


(1) [NiCl�

]�� sp� Tetrahedral Paramagnetic

(2) [Ni(CO)�

] sp� Tetrahedral Diamagnetic

(3) [Ni(CN)�

]�� dsp� Square planar Diamagnetic

(4) [CuCl�

]�� sp� Tetrahedral Paramagnetic

(5) [Cu(NH�)�

]�� dsp� (inner) Square planar Paramagnetic

(6) [Fe(H�O)

�]�� sp�d� (outer) Octahedral Paramagnetic

(7) [Fe(CN)�

]�� d�sp� (inner) Octahedral Paramagnetic

(8) [Fe(CO)�

] dsp� (inner) Trigonal bipyramidal Diamagnetic

(9) [Cr(H�O)

�]�� sp�d� (outer) Octahedral Paramagnetic

(10) [Co(NH�)�

]�� d�sp� (inner) Octahedral Diamagnetic

■

★ Q. 82. What are high spin and low spin Complexes ? (2 marks)

Ans.

(1) High spin complex (HS) :

(i) The complex which has greater number of unpaired electrons and hence higher value of resultant spin and

magnetic moment is called high spin (or spin free) or HS complex.

(ii) It is formed with weak field ligands and the complexes have lower values for crystal field splitting energy

(CFSE), �o.

(iii) The paramagnetism of HS complex is larger.

(2) Low spin complex (LS) :

(i) The complex which has the least number of unpaired electrons or all electrons paired and hence the lowest

(or no) resultant spin or magnetic moment is called low spin (or spin paired) or LS complex.

(ii) It is formed with strong field ligands and the complexes have higher values of crystal field splitting

energy (�o).

(iii) Low spin complex is diamagnetic or has low paramagnetism. ■


Table 9.5 : d-orbital diagrams for high spin and low spin complexes

d-orbital

electronic

configuration

High spin Low spin

d� eg

t�g

d� eg

t�g

d� eg

t�g

d� eg

t�g

(Only the electronic configurations d� to d� render the high spin and low spin complexes)

Try this...(Textbook page 206)

Q. Sketch qualitatively crystal field d orbital energy level diagrams for each of the following complexes :

(a) [Ni(en)�]�� (b) [Mn(CN)

�]�� (c) [Fe(H

�O)

�]��

Predict whether each of the complexes is diamagnetic or paramagnetic.

Ans.

(a) The complex ion, [Ni(en)�]�� is octahedral.

��Ni [Ar] 3d� 4s�

Ni�� [Ar] 3d� 4s�

Since en is a strong ligand there is pairing of electrons.

Number of unpaired electronsn2 in t�g orbitals

Magnetic moment��n(n�2)

�2(2�2)2.83 B.M.

The complex ion is paramagnetic.


(b) The complex ion [Mn(CN)�]�� is octahedral.

��Mn [Ar] 3d� 4s�

Mn�� [Ar] 3d� 4s�

Since CN� is a strong ligand there is pairing of electrons.

Number of unpaired electronsn2 in t�g orbitals


�2(2�2)2.83 B.M.


(c) The complex ion [Fe(H�O)

�]�� is octahedral.

��Fe [Ar] 3d� 4s�

Fe�� [Ar] 3d� 4s�

Since H�O is a weak ligand, there is no pairing

of electrons.

Number of unpaired electronsn4 in t�g

and eg orbitals.


�4(4�2)

4.90 B.M.


★ Q. 83. Give valence bond description for the bonding in the complex [VCl�]�. Draw box diagrams for free

metal ion. Which hybrid orbitals are used by the metal ? State the number of unpaired electrons. (3 marks)

Ans.��

V [Ar] 3d� 4s�

V�� [Ar] 3d� 4s�

Since Cl� is a weak ligand, there is no pairing of electrons.

Number of unpaired electrons2

Type of hybridisationsp�


Geometry of complex ionTetrahedral

The complex ion is paramagnetic. ■

★ Q. 84. (A) Draw qualitatively energy-level diagram showing d-orbital splitting in the octahedral environment.

(B) Predict the number of unpaired electrons in the complex [Fe(CN)�]��. (C) Is the complex diamagnetic

or paramagnetic ? (D) Is it coloured ? Explain. (4 marks)

Ans. (A) d-orbital splitting in the octahedral environment :

(B) [Fe (CN)�]�� is an octahedral complex.

(C) Since CN� is a strong ligand, there is pairing of electrons and the complex is diamagnetic.

(D) The complex exists as lemon yellow crystals.

(In the complex all electrons in t�g are paired and requires high radiation energy for excitation.) ■

Q. 85. Write a note on colour in coordination compounds. (3 marks)

Ans.

(1) A large number of coordination compounds show wide range of colours due to d-d transition of electron and this

can be explained by crystal field theory (CFT).

(2) The complex absorbs the light in one visible region (400 nm to 700 nm) and transmits the light in different visible

region giving complementary colour.

(3) Consider an octahedral purple coloured complex of [Ti(H�O)

�]�� which absorbs green light and transmits purple

colour. Similarly [Cu(H�O)

�]�� absorbs the light in the red region of radiation spectrum and transmits in the blue

region, hence the complex appears blue.

(4) The absorption of light arises due to d-d transition of electron from lower energy level ( t�g) to higher energy level

(eg) in octahedral complex.

(5) The energy required for transition depends upon crystal field splitting energy ��. If �

��E, then the energy of an

absorbed photon (h�) is

� Eh�hc

where , � and c are wavelength, frequency and velocity of the absorbed light.

(6) Higher the magnitude of ��

or �E, higher is the frequency or lower is the wavelength of the absorbed radiation.

(7) Since ��

depends upon nature of metal atom or ion, its oxidation state, nature of ligands and the geometry of the

complex, different coordination compounds have different colours. ■

Q. 86. Explain the purple colour of the complex, [Ti(H�O)

�]�� with the help of crystal field theory. (3 marks)

Ans.

(1) [Ti(H�O)

�]�� is an octahedral complex, oxidation state of titanium is �3 (Ti��) and the coordination

number is 6.

(2) Electronic configuration��

Ti [Ar] � 3d� 4s�

Electronic configurationTi�� [Ar] � 3d OR


(3) According to crystal field theory, 3d orbitals undergo crystal field splitting giving higher energy eg, two orbitals and

lower energy t�g , three orbitals.

Fig. 9.14 : Crystal field splitting in [Ti(H�O)

�]��

(4) The crystal field splitting energy (CFSE), ��

is found to be 3.99�10� � J/ion from the spectrochemical studies.

(5) The absorption of radiation of wavelength or frequency � results in the transition of one unpaired electron from

t�g orbital to eg orbital.

t 2g eg� ----E t��g e g

Lower energy configuration Higher energy configuration

Fig. 9.15 : Photon absorption and the absorption spectrum of [Ti(H�O)

�]��

(6) The wavelength of the absorbed radiation will be,

�� Eh�

hC

Js ms�

� hC�E

6.63�10��3�10�

3.99�10� �

J / ion

4.98�10�� m 498 nm

(7) Hence the complex [Ti(H�O)

�]�� absorbs the green radiation of wavelength 498 nm in the visible region and

transmits the complementary purple light. Therefore the complex is purple coloured. ■

Q. 87. An octahedral complex absorbs the radiation of wavelength 620 nm. Find the crystal field splitting energy.

(2 marks)

Ans. Crystal field splitting energy ��

is given by,

Js ms�

�� E

hC

6.63�10��3�10�

620�10��m

3.2�10� � J ■


Unit

9.10 Applications of coordination compounds

Q. 88. What are the applications of coordination compounds? (1 mark for each application)

Ans.

(1) In biology : Several biologically important natural compounds are metal complexes which play an important role

in number of processes occurring in plants and animals.

For example, chlorophyll in plants is a complex of Mg�� ions, haemoglobin in blood is a complex of iron, vitamin

B �

is a complex of cobalt.

(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more

stable, more effective and can be assimilated easily.

For example, platinum complex [Pt(NH�)�Cl

�] known as cisplatin is effectively used in cancer treatment.

EDTA is used to treat poisoning by heavy metals like lead.

(3) To estimate hardness of water :

(1) The hardness of water is due to the presence Mg�� and Ca�� ion in water.

(2) The strong field ligand EDTA forms stable complexes with Mg�� and Ca��. Hence these ions can be removed

by adding EDTA to hard water.

Similarly these ions can be selectively estimated due to the difference in their stability constants.

(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to

supply continuously the metal ions in small amounts.

For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For

example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)�] and K [Au(CN)

�] are used.

■

★Q. 89. Mention two applications of coordination compounds. (2 marks)

Ans. For answer refer to Q. 88. ■

Activity : (Textbook page 209)

(1) The reaction of chromium metal with H�SO

�in the absence of air gives blue solution of chromium ion.

Cr(s)�2H�(aq) � Cr��(aq)�H�(s)

Cr�� forms octahedral complex with H�O ligands.

(a) Write formula of the complex.

(b) Describe bonding in the complex using CFT and VBT.

Draw crystal field splitting and valence bond orbital diagrams.

(2) Reaction of complex [Co(NH�)�(NO

�)�] with HCl gives a complex

[Co(NH�)�H

�OCl

�]� in which two chloride ligands are trans to one another.

(a) Draw possible stereoisomers of starting material

(b) Assuming that NH�

groups remain in place, which of two starting isomers would give the observed

product ?

(Note : Students are supposed to carry out these activity on their own.)


................................................................................................................................................................................................................................................

Q. 90. Select and write the most appropriate answer

from the given alternatives for each subquestion :

(1 mark each)

★ 1. The oxidation state of cobalt ion in the complex

[Co(NH�)�Br]SO

�is

(a) �2 (b) �3

(c) �1 (d) �4

★ 2. IUPAC name of the complex [Pt(en)�(SCN)

�]�� is

(a) bis(ethylenediamine dithiocyanatoplatinum

(IV) ion

(b) bis(ethylenediamine) dithiocyantoplatinate

(IV) ion

(c) dicyanatobis(ethylenediamine)platinate(IV)

ion

(d) bis(ethylenediammine)dithiocynatoplatinate

(IV) ion

★ 3. Formula for the compound sodium hexacynofer-

rate (III) is

(a) [NaFe(CN)�] (b) Na

�[Fe(CN)

�]

(c) Na[Fe(CN)�] (d) Na

�[Fe(CN)

�]

★ 4. Which of the following complexes exist as cis and

trans isomers ?

(1) [Cr(NH�)�Cl

�]�

(2) [Co(NH�)�Br]��

(3) [PtCl�Br

�]�� (square planar)

(4) [FeCl�(NCS)

�]�� (tetrahedral)

(a) 1 and 3 (b) 2 and 3

(c) 1 and 2 (d) 4 only

★ 5. Which of the following complexes are chiral ?

(1) [Co(en)�Cl

�]� (2) [Pt(en)Cl

�]

(3) [Cr(C�O

�)�]�� (4) [Co(NH

�)�Cl

�]�

(a) 1 and 3 (b) 2 and 3

(c) 1 and 4 (d) 2 and 4

★ 6. On the basis of CFT predict the number of un-

paired electrons in [CrF�]�� .

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

★ 7. When an excess of AgNO�

is added to the complex

one mole of AgCl is precipitated. The formula of

the complex is

(a) [CoCl�(NH

�)�]Cl

(b) [CoCl(NH�)�]Cl

�(c) [CoCl

�(NH

�)�]

(d) [Co(NH�)�]Cl

�

★ 8. The sum of coordination number and oxidation

number of M in [M(en)�C

�O

�]Cl is

(a) 6 (b) 7 (c) 9 (d) 8

9. The coordination number of cobalt in the complex

[Co(en)�Br

�]Cl

�is

(a) 4 (b) 5 (c) 6 (d) 7

10. EDTA combines with cations to form

(a) chelates

(b) polymers

(c) clathrates

(d) non-stoichiometric compounds

11. Which one of the following compounds can exhibit

coordination isomerism ?

(a) [Co(en)�Cl

�]Br

(b) [Co(NH�)�] [Cr(CN)

�]

(c) [Co(en)�]Cl

�(d) [Co(NH

�)�NO

�]Cl

�

12. Which of the following compounds can exhibit

linkage isomerism ?

(a) [Co(en)�]Cl

�(b) [Co(en)

�Cl

�]Cl

(c) [Co(en)�NO

�Br]Cl (d) [Co(NH

�)�Cl]Br

�

13. Oxidation number of cobalt in K[CoCl�] is

(a) �1 (b) 1 (c) �3 (d) 3

14. The correct structure of [Cr(H�O)

�]�� is ...

(a) octahedral

(b) tetrahedral

(c) square pyramidal

(d) trigonal bipyramidal

15. Amongst the following ions which one has the

highest paramagnetism ?

(a) [Cr(H�O)

�]�� (b) [Fe(H

�O)

�]��

(c) [Cu(H�O)

�]�� (d) [Zn(H

�O)

�]��


................................................................................................................................................................................................................................................

16. The geometry of [Ni(CN)�]�� and [NiCl

�]�� are

(a) both tetrahedral

(b) both square planar

(c) tetrahedral and square planar respectively

(d) square planar and tetrahedral respectively

17. The complex cis-[Pt(NH�)�Cl

�] is used in treat-

ment of cancer under the name.

(a) Aspirin (b) Eqanil

(c) cisplatin (d) transplatin

18. [Co(NH�)�]�� is an ....... orbital complex and is

...... in nature.

(a) inner, paramagnetic

(b) inner, dimagnetic

(c) outer, paramagnetic

(d) outer, dimagnetic

19. The IUPAC name of [Ni(Co)�] is

(a) tetra carbonyl nickel (O)

(b) tetra carbonyl nickel (II)

(c) tetra carbonyl nickelate (O)

(d) tetra carbonyl nickelate (II)

20. The number of ions produced by the complex

[Co(NH�)�Cl

�] Cl is

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

21. The dimagnetic species is

(a) [Ni(CN)�]�� (b) [NiCl

�]��

(c) [CoCl�]�� (d) [CoF

�]��

22. Which one of the following is an inner orbital

complex as well as dimagnetic in behaviour

(Atomic no. Zn30, Cr24, Co27, Ni28)

(a) [Zn(NH�)�]�� (b) [Cr(NH

�)�]��

(c) [Co(NH�)�]�� (d) [Ni(NH

�)�]��

23. Among [Ni(Co)�], [Ni(CN)

�]��, [NiCl

�]��

Species, the hybridisation states at the Nickel atom

are respectively

(a) sp�, dsp�, sp�

(b) sp�, dsp�, dsp�

(c) dsp�, sp�, sp�

(d) sp�, sp�, dsp�

24. The strongest ligand in the following is

(a) CN� (b) Br� (c) HO� (d) F�

25. Magnetic moment of (NH�)�

(MnBr�) is ...... BM

(a) 5.91 (b) 4.91 (c) 3.91 (d) 2.91

26. The complex which violates EAN rule is

(a) Fe(CO)�

(b) [Fe(CN)�]��

(c) Ni(CO)�

(d) [Zn(NH�)�]Cl

�

27. EDTA is a ligand of the type

(a) bidentate (b) tridentate

(c) tetradentate (d) hexadentate

28. The cationic complex among the following is

(a) K�[Fe(CN)

�] (b) Ni(CO)

�(c) K

�HgI

�(d) [Co(NH

�)�]Cl

�

29. If Z is the atomic number of a metal, X is number

of electrons lost forming metal ion and Y is the

number of electrons from the ligands then EAN is

(a) Z�X�Y (b) XZ�Y

(c) ZX�Y (d) X�ZY

30. Octahedral complex has hybridisation,

(a) dsp� (b) d�sp� (c) dsp� (d) d�sp�

31. Inner complex has hybridisation,

(a) d�sp� (b) sp�d� (c) sp�d (d) sp�d�

32. The number of unpaired electrons in

[Co(NH�)�]�� is

(a) 0 (b) 1 (c) 2 (d) 4

33. The number of unpaired electrons in [NiCl�]��

and [Ni(CN)�]�� are respectively,

(a) 2, 2 (b) 2, 0 (c) 0, 0 (d) 1, 2

34. Among the following complexes, the highest

magnitude of crystal field stabilisation energy will

be for

[Co(H�O)

�]��, [Co(CN)

�]��, [Co(NH

�)�]��,

[CoF�]��

(a) [Co(H�O)

�]�� (b) [Co(CN)

�]��

(c) [Co(NH�)�]�� (d) [CoF

�]��

35. The number of unpaired electrons in a low spin

octahedral complex ion of d� is

(a) 0 (b) 1 (c) 2 (d) 3


................................................................................................................................................................................................................................................

36. The number of unpaired electrons in a high spin

octahedral complex ion of d� is

(a) 0 (b) 1 (c) 2 (d) 3

37. Ligand used in the estimation of hardness of water

is

(a) EDTA (b) DBG (c) chloride (d) bromo

38. Which of the following complexes will give a

white precipitate on treatment with a solution of

barium nitrate ?

(a) [Cr(NH�)�SO

�] Cl

(b) [Co(NH�)�Cl

�] NO

�(c) [Cr(NH

�)�Cl

�] SO

�(d) [CrCl

�(H

�O)

�] Cl

39 What is effective atomic number of Fe (z26) in

[Fe(CN)�]�� ?

(a) 12 (b) 30 (c) 26 (d) 36

40. Cisplatin compound is used in the treatment of

(a) malaria (b) cancer

(c) AIDS (d) yellow fever

Answers

1. (b) �3

2. (a) bis(ethylenediamine dithiocyanatoplatinum (IV)

ion

3. (d) Na�[Fe(CN)

�]

4. (a) 1 and 3

5. (a) 1 and 3

6. (c) 3

7. (a) [CoCl�(NH

�)�]Cl

8. (c) 9

9. (c) 6

10. (a) chelates

11. (b) [Co(NH�)�] [Cr(CN)

�]

12. (c) [Co(en)�NO

�Br]Cl

13. (b) 1

14. (a) octahedral

15. (b) [Fe(H�O)

�]��

16. (d) square planar and tetrahedral respectively

17. (c) cisplatin

18. (b) inner, dimagnetic

19. (a) tetra carbonyl nickel (O)

20. (b) 2

21. (a) [Ni(CN)�]��

22. (c) [Co(NH�)�]��

23. (a) sp�, dsp�, sp�

24. (a) CN�

25. (a) 5.91

26. (b) [Fe(CN)�]��

27. (d) hexadentate

28. (d) [Co(NH�)�]Cl

�29. (c) ZX�Y

30. (d) d�sp�

31. (a) d�sp�

32. (a) 0

33. (b) 2, 0

34. (b) [Co(CN)�]��

35. (b) 1

36. (d) 3

37. (a) EDTA

38. (c) [Cr(NH�)�Cl

�] SO

�39. (d) 36

40. (b) cancer



10. HALOGEN DERIVATIVES

10 HALOGEN DERIVATIVES

....

....

....

CHAPTER OUTLINE

Page No.

Important Definitions and Reactions … 51

Exercises

10.1 Classification of halogen derivatives … 53

10.1.1 Classification of monohalogen compounds … 54

10.2 Nomenclature of halogen derivatives … 56

10.3 Methods of preparation of alkyl halides/aryl halides … 63

10.3.1 From alcohol … 63

10.3.2 From hydrocarbon … 66

10.3.3 Halogen exchange … 70

10.3.4 Electrophilic substitution … 71

10.3.5 Sandmeyer’s reaction … 71

10.4 Physical properties … 71

10.4.1 Nature of intermolecular forces … 71

10.4.2 Boiling Point … 71

10.4.3 Solubility … 71

10.5 Optical isomerism in halogen derivatives … 73

10.5.1 Chiral atom and molecular chirality … 73

10.5.2 Plane polarized light … 73

10.5.3 Optical activity … 73

10.5.4 Enantiomers … 73

10.5.5 Representation of configuration of molecules … 77

10.6 Chemical properties … 78

10.6.1 Laboratory test of haloalkanes … 78

10.6.2 Nucleophilic substitution reaction of haloalkanes … 78

10.6.3 Mechanism of SN reaction … 84

10.6.4 Factors influencing SN1 and SN2 mechanism … 87

10.6.5 Elimination reaction : Dehydrohalogenation … 91

10.6.6 Reaction with active metals … 96

10.6.7 Reaction of haloarenes … 98

10.7 Uses and environmental effects of some polyhalogen compounds … 108

10.7.1 Uses of some polyhalogen compounds … 108

10.7.2 Environmental effects of some polyhalogen compounds … 110

Multiple Choice Questions … 111

10. HALOGEN DERIVATIVES 51

1. Classification of Halogen derivatives :

Halogen derivatives

!!D !D !D !D

Haloalkanes Haloalkenes Haloalkynes Haloarenes

Halogen derivatives

!!D !D !D

Monohalogen compounds Dihalogen compounds Trihalogen compounds

Alkyl halides or Haloalkanes

!!D !D !D

Primary alkyl halide Secondary alkyl halide Tertiary alkyl halide

Alkyl halides (Based on halogen atom bonded to sp�, sp� and sp hybridised carbon)

!!D !D !D !D !D

Allylic halide Benzylic halide Vinylic halides Haloalkyne Aryl halides

2. Preparation and Reactions of alkyl halides :Preparation methods Reactions

Hx

R�OHPCl

�SOCl

�

AlkanesX

�lightIIIIIIIIIIIIIIIIIJ RX IIIIIIIIIIJ

Alkyl halide

AlkenesHXIIIIII

Alkyl halide NaI

aq. NaOH or aq. KOHIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIJ R�OH alcohols

R��ONa�

IIIIIIIIIIIIIIIIIIJ R�OR� Ethers

R�COOAgIIIIIIIIIIIIIIIIIIJ R�COOR� Ester

NH�

excessIIIIIIIIIIIIIIIIIIJ R�NH

�Amine

aq. KCNIIIIIIIIIIIIIIIIIIJ R�CN Alkyl cyanide

AgCNIIIIIIIIIIIJ R�NC Alkyl isocyanide

KO�N�OIIIIIIIIIIIIIIIIIIIIIJ R�O�N�O Alkyl nitrite

Ag�O�N�OIIIIIIIIIIIIIIIIIIIIIIIIIIJ R�N

� O

O�Nitroalkane

Alc. KOHIIIIIIIIIIIIIIIIIJ Olefins (alkenes)

Mg, etherIIIIIIIIIIIIIIIIIJ RMgX (Grignard reagent)

Na, dry etherIIIIIIIIIIIIIIIIIIIIIIJ R�R Higher alkane (Wurtz reaction)


3. Preparation of haloarenes :

4. Reactions of chlorobenzene :

(i)

(ii)

(iii)

5. Uses and environmental effects :(1) Dichloromethane (CH

�Cl

�) (2) Chloroform (CHCl

�) (3) Carbon tetrachloride (CCl

�)

(4) Iodoform (CHI�) (5) Freons (CCl

�F�) (6) Dichlorodiphenyltrichloro ethane (DDT)

(★ ) Indicates questions from the textbook.

Unit

10.1 Classification of halogen derivatives

Q. 1. What are halogen derivatives of hydrocarbons ? (1 mark)

Ans. The replacement of hydrogen atom/s in aliphatic or aromatic hydrocarbons by halogen atom/s results in the

formation of halogen derivatives of hydrocarbons. ■

Q. 2. How are halogen derivatives of hydrocarbons classified ? (3 marks)

Ans. Halogen derivatives of alkane are classified as :

Halogen derivatives

!!D !D

Monohalogen derivatives

(Alkyl halides)

Polyhalogen derivatives

!!D !D !D

Dihalogen

derivatives

!

Trihalogen

derivatives

Tetrahalogen

derivatives

!D !D

Vicinal

dihalides

Geminal

dihalides


(1) Monohalogen derivative (or alkyl halide) : It is a halogen derivative of an alkane in which one hydrogen atom is

replaced by one halogen atom and it is also called alkyl halide. E.g. C�H

�Br.

(2) Polyhalogen derivatives : These are halogen derivatives in which more than one hydrogen atoms of an alkane

are substituted by corresponding number of halogen atoms.

They are classified as follows :

(i) Dihalogen derivatives : The compounds formed by the substitution of two hydrogen atoms of an alkane by

two halogen atoms are called dihalogen derivatives. E.g. CH�Cl

�.

They are further classified as :

(a) Vicinal dihalides : CH��CH

�� Cl Cl

(Two halogen atoms on vicinal or adjacent carbon atoms)

(b) Geminal dihalides : CH��CHCl

�(Two halogen atoms on the same carbon atom)

(ii) Trihalogen derivatives : The compounds formed by the substitution of three hydrogen atoms of an alkane

by three halogen atoms are called trihalogen derivatives. E.g. CHCl�, CH

��Cl

�CH�

Cl

�CH��

Cl

,

(iii) Tetrahalogen derivatives : The compounds formed by the substitution of four hydrogen atoms of an alkane

by four halogen atoms are called tetrahalogen derivatives. E.g. CCl�. ■

Unit

10.1.1 Classification of monohalogen compounds

Q. 3. What are alkyl halides ? How are they classified ? (3 marks)

Ans. The compound formed by the replacement of one hydrogen atom in an alkane by a halogen atom is called an alkyl

halide. The halogen atom is bonded to sp� hybridised carbon. Alkyl halides are classified into the following three

classes depending on the type of the carbon to which halogen atom is bonded.

(1) Primary (1°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a primary carbon atom is called

primary alkyl halide.

[Primary (1°) carbon atom i.e., the carbon atom which is attached to only one carbon atom.]

CH�CH

�–°

CH�

–Cl CH�

–

CH��

CH–°

CH�

–Br CH�

–

CH��

C—�

CH�

°

CH�

–Br

ethyl chloride isobutyl bromide

neopentyl bromide

They are represented by the general formula R–°

CH�

–X.

(2) Secondary (2°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a secondary carbon atom is

called secondary alkyl halide. [Secondary (2°) carbon i.e., the carbon atom which is attached to two other carbon

atoms.]

CH�

–�°

CH–CH��

Br

CH�

–CH�

–�°

CH–CH��

Cliso-propyl bromide sec-butyl chloride


They are represented by the general formula R–�°

CH–X�R�

(R and R� can be same or different)

(3) Tertiary (3°) alkyl halide : Alkyl halide in which halogen atom is bonded to a tertiary carbon atom is called

tertiary alkyl halide. [Tertiary (3°) carbon i.e., the carbon atom which is attached to three other carbon atoms.]

CH�

–

CH��°

C�CH

�

–Br CH�

–

C�H

��°

C –CH��

Cl

tert-butyl bromide tert-pentyl chloride

They are represented by the general formula R–

R��C–X�R�

(R, R� and R� may be same or different) ■

Q. 4. Explain the following :

(1) Alkyl halide or haloalkanes (2) Allylic halides (3) Benzylic halide (4) Vinylic halide

(5) Haloalkyne (6) Aryl halide or haloarenes. (1 mark each)

Ans.

(1) Alkyl halide or haloalkanes : In alkyl halides or haloalkanes the halogen atom is bonded to sp� hybridized carbon

which is a part of saturated carbon chain.

Example : CH��CH

��CH

��Br

n-proyl bromide

(2) Allylic halides : In allylic halides, halogen atom is bonded to a sp� hybridized carbon atom next to a

carbon-carbon double bond.

Example :

CH��CH�CH

��Br

Allyl bromide

(3) Benzylic halide : In benzylic halides, halogen atom is bonded to a sp� hybridized carbon atom which is further

bonded to an aromatic ring.

Example :

(4) Vinylic halides : In vinylic halides, halogen atom is bonded to a sp� hybridized carbon atom of aliphatic chain.

Vinylic halide is a haloalkene.

Example : CH��CH�Br

Vinyl bromide

(5) Haloalkyne : In haloalkynes, halogen atom is bonded to a sp hybridized carbon atom.

Example : CH � C�Cl

Chloro acetylene


(6) Aryl halides or haloarenes : In aryl halides, halogen atom is directly bonded to the sp� hybridized carbon atom of

aromatic ring.

Example :

■

Unit

10.2 Nomenclature of halogen derivatives

Q. 5. Give the IUPAC names of the following : (1 mark each)

Ans.Compound Ans. IUPAC names

(1) CH�

– Br Bromomethane

(2) CH�

– CH�

– Br Bromoethane

(3) CH�

– CHCl�

1,1-Dichloroethane

(4) Cl – CH�

– CH�

– Cl 1,2-Dichloroethane

(5) CH�Cl

�Dichloromethane

(6) Cl – CH�

–� CH�

–� CH�

– Cl 1,3-Dichloropropane

(7) CHCl�

Trichloromethane

(8) CHI�

Triiodomethane

(9) (CH�)�C – Cl OR

CH��

CH��C�Cl

�CH

�

2-Chloro-2-methylpropane

(10)4

CH�

–3

CH�

–2

CH�

–1

CH�

– Br 1-Bromobutane

(11)1

CH�

–2

CH –3

CH�

–4

CH�

–5

CH�

2-Chloropentane�

Cl

(12) (CH�)�CH – CH

�– Cl OR CH

��

CH��

CH�CH��Cl 1-Chloro-2-methylpropane

(13) CH�

– Cl�

(13) CH�

– CH – CH�

2-Chloropropane

(14)1

CH�

–2

CH –3

CH –4

CH�

–5

CH�

3-Bromo-2-chloropentane� �Cl Br

(15) CH�

– CH – CH – CH�

2,3-Dichlorobutane� �Cl Cl


Compound IUPAC names

(16) CH�


2-Chloro-3-methylbutane� �

Cl CH�

(17) Cl –3

CH�

–2

CH�

–1

CH�

– Br 1-Bromo-3-chloropropane

(18) CH�

– CH�

–

CH��

C – CH�

– Cl�CH

�

1-Chloro-2, 2-dimethylbutane

(19) sec-butyl chloride CH�

– CH�

– CH – CH�

2-Chlorobutane�

Cl

(20) tert-butyl bromide H�C –

CH��

C–Br�CH

�

2-Bromo-2-methylpropane

(21) n-Butyl bromide CH��CH

��CH

��CH

��Br 1-Bromobutane

(22) sec-Butyl bromide CH��CH�

�Br

CH��CH

�2-Bromobutane

(23) Tertiary butyl iodide CH��

CH��

C��CH

�

I 2-Iodo-2-methylpropane

(24) neo pentyl bromide CH�

–

CH��

C–CH�

–Br�CH

�

1-Bromo-2, 2-dimethyl propane

(25) CH��CH–Cl Chloroethene

(26) 1,3,5-tribromobenzene

(27) Bromophenyl methane

■


Q. 6. Draw the structures of the following compounds : (1 mark each)

Compound Ans. Structures

(1) Dibromomethane Br�

H�C��H

Br (CH�Br

�)

(2) 3-Bromopentane CH�

– CH�

– CH – CH�

– CH��

Br

CH�

– Br�

(3) 2-Bromo-2-methylpropane CH�

– C – CH��

CH�

CH�

– Cl�

(4) 2,2-Dichloropropane CH�

– C – CH��

Cl

(5) Iodoethane CH�

– CH�

– I

(6) 1,1-Dibromopropane CH�

– CH�

– CHBr�

(7) 1,1,1-Trichloroethane CH�

– CCl�

(8) 1,2-Dichloropropane Cl – CH�

– CH – CH��

Cl

(9) 1-Bromo-4-methylpentane Br –1

CH�

–2

CH�

–3

CH�

–4

CH –5

CH��

CH�

(10) 2-Chlorobutane CH�

– CH – CH�

– CH��

Cl

(11) 1,2-Diiodoethane I – CH�

– CH�

– I

(12) Tetrachloromethane CCl�

(13) 1,2,3-Trichloropropane Cl – CH�

– CH – CH�

– Cl�Cl

(14) 1-Iodobutane CH�

– CH�

– CH�

– CH�

– I

(15) n-Butyl bromide CH�

– CH�

– CH�

– CH�

– Br

(16) isobutyl bromide CH�

– CH – CH�Br

�CH

�


Compound Structures

CH�

– CH��

(17) 1-Chloro-2, 2-dimethylpropane CH�

– C – CH�

– Cl�CH

�

CH�

– C – CH�

– C –CH�

CH��

(18) 1-Bromo-2, 2, 4, 4-tetramethylpentane5

CH�

–4

C –3

CH�

–2

C –1

CH�

– Br� �CH

�CH

�CH

�– CH

��(19) 2-Chloro-2-methylpropane CH

�– C – CH

��Cl

CH�

– CH��

(20) 1-Chloro-3, 3-dimethylbutane CH�

– C – CH�

– CH�

– Cl�CH

�

CH�

– C – CH�

– C –CH�

CH��

(21) 1-Iodo-2, 2, 4, 4-tetramethylpentane CH�

– C – CH�

– C – CH�

– I� �CH

�CH

�

CH�

– Br�

(22) tertiary butyl bromide CH�

– C – CH��

CH�

CH�

– I�

(23) isopropyl iodide CH�

– CH – CH�

(24) sec-butyl chloride CH�

– CH – CH�

– CH��

Cl

(25) 3-Bromo-2-methyl pentane CH�

–CH�

–CH–�

Br

CH��

CH–CH�

(26) 2-Bromo-3-ethyl-2-methyl hexane CH�

–

CH��

C—–�Br

C�H

��CH–CH

�–CH

�–CH

�

(27) 1-Chlorobutane CH�

–CH�

–CH�

–CH�

–Cl


Compound Structures

(28) p-dichlorobenzene

(29) 1,1,1-trichloroethane CH�

–

Cl�C–Cl�Cl

■

Q. 7. Write the structure of –

(a) 3-chloro-3-ethylhex-1-ene (b) 1-Iodo-2, 3-dimethylbutane (c) 1, 3, 5-tribromobenzene (3 marks)

Ans.Cl�

(a) 3-Chloro-3-ethylhex-1-ene : CH�

– CH�

– H�C– C– CH�CH

��CH

�– CH

�

CH�

CH��

(b) 1-Iodo-2, 3-dimethylbutane : CH�


– I

(c) 1, 3, 5-Tribromobenzene :

■

★ Q. 8. Write IUPAC name of the following compounds. (1 mark each)

Compound Ans. IUPAC names

(1) CH��CH�C �

�CH

�

CH�Br�

CH�

4-Bromo-3, 4 dimethyl but-2-ene

(2) CH��CH�

�Cl

CH�CH��CH

��CH

�

2-Chloro-3-methyl pentane

(3) 1-Chloro-4-ethyl cyclohexane


Compound Ans. IUPAC names

(4) 1,4-Dichloro-2-methyl benzene

■

Q. 9. Write structures of

(a) 2-iodo-3-methyl pentane (b) 3-chlorohexane

(c) 1-chloro-2, 2-dimethyl propane (d) 1-chloro-4-ethyl cyclohexane. (1 mark each)

Ans. (a) H�C–HC–CH–H

�C–CH

�� I CH

�2-iodo-3-methyl pentane

(b) H�C–H

�C–HC–H

�C–H

�C–CH

��Cl

3-chlorohexane

(c) H�C–

CH��

C–CH�

–Cl�CH

�1-chloro-2, 2-dimethyl propane

(d)

1-chloro-4-ethyl cyclohexane

Q. 10. Write the possible isomers of monochloroderivatives of 2,3-Dimethylbutane and write their IUPAC

names. (2 marks)

Ans. The given parent hydrocarbon has molecular formula, CH

�. The monochloroderivative of this compound has

molecular formula CH

�Cl.

The parent hydrocarbon is,

CH��CH�CH�CH

�2,3-Dimethylbutane

� �CH

�CH

�Hence the structures of isomers of monochloroderivative are,

(1) CH��CH�CH�CH

�Cl 1-Chloro-2, 3-dimethylbutane

� �CH

�CH

�

(2) CH�

–

Cl�C � CH – CH

�� CH

�CH

�

2-Chloro-2, 3-dimethylbutane

Hence CH

�Cl has two isomers. ■


Q. 11. Write structures and IUPAC names of all possible isomers of C�H

Br and classify them as 1° / 2° / 3°.

(1 mark each)Ans. C

�H

Br is a monohalogen derivative.

Isomer IUPAC name Type

Br1°�

(1) CH��CH

��CH

��CH

��CH

�1-Bromopentane (1°) Primary alkyl halide

Br�

(2) CH��CH�CH

��CH

��CH

�2°2-Bromopentane (2°) Secondary alkyl halide

Br�

(3) CH��CH

��CH�CH

��CH

�2°3-Bromopentane (2°) Secondary alkyl halide

Br�

(4) CH��CH�CH

��CH

�1° �

CH�

1-Bromo-2-methylbutane (1°) Primary alkyl halide

Br� 3°

(5) CH��C�CH

��CH

��CH

�

2-Bromo-2-methylbutane (3°) Tertiary alkyl halide

Br�

(6) CH��CH�CH�CH

�� 2°CH

�

2-Bromo-3-methylbutane (2°) Secondary alkyl halide

1°(7) CH

��CH�CH

��CH

��Br

�CH

�

1-Bromo-3-methylbutane (1°) Primary alkyl halide

CH�� 1°

(8) CH��C�CH

��Br

�CH

�

1-Bromo-2, 2-dimethylpropane (1°) Primary alkyl halide

■



Write IUPAC names of the following :

(i) CH��CH�

�Br

CH�

(ii) CH��CH�CH

�I

�CH

�

(iii) CH��CH�CH�CH

�Cl

(iv) CH��C � C�CH

��Br (v) (vi)

Ans. (i) CH��CH�

�Br

CH�

2-Bromopropane

(ii) CH��CH�CH

�I

�CH

�

1-Iodo-2-methyl propane

(iii) CH��CH�CH�CH

��Cl 1-Chloro-but-2-ene

(iv) CH��C � C�CH

��Br 1-Bromo-but-2-yne

(v) Bromobenzene

(vi) 1, 4-Dibromobenzene (p-dibromobenzene)

Unit

10.3 Methods of preparation of alkyl halides/aryl halides

10.3.1 From alcohol

Q. 12. How are following compounds obtained from alcohols :

(1) ethyl chloride C�H

�Cl (2) isopropyl chloride (CH

��CHCl�CH

�) (3) tert-butyl chloride (CH

�)�

–Cl ?

(1 mark each)

Ans. Alcohols in the presence of Lucas reagent which is a solution of concentrated HCl and ZnCl�

form alkyl halides.

Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols.

(1) C�H

��OH � HCl

anhydrous ZnCl�

�J C

�H

��Cl � H

�O

ethyl alcoholconc.

chloroethane

(2) CH��CH

�OH

�CH�

� HClanhydrous ZnCl

��

J CH��CH

�Cl

�CH�

� H�O

isopropyl alcohol

conc.

isopropyl chloride


(3) Tertiary alcohols don’t need ZnCl�

to react with HCl.

CH��

CH��

C��CH

�

OH � HCl J CH��

CH��

C��CH

�

Cl � H�O

tert-butyl alcohol tert-butyl chloride

The order of reactivity of alcohols with a given halo acid is 3°�2°�1°. ■

Do you know ?(Textbook page 212)

Zinc chloride is a Lewis acid and consequently can coordinate with the alcohol, weakening R-O bond. Mixture of

concentrated HCl and anhydrous ZnCl�

is called Lucas reagent.

Q. 13. How are following compounds prepared from alcohols :

(1) ethyl bromide (C�H

�Br) (2) isopropyl bromide (CH

��CHBr�CH

�)

(3) tert-butyl bromide (CH�)�C–Br ? (1 mark each)

Ans.

(1) Ethyl alcohol on heating with conc. hydrobromic acid (48%) forms ethyl bromide. OR

When ethyl alcohol is treated with a mixture of NaBr and H�SO

�, ethyl bromide is formed. Here HBr is generated

in situ.

C�H

��OH � HBr

NaBr�conc. H�SO

�IIIIIIIIIIIIIIIIIIIIIIIIJ C�H

��Br � H

�O

ethyl alcohol ethyl bromide

(2) Isopropyl alcohol, on reaction with NaBr and dil. H�SO

�forms isopropyl bromide.

CH��CH

�OH

�CH�

� HBrNaBr�H

�SO

�conc.

IIIIIIIIIIIIIIIIIIJ CH��CH�

�Br

CH�

� H�O

isopropyl alcohol isopropyl bromide

(3) Tertiary alcohol on reaction with sodium bromide and dil. H�SO

�forms tert-butyl bromide.

CH�

–

CH��

C–OH � HBr�CH

�

NaBr�dil. H�SO

�IIIIIIIIIIIIIIIIIIIIIIJ CH�

–

CH��

C–Br � H�O

�CH

�tert-butyl alcohol tert-butyl bromide ■

Q. 14. How is ethyl iodide obtained from ethyl alcohol ? (1 mark)

Ans. When ethyl alcohol is treated with sodium or potassium iodide in 95% phosphoric acid, ethyl iodide is formed.

Here HI is generated in situ.

CH��CH

��OH�HI

NaI/H�PO

��

CH��CH

��I�H

�O

ethyl alcohol ethyl iodide ■

Q. 15. How will you prepare the following : (1 mark each)

(1) Ethyl chloride (chloroethane) from ethyl alcohol using (i) PCl�

(ii) PCl�

and (iii) SOCl�.

Ans.

(i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.

3CH�CH

�OH � PCl

�

�IIIIIJ 3CH

�CH

�Cl � H

�PO

�ethyl alcohol ethyl chloride phosphorus acid


(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

CH�CH

�OH � PCl

�

�IIIIIJ CH

�CH

�Cl � POCl

�� HCl

ethyl alcohol ethyl chloride phosphoryl chloride

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethyl chloride is formed.

The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.

CH�CH

�OH � SOCl

�

pyridine

refluxIIIIIIIIIJ CH

�CH

�Cl � SO

�� HCl �

ethyl alcohol ethyl chloride

(2) Isopropyl chloride (2-chloropropane) from isopropyl alcohol using (i) PCl�

(ii) PCl�

(iii) SOCl�.

Ans. When isopropyl alcohol is refluxed with phosphorus trichloride, isopropyl chloride is formed.

(i) 3CH�

–CH–CH��PCl

�

�IIIIIJ 3CH

�–CH–CH

��H

�PO

�phosphorus acid� �

OH Cl

isopropyl alcohol isopropyl chloride

When isopropyl alcohol is refluxed with phosphorus pentachloride, isopropyl chloride is formed.

(ii) CH�

–CH–CH��PCl

�

�IIIIIJ CH

�–CH–CH

��POCl

��HCl

� �OH Cl


When isopropyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, isopropyl chloride is formed.

The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.

(iii) CH�

–CH–CH��SOCl

�

pyridine

refluxIIIIIIIIIJ CH

�–CH–CH

��SO

��HCl �

� �OH Cl


(3) Ethyl bromide (bromoethane) from ethyl alcohol.

Ans. When ethyl alcohol is treated with a mixture of red phosphorus and bromine or hydrobromic acid (phosphorus

tribromide is generated in situ), ethyl bromide is formed.

2P � 3Br�

IIIIIJ 2PBr�

3CH�CH

�OH �PBr

�

�IIIIIJ 3CH

�CH

�Br � H

�PO

�ethyl alcohol ethyl bromide

(4) Ethyl iodide (iodo ethane) from ethyl alcohol.

Ans. When ethyl alcohol is heated with a mixture of red phosphorus and iodine, (phosphorus triiodide is generated in

situ), ethyl iodide is formed.

2P � 3I�IIIIIJ 2PI

�3CH

�CH

�OH � PI

�IIIIIJ 3CH

�CH

�I�H

�PO

�ethyl alcohol ethyl iodide ■


Unit

10.3.2 From hydrocarbon

Q. 16. Explain halogenation of methane. (2 marks)

Ans. Halogenation : A reaction of alkanes with halogens (Cl�, Br

�, I

�) in the presence of appropriate conditions

forming a mixture of alkyl halides.

(1) Chlorination :

CH��Cl

�

UV lightIIIIIIIIIIIIJ CH

��Cl � HCl

(excess)

methane

methyl

chloride

(major product)

(2) When excess of chlorine is used, tetrachloro methane, a major product is obtained. When excess of methane is

used, chloromethane, a major product is obtained. The order of reactivity of halogens towards alkane is

F��Cl

��Br

��I

�.

CH��Cl

�

UV light

or �IIIIIIIIIIIIIIJ CH

�Cl�CH

�Cl

��CHCl

�� CCl

�� HCl

excess methyl

chloride

methylene

dichloride

chloroform Carbon tetrachloride

(major product)

(3) Iodination :

CH��I

�� CH

�–I�HI

methyl iodide

However, iodination reaction is a reversible reaction. HI being a strongest reducing agent reduces methyl iodide

back to methane.

(4) Fluorination : A reaction of alkane with fluorine is explosive and also hydrofluoric acid is poisonous and

corrosive. Hence, alkyl fluorides are not prepared by halogenation of alkane. ■

Q. 17. Predict the possible products of the following reaction :

(1) Bromination of propane (2) Bromination of n-butane (3) Bromination of 2-Methyl propane.

(3 marks)

Ans.

(1) Bromination of propane

CH��CH

��CH

�

Br�

UV lightIIIIIIIIIIIJ CH

��CH

��CH

��Br � CH

��CH�CH

��BrPropane 1-Bromopropane (3%)

2-Bromo propane (97%)(2) Bromination of n-butane

H�C�CH

��CH

��CH

�

Br�

UV lightIIIIIIIIIIIJ H

�C�CH

��CH

��CH

��Br � H

�C�CH

��CH�CH

��Brn-Butane

1-Bromobutane (2%)2-Bromobutane (98%)

(3) Bromination of 2-Methyl propane

H�C�

CH��

CH�CH�

Br�

UV lightIIIIIIIIIIIJ H

�C�

CH��

CH�CH��Br � H

�C�

CH��

C��Br

CH�

2-Methylpropane 1-Bromo-2-methylbutane

(trace) 2-Bromo-2-methylpropane (over 99%)■



Some times during replacement of – OH by – X, alcohols tend to undergo rearrangement. This tendency can be

minimized by use of phosphorous halides. Straight chain primary alcohols react with phosphorous trihalide to give

unrearranged alkyl halides.

Q. 18. How are following compounds prepared by halogenation of ethane :

(1) Chloroethane (2) Bromoethane (3) Iodoethane ? (1 mark each)

Ans.

(1) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorine in the presence of

diffused sunlight or U.V. light or at high temperature, chloroethane is obtained.

C�H

�Cl

�

diffusedsunlight

IIIIIIIIIIIIJ C�H

��Cl � HCl

(excess)ethane

Chloroethane

(2) Bromination of ethane : When ethane is heated with Br�

in the presence of anhydrous AlBr�, bromoethane is

obtained.

C�H

�Br

�

Anhydrous AlBr�IIIIIIIIIIIIIIIIIIIIIIJ C

�H

��Br � HBr

ethane Bromoethane

(3) Iodination : When ethane is reacted with I�

in the presence of suitable oxidising agents like–HgO or HIO�

or

dilute HNO�

iodoethane is obtained.

2 C�H

�2I

��HgO � 2C

�H

��I � HgI

�� H

�O

ethane mercuric Iodoethaneoxide

5C�H

�2I

��HIO

�� 5C

�H

��I � 3H

�O

ethane iodic acid Iodoethane

8C�H

�4I

��HNO

�� 8 C

�H

��I � 3H

�O � NH

�ethane Iodoethane ■

Q. 19. Explain :

(1) Halogenation of alkanes is not a good method of preparation of alkyl halides. (2 marks)

Ans.

(1) Direct fluorination of alkanes is highly exothermic, explosive and invariably leads to polyfluorination and

decomposition of the alkanes. It is difficult to control the reaction.

(2) Direct iodination of alkanes is highly reversible and difficult to carry out.

(3) In direct chlorination and bromination, the reaction is not selective. It can lead to different isomeric

monohalogenated alkanes (alkyl halides) as well as polyhalogenated alkanes.

Hence, halogenation of alkanes is not a good method of preparation of alkyl halides.

(2) Direct iodination of alkanes is not possible. (2 marks)

Ans.

(1) Direct iodination of alkanes using iodine is highly reversible.

RH � I�

� RI � HI

(2) Hydroiodic acid HI being strong reducing agent, it reduces RI to alkane RH.


(3) The reaction takes place only in the presence of a suitable oxidizing agent like HgO, HIO�

or dilute HNO�

which decomposes HI.

Hence, direct iodination of alkanes is not possible.

(4) 5C�H

�2I

��HIO

�� 5C

�H

��I�3H

�O

ethane iodoethane ■

Q. 20. How are following compounds obtained from alkenes :

(1) C�H

�Cl (2) C

�H

�Br (3) CH

��CH

�I

�CH�

(4) CH�

–CH�

–CH�I

–CH�

? (1 mark each)

Ans.

(1) Ethene on reaction with hydrogen chloride forms C�H

�Cl.

H�C�CH

��HCl � CH

��CH

��Cl

Ethene Chloroethane

(2) Ethene on reaction with hydrogen bromide forms C�H

�Br.

H�C�CH

��HBr � CH

��CH

��Br

Ethene Bromoethane

(3) Propene on reaction with hydrogen iodide forms CH��CHI�CH

�.

CH��CH�CH

��HI � CH

��CH�CH

��I

propene

2-Iodopropane

(4) but-2-ene on reaction with hydrogen iodide forms 2-iodobutane.

CH�

–CH�CH–CH��HI IIIIIJ CH

�–CH

�–CH–CH

�but-2-ene

�I 2-iodobutane ■

Q. 21. State and explain Markovnikov’s rule. (2 marks)

Ans. Markovnikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of

the reagent gets attached to that carbon atom of the double bond which carries less number of hydrogen atoms.

Example : Addition of HBr to unsymmetrical alkene like propene gives two products.

CH�

–CH�CH��HBr IIIII

� CH�

–CH–CH�

(Major product)�Br

IsopropylbromidePropene

� CH�

–CH�

–CH�

–Br (Minor product)

n-propylbromide

Isopropyl bromide is the major product, since the negative part (Br�) of HBr is attached to carbon atom of a

double bond with less number of hydrogen atoms. ■

Q. 22. Explain peroxide effect. OR Write a note on Kharasch-Mayo effect. OR

Explain the addition of HBr to (unsymmetrical alkene) propene in the presence of benzoyl peroxide.

(2 marks)

Ans. The addition of HBr to an unsymmetrical alkene (propene) in the presence of benzoyl peroxide takes place in the


opposite orientation to that of Markovnikov’s rule and this is known as Kharasch-Mayo effect or peroxide effect or

Anti-Markovnikov addition.

H�C�CH�CH

��H�Br

(CH

�CO)

�O

�Benzoyl peroxideIIIIIIIIIIIIIIIIIIIIIJ H

�C�CH

��CH

��Br

Propene 1-Bromopropane ■

Q. 23. Write the structure of alkyl halide obtained by the action hydrogen bromide on 2-Methylprop-

1-ene in the presence of peroxide. (1 mark)

Ans. In the presence of peroxide, HBr to 2-Methyl prop-1-ene forms 1-Bromo-2-methylpropane.

H�C�

CH��

C�CH��H�Br

(CH

�CO)

�O

�Benzoyl peroxideIIIIIIIIIIIIIIIIIIIIJ H

�C�

CH��

C��

H

CH��Br

Z-Methylprop-1-ene

1-Bromo-2-methylpropane ■


How will you obtain 1-bromo-1-methylcyclohexane from alkene ?

Write possible structures of alkene and the reaction involved.

Answer :

(1) (2)


Rewrite the following reaction by filling the blanks :

(1) CH��CH�CH

��HBr � ……�……

(major) (minor)

(2) (CH�)�C�CHCH

��HBr

peroxideIIIIIIIIIIIIJ ……�……

(major) (minor)

(3) CH��CH�CH

��HBr

peroxideIIIIIIIIIIIIJ ……�……

(major) (minor)

Ans.

(1) CH��CH�CH

��HBr � CH

��CH�

�Br

CH��CH

��CH

��CH

��Br

Propene

isopropyl bromide

(major)

n-propyl bromide

(minor)

(2) CH��

CH��

C�CH�CH��HBr

peroxideIIIIIIIIIIIIJ CH

��

CH��

CH� CH�CH��

Br

� CH��

CH��

C��

Br

CH��CH

�2-methyl but-2-ene

2-bromo-3-methyl butane

(major)

2-bromo-2-methyl butane

(minor)


(3) CH��CH�CH

��HBr


��CH

��CH

��Br � CH

��CH�CH

�Propene n-propyl bromide

(major)

isopropyl bromide

(minor)


Alkenes form addition product, vicinal dihalide, with chlorine or bromine usually in inert solvent like CCl�

at room

temperature.

(X�Cl, Br)


When alkenes are heated with Br�

or Cl�

at high temperature, hydrogen atom of allylic carbon is substituted with

halogen atom giving allyl halide.

CH��CH�CH

��Cl

�� CH

��CH�CH

�Cl�HCl

Unit

10.3.3 Halogen exchange

Q. 24. How are alkyl iodides prepared from alkyl chlorides/bromides ? (2 marks)

Ans. Alkyl iodide is prepared by treating alkyl chloride or alkyl bromide with sodium iodide, in the presence of dry

acetone, sodium chloride or sodium bromide precipitates from the solution and can be separated by filtration.

This reaction is known as Finkelstein reaction.

RX�NaIdry etherIIIIIIIIIIJ RI�NaX

R–Cl � NaIdry acetoneIIIIIIIIIIIIIIIJ RI � NaCl

alkyl chloride alkyl iodide

R–Br � NaIdry acetoneIIIIIIIIIIIIIIIJ RI � NaBr

alkyl bromide alkyl iodide

C�H

�Br � NaI

dry acetoneIIIIIIIIIIIIIIIJ C

�H

�I � NaBr

ethyl bromide ethyl iodide ■

Q. 25. How are alkyl fluorides prepared with alkyl chlorides / alkyl bromides ? (2 marks)

Ans. When alkyl chloride or alkyl bromide is heated with metallic fluorides like AgF, CaF�, CoF

�or Hg

�F�, alkyl

fluoride is formed.

R–X�AgF�

IIIIIJ R–F�AgX

CH�

–Cl � AgF�IIIIIJ CH

�F � AgCl

methyl methylchloride fluoride

This reaction is known as Swarts reaction. ■


Units

10.3.4 Electrophilic Substitution

10.3.5 Sandmeyer’s reaction

Q. 26. Explain the preparation of haloarenes using electrophilic substitution. (1 mark)

Ans. When arene is treated with chlorine or bromine in dark at ordinary temperature in the presence of lewis acid as a

catalyst like Fe, FeCl�

or anhydrous AlCl�, aryl chloride or aryl bromide is formed.

When toluene is brominated in dark at ordinary temperature in the presence of iron, a mixture of ortho and para

bromo tolerene is obtained.

Ortho and para isomers can be easily separated as there is large difference in melting points of ortho and para

isomers. ■

Q. 27. Write a note on Sandmeyer’s reaction. (3 marks)

Ans. Aryl halides are most commonly prepared by replacement of nitrogen of diazonium salt. The replacement of

diazonium group by –Cl or –Br using cuprous salt is called Sandmeyer’s reaction. When a primary aromatic amine

(like aniline) suspended in cold HCl, is treated with sodium nitrite, a diazonium salt (benzene diazonium chloride)

is formed. When diazonium salt is treated cuprous chloride or cuprous bromide, aryl halide (chlorobenzene or

bromobenzene) is formed.

When benzene diazonium salt is mixed with potassium iodide, iodobenzene is formed.

■

Units

10.4 Physical properties

10.4.1 Nature of intermolecular forces

10.4.2 Boiling point

10.4.3 Solubility


Q. 28. Give reasons : (2 marks each)

★ (1) Alkyl halides through polar are immiscible with water.

Ans.

(1) In alkyl halide, the halogen atom is more electronegative than carbon atom, the C�X bond is polar.

(2) Though alkyl halide is polar, it is insoluble in water because alkyl halide is not able to form hydrogen bonds with

water. Attraction between alkyl halide molecule is stronger than attraction between alkyl halide and water.

(2) C–F bond in CH�F is the strongest bond and C–I bond in CH

�I is the weakest bond. Explain.

Ans.

(1) Methyl fluoride (CH�F) is highly polar molecule and has the shortest C–F bond length (139 pm) and the strongest

C–F bond due to greater overlap of orbitals of the same principal quantum number i.e., overlap of 2sp� orbital of

carbon with 2pz orbital of fluorine.

(2) Methyl iodide (CH�I) is much less polar and has the longest (C–I) bond length (214 pm) and the weakest C–I bond

due to poor overlap of 2sp� orbital carbon with 5pz orbital of iodine i.e., 2sp� orbital of carbon cannot penetrate into

larger p-orbitals.

(3) The boiling point of alkyl iodide is higher than that of alkyl fluoride.

Ans. For a given alkyl group, the boiling point increases with increasing atomic mass of the halogen, because magnitude

of van der Waals force increases with increase in size and mass of halogen. Therefore, boiling point of alkyl iodide

is higher than that of alkyl fluoride.

(4) The boiling point of isopropyl bromide is lower than that of n-propyl bromide.

Ans. For isomeric alkyl halides (isopropyl bromide and n-propyl bromide), the boiling point decreases as the branching

increases, surface area decreases on branching and van der Waals forces decrease, therefore, the boiling point of

isopropyl bromide is lower than that of n-propyl bromide.

(5) p-Dichlorobenzene � � has m.p. higher than those of o-and m-isomers.

Ans. p-Dichlorobenzene has higher melting point than those of o-and m-isomers. This is because of its symmetrical

structure which can easily fits in crystal lattice. As a result intermolecular forces of attraction are stronger and

therefore greater energy is required to overcome its lattice energy. ■

★ Q. 29. Arrange the following in the increase order of boiling points :

(a) 1-Bromopropane (b) 2-Bromopropane (c) 1-Bromobutane (d) 1-Bromo-2-methylpropane (2 marks)

Ans. 1-Bromo-2-methylpropane, 2-Bromopropane, 1-Bromopropane, 1-Bromo butane ■


Arrange the following compounds in order of increasing boiling points : bromoform, chloromethane,

dibromomethane, bromomethane.

Ans. The comparative boiling points of halogen derivatives are mainly related with van der Waals forces of

attraction which depend upon the molecular size. In the present case all the compounds contain only one carbon.

Thus the molecular size depends upon the size of halogen and number of halogen atoms present.

Thus increasing order of boiling point is,

CH�Cl � CH

�Br � CH

�Br

�� CHBr

�


Units10.5 Optical isomerism in halogen derivatives

10.5.1 Chiral atom and molecular chirality10.5.2 Plane polarized light10.5.3 Optical activity10.5.4 Enantiomers


(1) Make a three-dimensional model of 2-chlorobutane.(2) Make another model which is a mirror image of the first model.(3) Try to superimpose the two models on each other.(4) Do they superimpose on each other exactly ?(5) Comment on whether the two models are identical or not.Ans. (1) (2) and (3)

Fig : Nonsuperimposable mirror images

(4) Two models are nonsuperimposable mir ror images of each other called enantiomers.

(5) Two enantiomers are identical. Theyhave the same physical properties (such as melting points, boiling points,densities refractive index). They also have identical chemical properties. The magnitude of their opticalrotation is equal but the sign of optical rotation is opposite.


Nicol prism is a special type of prism made from pieces of calcite, a crystalline form of CaCO�, arranged in a

specific manner. Nicol prism is also called polarizer.

Q. 30. Define the following : (1 mark each)Ans.(1) Monochromatic light : It consists of rays of single wavelength vibrating in different planes perpendicular to the

direction of propagation of the light.

(2) Plane polarized light : A light having oscillations only in oneplane perpendicular to direction of propagation of light is knownas plane polarized light.

Fig. Plane polarized light

Fig. Rotation of plane polarized lightdue to optically active substance


(3) Optical isomerism : The steroisomerism in which the isomers have different spatial arrangements of

groups/atoms around a chiral atom is called optical isomerism.

(4) Optical activity : The property of a substance by which it rotates plane of polarization of incident plane polarized

light is known as optical activity.

(5) Optically active compound : The compound which rotate the plane of plane polarized light is called optically

active compound.

(6) Enantiomers : The optical isomers which are non-superimposable mirror images of each other are called

enantiomers or enantiomorphs or optical antipodes.

Example : 2-chlorobutane, lactic acid

(7) Chiral carbon atom : Carbon atom in a molecule which carries four different groups/atoms is called chiral carbon

atom.

Chiral atom in a molecule is marked with asterisk ( * )

For example : C*

in lactic acid

H�

CH��

C*�OH

�COOH

(8) Chiral molecule : When a molecule contains one chiral atom, it acquires a unique property i.e. it is non-

superimposable with its mirror image is said to be chiral molecule.

(9) Chirality : The relationship between a chiral molecule and its mirror image is similar to the relationship between

left and right hands. Therefore it is called handedness or chirality.

(10) Dextrorotatory substance or d-Isomer : An optically active substance (or isomer) which rotates the plane of a

plane polarized light to the right hand side (RHS) is called dextrorotatory substance (or isomer) and denoted by d or

(�) sign.

(11) Laevorotatory substance or l-Isomer : An optically active substance (or isomer) which rotates the plane of a

plane polarized light to the left hand side (LHS) is called laevorotatory substance (or isomer) and denoted by l or

(�) sign.

(12) Racemic mixture or Racemate : A mixture containing equimolar quantities of dextro (d) and laevo (l) optical

isomers which is optically inactive due to molecules of one enantiomer is cancelled by equal and opposite optical

rotation due to molecules of the other enantiomer is called a racemic mixture or racemate. It is represented as (dl )

or (�). ■

Q. 31. Calculate the number of isomers for 2-chlorobutane. (1 mark)

Ans. The number of optical isomers possible for a compound is 2n where n�number of asymetric carbon atoms.

As n�1 for 2-chlorobutane, 2n�2�2. Hence, it has two optical isomers. ■

Q. 32. How many optical isomers are possible for C�H

Cl ? (1 mark)

Ans. The number of optical isomers : 3. ■

Q. 33. How many optical isomers are possible for glucose ? (1 mark)

Ans. The number of optical isomers : 16. ■


The phenomenon of optical isomerism in organic compounds was observed first and its origin in molecular

chirality was recognized later



● Optical activity is an experimentally observable property of compounds. Chirality is a description of molecular

structure. Optical activity is the consequence of chirality.

● Molecules which contain one chiral atom are chiral, that is, they are nonsuperimposable on their mirror image.

● The two non-superimposable mirror image structures are called pair of enantiomers.

● Enantiomers have equal and opposite optical rotation. Thus, enantiomers are a kind of optical isomers.

Q. 34. Draw the structures and indicate the chiral carbon atoms in (1) Lactic acid (2) 2-Chlorobutane.

(2 marks each)Ans.

(1) In lactic acid structure, CH�

–*CH – COOH, the starred carbon atom is chiral carbon atom as it is attached to

the starred�OH

four different substituents, COOH, OH, CH�

and H.

(2) In 2-chlorobutane structure, CH�

–

Cl�CH – CH

�– CH

�*

the starred carbon atom is chiral carbon atom as it is

attached to four different substituents, – CH�

– CH�

(ethyl), CH�

(methyl), Cl and H.

(The carbon atom marked with an asterisk (*) is an asymmetric carbon atom.) ■

Q. 35. Identify chiral and achiral molecules. (1 mark each)

(1) CH�

–CH–CH�

–CH��

Br

(2) CH�

–CHBr�

(3) CH�

–CH(OH)Br (4)

Ans. Chiral Ans. Achiral Ans. Chiral Ans. Chiral

■

★ Q. 36. Explain optical isomerism in 2-chlorobutane. (3 marks)

Fig. Enantiomers of 2-Chlorobutane(with structural and Fischer’s projection formula)

Ans.

(1) 2-Chlorobutane contains an asymmetric.

CH�

–

Cl�CH–CH

�–CH

�*carbon atom (the starred

carbon atom) which is attached to four different

groups, i.e., ethyl (– CH�

– CH�), methyl (CH

�),

chloro (Cl) and hydrogen (H) groups.

(2) Two different arrangements of these groups

around the carbon atom are possible as shown in

the figure. Hence, it exists as a pair of enanti-

omers. The two enantiomers are mirror images of

each other and are not superimposable.

(3) One of the enantiomers will rotate the plane of

plane-polarized light to the left hand side and is

called the laevorotatory isomer (l-isomer). The

other enantiomer will rotate the plane of plane-polarized light to the right hand side and is called the dextrorotatory

isomer (d-isomer).


(4) Equimolar mixture of the d- and the l-isomers is optically inactive and is called the racemic mixture or the racemate

(dl-mixture). The optical inactivity of the racemic mixture is due to external compensation. ■

Q. 37. Complete the following reactions and explain optical activity of the products formed : (1 mark each)

(i) Pent-1-ene with HBr

Ans. CH�

–CH�

–CH�

–CH�CH��HBr IIIIIJ CH

�–CH

�–CH

�–CH–CH

��BrPent-1-ene

2-bromopentane

CH�

–CH�

–H�C–

H�C–CH

��Br

2-bromopentane

(Optically active)(ii) Pent-2-ene with HBr

Ans. CH�

–CH�

–CH�CH–CH��HBr IIIIIJ CH

�–CH

�–CH–CH

�–CH

��Br

Pent-2-ene

3-bromopentane

H�C–H

�C–

H�C–CH

�–CH

��Br

3-bromopentane

(Optically inactive) ■

Q. 38. CH

�(A) on treatment with HCl produced a compound Y. Which is optically active, what is structure A?

(2 marks)

Ans. CH�

–CH�

–CH�CH–CH�

–CH��HCl IIIIIJ CH

�–CH

�–CH–CH

�–CH

�–CH

��Clhex-3-ene

3-chloro hexane

The structure of A�CH�

–CH�

–CH�CH–CH�

–CH�

hex-3-ene

Optically active compound�CH�

–CH�

–CH–CH�

–CH�

–CH��

Cl3-chloro hexane ■

★ Q. 39. Identify chiral molecule/s from the following : (1 mark each)

(a) CH��CH�

�OH

CH��CH

�(b) CH

��CH

��CH�

�Br

CH��CH

�

(c) CH��CH

��CH

��CH

�Br (d) CH

��CH�

�CH

�

CH��CH

�


Ans. Chiral molecule

(a) CH��

H�

C��

OH

CH��CH

�

■

Q. 40. A racemic mixture is optically inactive. Explain. (2 marks)

Ans.

(1) A racemic mixture contains equimolar (or equimolecular) quantities of the dextrorotatory (d-) and laevorotatory

(l-) isomers (enantiomers) of a compound.

(2) The d-enantiomer rotates the plane of plane-polarized light to the right, while the l-enantiomer rotates the same to

the left to the same extent.

(3) The quantities of the d- and l-enantiomers being the same, both the rotations are of the same magnitude, but of

opposite directions. Hence, they cancel each other.

Hence, a racemic mixture is optically inactive.

(4) It is represented as dl or (�). Example : (�) lactic acid ■

Unit

10.5.5 Representation of configuration of molecules

Q. 41. Explain fischer projection formula with illustration. OR

Write a note on fischer projection formula. (3 marks)

Ans. Fischer projection formula or cross formula : The three dimensional (3–D) view of a molecule is presented on

plane of paper. A Fischer projection formula can be drawn by visualizing the main carbon chain verical in the

molecule. Each carbon on the vertical chain is represented by a cross.

Conventionally the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines

represent the bonds going below the carbon. ■

Q. 42. Explain Wedge formula with illustration OR Write a note on Wedge formula. (3 marks)

Ans. Wedge formula : When a tetrahedral carbon is imagined to be

present in the plane of paper all the four bonds at this carbon cannot

lie in the same plane. The bonds in the plane of paper are represented

by normal lines, the bonds projecting above the plane of paper are

represented by solid wedges (or simply by bold lines) while bonds

going below the plane of paper are represented by broken wedges

(or simply by broken lines).■



(1) Draw structures of enantiomers of lactic acid �CH

��CH�

�OH

COOH

� using Fischer projection formulae.

(2) Draw structures of enantiomers of 2-bromobutane using wedge formula.

Ans.

(1) Fischer projection formula : CH��CH�

�OH

COOH

(2) Wedge formula : 2-bromobutane

Units

10.6 Chemical properties

10.6.1 Laboratory test of haloalkanes

10.6.2 Nucleophilic substitution reaction of haloalkanes

Q. 43. Give a laboratory test to confirm the presence of halogen in the original organic compound. (1 mark)

Ans. Haloalkanes are of neutral type in aqueous medium. On warming with aqueous sodium or potassium hydroxide the

covalently bonded halogen in haloalkane is converted to halide ion.

R�X�OH�� IIIIIJ R�OH�X��

When this reaction mixture is acidified by adding dilute nitric acid and silver nitrate solution is added a precipitate

of silver halide is formed which confirms presence of halogen in the original organic compound.

Ag�� (aq)�X�� (aq) IIIIIIIJ AgX� (s)silver halide ■

Q. 44. Define the following : (1 mark each)

(1) Mechanism of a reaction : It is a step by step description of exactly how the reactants are transformed into

products in as much details as possible.

(2) Substitution reaction : When a group bonded to a carbon in a substrate is replaced by another group to get a

product with no change in state of hybridization of that carbon, the reaction is called substitution reaction. ■


Q. 45. Describe the action of aqueous KOH (or NaOH) on :(1) ethyl bromide (2) isopropyl bromide (3) tert-butyl chloride (4) methyl bromide(5) 2-chlorobutane. (1 mark each)

Ans.(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol

is formed. The reaction is called a hydrolysis reaction.

CH�

– CH�

– Br� KOHboil

J CH�

– CH�

– OH�KBr

ethyl bromide(aq)

ethanol

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide,

isopropyl alcohol is formed.

CH�

– CH – CH�

� KOH�Br

boilJ

CH�

– CH – CH�

� KBr�OH

isopropyl bromide

(aq)

isopropyl alcohol

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is

formed.

CH��

CH�

– C�CH

�

– Cl � KOHboil

J

CH��

CH�

– C�CH

�

– OH � KCl

tert-butyl chloride

(aq)

tert-butyl alcohol

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl

alcohol (methanol).

CH�

– Br � KOHboil

J CH�

– OH � KBr

methyl bromide(aq)

methyl alcohol

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

CH�

–

Cl�CH – CH

�– CH

�� KOH

boilJ CH

�–

OH�CH – CH

�– CH

�� KCl

2-Chlorobutane (aq) Butan-2-ol ■

Q. 46. Describe the action of sodium ethoxide on (1) ethyl bromide (2) methyl bromide : OR

Write a note on Williamson’s synthesis. (2 marks) OR How are ethers prepared from alkyl halides ?(2 marks)

Ans. Williamson’s synthesis : When an alkyl halide (R�X) is heated with sodium alkoxide (R�O�Na), an ether

is obtained. In this reaction halide (–X) of alkyl halide is replaced by an alkoxy group (–OR). This reaction is

known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed

(or unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

2R�OH � 2Na IIIIIJ 2R�O� �

Na � H�

E.g. 2C�H

�OH � 2Na IIIIIJ 2 C

�H

�O� �

Na � H�

(1) Simple (symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated,

symmetrical ether is obtained.

R�O�Na � RX�IIIIJ R�O�R � NaX

e.g., When ethyl bromide is heated with sodium ethoxide, diethyl ether is formed.


C�H

��O�Na � C

�H

��Br

�IIIIJ C

�H

��O�C

�H

�� NaBr

sodium ethoxide Bromoethane diethyl ether

(2) Mixed (unsymmetrical) ether : When an alkyl halide and sodium alkoxide having different alkyl groups are

heated, unsymmetrical ether is obtained.

R�O�Na � R��X�IIIIJ R�O�R� � NaX

When methyl bromide is heated sodium ethoxide, ethyl methyl ether is formed.

C�H

��O�Na � CH

��Br

�IIIIJ C

�H

��O�CH

�� NaBr

sodium ethoxide methylbromide

ethyl methyl ether

When ethyl bromide is heated with sodium methoxide, ethyl methyl ether is formed.

CH�

–O–Na�C�H

�Br

�IIIIJ C

�H

�–O–CH

��NaBr

Sodium methoxideethylbromide

ethyl methyl ether ■

Q. 47. What is the action of silver salt of carboxylic acid on alkyl halide ? (1 mark)

Ans. When an alkyl halide (R�X) is heated with silver salt of carboxylic acid (R�COOAg), an ester is obtained.

R�COOAg�R�X �alcoholIIIIIIIIIJ R�COOR��Agx

silver salt ofcarboxylic acid

Ester■

Q. 48. Describe the action of alcoholic silver acetate on (1) methyl bromide (2) ethyl bromide. (1 mark each)

Ans.

(1) Methyl bromide : When methyl bromide is heated with an alcoholic silver acetate, methyl acetate is formed.

CH�

– COOAg�CH�

– Br�

alcoholIIIIIIIIIIJ CH

�– COO – CH

��AgBr

silver acetate methyl acetate

(2) Ethyl bromide : When ethyl bromide is heated with an alcoholic silver acetate, ethyl acetate is formed.

CH�

– COOAg� CH�

– CH�

– Br�

alcoholIIIIIIIIIIJ CH

�– COO�CH

��CH

��AgI

silver acetate ethyl acetate ■

Q. 49. What is the action of alcoholic silver propionate on ethyl bromide ? (1 mark)

Ans. When ethyl bromide is heated with an alcoholic silver propionate, ethyl propionate is formed.

C�H

�–Br�C

�H

�COOAg

�

alcoholIIIIIIIIJ C

�H

�–COO–C

�H

��AgBr ■

Q. 50. Describe the action of excess of ammonia on (1) ethyl bromide (2) n-propyl bromide. (1 mark each)

Ans.

(1) Ethyl bromide : When ethyl bromide is boiled under pressure with an excess of alcoholic ammonia, ethylamine

(ethanamine) is formed. This is known as ‘ammonolysis’ of ethyl bromide.

CH�

– CH�

– Br � NH�

413 K

pressureIIIIIIIIIIIIJ CH

�– CH

�– NH

�� HBr

ethyl bromide(alc.)

(excess)ethyl amine

(2) n-propyl bromide : When n-propyl bromide is boiled under pressure with an excess of ammonia, n-propyl amine

(propanamine) is formed.


CH�

– CH�

– CH�

– Br�NH�

413 K

pressureIIIIIIIIIIIIJ CH

�– CH

�– CH

�– NH

��HBr

n-propyl bromide (alc.)

(excess)n-propyl amine

■

Q. 51. What is ammonolysis ? Give a suitable example for the reaction. (2 marks)

Ans. When an alkyl halide is boiled under pressure with an excess of alcoholic solution of ammonia (NH�),

corresponding (primary amine) alkyl amine is formed. This reaction is known as ammonolysis of alkyl halide.

R�X � NH�

pressureJ R�NH

�� HX

alkyl halide (alc.)(excess)

alkyl amine(primary amine)

Further refer to Q. 45. ■

Q. 52. Describe the action of aqueous alcoholic potassium cyanide on (1) ethyl bromide (2) methyl iodide.

(2 marks)Ans.

(1) Ethyl bromide : When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide

in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.

CH�

– CH�

– Br � KCNboil

J CH�

– CH�

– CN � KBrethyl bromide ethyl cyanide(alc.)

(2) Methyl iodide : When methyl iodide is boiled with alcoholic solution of potassium cyanide, methyl cyanide is

formed.

CH�

–I�KCNboilIIIIIJ CH

�–CN�KI

(alc.) methyl cyanide ■

Q. 53. Describe the action of alcoholic silver cyanide on (1) ethyl bromide (2) methyl chloride. OR

Explain isocyanide reaction of (1) ethyl bromide (2) methyl chloride. (2 marks)

Ans.

(1) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl isocyanide is formed.

CH�

– CH�

– Br�AgCNheat

J CH�

– CH�

– NC�AgBr

ethyl bromide ethyl isocyanide (ethyl carbylamine)(alc.)

(2) Methyl chloride : When methyl chloride is heated with alcoholic silver cyanide, methyl isocyanide is formed.

CH�

– Cl�AgCNheat

J CH�

– NC�AgCl

methyl chloride methyl isocyanide(alc.)

The above reactions (1) and (2) are called isocyanide reaction. ■

Q. 54. Describe the action of potassium nitrite on (i) ethyl bromide. (ii) methyl chloride. (2 marks)

Ans.

(1) Ethyl bromide : When ethyl bromide is treated with potassium nitrite, ethyl nitrite is formed.

CH��CH

��Br � K

�

O�

�N�O IIIIIJ CH��CH

��O�N�O�KBr

ethyl bromide potassium

nitrite

ethyl nitrite

(2) Methyl chloride : When methyl chloride is treated with potassium nitrite, methyl nitrite is formed.

CH��CH

��Cl � K

�

O�

�N�O IIIIIJ CH�� O�N�O�KCl

methyl chloride potassium

nitrite

methyl nitrite■


Q. 55. Describe the action of silver nitrite on (1) ethyl chloride (2) n-propyl bromide. (2 marks)

Ans.

(1) Ethyl chloride : When ethyl chloride is treated with silver nitrite, nitroethane is obtained.

(2) n-Propyl bromide : When n-propyl bromide is treated with silver nitrite, nitropropane is obtained.

■


Cyanide ion is capable of attacking through more than one site (atom).

Such nucleophiles are called ambident nucleophiles. KCN is predominantly ionic (K��C�� N) and provides

cyanide ions. Both carbon and nitrogen are capable of donating electron pair. C�C Bond being stronger than C–N

bond, attack occurs through carbon atom of cyanide group forming alkyl cyanides as major product. However

AgCN (Ag�C � N) is mainly covalent compound and nitrogen is free to donate pair of electron. Hence attack

occurs through nitrogen resulting in formation of isocyanide.

Another ambident nucleophile is nitrite ion, which can attack through ‘O’ or ‘N’.


Alkyl halides when treated with alcoholic solution of silver nitrite give nitroalkanes whereas with sodium nitrite

they give alkyl nitrites. Explain.

Ans. Nitrite ion is an ambident nucleophile, which can attack through ‘O’ or ‘N’.

Both nitrogen and oxygen are capable of donating electron pair. C�N bond, being stronger than N�O bond,

attack occurs through C atom from alkyl halide forming nitroalkane.

However, sodium nitrite (NaNO�) is an ionic compound and oxygen is free to donate pair of electrons. Hence,

attack occurs through oxygen resulting in the formation of alkyl nitrite.

★ Q. 56. Write structure and IUPAC name of the major product in each of the following reaction. (1 mark each)

Ans. Structure and IUPAC name

(1) CH��CH�

�CH

�

CH�Cl�NaI

AcetoneIIIIIIIIIIIIJ CH

��CH�

�CH

�

CH�I�NaCl

1-Iodo-2-methyl propane


(2) CH��CH

�Br�SbF

�IIIIIIIJ F

��

CH�

�CH

��Br

1,1-difluoro-2-bromoethane

(3) CH��CH�

�CH

�

CH�CH��HBr

PeroxideIIIIIIIIIIIIJ CH

��CH�

�CH

�

CH��CH

��Br

1-Bromo-3-methyl butane

(4)

(5)

■

Q. 57. How will you convert the following : (1 mark each)

(1) Ethyl bromide to ethanol.

Ans. C�H

�Br � NaOH

�IIIIJ C

�H

�OH � NaBr

ethyl bromide ethanol

(2) Ethyl bromide to propane nitrile.

Ans. C�H

�Br � KCN

�IIIIJ C

�H

�CN � KBr

ethyl bromide (alc.) propane nitrile

(3) Ethyl bromide to ethyl amine.

Ans. C�H

�Br � NH

�

�

pressureIIIIIIIIIIIJ C

�H

�NH

�� HBr

ethyl bromide (alc.) ethyl amine

(4) Ethyl bromide to ethyl acetate.

Ans. CH�COOAg � C

�H

�Br

�IIIIJ CH

�COOC

�H

�� AgBr

silver acetate ethyl bromide ethyl acetate

(5) Ethyl bromide to ethyl isocyanide.

Ans. C�H

�Br � AgCN

�IIIIJ C

�H

�NC � AgBr

ethyl bromide (alc.) ethyl isocyanide

(6) Ethyl bromide to ethyl methyl ether.

Ans. C�H

�Br � CH

�–ONa

�IIIIJ C

�H

�–O–CH

�� NaBr

ethyl bromide ethyl methyl ether

(7) Ethyl bromide to n-butane.

Ans. C�H

�Br � Na � C

�H

�Br

dry etherIIIIIIIIIIIIJ CH

��CH

��CH

��CH

�� 2NaBr

ethyl bromide n-butane

(8) Ethyl bromide to Ethyl magnesium bromide.

Ans. C�H

�Br � Mg

dry etherIIIIIIIIIIIIJ C

�H

�MgBr

ethyl bromide ethyl magnesium bromide ■


Unit

10.6.3 Mechanism of SN reaction

Q. 58. Define the following : (1 mark each)

Ans.

(1) Nucleophilic bimolecular reaction (SN�) : The substitution reaction in which a nucleophile reacts with the

substrate and the rate of the reaction depends on the concentration of the substrate and the nucleophile is called a

nucleophilic bimolecular reaction.

Example : CH�Br � OH� � CH

�OH � Br�

substrate nucleophile

(2) SN reaction : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the

reaction depends only on the concentration of the substrate is called nucleophilic unimolecular or first order

reaction or SN reaction. ■

Q. 59. Explain, the mechanism of alkaline hydrolysis (reaction with aqueous KOH) of tert-butyl bromide

(2-Bromo-2-methylpropane) with energy profile diagram. (4 marks) OR

Explain only reaction mechanism for alkaline hydrolysis of tert-butyl bromide. (2 marks)

Ans.

(i) Consider alkaline hydrolysis of tert-butyl bromide (2-Bromo-2-methylpropane) with aqueous NaOH or KOH.

CH�

–

CH��

C – Br�CH

�

� OH� IIIIIIJ CH�

–

CH��

C – OH�CH

�

� Br�

tert-butyl bromide nucleophile tert-butyl alcohol

(substrate)

(ii) Kinetics of the reaction : Due to steric hindrance of voluminous three methyl groups around carbon, nucleophile

OH� cannot attack carbon atom directly. Hence, the reaction takes place in two steps.

Step I : This involves heterolytic fission of C�Br covalent bond in the substrate forming carbocation and Br�.

This is a slow process.

CH�

–

CH��

C – Br�CH

�

slow

RDSCH

�–

CH��

C�

�CH

�

� Br�

Carbonium ion

Step II : This step involves attack of nucleophile OH� or carbocation forming C�OH bond and product tert-butyl

alcohol. Since it involves ionic charge neutralisation, it is a fast step.

CH�

–

CH��

C�

�CH

�

� OH� fastIIIIIIJ CH

�–

CH��

C – OH�CH

�Rate Determining Step (R.D.S.) : Since the first step is a slow step, it is R.D.S., and therefore the rate of the

reaction depends on the concentration of only one reactant, (CH�)�

C�Br.

Rate�R�k [(CH�)�

C�Br] where k is a rate constant of the reaction.

SN reaction : The reaction between tert.butyl bromide and hydroxide ion to form tert.butyl alcohol follows

a first – order kinetics. The rate of this reaction depends only on the concentration of one substance (tert-butyl


bromide) and is independent of concentration of alkali added. It is unimolecular first (1st) order Nucleophilic

Substitution reaction denoted as SN reaction.

Stereochemistry and mechanism of the reaction : The reaction takes place in two steps and both the steps

involve formation of transition states (T.S.).

T.S. – I for first step :

Fig. : Formation of carbocation intermediate

In this transition state, C�Br bond is partially broken, so that carbon atom carries partial positive charge

(��) and Br carries partial negative charge (��) which further breaks forming carbocation and Br�. Tert-butyl

cation (carbocation) has a planar structure and the CH�

–C–CH�

bond angle is 120°. It is the intermediate of the

reaction. It is unstable. In this step, hybridisation of carbon atom changes from sp� (tetrahedral geometry) to sp�

(planar geometry).

T.S. – II for second step :

Fig. : SN mechanism

In this transition state, C�OH bond is partially formed so that carbon atom carries partial positive charge

(�� ) and OH carries partial negative charge (�� ) which further forms tert-butyl alcohol.

Formation of a racemic mixture : Since OH� has equal probability of the attack on carbocation from

frontside and from backside, the products obtained are equal. In case of optical active alkyl halide, a racemic

mixture is obtained. ■

Q. 60. Discuss SN� mechanism of methyl bromide using aqueous KOH. Draw energy profile diagram. OR

Discuss the mechanism of alkaline hydrolysis of methyl bromide or Bromomethane. (4 marks)

Ans.

(1) Consider alkaline hydrolysis of methyl bromide (Bromomethane), CH�Br with aqueous NaOH or KOH.

CH��Br � OH� � CH

��OH � Br�

Bromomethane nucleophile Methanolsubstrate

(2) Stereochemistry and Kinetics of the reaction (R.D.S.) : This hydrolysis reaction takes place only in one step

which is a rate determining step i.e. R.D.S. The rate of hydrolysis reaction depends on the concentration of CH�Br

and OH� which are present in the R.D.S. of the reaction.

Rate�R�k [CH�Br] [OH�]

where k is rate constant of the reaction.


SN� reaction : The reaction between methyl bromide and hydroxide ion to form methanol follows a second orderkinetics, since the rate of the reaction depends on the concentrations of two reacting species, namely methylbromide and hydroxide ion it is bimolecular second order (2nd) Nucleophilic Substitution reaction denoted by SN�.

(3) Mechanism of the reaction :(i) It is a single step mechanism. The reaction takes place in the following steps :

Fig. : Backside attack of nucleophile in SN� mechanism

(ii) Backside attack of the nucleophile : Nucleophile, OH� attacks carbon atom of CH�Br from back side

i.e. from opposite side to that of the leaving group i.e. Br� to experience minimum steric repulsion andelectrostatic repulsion between the incoming nucleophile (OH�) and leaving Br�.

(iii) Transition state : When a nucleophile, OH� approaches carbon atom of CH�Br, the potential energy of the

system increases until a transition state (T.S.) of maximum potential energy is formed in which C�Br bond ispartially broken and C�OH bond is partially formed. The negative charge is equally shared by bothincoming nucleophile –OH� and outgoing, leaving group –Br�. (Thus, the total negative charge is diffused.)

(iv) In CH�Br, carbon atom is sp�-hybridized and CH

�Br molecule is tetrahedral. The hybridisation of carbon

atom changes to sp� hybridisation. The transition state contains pentacoordinate carbon having three � (sigma)bonds in one plane making bond angles of 120° with each other i.e., H

, H

�and H

�atoms lie in one plane

while two partial covalent bonds containing Br and OH lie collinear and on opposite sides perpendicular to theplane.

(v) Inversion of configuration : The transition state decomposes fast by the complete breaking of the C–Brbond and the new C–OH bond is formed on the other side. The breaking of C–Br bond and the formation ofC–OH bond take place simultaneously. The energy required to break the C–Br bond is partly obtained fromthe energy released when C–OH bond is formed. The formation of product CH

�OH is accompanied by

complete or 100% inversion of configuration forming again sp�-hybridized carbon atom giving tetrahedralCH

�OH molecule. But in this structure the positions of H

�and H

�atoms in the reactant (CH

�Br) and in

product are on the opposite side. This inversion of configuration is called Walden inversion. ■


● Draw the Fischer projection formulae of two products obtained when compound (A) reacts with OH�� by

SN1 mechanis.

H�C�

n–

C�H

��C�Br�

C�H

(A)

Ans. CH��

n–

C�H

��C�Br�C

�H

� OH�� IIIIIIJ H�C�

n–

C�H

��C�OH�

C�H

� HO�

n–

C�H

��C�CH

��C

�H

I II

(Racemic mixture)


● Draw the Fischer projection formula of the product formed when compound (B) reacts with OH�� by SN2

mechanism.

H�

CH��

C�Cl (B)�

C�H

�

Ans. H�

CH��

C�Cl�

C�H

�

�OH�� IIIIIIIJ HO�

CH��

C�H�

C�H

�

�Cl�

Unit

10.6.4 Factors influencing SN1 and SN2 mechanism

Q. 61. Discuss the factors influencing SN1 and SN2 mechanism. (4 marks)

Ans.

(1) Nature of substrate : SN2 : The transition state (T.S.) of SN2 mechanism � is pentacoordinate, it is crowded. As a

result SN2 mechanism is favoured in primary halides and least favoured in tertiary halides.

SN1 : A planar carbocation intermediate is formed in SN1 reaction. Bulky alkyl groups can be easily accommodated

in planar carbocation and it has no steric crowding. As a result SN1 mechanism is favoured in tertiary halides and

least favoured in primary halides.

The carbocation intermediate is stabilized by�effect of alkyl substituents and also by hyperconjugation

effect of alkyl substituents containing -hydrogens. As a result, SN1 mechanism is favoured in tertiary halides and

least favoured in primary halides.

SN1 rate increases

CH��X 1° 2° 3°

IIIIIIIIIIIIIIIIIIIIIIIIIJ

SN2 rate increasesZIIIIIIIIIIIIIIIIIIIIII

Thus, tertiary alkyl halides undergo nucleophilic substitution by SN1 mechanism while primary halides follow SN2

mechanism.

(2) Nucleophilicity of the reagent : A strong nucleophile attacks the substrate faster and favours SN2 mechanism. The

rate of SN1 mechanism is independent of the nature of nucleophile. Nucleophile does not react in the 1st step (slow

step) of SN1. Nucleophile reacts fast after the carbocation intermediate is formed.

(3) Solvent polarity : (1) SN1 reaction proceeds more rapidly in polar protic solvents than in aprotic solvent. Polar

protic solvent decreases the rate of SN2 reaction. (2) In SN2 mechanism, rate depends on substrate as well as

nucleophile. A polar solvent stabilizes nucleophile by solvation. Thus solvent deactivates the nucleophile by

stabilizing it. Hence, aprotic solvents or solvent of low polarity will favour SN2 mechanism. ■



Allylic and benzylic halides show high reactivity towards the SN1 mechanism than other primary alkyl

halides. Explain. (2 marks)

Ans. In allylic and benzylic halide, the carbocation formed undergoes stabilization through the resonance. Hence,

allylic and benzylic halides show high reactivity towards the SN1 reaction. The resonating structures are

Resonance stabilization of allylic carbocation

Resonance stabilization of benzylic carbocation


1. A negatively charged nucleophile is more powerful than its conjugate acid. For example, R�O�� is better

nucleophile than R�OH.

2. When donor atoms are from same period of periodic table, nucleophilicity decreases from left to right in a

period. For example, H�O��

is less powerful nucleophile than NH�.

3. When donor atoms are from same group of the periodic table, nucleophilicity increases down the group. For

example, 1�� is better nucleophile than Cl��.


Which of the following two compounds would react faster by SN2 mechanism and Why ?

CH��CH

��CH

��CH

�Cl CH

��CH�

�Cl

CH��CH

�

1-Chlorobutane 2-Chlorobutane

Answer : In SN2 mechanism, a pentacoordinate T.S. is involved. The order of reactivity of alkyl halides towards

SN2 mechanism is,

Primary � Secondary � Tertiary, (due to increasing crowding in T.S. from primary to tertiary halides.

1-Chlorobutane being primary halide will react faster by SN2 mechanism, than the secondary halide

2-chlorobutane.)


★ Q. 62. Distinguish between SN1 reaction and SN2 reaction : (2 marks)

Ans. Factor SN1 SN2

(1) Number of steps Two steps One step

(2) Molecularity/Order Unimolecular/1st order Bimolecular/2nd order

(3) Reaction rate Depends upon concentration of one reacting

species

Depends upon concentration of two

reacting species

(4) Attack of a nucleo-

phile

Back side attack and front side attack on a

substrate with equal probability

Only back side attack on a substrate

(5) Transition state Two steps, two transition states One step, one transition state

(6) Type of substrate Mainly tertiary (3°) substrate Mainly primary (1°) substrate

(7) Stereochemistry 50% inversion and 50% retention of configur-

ation

100% inversion of configuration

(8) Enantiomer Forms racemic mixture Forms opposite enantiomer

(9) Solvent Polar solvent favourable Nonpolar solvent favourable

(10) Energy of activation Two values of energies of activation One value of energy activation

(11) Intermediate Carbocation intermediates No intermediate

(12) Nucleophile Weak nucleophile favourable Strong nucleophile favourable

(13) Order of reactivity

in alkyl halides

Tertiary � Secondary � Primary Primary � Secondary � Tertiary

■

Q. 63. How does relative reactivity for alkaline hydrolysis with respect to SN2 and SN1 vary in the following alkyl

halides :

(1) Bromomethane (2) Bromoethane (3) 2-Bromopropane (4) 2-Bromo-2-methylpropane?

(2 marks each)

Ans.

(A) Relative reactivity for SN2 mechanism decreases in the order of :

Bromomethane � Bromoethane � 2-Bromopropane � 2-Bromo-2-methyl propane

CH�Br � CH

��CH

��Br � CH

��CHBr�CH

�� CH

�–

CH��

C – Br�CH

�

(B) Relative reactivity for SN1 mechanism decreases in the order of :

2-Bromo-2-methylpropane � 2-Bromopropane � Bromoethane � Bromomethane

CH�

–

CH��

C – Br�CH

�

� CH��CHBr�CH

�� CH

��CH

��Br � CH

�Br

■


Q. 64. Explain with reason the relative order of reactivity of 1°/2°/3° alkyl halides by SN1 mechanism. (2 marks)

Ans. In alkaline hydrolysis of an alkyl halide by SN1 mechanism, the formation of carbocation as an intermediate

product is involved.

The increasing order of a stability of carbocation is,

(1°) primary � (2°) secondary � (3°) tertiary

CH��CH

�� CH

��CH�CH

�� CH

��–

CH��

C�CH

�(less stable) (most stable)

The stability order for carbocation is 3° � 2° � 1°.

Therefore the increasing order of reactivity by SN1 mechanism of alkyl halides is

(1°) primary � (2°) secondary � (3°) tertiary ■

Q. 65. Which one of the following is more easily hydrolysed in SN1 and SN2 reaction by aqueous KOH,

CH

�CHCl C

H

�and C

H

�CH

�Cl ? (1 mark each)

Ans. In SN1 reaction CH

�CHCl C

H

�will be more easily hydrolysed than C

H

�CH

�Cl

In SN2 reaction CH

�CH

�Cl will be more easily hydrolysed than C

H

�CHCl C

H

�. ■

★ Q. 66. Which one compound from the following pairs would undergo SN2 faster ? (2 marks)

(1)

(2) CH�CH

�CH

�I and CH

�CH

�CH

�Cl

Ans.

(1) Since is a primary halide it undergoes SN2 reaction faster than .

(2) Since iodine is a better leaving group than chloride, 1-iodo propane (CH�CH

�CH

�I) undergoes SN2 reaction faster

than 1-chloropropane (CH�CH

�CH

�Cl). ■

Q. 67. Choose the member that will react faster than the following pairs by SN1 mechanism. (2 marks each)

(1) 1-bromo-2, 2-dimethyl propane or 2-bromopropane.

CH�

–

CH��

C—�CH

�

CH�

–Br CH�

–

CH–CH��

Br

1-bromo-2, 2-dimethyl propane 2-bromopropane

Ans. The reactivity of SN1 reaction depends on the steric hindrance, in 2-bromopropane, -carbon atom is attached to

two methyl groups suffers greater steric hindrance to nucleophilic attack than 1-bromo-2, 2-dimethyl propane.

Hence, 2-bromopropane react faster by SN mechanism.

(2) 2-Iodo-2-methyl butane or 2-iodio-3-methyl butane.

CH�

–CH�

–

CH��

C–CH��

I

CH�

–

CH��

CH–CH–CH��

I

2-Iodo-2-methyl butane 2-Iodo-3-methyl butane


Ans. Since, 2-Iodo-2-methyl butane is a tertiary alkyl halide, it undergoes SN reaction faster than 2-iodo-3-methyl

butane.

(3) 1-Chloro propane or 2-chloropropane.

CH�

–CH�

–CH�

–Cl CH�

–CH–CH��

Cl1-chloropropane

2-chloropropane

Ans. Since, 2-chloropropane is a secondary alkyl halide, it undergoes SN reaction faster than 1-chloropropane.

(4) 2-Iodo-2-methyl butane or tert-butyl chloride.

CH�

–

I�C–CH

�–CH

��CH

�

CH�

–

CH��

C–Cl�CH

�2-Iodo-2-methyl butane tert-butyl chloride

Ans. Since, iodine is a better leaving group than chloride 2-iodo-2-methyl butane undergo SN reaction faster than

tert-butyl chloride. ■

★ Q. 68. Observe the following and answer the questions given below :

CH��CH� X�

�

:�� CH��CH�

��X�

:(1) Name the type of halogen derivative

(2) Comment on the bond length of C�X bond in it

(3) Can react by SN1 mechanism ? Justify your answer.

Ans.

(1) Vinyl halide

(2) C�X bond length shorter in vinyl halide than alkyl halide. Vinyl halide has partial double bond character due to

resonance.

In vinyl halide, carbon is sp� hybridised. The bond is shorter and stronger and the molecule is more stable.

(3) Yes, It reacts by SN1 mechanism. SN1 mechanism involves formation of carbocation intermediate. The vinylic

carbocation intermediate formed is resonance stabilized, hence SN mechanism is favoured. ■

Unit

10.6.5 Elimination reaction : Dehydrohalogenation

Q. 69. Write a note on elimination reaction. OR Explain dehydrohalogenation reaction. (2 marks)

Ans. When alkyl halide having at least one -hydrogen is boiled with alcoholic solution of potassium hydroxide (KOH),

an alkene is formed due to elimination of hydrogen atom from -carbon and halogen atom from -carbon, is called

dehydrohalogenation.

�

H�

C��

�

X�

C�

� � KOH�

J �C��C

�� KX�H

�O

(alc.)

For example,

C

H��C

H��Br � KOH

�J H

�C�CH

��KBr�H

�O

ethyl bromide (alc.) ethylene


Tertiary butyl bromide when heated with alcoholic solution of potassium hydroxide forms isobutylene.

CH�

–

CH��

C–

CH��

�Br

KOH(alc.)

�IIIIIJ CH

�–

CH��

C�CH��KBr�H

�O

isobutylene

This reaction is called -elimination (or 1,2-elimination) reaction as it involves elimination of halogen and a

-hydrogen atom.

As hydrogen and halogen is removed in this reaction it is also known as dehydrohalogenation reaction. ■


The carbon bearing halogen is commonly called -carbon (alpha carbon) and any carbon attached to -carbon is

-carbon (beta carbon). Hydrogens attached to -carbon are -hydrogens.

Q. 70. Describe the action of alcoholic potassium hydroxide (alc. KOH) on

(1) ethyl bromide (2) n-propyl bromide (3) isopropyl bromide (4) tert-butyl chloride. (1 mark each)

Ans.

(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic

alkali), ethene (gas) is formed by the dehydrobromination reaction.

CH�

– CH�

– Br � KOHheat

J CH��CH

�� KBr � H

�O

ethyl bromide (alc.) ethene

(2) n-Propyl bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

CH�

–CH�

–CH�

–Br�KOHheat

J CH�

– CH�CH��KBr�H

�O

n-propyl bromide (alc.) propene

(3) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide,propene is formed.

CH�

– CH – CH�� KOH

heatJ CH

�– CH�CH

�� KBr � H

�O

� (alc.) propeneBr

isopropyl bromide

(4) Tert-butyl chloride : When tert-butyl chloride (2-chloro-2-methyl propane) is heated with alcoholic KOH,2-methyl propene is formed.

CH��

CH�

– C�CH

�

– Cl�KOHheat

J

CH��

CH�

– C�CH��KBr�H

�O

tert-butyl chloride

2-methyl propene(alc.)

■

Q. 71. Describe the action of alc.KOH on 2-bromobutane. (1 mark)When 2-bromobutane is boiled with alc.KOH on 2-bromobutane, a mixture of but-1-ene and but-2-ene isformed.


Ans.

CH��

CH��

CH

�Br

��

CH�

(2-bromobutane)

alc. KOH � �

� �loss of �-hydrogen loss of -hydrogen

HC��CH

��CH�CH

�(But-1-ene)


�(But-2-ene) ■

Q. 72. Explain Saytzeff’s rule with suitable example. (2 marks)

Ans. Saytzeff’s rule : In dehydrohalogenation reaction the preferred product is that alkene which has the greater

number of alkyl groups attached to the doubly bonded carbon atoms.

Hence the number of alkyl substituents on doubly bonded carbon atoms increases, the stability of the alkene

giving its major products.

Hence the increasing stability of alkenes is,

CH��CH

�� CH

��CH�R � R�CH�CH�R �

RR

C�CH�R �RR

C�CRR

Monosubstitutedalkene

Disubstitutedalkene

Trisubstitutedalkene

Tetrasubstitutedalkene

There are two types of hydrogens (

and �) therefore two alkenes are expected.

2

CH�

–

CH–�

CH�

–CH��2KOH

�IIIIJ H

�C–CH�CH–CH

��H

�C�CH–CH

�–CH

��2KBr�2H

�O

� (alc.) But-2-ene (Disubstituted alkene) (80%) But-1-ene (20%)Cl2-chlorobutane (Monosubstituted alkene) ■


Elimination versus substitution :

Alkyl halides undergo substitution as well as elimination reaction. Both reactions are brought about by basic

reagent, hence there is always a competition between these two reactions. The reaction which actually

predominates depends upon following factors.

(a) Nature of alkyl halides : Tertiary alkyl halides prefer to undergo elimination reaction whereas primary alkyl

halides prefer to undergo substitution reaction.

(b) Strength and size of nucleophile : Bulkier electron rich species prefers to act as base by abstracting proton,

thus favours elimination. Substitution is favoured in the case of comparatively weaker bases, which prefer to

act as nucleophile.

(c) Reaction conditions : Less polar solvent, high temperarture favours elimination whereas low temperature,

polar solvent favours substitution reaction.


Q. 73. Answer the following :

★ (1) HCl is added to a hydrocarbon ‘A’ (C�H

�) to give a compound ‘B’ which on hydrolysis with aqueous alkali

forms tertiary alcohol ‘C’(C�H

�O). Identify ‘A’, ‘B’ and ‘C’. (3 marks)

Ans. CH��

CH��

C�CH��HCl IIIIIJ CH

��

CH��

C�CH��

Cl

(A)

2-methyl propene

(B)

tert.butyl chloride

CH��

CH��

C�CH��aq.KOH

�Cl

�IIIIIIIIJ H

�C�

CH��

C�CH��

OH

(B) tert. butyl alcohol

(C)

A�CH��

CH��

C�CH�

B�CH��

CH��

C�CH��

Cl

C�CH��

CH��

C�OH�

CH�

★ (2) Complete the following reaction sequences by writing the structural formulae of the organic compounds

‘A’, ‘B’ and ‘C’.

(1) 2-Bromobutanealc.KOH

IIIIIIIIIIIIIJ ABr

�

IIIIIIIIIIIIJ BNaNH

�

IIIIIIIIIIIIJ C (3 marks)

(2) Isopropyl alcohol�

PBr�

IIIIIIIIIJ ANH

�excess

IIIIIIIIIIIIIIIJ B (2 marks)

Ans.

(1) CH��CH�

�Br

CH��CH

�

alc. KOH

�IIIIIIIIIIIIIIJ CH

��CH�CH�CH

��CH

��CH

��CH�CH

��2KBr�2H

�O

2-Bromobutane

80%

But-2-ene

20%

But-1-ene

(A)


��Br

�IIIIIIIIIIIJ CH

��CH�

�Br

CH��

Br

CH�

2,3-Dibromobutane

(B)

CH��CH�

�Br

CH��

Br

CH��2NaNH

�IIIIIIIIJ CH

��CH�

�NH

�

CH��

NH�

CH��2NaBr

2,3-diaminobutane

(C)

A�CH��CH�CH�CH

�

B�CH��CH�

�Br

CH��

Br

CH� C�CH

��CH�

�NH

�

CH��

NH�

CH�


(2) CH��CH�

�OH

CH�

�

PBr�

IIIIIIIIIIIJ CH��CH�

�Br

CH�

NH�

excessIIIIIIIIIIIJ CH

��CH�

�NH

�

CH�

(A)

isopropyl bromide

(B)

isopropyl amine

A�CH��CH�

�Br

CH isopropyl bromide

B�CH��CH�

�NH

�

CH�

isopropyl amine ■

★ Q. 74. Convert the following :

(1) Propene to propan-1-ol (2 marks)

Ans. CH��CH�CH

��HBr


��CH

��CH

�

Propene n-propyl bromide

CH��CH

��CH

��Br�aq. KOH

�IIIIIIIJ CH

��CH

��CH

��OH�KBr

Propan-1-ol

(2) Benzyl alcohol to benzyl cyanide (2 marks)

Ans.

(3) Ethanol to propane nitrile. (2 marks)

Ans. CH��CH

��OH�HBr IIIIIIJ CH

��CH

��Br�H

�O

Ethanol (48%) Ethyl bromide

CH��CH

��Br�KCN

alc.IIIIIIIIJ CH

��CH

��CN�KBr

Propane nitrile

(4) But-1-ene to n-butyl iodide. (2 marks)

Ans. 6CH��CH

��CH�CH

��B

�H

IIIIIIIJ 2 (CH

��CH

��CH

��CH

�)�

B

But-1-ene Tributyl borane

(CH��CH

��CH

��CH

�)�

B�3H�O

�

OH�

H�O

IIIIIIIJ 3CH��CH

��CH

��CH

��OH�B (OH)

�Butan-1-ol

3CH��CH

��CH

��CH

��OH�PI

�

red P/I�

�IIIIIIIIIIIIJ 3CH

��CH

��CH

��CH

�I�H

�PO

�

n-butyl iodide


(5) 2-Chloropropane to propan-1-ol. (3 marks)

Ans. CH��CH�

�Cl

CH��alc.KOH

�IIIIIJ CH

��CH�CH

��KCl�H

�O

Propene

6CH��CH�CH

��B

�H

IIIIIIJ 2 (CH

��CH

��CH

�)�

B

Propene tripropyl borane

(CH��CH

��CH

�)�

B�3H�O

�

OH�

H�O

IIIIIIIIJ 3CH��CH

��CH

��OH�B(OH)

�Propan-1-ol

(6) tert. butyl bromide to isobutyl bromide. (2 marks)

Ans. CH��

CH��

C��

CH�

Br�alc.KOH�

IIIIIIIJ CH��

CH��

C�CH��KBr�H

�O

tert. butyl bromide

2-Methyl prop-1-ene

CH��

CH��

C�CH��HBr


��

CH��

C�CH�Br

isobutyl bromide

(7) Aniline to chlorobenzene. (2 marks)

Ans.

(8) Propene to 1-nitropropane. (2 marks)


��HBr


��CH

��CH

��Br

Propene 1-bromopropane

CH��CH

��CH

��Br�AgNO

�

�

�NaClIIIIIIIIIJ CH

��CH

��CH

��NO

��AgBr(s)

1-nitropropane ■

Unit

10.6.6 Reaction with active metals

Q. 75. What is a Grignard reagent ? (2 marks)

Ans. Grignard reagent : An organometallic compound in which the divalent magnesium is directly linked to an alkyl

group (R�) and a halogen atom (X), and has general formula R�Mg�X is called Grignard reagent. OR

When alkyl halide is treated with magnesium in dry ether as solvent, it gives alkyl magnesium halide. It is known as

Grignard reagent.


R�X�Mgdry ether

IIIIIIIIIIIIJ R�Mg�X

alkyl magnesium halide

(Grignard reagent)

E.g.��CH

��

��Mg�

��I

The carbon-magnesium bond is highly polar and magnesium-halogen bond is in ionic in nature. Grignard

reagent is highly reactive. It is an important reagent and used in the preparation of a large number of organic

compounds. ■

Q. 76. How is Grignard reagent prepared ? (2 marks)

Ans. Grignard reagent is an alkyl magnesium halide, R�Mg�X obtained by the reaction of alkyl halide R�X with

magnesium (Mg) in dry ether.

R�X � Mgdry etherIIIIIIIIIIIJ

��R �

��Mg�

��X

alkyl halide alkyl magnesium halide

When an alkyl halide like CH�I is added from a dropping funnel to a flask containing pieces of pure Mg in pure

and dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH��Mg�I is formed.

CH��I � Mg

dry etherIIIIIIIIIIIJ

��CH

��

��Mg�

��I OR (

��CH

��

��Mg�

��I )

methyl iodide methyl magnesium iodide

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesium iodide.

C�H

�I�Mg

dry etherIIIIIIIIIIIJ C

�H

�–Mg–I OR (

��C�H

��

2��Mg�

��I )

ethyl magnesiumiodide ■

Q. 77. Write a note on Grignard reagent. (2 marks)

Ans. For the answer refer to Qs. 70 and 71. ■


Carbon-magnesium bond in Grignard reagent is a polar covalant bond. The carbon pulls electrons from the

electropositive magnesium. Hence carbon in Grignard reagent has negative polarity and acts as a nucleophite.

Victor Grignard received Nobel Prize in 1912 for synthesis and study of organomagnesium compounds. Grignard

reagent is a very versatile reagent used by organic chemist. Vinyl and aryl halides also form Grignard reagent.

Q. 78. Describe the action of water on (1) methyl magnesium iodide (2) ethyl magnesium iodide. (2 marks)

Ans.

(1) Methyl magnesium iodide : When methyl magnesium iodide is treated with water, methane is obtained

CH��Mg�I�H�OH IIIIIIJ CH

��Mg

OHI

Methane

(2) Ethyl magnesium iodide : When ethyl magnesium iodide is treated with water, ethane is obtained.

CH��CH

��Mg�I�H�OH IIIIIIIJ CH

��CH

��Mg

OHI

Ethane ■


Q. 79. Describe the action of ammonia on (1) ethyl magnesium bromide (2) n-propyl magnesium chloride.

(2 marks)

Ans.

(1) Ethyl magnesium bromide : When ethyl magnesium bromide is treated with ammonia, ethane is formed.

CH��CH

��Mg�Br�H�NH

�IIIIIIIJ CH

��CH

��Mg

BrNH

�ammonia Ethane

(2) n-Propyl magnesium chloride : When n-propyl magnesium chloride is treated with ammonia, propane is formed.

CH��CH

��CH

��Mg�Br�H�NH

�IIIIIIIJ CH

��CH

��CH

��Mg

BrNH

�ammonia Propane ■

Q. 80. Explain Wurtz reaction. (2 marks) OR Explain the action of sodium with alkyl halides. (2 marks)

Ans.

(1) When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher alkane is formed. This

is called Wurtz reaction or Wurtz coupling reaction.

R – X � 2Na � X – RetherIIIIIIIJ R – R � 2NaX

alkane

(2) In this reaction the alkyl radicals from two molecules of the reacting alkyl halide combine or couple to form the

higher alkane.

(3) Thus, methyl bromide reacts with sodium in ether to form ethane (C�H

), while ethyl bromide under the same

conditions forms n-butane (C�H

�).

2 CH�

– Br � 2NaetherIIIIIIIIJ CH

�– CH

�� 2NaBr

methyl bromide ethane

2 CH�

– CH�

– Br � 2NaetherIIIIIIIIJ CH

�– CH

�– CH

�– CH

�� 2NaBr

ethyl bromide n-butane

(Zinc metal can also be used in place of sodium.)

(4) If a mixture of two different alkyl halides is treated with Na in dry ether, then a mixture of alkanes is obtained

called self coupling products. For example, a mixture of CH�Br and C

�H

�Br gives propane along with C

�H

and

C�H

�.

CH�Br � 2Na � C

�H

�Br

dry etherJ C

�H

� C

�H

�� C

�H

�� NaBr

methyl bromide major major minor ■

Unit

10.6.7 Reaction of haloarenes

Q. 81. Explain the reaction of haloarene with alkyl halide and sodium metal. OR

Write a note on Wurtz–Fittig reaction. (2 marks)

Ans. When an alkyl halide and an aryl halide is treated with sodium metal in dry ether the corresponding alkylarene

(alkyl benzene) is formed. The reaction is known as Wurtz-Fittig reaction. This reaction allows alkylation of alkyl

halides.


■

Q. 82. Describe the action of aryl halide on sodium metal. (1 mark)

Ans. Aryl halide reacts with sodium metal in dry ether, biphenyl is formed. This reaction is known as Fittig reaction.

■


Conversion of chlorobenzene to phenol by aqueous sodium hydroxide requires high temperature of

about 623K and high pressure. Explain.

Ans. Due to partial double bond character in chlorobenzene, the bond cleavage in chlorobenzene is difficult and

are less reactive. Hence, during the conversion of chlorobenzene to phenol by aq. NaOH requires high

temperature & high pressure.

Electrophilic substitution (SE) in aryl halides

Q. 83. Identify the product A of following reaction. (Textbook page 228)

(1 mark)

Ans.

■


The �I effect of Cl is more powerful than its �R effect. Therefore Cl is o-/p- directing but ring deactivating

group.



Occurrence of nucleophilic substitution in p-nitrochlorobenzene can be explained on the basis of resonance

stabilization of the intermediate.

The resonance structure (III) shows that the electron withdrawing nitro group (�NO�) in the p-position

extends the conjugation. As a result, the intermediate carbanion is better stabilized which favours nucleophilic

substitution reaction.

Q. 84. Explain the following substitution reactions of chlorobenzene :

(1) Halogenation (2) Nitration (3) Sulphonation. (1 mark each)

Ans.

(1) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrous ferric chloride, a mixture

of ortho and para-dichlorobenzene (major product) is formed.

(2) Nitration : When chlorobenzene is heated with nitrating mixture (conc. nitric acid�conc. sulphuric acid) a

mixture of 1-chloro-4-nitro benzene (major product) and 1-chloro-2-nitrobenzene is formed.

(3) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixture of 4-chlorobenzene

sulphonic acid (major product) and 2-chlorobenzene sulphonic acid is formed.

■


Q. 85. Describe the action of the following on chlorobenzene :

(1) Methyl chloride in the presence of anhydrous AlCl�

(2) Acetyl chloride in the presence of anhydrous AlCl�. (1 mark each)

Ans.

(1) Methyl chloride in the presence of anhydrous AlCl�

: When chlorobenzene is treated with methyl chloride in

the presence of anhydrous AlCl�, a mixture of 1-chloro-4-methyl benzene (major product) and 1-chloro-2-methyl

benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s

alkylation.

(2) Acetyl chloride in the presence of anhydrous AlCl�

: When chlorobenzene is reacted with acetyl chloride in the

presence of anhydrous AlCl�, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is

formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

■

Q. 86. Write a note on Friedel Craft’s reaction. (2 marks)

Ans. For answer refer to Q. 85. ■

Q. 87. Give reasons : ★ (1) Haloarenes are less reactive than haloalkanes. (3 marks)

Ans. Haloarenes (Aryl halides) are less reactive than (alkyl halides) haloalkanes due to the following reasons :

(1) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with �-electrons of the

benzene ring. The delocalization of these electrons C-Cl bond acquires partial double bond character.

Due to partial double bond character of C-Cl bond in aryl halides, the bond cleavage in haloarene is difficult and

are less reactive. On the other hand, in alkyl halides, carbon is attached to chlorine by a single bond and it can be

easily broken.

(2) Aryl halides are stabilized by resonance but alkyl halides are not. Hence, the energy of activation for the

displacement of halogen from aryl halides is much greater than that of alkyl halides.


(3) Different hybridization state of carbon atom in C-X bond :

(i) In alkyl halides, the carbon of C-X bond is sp�-hybridized with less s-character and greater bond length of

178 pm, which requires less energy to break the C-X bond.

(ii) In aryl halides, the carbon of C-X bond is sp�-hybridized with more s-character and shorter bond length which

requires more energy to break C-X bond. Therefore, aryl halides are less reactive than alkyl halides.

(iii) Polarity of the C-X bond : In aryl halide C-X bond is less polar than in alkyl halides. Because sp�-hybrid

carbon of C-X bond has less tendency to release electrons to the halogen than a sp�-hybrid carbon in alkyl

halides. Thus halogen atom in aryl halides cannot be easily displaced by nucleophile.

(2) Aryl halides are extremely less reactive towards nucleophilic substitution reactions. (3 marks)

Ans. Aryl halides are extremely less reactive towards nucleophilic substitution reaction due to the following reasons :

(1) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with �-electrons of the

benzene ring. The delocalization of these electrons C-Cl bond acquires partial double bond character.

Due to partial double bond character of C-Cl bond in aryl halides, the bond cleavage in haloarene is difficult

and are less reactive towards nucleophilic substitution.

(2) Sp� hybrid state of C : Different hybridization state of carbon atom in C-X bond : In aryl halides, the carbon of

C-X bond is sp�-hybridized with more s-character and shorter bond length of 169 pm which requires more energy

to break C-X bond. It is difficult to break a shorter bond than a longer bond, in alkyl chloride (bond length 178 pm)

therefore, aryl halides are less reactive towards nucleophilic substitution reaction.

(3) Instability of phenyl cation : In aryl halides, the phenyl cation formed due to self ionisation will not be stabilized

by resonance which rules out possibility of SN1 mechanism. Also backside attack of nucleophile is blocked by the

aromatic ring which rules out SN2 mechanism. Thus cations are not formed and hence aryl halides do not undergo

nucleophilic substitution reaction easily.

(4) As any halides are electron rich molecules due to the presence of �-bond, they repel electron rich nucleophilic.

attack. Hence, aryl halides are less reactive towards nucleophilic substitution reactions. However, the presence of

electron withdrawing groups at o/p position activates the halogen of aryl halides towards substitution.

■

(3) Aryl halides undergo electrophilic substitution reactions slowly. (3 marks)

Ans. Aryl halides undergo electrophilic substitution reactions slowly and it can be explained as follows :

(1) Inductive effect : The strongly electronegative halogen atom withdraws the electrons from carbon, atom of the

ring, hence aryl halides show reactivity towards electrophilic attack.

(2) Resonance effect : The resonating structures of aryl halides show increase in electron density at ortho and para

position, hence it is o, p directing.


The inductive effect and resonance effect compete with each other. The inductive effect is stronger than

resonance effect. The reactivity of aryl halides is controlled by stronger inductive effect and o, p orientation is

controlled by weaker resonating effect.

The attack of electrophile (Y) on haloarenes at ortho and para positions are more stable due to formation of

chloronium ion. The chloronium ion formed is comparatively more stable than other hybrid structures of

carbonium ion.

★ (4) Reactions involving Grignard reagent must be carried out under anhydrous condition. (2 marks)

Ans.

(1) Grignard reagent (R Mg X) is an organometallic compound. The carbon-magnesium bond is highly polar and

magnesium halogen bond is in ionic in nature. Grignard reagent is highly reactive.

(2) The reactions of Grignard reagent are carried out in dry conditions because traces of moisture may spoil the

reaction and Grignard reagent reacts with water to produce alkane. Hence, reactions involving Grignard reagent

must be carried out under anhydrous condition.

★ (5) Alkyl halides are generally not prepared by free radical halogenation of alkane. (2 marks)

Ans.

(1) Free radical halogenation of alkane gives a mixture of all different possible Monohaloalkanes as well as

polyhalogen alkanes.

(2) In this method, by changing the quantity of halogen the desired product can be made to predominate over the other

products. Hence, alkyl halides are generally not prepared by free radical halogenation of alkane. ■

★ Q. 88. Complete the following reactions giving major product.

(1) CH��CH�CH

�

HBr

PeroxideIIIIIIIIIIIJ A

alc.KOHIIIIIIIIIIIIJ B (2 marks)

Ans. CH��CH

��CH

�

HBr

PeroxideIIIIIIIIIIIJ CH

��CH

��CH

��Br

alc.KOHIIIIIIIIIIIIJ CH

��CH�CH

��KBr�H

�O

(A) (B)

1-Bromopropane

(major product)

Propene

A�CH��CH

��CH

��Br

B�CH��CH�CH

�


(2) CH��CH�CH

��OH

Red P/Br�IIIIIIIIIIIIIIJ A

Ag�O/H

�O

IIIIIIIIIIIIIIJ B (2 marks)


��OH

Red P/Br�IIIIIIIIIIIIIIJ CH

��CH

�OH

�CH��Br

Ag�O/H

�O

IIIIIIIIIIIIIIJ CH��CH

�OH

�CH��OH

(A) (B)

A�CH��CH

�OH

�CH��Br, B�CH

��CH

�OH

�CH��OH

(3) CH��

CH��

C��

CH�

CH��Cl

Na/dry etherIIIIIIIIIIIIIIIJ A (1 mark)

Ans. CH��

CH��

C��

CH�

CH��Cl

Na/dry etherIIIIIIIIIIIIIIIJ CH

��

CH��

C��

CH�

CH��CH

��

CH��

C��

CH�

CH�

(A)

2,2,5,5�tetramethylhexane

A�CH��

CH��

C��

CH�

CH��CH

��

CH��

C��

CH�

CH�

(4) (1 mark)

Ans.

■

Q. 89. Name the reagent used to bring about the following conversions.

★ (1) Bromoethane to ethoxyethane ★ (2) 1-Chloropropane to 1 nitropropane ★ (3) Ethyl bromide to ethyl

isocyanide ★ (4) Chlorobenzene to biphenyl (5) Ethyl iodide to ethyl magnesium iodide (6) Ethyl chloride to

ethyl iodide (7) Toluene to a mixture of ortho and para bromo toluene. (1 mark each)


Ans.

Reactant Ans. Reagent Product

(1) Bromoethane�

Sodium alkoxideIIIIIIIIIIIIIIIIIIIJ ethoxy ethane

(2) 1-ChloropropaneSilver nitrite

alcoholIIIIIIIIIIIIIIIIIIIJ 1-nitropropane

(3) Ethyl bromideAlcohol

AgCNIIIIIIIIIIIIIIIIIIIJ Ethyl isocyanide

(4) ChlorobenzeneNa

dry etherIIIIIIIIIIIIIIIIIIIJ Biphenyl

(5) Ethyl iodideMg

dry etherIIIIIIIIIIIIIIIIIIIJ Ethyl magnesium iodide

(6) Ethyl chlorideNaI

acetoneIIIIIIIIIIIIIIIIIIIJ Ethyl iodide

(7) Toluene�Br�

Fe

darkIIIIIIIIIIIIIIIIIIIJ ortho and para bromo toluene

■

Q. 90. Convert 1-chlorobutane into the following compounds :

(1) butan-1-ol (2) 1-iodobutane (3) CH�

–CH�

–CH�

–CH�

–CN

(4) CH�

–CH�

–CH�

–CH�

–O–

O�C–CH

�. (1 mark each)

Ans.

(1) 1-Chlorobutane to butan-1-ol :

CH�

–CH�

–CH�

–CH�

–Cl�aq·KOH�IIIIJ CH

�–CH

�–CH

�–CH

�–OH�KCl

1-chlorobutane butan-1-ol

(2) 1-Chlorobutane to 1-iodobutane :

(i) CH�

–CH�

–CH�

–CH�

–Cl�aq·KOH�IIIIJ CH

�–CH

�–CH

�–CH

�–OH�KCl

1-chlorobutane butan-1-ol

(ii) 3CH�

–CH�

–CH�

–CH�

–OH�PI�

red P�I�IIIIIIIIIIIIIJ 3–CH

�–CH

�–CH

�–CH

�–I�H

�PO

�butan-1-ol 1-iodobutane

(3) 1-Chlorobutane to n-butyl cyanide (CH�

–CH�

–CH�

–CH�

–CN) :

CH�

–CH�

–CH�

–CH�

–Cl�KCNalcohol

�IIIIIIIIJ CH

�–CH

�–CH

�–CH

�–CN�KCl

1-chlorobutane potassium n-butyl cyanidecyanide

(4) 1-Chlorobutane to CH�

–CH�

–CH�

–CH�

–O–

O�C–CH

�:

CH�

–CH�

–CH�

–CH�

–Cl�Ag–O–

O�C–CH

�

�IIIIJ CH

�–CH

�–CH

�–CH

�–

O�C–O–CH

��AgCl .

1-chlorobutane silver acetate butyl acetate ■


Q. 91. Predict the expected product of substitution reactions : (1 mark each)

(1) Isobutyl chloride�sodium ethoxide.

Ans. CH�

–

CH��

CH–CH�

–Cl�NaOC�H

�IIIIIJ CH

�–

CH��

CH–CH�

–OC�H

��NaCl

Isobutyl chloride Sodium ethoxide Ethyl isobutyl ether

Product : CH�

–

CH��

CH–CH�

–OC�H

�ethyl isobutyl ether

(2) n-butyl chloride�sodium.

Ans. 2CH�

–CH�

–CH�

–CH�

–Cl�2Nadry etherIIIIIIIIIJ C

�H

��2NaCl

n-butyl chloride sodium n-octane

Product : C�H

�n-octane

(3) 1-chloropropane � aq. potassium hydroxide.

Ans. CH�

–CH�

–CH�

–Cl�aq. KOH�IIIIJ CH

�–CH

�–CH

�–OH�KCl

1-chloropropane propan-1-ol

Product : Propan-1-ol

(4) Aniline�NaNO�

/ HCl.

Ans. Product :

■

Q. 92. Write the products : (1 mark each)

(1) CH

�–CH

�–CH�CH

��HBr IIIIIJ C

H

�–CH

�–

Br�CH–CH

�

3-phenyl prop-1-ene 2-Bromo-3-phenyl propane

(Product)

(2) CH

�–CH�CH

��HBr

PeroxideIIIIIIIIIIIJ C

H

�–CH

�–CH

�–Br

Phenyl ethene 1-Bromo-2-phenyl ethane

(Product)

(3) CH�

–

CH��

C�

CH��

C–CH��HBr IIIIIJ CH

�–

CH��

C–CH�

–CH��

Br2,3-dimethyl but-2-ene

2-Bromo-2-methyl butane

(Product)


(4) CH�

–

CH��

C�CH–CH��HBr IIIIJ CH

�–

CH��

CH–CH–CH��

Br2-Methyl but-2-ene 2-Bromo-3-methyl butane

(Product)

(5)

■

Q. 93. Identify A and B in the following : (2 marks each)

(1)

Ans.

(2)

Ans.

(3) C�H

�OH

NaIIIIIJ A

C�H

�Br

IIIIIIIIIIJ B.

Ans. 2C�H

�OH�2Na

–H�

IIIIIIJ C�H

�ONa

C�H

�Br

IIIIIIIIIIJ C�H

�OC

�H

��NaBr

(A) C�H

�ONa (B) C

�H

�OC

�H

�Sodium ethoxide Diethyl ether

(4) CH�COOH

AgIIIIIJ A

C�H

�Br

IIIIIIIIIIJ B.

Ans. 2CH�OOH�2Ag

–H�

IIIIIIJ CH�COOAg � C

�H

�Br

�IIIIJ CH

�COOC

�H

��AgBr

(A) CH�COOAg (B) CH

�COOC

�H

�silver acetate ethyl acetate


(5) CH�

–CH�

–CH�

–ClNaIIIIIJ A

Cl�IIIIIIJ B.

Ans. 2CH�

–CH�

–CH�

–Cl�2Na–2NaCl�IIIIIIIIIIJ C

H

�

Cl�IIIIIIJ C

H

�Cl

(A) CH

�(B) C

H

�Cl

n-Hexane n-Hexyl chloride

(6) CH�

–CH�CH�

HBrIIIIIIJ A

alc. KOHIIIIIIIIIIIIIJ B.

Ans. CH�

–CH�CH��HBr IIIIIJ CH

�–CH–CH

��Br

alc. KOHIIIIIIIIIIIIJ CH

�–CH�CH

��KBr

(A) CH�

–CH–CH��

Br

(B) CH�

–CH�CH�

isopropyl bromide

propene

■

★ Q. 94. Match the pairs : (3 marks)

Column I Column II

(1) CH��CH�CH

��X

(2) CH��CH�CH

�X

(3) CH��CH�X

(a) vinyl halide

(b) alkyl halide

(c) allyl halide

(d) benzyl halide

(e) aryl halide

Ans. (1) CH��CH�CH

��X

– (b) alkyl halide

(2) CH��CH�CH

�X – (c) allyl halide

(3) CH��CH�X – (a) vinyl halide

■

Units

10.7 Uses and environmental effects of some polyhalogen compounds

10.7.1 Uses of some polyhalogen compounds

Q. 95. State the uses of the following compounds :

(1) Dichloromethane (CH�Cl

�) (2) Trichloromethane or Chloroform (CHCl

�)

(3) Tetrachloromethane or carbon tetrachloride (CCl�) (4) Iodoform (CHI

�)

(5) Freons (6) DDT (p, p�-Dichlorodiphenyl trichloroethane). (2 marks each)

Ans.


�) :

(1) Dichloromethane dissolves wide range of organic compounds, hence it is used as solvent for many chemical

reactions.

(2) It is used as a solvent as a paint remover and degreaser.

(3) It is used as propellant in aerosols and as a fumigant pesticide for grains and strawberries.

(4) It is used to decaffinate tea or coffee.

(2) Trichloromethane or Chloroform (CHCl�) :

(1) Chloroform in the production of chlorofluoromethane, freon refrigerant R-22.

(2) It is used as solvent in pharmaceuticals, pesticides, gums, fats, resins and dye industry.

(3) It is a good source of dichlorocarbene species.


(3) Tetrachloromethane or carbon tetrachloride (CCl�) :

(1) Carbon tetrachloride is used in the manufacture of refrigerants.

(2) It is used as a dry cleaning agent and as a pesticide for stored grains.

(3) It is very useful solvent for oils, fats and resins. It serves as a source of chlorine.

(4) Iodoform (CHI�) :

(1) Iodoform is used as antiseptic, dressing of wounds and sores.

(2) On small scale it is used as disinfectant.

(5) Freons :

(1) Freons are widely used as propellants in aerosol, products of food, cosmetics and pharmaceutical industries.

(2) Freons containing bromine in their molecules are used as fire extinguishers.

(3) They are used in aerosol insecticides, solvent for cleaning clothes and metallic surfaces. It is used as foaming

agents in the preparation of foamed plastics and in production of certain fluorocarbons.

(4) It is used as refrigerants and air conditioning purposes. ■


How do CFC destroy the ozone layer in the atmosphere ?

When ultraviolet radiation (UV) strikes CFC (CFC13) molecules in the upper atmosphere, the carbon-chlorine

bond breaks and produces highly reactive chlorine atom (Cl).

CFCl�IIIIIJ CFCl

��Cl

This reactive chlorine atom decomposes ozone (O�) molecule into oxygen molecule (O

�).

O��Cl IIIIIJ O

��ClO

ClO�O IIIIIJ O��Cl

One atom of chlorine can destroy up to 100,000 ozone molecules.

(6) p, p-Dichlorodiphenyl trichloroethane (DDT) :

(1) DDT is used as insecticide against malaria and typhus.

(2) It is used to kill various insects like housefly and mosquitoes. ■


DDT, the first chlorinated organic insecticides, was originally prepared in 1873, but it was not until 1939 that

Paul Muller of Geigy Pharmaceuticals in Switzerland discovered the effectiveness of DDT as an insecticide. Paul

Muller was awarded the Nobel Prize in Medicine and Physiology in 1948 for this discovery. The use of DDT

increased enromously on a worldwide basis after World War II, primarily because of its effectiveness against the

mosquito that spreads malaria and lice that carry typhus. Many species of insects developed resistance to DDT, and

it was also discovered to have a high toxicity towards fish. DDT is not metabolised very rapidly by animals;

instead, it is deposited and stored in the fatty tissues. The use of DDT was banned in the United States in 1973,

although it is still in use in some other parts of the world.


Unit

10.7.2 Environmental effects of some polyhalogen compounds

Q. 96. State the environmental effects of the following compounds :


�) (2) Trichloromethane or chloroform (CHCl

�)

(3) Tetrachloromethane or carbon tetrachloride (CCl�) (4) Iodoform (CHI

�)

(5) Freons (CCl�F�, CCl

�F, CHClF

�) (6) DDT (2 marks each)

Ans.


�) :

(1) Higher levels of dichloromethane in air causes nausea, numbness in fingers and toes, dizziness.

(2) Lower levels of dichloromethane causes impaired vision and hearing.

(3) Direct contact with eyes can damage cornea.

(2) Trichloromethane or chloroform (CHCl�) :

(1) When chloroform is exposed to air in the presence of sunlight, it slowly oxidised to phosgene, a poisonous

compound, therefore it is stored in dark, amber coloured bottles.

(2) Chloroform vapour when inhaled for a short time causes dizziness, headache and fatigue and if inhaled for a

long time affects central nervous system.

(3) Tetrachloromethane or carbon tetrachloride (CCl�) :

(1) Exposure to carbon tetrachloride causes eye irritation, damages nerve cells, vomiting sensation, dizziness,

unconciousness or death. Long exposure to chloroform may affect liver.

(2) When mixed with air it causes depletion of the ozone layer, which affects human skin leading to cancer.

(4) Iodoform (CHI�) : Iodoform has a strong smell. It causes irritation to skin and eyes. It may cause respiratory

irritation or breathing difficulty, dizziness, nausea, depression of central nervous system, visual disturbance.

(5) Freons (CCl�F�, CCl

�F, CHClF

�) :

(1) Freon as refrigerant causes ozone depletion.

(2) Freons have low toxicity and low biological activity.

(3) Freons from propane group are more toxic in nature.

(4) Regular large inhalation of freon results in breathing problems, organ damage, loss of consciousness.

(6) DDT :

(1) DDT is not readily metabolised by animals.

(2) It is deposited and stored in fatty tissues.

(3) Exposure to high doses of DDT may cause vomiting, tremors or shakiness.


(4) Laboratory animal studies showed adverse effect of DDT on liver and reproduction.

(5) DDT is a pressistent organic pollutant, readily absorbed in soils and tends to accumulate in the ecosystem.

(6) When dissolved in oil or other lipid, it is readily absorbed by the skin. It is resistant to metabolism. There is a

ban on use of DDT due to all these adverse effects. ■

Q. 97. What is the chemical name of freon ? (1 mark)

Ans. The chemical name of freon is Dichlorodifluoro methane.■

Q. 98. What is the chemical name of DDT ? (1 mark)

Ans. The chemical name of DDT is p, p�-Dichlorodiphenyltrichloroethane. ■

Activity : (Textbook page 233)

(1) Collect detailed information about Freons and their uses.

(2) Collect information about DDT as a persistent pesticide.

Reference books :

(1) Organic chemistry by Morrison, Boyd, Bhattacharjee, 7th edition, Pearson.

(2) Organic chemistry by Finar, Vol 1, 6th edition, Pearson

(Note : Students are supposed to do these activity on their own.)

Q. 99. Select and write the most appropriate answerfrom the given alternatives for each sub-question :

(1 mark each)★ 1. The correct order of increasing reactivity of C-X

bond towards nucleophile in the following

compounds is

(I) (II)

(CH�)�

C�X (CH�)�

CH�X

(III) (IV)

(a) I � II � III � IV (b) II � I � III � IV

(c) III � IV � II � I (d) IV � III � I � II

★ 2. CH��CH�CH

�

Hl

peroxideIIIIIIIIIIIIJ

The major product of the above reaction is,

(a) I�CH��CH�CH

�(b) CH

��CH

��CH

�I

(c) CH��CH

�I

�CH�

(d) CH��CH

�I

�CH��

OH

★ 3. Which of the following is likely to undergo

racemization during alkaline hydrolysis ?

CH��CH�

�Cl

C�H

�

(I) (II)

CH��

CH��

CH�

CH�Cl

(III) (IV)

(a) Only I (b) Only II

(c) II and IV (d) Only IV

★ 4. The best method for preparation of alkyl fluorides

is

(a) Finkelstein reaction

(b) Swartz reaction

(c) Free radical fluorination

(d) Sandmeyer’s reaction

★ 5. Identify the chiral molecule from the following :

(a) 1-Bromobutane

(b) 1, 1-Dibromobutane

(c) 2, 3-Dibromobutane

(d) 2-Bromobutane


★ 6. An alkyl chloride on Wurtz reaction gives 2, 2, 5,

5-tetramethylhexane. The same alkyl chloride on

reduction with zinc-copper couple in alcohol give

hydrocarbon with molecular formula C�H

�. What

is the structure of alkyl chloride

(a) CH��

CH��

C��

CH�

CH�Cl

(b) CH��

CH��

C��

Cl

CH�CH

�

(c) CH��CH

��CH�Cl

�CH

�

(d) CH��CH�

�CH

�

CH��

CH�

CHCl

★ 7. Butanenitrile may be prepared by heating

(a) Propanol with KCN

(b) butanol with KCN

(c) n-butyl chloride with KCN

(d) n-propyl chloride with KCN

★ 8. Choose the compound from the following that will

react fastest by SN1 mechanism

(a) 1-iodobutane

(b) 1-iodopropane

(c) 2-iodo-2 methylbutane

(d) 2-iodo-3-methylbutane

★ 9.

The product ‘B’ in the above reaction sequence is,

(a) (b)

(c) (d)

★ 10. Which of the following is used as source of

dichlorocarbene

(a) tetrachloromethane (b) chloroform

(c) iodoform (d) DDT

11. The rate of SN2 reaction depends on the concentra-

tion of

(a) only the substrate

(b) only the reagent

(c) both the substrate and the reagent

(d) neither the substrate nor the reagent

12. In SN2 reaction, the hydrolysis of alkyl halide

shows

(a) the retention of configuration

(b) the inversion of configuration

(c) both retention and inversion of configuration

(d) no change in the configuration

13. The one step exothermic reaction is

(a) SN1 (b) SN2 (c) SN (d) S2N

14. Which of the following is correct about SN2

mechanism ?

(a) Two step reaction

(b) Complete inversion of configuration

(c) Formation of carbonium ion

(d) Favoured by polar solvent

15. Which of the following is not a nucleophile ?

(a) Ammonia (b) Ammonium ion

(c) Primary amine (d) Secondary amine

16. Which of the following undergoes nucleophilic

substitution exclusively by SN2 mechanism ?

(a) ethyl chloride (b) isopropyl chloride

(c) chlorobenzene (d) benzyl chloride

17. Which of the following is most reactive towards

nucleophilic substitution reaction ?

(a) CH��CH–Cl (b) CH

�CH�CHCl

(c) CH

�Cl (d) ClCH

�–CH�CH

�18. The stability order of carbocation is

(a) 2°� 3°� 1° (b) 3°� 2°� 1°(c) 3°� 1°� 2° (d) 1°� 3°� 2°

19. CH�

– CH�

– BrAlc. KOH

�IIIIIIIIIIIIJ B

HBrIIIIIJ C

Na EtherIIIIIIIIIIJ D the compound D is :

(a) ethane (b) propane

(c) n-butane (d) n-pentane


20. Which of the following characteristic properties of

the enantiomers is correct?

(a) The enantiomers possess same physical and

chemical properties

(b) The enantiomers are optically active compounds

(c) The enantiomers have different optical rotations

(d) All of these

21. The optically inactive compound is

(a) glucose (b) lactic acid

(c) isopropyl alcohol (d) 2-bromo butane

22. A compound with the molecular formula

CH�OH(CHOH)

�CH

�OH has ……… optically

active forms

(a) 3 (b) 4 (c) 6 (d) 8

23. A racemic mixture consists of

(a) equal amount of d and l isomers

(b) unequal amounts of d and l isomers

(c) unknown amounts of d and l isomers

(d) only d isomers

24. Which of the following compounds is not optically

active?

(a) Lactic acid (b) Secondary butyl chloride

(c) n-propyl iodide (d) Glucose

25. Which of the following compounds shows optical

activity ?

(a) n-butyl chloride (b) isobutyl chloride

(c) sec-butyl chloride (d) t-butyl chloride

26. The major product of the following reaction is

27.

The above reaction is known as

(a) Wurtz-Fittig reaction (b) Friedel Craft’s reaction

(c) Sandmeyer’s reaction(d) Swarts reaction

28. Iodoform is used as

(a) an anaesthetic (b) an antiseptic

(c) an analgesic (d) an antibiotic

29. p, p�-dichlorodiphenyl trichloroethane is used as

(a) insecticide (b) anaesthetic

(c) antiseptic (d) refrigerant

30. The order of reactivity in nucleophilic substitution

reaction is

(a) CH�F � CH

�Cl � CH

�I � CH

�Br

(b) CH�F � CH

�Cl � CH

�Br � CH

�I

(c) CH�F � CH

�Br � CH

�Cl � CH

�I

(d) CH�I � CH

�Br � CH

�Cl � CH

�F

31. Racemate is

(a) optically active (b) optically dextro rotatory

(c) optically inactive (d) optically laevorotatory

32. The number of asymmetric carbon atoms in

glucose are

(a) 2 (b) 3 (c) 4 (d) 5

33. The geometry of carbonium ion is

(a) Tetrahedral (b) planar

(c) linear (d) pyramidal

34. In its nucleophilic substitution reaction, aryl halide

resembles

(a) Vinyl chloride

(b) allyl chloride

(c) Benzyl chloride

(d) ethyl chloride


35. The weakest C-Cl bond is present in

36. Which alkyl halide among the following com-

pounds has the highest boiling point ?

(a) (CH�)�CCI (b) CH

�CH

�CH

�CH

�Cl

(c) CH�CH

�CH

�Cl (d) CH

�CH(CH

�)CH

�Cl

37. It is difficult to break C-Cl bond in CH��CH-Cl

due to

(a) Hyper conjugation (b) Resonance

(c) Electromeric effect (d) Inductive effect

38. Which one of the following when heated with

metallic sodium will not give the corresponding

alkane ?

(a) CH�

–CH–Cl�CH

�

(b) CH�

–CH�

–CH�

–Cl

(c) CH�

–

CH��

C–Cl�CH

�

(d) CH�

–CH�

–CH�

–CH�

–Cl

39. The most reactive alkyl halide towards SN2 reac-

tion is

(a) CH�X (b) R

�CX (c) R

�CHX (d) RCH

�X

40. The number of electrons surrounding the carbon-

ium ion is

(a) 6 (b) 8 (c) 10 (d) 7

41. The lowest stability of carbocation among the

compounds

42. Carbon atom in methyl carbocation contains how

many pairs of electrons ?

(a) 8 (b) 4 (c) 3 (d) 5

43. The optically inactive compound is

(a) Glucose (b) Lactic acid

(c) 2-Chlorobutane (d) 2-Chloropropane

44. The hydrogen halide which does not obey

Markownikv rule in presence of a peroxide is

(a) HCl (b) HBr (c) HF (d) HI

45. Which one of the following is NOT used to prepare

alkyl halide from an alcohol ?

(a) SOCl�

(b) PCl�

(c) HCl�ZnCl�

(d) NaCl

46. The total number of electrons present in the central

carbon atom of a free radical is

(a) 7 (b) 8 (c) 9 (d) 6

47. In which of the following pairs both are

nucleophiles ?

(a) BF�, AlCl

�(b) NO�

�, Cl�

(c) CN�, NH�

(d) Br�, BCl�

48. Which one of the following alkane is NOT formed

in Wurtz reaction ?

(a) Methane (b) Ethane (c) Propane (d) Butane

49. Which of the following groups has highest priority

according to R, S convention?

(a) CH�OH (b) COOH

(c) COCH�

(d) COOCH�

50. The halogen atom in aryl halides is

(a) o- and p-directing (b) m-directing

(c) o, m and p-directing (d) only m-directing

51. Chlorobenzene can be obtained by benzene

diazonium chloride by

(a) Friedel Craft’s reaction (b) Wurtz reaction

(c) Gatterman’s reaction (d) Fittig reaction

52. Which of the following carbocations is least stable ?

(a) CH��CH

��CH

��CH

��

(a) CH��CH

�� C�

�CH

��CH

�

(b) CH��CH

��CH�CH

��CH

�


(c) CH��CH

��CH

�(d) CH

��CH

��CH�H

�(d) CH

��CH

��CH�C�CH

��CH

�

53. But-1-ene on reaction with HCl in the presence of

sodium peroxide yields

(a) n-butyl chloride (b) isobutyl chloride

(c) secondary butyl chloride (d) tertiary butyl chloride

54. Carbon tetrachloride is used as

(a) anaesthetic (b) antiseptic

(c) dry cleaning agent (d) fire extinguisher

55. Identify the product D in the following sequence of

reactions :

H�C – CH

�– CH

�– Cl

Alc. KOH

�IIIIIIIIIIIIJ B

HBrIIIIIJ C

NaEtherIIIIIIJ D

(a) 2, 2-dimethyl butane (b) 2, 3-dimethyl butane

(c) hexane (d) 2, 4-dimethylpentane

56. The preparation of alkyl fluoride from alkyl chlor-

ide, in presence of metallic fluorides is known as

(a) Williamson’s reaction

(b) Finkelstein reaction

(c) Swarts reaction

(d) Wurtz reaction

57. IUPAC name of the following compound is

(a) 3-Bromo-3, 4-dimethylheptane

(b) 3,4-dimethyl-3-bromoheptane

(c) 5-Bromo-4,5-dimethylheptane

(d) 4,5-dimethyl-5-bromoheptane

Answers

1. (d) IV � III � I � II 2. (b) CH��CH

��CH

��I

3. (a) Only I 4. (b) Swartz reaction 5. (d) 2-Bro-

mobutane 6. (a) CH��

CH��

C��

CH�

CH�Cl 7. (d) n-propyl

chloride with KCN 8. (c) 2-iodo-2-methyl butane

9. (d) 10. (b) chloroform 11. (c) both

the substrate and the reagent 12. (b) the inversion of

configuration 13. (b) SN2 14. (b) Complete inver-

sion of configuration 15. (b) Ammonium ion

16. (d) benzyl chloride 17. (d) ClCH�

–CH�CH�

18. (b) 3°� 2°� 1° 19. (c) n-butane 20. (d) All of

these 21. (c) isopropyl alcohol 22. (d) 8

23. (a) equal amount of d and l isomers

24. (c) n-propyl iodide 25. (c) sec-butyl chloride

26. (c) 27. (b) Friedel Craft’s reaction

28. (b) an antiseptic 29. (a) insecticide

30. (b) CH�F � CH

�Cl � CH

�Br � CH

�I

31. (c) optically inactive 32. (c) 4 33. (b) planar

34. (a) Vinyl chloride 35. (d)

36. (b) CH�CH

�CH

�CH

�Cl 37. (b) Resonance

38. (c) CH�–

CH��

C–Cl�CH

�

39. (a) CH�X 40. (a) 6

41. (a) 42. (b) 4 43. (d) 2-Chloropropane

44. (b) HBr 45. (d) NaCl 46. (a) 7

47. (c) CN�, NH�

48. (a) Methane

49. (d) COOCH�

50. (a) o- and p-directing

51. (c) Gatterman’s reaction

52. (c) CH��CH

��CH

�53. (c) secondary butyl chloride

54. (c) dry cleaning agent

55. (b) 2, 3-dimethyl butane 56. (c) Swarts reaction

57. (a) 3-Bromo-3, 4-dimethylheptane


E0289.pdf - Navneet eCatalogue

Documents

Transcript of E0289.pdf - Navneet eCatalogue