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Chemistry
Part II
Textbook for Class XI
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chemistry
ALL RIGHTS RESERVED
No part of this publication may be reproduced, stored in a retrieval system or
transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher. This book is sold subject to the condition that it shall not, by way of trade, be lent, re-sold, hired out or otherwise disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published. The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable. Publication Team
Head, Publication : Anup Kumar Rajput
Division
Chief Editor : Shveta Uppal
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OfficerChief Business
: Amitabh Kumar
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: Binoy Banerjee
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DIVISION, NCERT
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Phone : 0361-2674869 First Edition
March 2006 Phalguna 1927
ReprintedOctober 2006, November 2007,
January 2009, December 2009,
November 2010, January 2012,
November 2012, November 2013,
December 2014, December 2015,
February 2017, February 2018,
December 2018, September 2019,
August 2021 and November 2021
Revised Edition
October 2022 Kartika 1944
Reprinted
March 2024 Chaitra 1946
PD 360T SU
© National Council of Educational
Research and Training, 2006, 2022
` 120.00
Printed on 80 GSM paper with NCERT watermark
Published at the Publication Division
by the Secretary, National Council of
Educational Research and Training,
Sri Aurobindo Marg, New Delhi 110 016
and printed at Hi-Tech Graphics D-4/3,
Basement, Okhla Industrial Area, Phase-II, New Delhi -110 02011083 – C hemistry Part ii
Textbook for Class XIISBN 81-7450-494-X (Part I)
ISBN 81-7450-535-0 (Part II)
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Foreword
The National Curriculum Framework (NCF), 2005 recommends that children’s life at
school must be linked to their life outside the school. This principle marks a departure from the legacy of bookish learning which continues to shape our system and causes a gap between the school, home and community. The syllabi and textbooks developed on the basis of NCF signify an attempt to implement this basic idea. They also attempt to discourage rote learning and the maintenance of sharp boundaries between different subject areas. We hope these measures will take us significantly further in the direction of a child-centred system of education outlined in the National Policy on Education (1986). The success of this effort depends on the steps that school principals and
teachers will take to encourage children to reflect on their own learning and to pursue imaginative activities and questions.
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We must recognise that, given space, time and freedom, children generate new knowledge by engaging with the information passed on to them by adults. Treating the prescribed textbook as the sole basis of examination is one of the key reasons why other resources and sites of learning are ignored. Inculcating creativity and initiative is possible if we perceive and treat children as participants in learning, not as receivers of a fixed body of knowledge. These aims imply considerable change in school routines and mode of functioning. Flexibility in the daily time-table is as necessary as rigour in implementing the annual calender so that the required number of teaching days are actually devoted to teaching. The methods used for teaching and evaluation will also determine how effective this textbook proves for making children’s life at school a happy experience, rather than a source of stress or boredom. Syllabus designers have tried to address the problem of curricular burden by restructuring and reorienting knowledge at different stages with greater consideration for child psychology and the time available for teaching. The textbook attempts to enhance this endeavour by giving higher priority and space to opportunities for contemplation and wondering, discussion in small groups, and activities requiring hands-on experience. The National Council of Educational Research and Training (NCERT) appreciates
the hard work done by the textbook development committee responsible for this book. We wish to thank the Chairperson of the advisory group in science and mathematics, Professor J.V.
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Narlikar and the Chief Advisor for this book, Professor B.
L. Khandelwal for guiding the work of this committee. Several teachers contributed to the development of this textbook; we are grateful to their principals for making this possible. We are indebted to the institutions and organisations which have generously permitted us to draw upon their resources, material and personnel. We are especially grateful to the members of the National Monitoring Committee, appointed by the Department of Secondary and Higher Education, Ministry of Human Resource Development under the Chairpersonship of Professor Mrinal Miri and Professor G.P. Deshpande, for their valuable time and contribution. As an organisation committed to systemic reform and continuous improvement in the quality of its products, NCERT welcomes comments and suggestions which will
enable us to undertake further revision and refinement. Director
New Delhi National Council of Educational
20 December 2005 Research and Training
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Rationalisation of Content in the textbook
In view of the COVID-19 pandemic, it is imperative to reduce
content load on students. The National Education Policy 2020, also emphasises reducing the content load and providing opportunities for experiential learning with creative mindset. In this background, the NCERT has undertaken the exercise to rationalise the textbooks across all classes. Learning Outcomes already developed by the NCERT across classes have been taken into consideration in this exercise.Contents of the textbooks have been rationalised in view of the following:• Overlapping with similar content included in other subject areas in the same class
• Similar content included in the lower or higher class in the same subject
• Difficulty
level
• Content, which is easily accessible to students without
much
interventions from teachers and can be learned by children through self-learning or peer-learning
• Content, which is irrelevant in the present context
This present edition, is a reformatted version after carrying out
the changes given above.
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textbook develoPment Committee
ChairPerson , advisory GrouP For textbooks in sCienCe and mathematiCs
J.V. Narlikar, Emeritus Professor, Chairman, Advisory Committee, Inter
University Centre for Astronomy and Astrophysics (IUCCA), Ganeshbhind,
Pune University, Pune
ChieF advisor
B.L. Khandelwal, Professor (Retd. ), Emeritus Scientist, CSIR; Emeritus Fellow, AICTE
and formerly Chairman, Department of Chemistry, Indian Institute of Technology,
New Delhi
members
A. S. Brar, Professor, Indian Institute of Technology, Delhi
Anjni Koul, Lecturer, DESM, NCERT, New Delhi
H.O. Gupta, Professor, DESM, NCERT, New DelhiI.P. Aggarwal, Professor, Regional Institute of Education, NCERT, BhopalJaishree Sharma, Professor, DESM, NCERT, New DelhiM. Chandra, Professor, DESM, NCERT, New Delhi
Poonam Sawhney, PGT (Chemistry), Kendriya Vidyalaya, Vikas Puri, New DelhiR.K. Parashar, Lecturer, DESM NCERT, New Delhi
S.K. Dogra, Professor, Dr. B.R. Ambedkar Centre for Biomedical Research Delhi
University, Delhi
S.K.
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Gupta, Reader, School of Studies in Chemistry, Jiwaji University, GwaliorSadhna Bhargava, PGT (Chemistry), Sardar Patel Vidyalaya, Lodhi Estate, New DelhiShubha Keshwan, Headmistress, Demonstration School, Regional Institute of
Education, NCERT, Mysore
Sukhvir Singh, Reader, DESM, NCERT, New DelhiSunita Malhotra, Professor, School of Sciences, IGNOU, Maidan Garhi, New Delhi
V.K. Verma, Professor (Retd.) Institute of Technology, Banaras Hindu University,
Varanasi
V.P. Gupta, Reader, Regional Institute of Education, NCERT, Bhopal
member -Coordinator
Alka Mehrotra, Reader, DESM, NCERT, New Delhi
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Acknowledgements
The National Council of Educational Research and Training acknowledges the valuable
contributions of the individuals and organisations involved in the development of Chemistry textbook for Class XI. It also acknowledges that some useful material from the reprint editions (2005) of Chemistry textbooks has been utilised in the development of the present textbook. The following academics contributed effectively for editing, reviewing, refining and finalisation of the manuscript of this book: G.T. Bhandage, Professor, RIE, Mysuru; N. Ram, Professor, IIT, New Delhi; R. Sindhu,
Reader, RIE (NCERT), Bhopal; Sanjeev Kumar, Reader, Desh Bandhu College, Kalkaji, New Delhi; Shampa Bhattacharya, Reader, Hans Raj College, Delhi; Vijay Sarda, Reader, Zakir Husain College, New Delhi. K.K. Arora, Reader, Zakir Husain College,
New Delhi; Shashi Saxena, Reader, Hans Raj College, Delhi; Anuradha Sen, Apeejay School, Sheikh Sarai, New Delhi; C. Shrinivas, PGT, Kendriya Vidyalaya, Pushp Vihar,
New Delhi; D.L. Bharti, PGT, Ramjas School, Sector IV, R.K. Puram, New Delhi;
Ila Sharma, PGT, Delhi Public School, Dwarka, Sector-B, New Delhi; Raj Lakshmi Karthikeyan, Head (Science), Mother’s International School, Sri Aurobindo Marg,
New Delhi; Sushma Kiran Setia, Principal, Sarvodaya Kanya Vidyalaya, Hari Nagar (CT), New Delhi; Nidhi Chaudray, PGT, CRPF Public School, Rohini, Delhi; and Veena Suri, PGT, Bluebells School, Kailash, New Delhi.
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We are thankful to them. We express gratitude to R.S. Sindhu, Professor (Retd. ), DESM, NCERT, New
Delhi, for editing, reviewing and refining the textbook right from the initial stage. We are also grateful to Ruchi Verma, Associate Professor, DESM, NCERT, New
Delhi; Pramila Tanwar, Associate Professor, DESM, NCERT, New Delhi; R.B. Pareek, Associate Professor, RIE, Ajmer and A.K. Arya, Associate professor, RIE, Ajmer, for reviewing and refining the content of the textbook. Special thanks are due to M. Chandra, Professor and Head, DESM, NCERT for
her support. The Council also gratefully acknowledges the contribution of Surendra Kumar
and Hari Darshan Lodhi DTP Operator; Subhash Saluja, Ramendra Kumar Sharma
and Abhimanyu Mohanty, Proof Readers; Bhavna Saxena, Copy Editor and Deepak
Kapoor, Incharge, Computer Station, in shaping this book. The contributions of
the Publication Department in bringing out this book are also duly acknowledged.
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Contents
Foreword iii
rationalisation oF Content in the textbook v
Unit 7 Redox Reactions 235
7.1 Classical Idea of Redox Reactions-Oxidation and Reduction Reactions 235
7.2 Redox Reactions in Terms of Electron Transfer Reactions 237
7.3 Oxidation Number 239
7.4 Redox Reactions and Electrode Processes 249
Unit 8 Organic Chemistry – Some Basic Principles and Techniques 256
8.1 General Introduction 256
8.2 Tetravalence of Carbon: Shapes of Organic Compounds 257
8.3 Structural Representations of Organic Compounds 258
8.4 Classification of Organic Compounds 261
8.5 Nomenclature of Organic Compounds 262
8.6 Isomerism 270
8.7 Fundamental Concepts in Organic Reaction Mechanism 271
8.8 Methods of Purification of Organic Compounds 278
8.9 Qualitative Analysis of Organic Compounds 284
8.10 Quantitative Analysis 285
Unit 9 Hydrocarbons 295
9.1 Classification 295
9.2 Alkanes 296
9.3 Alkenes 306
9.4 Alkynes 314
9.5 Aromatic Hydrocarbon 318
9.6 Carcinogenicity and Toxicity 325
Answers 328
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Contents oF Chemistry Part i
unit 1 Some BaSic conceptS of chemiStry 1
unit 2 Structure of atom 29
unit 3 c laSSification of elementS and periodicity in propertieS 74
unit 4 c hemical Bonding and molecular Structure 100
unit 5 t hermodynamicS 136
unit 6 e quiliBrium 168
a ppendiceS 215
a nSwer to Some Selected queStionS 229
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redox reactions
Chemistry deals with varieties of matter and change
of one kind of matter into the other. Transformation of
matter from one kind into another occurs through the various types of reactions. One important category of such reactions is Redox Reactions. A number of phenomena, both physical as well as biological, are concerned with redox reactions. These reactions find extensive use in pharmaceutical, biological, industrial, metallurgical and agricultural areas. The importance of these reactions is apparent from the fact that burning of different types of fuels for obtaining energy for domestic, transport and other commercial purposes, electrochemical processes for extraction of highly reactive metals and non-metals, manufacturing of chemical compounds like caustic soda, operation of dry and wet batteries and corrosion of metals fall within the purview of redox processes. Of late, environmental issues like Hydrogen Economy (use of
liquid hydrogen as fuel) and development of ‘Ozone Hole’
have started figuring under redox phenomenon. 7.1 CLASSICAL
IDEA OF REDOX REACTIONS –
OXIDATION AND REDUCTION REACTIONS
Originally, the ter
m oxidation was used to describe the
addition of oxygen to an element or a compound. Because
of the presence of dioxygen in the atmosphere (~20%), many elements combine with it and this is the principal reason why they commonly occur on the earth in the
form of their oxides. The following reactions represent oxidation processes according to the limited definition of oxidation: 2 Mg (s) + O
2 (g) → 2 MgO (s) (7.1)
S (s) + O2 (g) → SO2 (g) (7.2)After studying this unit you will be able to
•
identify redox reactions as
a class
of reactions in which oxidation
and reduction reactions occur simultaneously;
•
define the terms oxidation, reduction, oxidant (oxidising
agent) and reductant (reducing agent);
•
explain mechanism of redox reactions by electr
on transfer
process;
• use the concept of oxidation number to identify oxidant and r
eductant in a reaction;
• classify r
edox reaction into
combination (synthesis), decomposition, displacement and disproportionation reactions;
•
suggest a comparative order among various reductants and oxidants;
•
bala
nce chemical equations
using (i) oxidation number
(ii) half reaction method;
• learn the concept of redox r
eactions in terms of electrode
processes.UNIT 7
REDOX REACTIONS
Where t here is oxidation, there is always reduction –
Chemistry is essentially a study of redox systems.
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In reactions (7.1) and (7.2), the elements
magnesium and sulphur are oxidised on
account of addition of oxygen to them. Similarly, methane is oxidised owing to the addition of oxygen to it. CH
4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) (7.3)
A careful examination
of reaction (7.3) in
which hydrogen has been replaced by oxygen
prompted chemists to reinterpret oxidation in terms of removal of hydrogen from it and, therefore, the scope of term oxidation was broadened to include the removal of hydrogen from a substance. The following illustration is another reaction where removal of hydrogen can also be cited as an oxidation reaction. 2 H
2S(g) + O2 (g) → 2 S (s) + 2 H2O (l) (7.4)
As knowledge of chemists grew, it was
natural to extend the term oxidation for
reactions similar to (7.1 to 7.4), which do
not involve oxygen but other electronegative elements. The oxidation of magnesium with fluorine, chlorine and sulphur etc. occurs according to the following reactions :
Mg (s) + F
2 (g) → MgF2 (s) (7.5)
Mg (s) + Cl2 (g) → MgCl2 (s) (7.6)
Mg (s) + S (s) → MgS (s) (7.7)
Incorporating the reactions (7.5 to
7.7) within the fold of oxidation reactions
encouraged chemists to consider not only
the removal of hydrogen as oxidation, but also the removal of electropositive elements as oxidation. Thus the reaction :
2K
4 [Fe(CN)6](aq) + H2O2 (aq) → 2K3[Fe(CN)6](aq)
+ 2 KOH (aq)
is interpreted as oxidation due to the removal of electropositive element potassium from potassium ferrocyanide before it changes to potassium ferricyanide. To summarise, the term “oxidation” is defined as the addition
of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance. In the beginning, reduction was considered
as removal of oxygen from a compound.
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However, the term reduction has been broadened these days to include removal
of oxygen/electronegative element from a substance or addition of hydrogen/electropositive element to a substance. According to the definition given above,
the following are the examples of reduction processes:
2 HgO (s)
2 Hg (l) + O2 (g) (7.8)
(removal of oxygen from mercuric oxide )
2 FeCl3 (aq) + H2 (g) → 2 FeCl2 (aq) + 2 HCl(aq)
(7.9)
(removal of electronegative element, chlorine
from ferric chloride)
CH2 = CH2 (g) + H2 (g) → H3C – CH3 (g) (7.10)
(addition of hydrogen)2HgCl
2 (aq) + SnCl2 (aq) → Hg2Cl2 (s)+SnCl4 (aq)
(7.11)
(addition of mercury to mercuric chloride)
In reaction (7.11) simultaneous oxidation
of stannous chloride to stannic chloride is
also occurring because of the addition of electronegative element chlorine to it. It was soon realised that oxidation and reduction always occur simultaneously (as will be apparent by re-examining all the equations given above), hence, the word “redox” was coined for this class of chemical reactions. Problem 7.1
In the reactions given below, identify
the species undergoing oxidation and reduction:
(i) H
2S (g) + Cl2 (g) → 2 HCl (g) + S (s)
(ii) 3Fe3O4 (s) + 8 Al (s) → 9 Fe (s)
+ 4Al2O3 (s)
(iii) 2 Na (s) + H2 (g) → 2 NaH (s)
Solution (i) H
2S is oxidised because a more
electronegative element, chlorine is added
to hydrogen (or a more electropositive element, hydrogen has been removed
from S). Chlorine is reduced due to addition of hydrogen to it. (ii) Aluminium is oxidised because
oxygen is added to it. Ferrous ferric oxide
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redox reactions
(Fe3O4) is reduced because oxygen has
been removed from it. (iii) With the careful application of the
concept of electronegativity only we may infer that sodium is oxidised and hydrogen is reduced. Reaction (iii) chosen here prompts us to
think in terms of another way to define redox reactions. 7.2
REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS
We have already lear
nt that the reactions
2Na(s) + Cl2(g) → 2NaCl (s) (7.12)
4Na(s) + O2(g) → 2Na2O(s) (7.13)
2Na(s) + S(s) → Na2S(s) (7.14)
are redox reactions because in each of these r
eactions sodium is oxidised due to the addition
of either oxygen or more electronegative element to sodium.
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Simultaneously, chlorine, oxygen and sulphur are reduced because to each of these, the electropositive element sodium has been added. From our knowledge of chemical bonding we also know that sodium chloride, sodium oxide and sodium sulphide are ionic compounds and perhaps better written as Na
+Cl– (s), (Na+)2O2–(s), and (Na+)2
S2–(s). Development of charges on the species
produced suggests us to rewrite the reactions (7.12 to 7.14) in the following manner : For convenience, each of the above
processes can be considered as two separate steps, one involving the loss of electrons and the other the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride. 2 Na(s) → 2 Na
+(g) + 2e–
Cl2(g) + 2e– → 2 Cl–(g)
Each of the above steps is called a half
reaction, which explicitly shows involvement
of electrons. Sum of the half reactions gives the overall reaction :
2 Na(s) + Cl
2 (g) → 2 Na+ Cl– (s) or 2 NaCl (s)
Reactions 7.12 to 7.14 suggest that half
reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions. It may not be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species as per the classical idea and their interplay in electron-transfer change. In reactions (7.12 to 7.14) sodium, which is oxidised, acts as
a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine, oxygen and sulphur are reduced and act as oxidising agents because these accept electrons from sodium.
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To summarise, we may mention that
Oxidation : Loss of electron(s) by any species.Reduction : Gain of electron(s) by any species. Oxidising agent : Acceptor of electron(s).Reducing agent : Donor of electron(s). Problem 7.2 Justify that the reaction:2 Na(s) + H
2(g) → 2 NaH (s) is a redox
change.SolutionSince in the above reaction the compound
formed is an ionic compound, which may also be represented as Na
+H– (s), this
suggests that one half reaction in this process is :
2 Na (s)
→ 2 Na+(g) + 2e–
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and the other half reaction is:
H2 (g) + 2e– → 2 H–(g)
This splitting of the reaction under
examination into two half reactions automatically reveals that here sodium is oxidised and hydrogen is reduced, therefore, the complete reaction is a redox change. 7.2.1
Competitive Electron Transfer Reactions
Place a strip of metallic zinc in an aqueous solution of copper nitrate as shown in Fig. 7.1, for about one hour. Y
ou may notice
that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn
2+ ions
among the products can easily be judged when the blue colour of the solution due to Cu
2+ has disappeared. If hydrogen sulphide
gas is passed through the colourless solution containing Zn
2+ ions, appearance of white zinc
sulphide, ZnS can be seen on making the solution alkaline with ammonia. The reaction between metallic zinc and the
aqueous solution of copper nitrate is :
Zn(s) + Cu
2+ (aq) → Zn2+ (aq) + Cu(s) (7.15)
In reaction (7.15), zinc has lost electrons
to form Zn2+ and, therefore, zinc is oxidised. Evidently, now if zinc is oxidised, releasing
electrons, something must be reduced, accepting the electrons lost by zinc.
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Copper ion is reduced by gaining electrons from the zinc. Reaction (7.15) may be rewritten as : At this stage we may investigate the state
of equilibrium for the reaction represented by equation (7.15). For this purpose, let us place a strip of metallic copper in a zinc sulphate solution. No visible reaction is noticed and attempt to detect the presence of Cu
2+ ions
by passing H2S gas through the solution to
produce the black colour of cupric sulphide, CuS, does not succeed. Cupric sulphide has such a low solubility that this is an extremely sensitive test; yet the amount of Cu
2+ formed
cannot be detected. We thus conclude that the state of equilibrium for the reaction (7.15) greatly favours the products over the reactants. Let us extend electron transfer reaction
now to copper metal and silver nitrate solution in water and arrange a set-up as shown in
Fig. 7.2. The solution develops blue colour due to the formation of Cu
2+ ions on account
of the reaction:
Fig. 7.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.
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(7.16)
Here, Cu(s) is oxidised to Cu2+(aq) and
Ag+(aq) is reduced to Ag(s). Equilibrium greatly
favours the products Cu2+ (aq) and Ag(s). By way of contrast, let us also compare
the reaction of metallic cobalt placed in nickel sulphate solution. The reaction that occurs here is :
(7.17)
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redox reactions
Fig. 7.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker. At equilibrium, chemical tests reveal that both
Ni2+(aq) and Co2+(aq) are present at moderate
concentrations. In this case, neither the reactants [Co(s) and Ni
2+(aq)] nor the products
[Co2+(aq) and Ni (s)] are greatly favoured. This competition for release of electrons
incidently reminds us of the competition for release of protons among acids. The similarity suggests that we might develop a table in which metals and their ions are listed on the basis of their tendency to release electrons just as we do in the case of acids to indicate the strength of the acids. As a matter of fact we have already made certain comparisons.
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By comparison we have come to know that zinc releases electrons to copper and copper releases electrons to silver and, therefore, the electron releasing tendency of the metals is in the order: Zn>Cu>Ag. We would love to make our list more vast and design a metal activity series or electrochemical series. The competition for electrons between various metals helps us to design a class of cells, named as Galvanic cells in which the chemical reactions become the source of electrical energy. We would study more about these cells in Class XII. 7.3
OXIDATION NUMBER
A less obvious example of electron transfer is
realised when hydrogen combines with oxygen
to form water by the reaction:
2H2(g) + O2 (g) → 2H2O (l) (7.18)
Thoug
h not simple in its approach, yet
we can visualise the H atom as going from a
neutral (zero) state in H2 to a positive state in
H2O, the O atom goes from a zero state in O2
to a dinegative state in H2O. It is assumed that
there is an electron transfer from H to O and
consequently H2 is oxidised and O2 is reduced. However, as we shall see later, the charge
transfer is only partial and is perhaps better described as an electron shift rather than a complete loss of electron by H and gain by O. What has been said here with respect to equation (7.18) may be true for a good number of other reactions involving covalent compounds. Two such examples of this class of the reactions are:
H
2(s) + Cl2(g) → 2HCl(g) (7.19)
and,CH
4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g) (7.20)
In order to keep track of electron shifts
in chemical reactions involving formation
of covalent compounds, a more practical method of using oxidation number has been developed. In this method, it is always assumed that there is a complete transfer of electron from a less electronegative atom to a more electonegative atom.
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For example, we rewrite equations (7.18 to 7.20) to show charge on each of the atoms forming part of the reaction :
0 0 +1 –2
2H2(g) + O2(g) → 2H2O (l) (7.21)
0 0 +1 –1
H2 (s) + Cl2(g) → 2HCl(g) (7.22)
–4+1 0 +4 –1 +1 –1
CH4(g) + 4Cl2(g) → CCl4(l) +4HCl(g) (7.23)
It may be emphasised that the assumption
of electron transfer is made for book-keeping purpose only and it will become obvious at a later stage
in this unit that it leads to the
simple description of redox reactions. Oxidation number denotes the oxidation
state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair
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in a covalent bond belongs entirely to more
electronegative element. It is not always possible to remember or
make out easily in a compound/ion, which element is more electronegative than the other. Therefore, a set of rules has been formulated to determine the oxidation number of an element in a compound/ion. If two or more than two atoms of an element are present in the molecule/ion such as Na
2S2O3/Cr2O7 2–, the oxidation number of the
atom of that element will then be the average of the oxidation number of all the atoms of that element. We may at this stage, state the rules for the calculation of oxidation number. These rules are:
1. In elements, in the free or the uncombined
state, each atom bears an oxidation number
of zero. Evidently each atom in
H2, O2, Cl2, O3, P4, S8, Na, Mg, Al has the
oxidation number zero. 2.
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For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Na
+ ion has an oxidation
number of +1, Mg2+ ion, +2, Fe3+ ion, +3,
Cl– ion, –1, O2– ion, –2; and so on. In their
compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds. 3. The oxidation number of oxygen in most
compounds is
–2. However, we come
across two kinds of exceptions here. One arises in the case of peroxides and superoxides, the compounds of oxygen in which oxygen atoms are directly linked to each other. While in peroxides (e.g., H
2O2,
Na2O2), each oxygen atom is assigned an
oxidation number of –1, in superoxides (e.g., KO
2, RbO2) each oxygen atom is
assigned an oxidation number of –(½). The second exception appears rarely, i.e.
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when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF
2) and dioxygen difluoride (O2F2), the
oxygen is assigned an oxidation number of +2 and +1, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number
would now be a positive figure only. 4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing
two elements). For example, in LiH, NaH, and CaH
2, its oxidation number is –1. 5. In all its compounds, fluorine has an oxidation number of
–1. Other halogens (Cl,
Br, and I) also have an oxidation number of –1, when they occur as halide ions in their compounds. Chlorine, bromine and iodine when combined with oxygen, for example in oxoacids and oxoanions, have positive oxidation numbers.
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6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum
of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, (CO
3)2– must equal –2. By the application of above rules, we can
find out the oxidation number of the desired element in a molecule or in an ion. It is clear that the metallic elements have positive oxidation number and nonmetallic elements have positive or negative oxidation number. The atoms of transition elements usually display several positive oxidation states. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups. Thus, it implies that the highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table. In the third period, the highest value of oxidation number changes from 1 to 7 as indicated below in the compounds of the elements.
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A term that is often used interchangeably
with the oxidation number is the oxidation state. Thus in CO
2, the oxidation state of
carbon is +4, that is also its oxidation number and similarly the oxidation state as well as oxidation number of oxygen is – 2. This implies that the oxidation number denotes the
oxidation state of an element in a compound. Unit 7.indd 240 10/10/2022 10:37:04 AM
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redox reactions
The oxidation number/state of a metal in a
compound is sometimes presented according to the notation given by German chemist, Alfred Stock. It is popularly known as Stock notation. According to this, the oxidation
number is expressed by putting a Roman numeral representing the oxidation number in parenthesis after the symbol of the metal in the molecular formula. Thus aurous chloride and auric chloride are written as Au(I)Cl and Au(III)Cl
3. Similarly, stannous chloride and
stannic chloride are written as Sn(II)Cl2 and
Sn(IV)Cl4. This change in oxidation number
implies change in oxidation state, which in turn helps to identify whether the species is present in oxidised form or reduced form. Thus, Hg
2(I)Cl2 is the reduced form of Hg(II) Cl2.
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Problem 7.3
Using Stock notation, represent the
following compounds :HAuCl4, Tl2O, FeO,
Fe2O3, CuI, CuO, MnO and MnO2. Solution
By applying various rules of calculating
the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:
HAuCl
4 → Au has 3
Tl2O → Tl has 1
FeO → Fe has 2
Fe2O3 → Fe has 3
CuI → Cu has 1
CuO → Cu has 2
MnO → Mn has 2
MnO2 → Mn has 4
Therefore, these compounds may be
r
epresented as:
HAu(III)Cl4, Tl2(I)O, Fe(II)O, Fe2(III)O3,
Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O2.The idea of oxidation number has been
invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:
Oxidation: An increase in the oxidation
number of the element in the given substance. Reduction
: A decrease in the oxidation
number of the element in the given substance.Oxidising
agent: A reagent which can
increase the oxidation number of an element
in a given substance. These reagents are called as oxidants also. Reducing agent: A reagent which lowers the
oxidation number of an element in a given
substance. These reagents are also called as reductants. Redox reactions: Reactions which involve
change in oxidation number of the interacting
species. Problem 7.4
Justify that the reaction:2Cu
2O(s) + Cu2S(s) → 6Cu(s) + SO2(g)
is a redox reaction. Identify the species
oxidised/reduced, which acts as an oxidant and which acts as a reductant.
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Solution
Let us assign oxidation number to each
of the species in the reaction under examination. This results into:
+1 –2 +1 –2 0 +4 –2
2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2
We therefore, conclude that in this reaction copper is reduced from +1 state
to zero oxidation state and sulphur is oxidised from –2 state to +4 state. The above reaction is thus a redox reaction. Group 1 2 13 14 15 16 17
Element Na Mg Al Si P S Cl
Compound NaCl MgSO4AlF3SiCl4P4O10SF6HClO
Highest oxidation number state of the group element+1 +2 +3 +4 +5 +6 +7
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Further, Cu2O helps sulphur in Cu2S to
increase its oxidation number, therefore,
Cu(I) is an oxidant; and sulphur of Cu2S
helps copper both in Cu2S itself and
Cu2O to decrease its oxidation number;
therefore, sulphur of Cu2S is reductant. 7.3.1 Types of Redox Reactions
1. Combination reactions
A combination reaction may be denoted in
the manner:
A + B
→ C
Either A and B or both A and B must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:
0 0 +4 –2
C(s) + O2 (g) CO2(g) (7.24)
0 0 +2 –3
3Mg(s) + N2(g) Mg3N2(s) (7.25)
–4+1 0 +4 –2 +1 –2
CH4(g) + 2O2(g) CO2(g) + 2H2O (l)
2. Decomposition reactions
Decomposition reactions are the opposite
of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.
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Examples of this class of reactions are:
+1 –2 0 0
2H2O (l) 2H2 (g) + O2(g) (7.26)
+1 –1 0 0
2NaH (s) 2Na (s) + H2(g) (7.27)
+1 +5 –2 +1 –1 0
2KClO3 (s) 2KCl (s) + 3O2(g) (7.28)
It may carefully be noted that there is no
change in the oxidation number of hydr
ogen
in methane under combination reactions and that of potassium in potassium chlorate in reaction (7.28). This may also be noted here that all decomposition reactions are not
redox reactions. For example, decomposition
of calcium carbonate is not a redox reaction. +2 +4 –2 +2 –2 +4 –2
CaCO3 (s) CaO(s) + CO2(g)
3. Displacement reactions
In a displacement reaction, an ion (or an
atom)
in a compound is replaced by an ion
(or an atom) of another element. It may be denoted as:
X + YZ → XZ + Y
Displacement reactions fit into two categories: metal displacement and non-metal displacement. (a) Metal dis
placement: A metal in a
compound can be displaced by another metal
in the uncombined state. We have already discussed about this class of the reactions under section 7.2.1. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores. A few such examples are:
+2 +6 –2 0 0 +2 +6 –2
CuSO4(aq) + Zn (s) → Cu(s) + ZnSO4 (aq)
(7.29)
+5 –2 0 0 +2 –2
V2O5 (s) + 5Ca (s) 2V (s) + 5CaO (s)
(7.30)
+4 –1 0 0 +2 –1
TiCl4 (l) + 2Mg (s) Ti (s) + 2 MgCl2 (s)
(7.31)
+3 –2 0 +3 –2 0
Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s)
(7.32)
In each case, the reducing metal is a
better reducing agent than the one that is being reduced which evidently shows more capability to lose electrons as compared to the one that is reduced.
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(b) Non-metal displacement: The non-
metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. Unit 7.indd 242 10/10/2022 10:37:04 AM
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redox reactions
All alkali metals and some alkaline earth
metals (Ca, Sr, and Ba) which are very good
reductants, will displace hydrogen from cold water. 0 +1 –2 +1 –2 +1 0
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
(7.33)
0 +1 –2 +2 –2 +1 0
Ca(s) + 2H2O(l) → Ca(OH)2 (aq) + H2(g)
(7.34)
Less active metals such as magnesium and
iron react with steam to produce dihydrogen gas:
0 +1 –2 +2 –2 +1 0
Mg(s) + 2H2O(l) Mg(OH)2(s) + H2(g)
(7.35)
0 +1 –2 +3 –2 0
2Fe(s) + 3H2O(l) Fe2O3(s) + 3H2(g) (7.36)
Many metals, including those which do not
react with cold water, are capable of displacing
hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals. A few examples for the displacement of hydrogen from acids are:
0 +1 –1 +2 –1 0
Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g)
(7.37)
0 +1 –1 +2 –1 0
Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)
(7.38)
0 +1 –1 +2 –1 0
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
(7.39)
Reactions (7.37 to 7.39) are used to
prepare dihydrogen gas in the laboratory. Here, the reactivity of metals is reflected in the rate of hydrogen gas evolution, which is the slowest for the least active metal Fe, and the fastest for the most reactive metal, Mg. Very less active metals, which may occur in the native state such as silver (Ag), and gold (Au) do not react even with hydrochloric acid. In section (7.2.1) we have already discussed
that the metals – zinc (Zn), copper (Cu) and silver (Ag) through tendency to lose electrons show their reducing activity in the order Zn> Cu>Ag. Like metals, activity series also exists
for the halogens.
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The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table. This implies that fluorine is so reactive that it can replace chloride, bromide and iodide ions in solution. In fact, fluorine is so reactive that it attacks water and displaces the oxygen of water :
+1 –2 0 +1 –1 0
2H2O (l) + 2F2 (g) → 4HF(aq) + O2(g) (7.40)
It is for this reason that the displacement
reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution. On the other hand, chlorine can displace bromide and iodide ions in an aqueous solution as shown below:
0 +1 –1 +1 –1 0
Cl2 (g) + 2KBr (aq) → 2 KCl (aq) + Br2 (l)
(7.41)
0 +1–1 +1 –1 0
Cl2 (g) + 2KI (aq) → 2 KCl (aq) + I2 (s)
(7.42)
As Br2 and I2 are coloured and dissolve in CCl4,
can easily be identified from the colour of the solution. The above reactions can be written in ionic form as:
0 –1 –1 0
Cl2 (g) + 2Br– (aq) → 2Cl– (aq) + Br2 (l) (7.41a)
0 –1 –1 0
Cl2 (g) + 2I– (aq) → 2Cl– (aq) + I2 (s) (7.42b)
Reactions (7.41) and (7.42) form the basis
of identifying Br– and I– in the laboratory
through the test popularly known as ‘Layer
Test’. It may not be out of place to mention here that bromine likewise can displace iodide ion in solution:
0 –1 –1 0
Br2 (l) + 2I – (aq) → 2Br– (aq) + I2 (s) (7.43)
The halogen displacement reactions have
a dir
ect industrial application. The recovery
of halogens from their halides requires an
oxidation process, which is represented by:
2X– → X2 + 2e– (7.44)
here
X denotes a halogen element. Whereas
chemical means are available to oxidise Cl–,
Br– and I–, as fluorine is the strongest oxidising
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agent; there is no way to convert F– ions to F2
by chemical means. The only way to achieve
F2 from F– is to oxidise electrolytically, the
details of which you will study at a later stage. 4.
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Disproportionation reactions Disproportionation reactions are a special type
of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction. The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation. +1 –1 +1 –2 0
2H2O2 (aq) → 2H2O(l) + O2(g) (7.45)
Here the oxygen of
peroxide, which is present
in –1 state, is converted to zero oxidation state in O
2 and decreases to –2 oxidation state in
H2O. Phosphorous, sulphur and chlorine
undergo disproportionation in the alkaline medium as shown below :
0 –3 +1
P4(s) + 3OH–(aq)+ 3H2O(l) → PH3(g) + 3H2PO2–
(aq)
(7.46)
0 –2 +2
S8(s) + 12 OH– (aq) → 4S2– (aq) + 2S2O32–(aq)
+ 6H2O(l)
(7.47)
0 +1 –1
Cl2 (g) + 2 OH– (aq) → ClO– (aq) + Cl– (aq) +
H2O (l)
(7.48)
The reaction (7.48)
describes the
formation of household bleaching agents. The hypochlorite ion (ClO
–) formed in the
reaction oxidises the colour-bearing stains of the substances to colourless compounds. It is of interest to mention here that whereas
bromine and iodine follow the same trend as exhibited by chlorine in reaction (7.48), fluorine shows deviation from this behaviour when it reacts with alkali. The reaction that takes place in the case of fluorine is as follows:
2 F
2(g) + 2OH–(aq) → 2 F–(aq) + OF2(g) + H2O(l)
(7.49)
(It is to be noted with care that fluorine in
reaction (7.49) will
undoubtedly attack water
to produce some oxygen also).
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This departure shown by fluorine is not surprising for us as we know the limitation of fluorine that, being the most electronegative element, it cannot exhibit any positive oxidation state. This means that among halogens, fluorine does not show a disproportionation tendency. Problem 7.5
Which of the following species, do not
show disproportionation reaction and why ? ClO
–, ClO2–, ClO3– and ClO4 –
Also write reaction for each of the species
that disproportionates. Solution Among the oxoanions of chlorine listed
above, ClO
4– does not disproportionate
because in this oxoanion chlorine is present in its highest oxidation state that is, +7. The disproportionation reactions for the other three oxoanions of chlorine are as follows:
+1 –1 +5
3ClO– → 2Cl– + ClO–
3
+3 +5 –1
6 ClO2–
4ClO3– + 2Cl–
+5 –1 +7
4ClO–
3 → Cl– + 3 ClO4–
Problem 7.6 Suggest a scheme of classification of the
following redox reactions
(a) N
2 (g) + O2 (g) → 2 NO (g)
(b) 2Pb(NO3)2(s) → 2PbO(s) + 4 NO2 (g) +
O2 (g)
(c) NaH(s) + H2O(l) → NaOH(aq) + H2 (g)
(d) 2NO2(g) + 2OH–(aq) → NO2–(aq) +
NO3– (aq)+H2O(l)
Unit 7.indd 244 11/10/2022 15:18:43
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redox reactions
Solution
In reaction (a), the compound nitric
oxide is formed by the combination of the elemental substances, nitrogen and oxygen; therefore, this is an example of combination redox reactions. The reaction (b) involves the breaking down of lead nitrate into three components; therefore, this is categorised under
decomposition redox reaction. In
The Paradox of Fractional Oxidation Number
Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction. Examples are:
C
3O2 [where oxidation number of carbon is (4/3)],
Br3O8 [where oxidation number of bromine is (16/3)]
and Na2S4O6 (where oxidation number of sulphur is 2.5). We know that the idea of fractional oxidation number is unconvincing to us, because electrons
are never shared/transferred in fraction.
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Actually this fractional oxidation state is the average
oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states. Structure of the species C
3O2, Br3O8 and S4O62– reveal the following bonding situations:
+2 0 +2
O = C = C*= C = O
Structure of C3O2
(carbon suboxide)
Structure of Br3O8 (tribromooctaoxide) Structure of S4O62– (tetrathionate ion)
The element marked with asterisk in each species is exhibiting the different oxidation
state (oxidation number) from rest of the atoms of the same element in each of the species. This reveals that in C
3O2, two carbon atoms are present in +2 oxidation state each, whereas
the third one is present in zero oxidation state and the average is 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br
3O8, each
of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in +4 oxidation state. Once again the average, that is different from reality, is
16/3. In the same fashion, in the species S
4O62–, each of the two extreme sulphurs exhibits oxidation
state of +5 and the two middle sulphurs as zero. The average of four oxidation numbers of sulphurs of the S
4O62– is 2.5, whereas the reality being + 5,0,0 and +5 oxidation number respectively for
each sulphur. We may thus, in general, conclude that the idea of fractional oxidation state should be taken
with care and the reality is revealed by the structures only. Further, whenever we come across with fractional oxidation state of any particular element in any species, we must understand that this is the average oxidation number only.
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In reality (revealed by structures only), the element in that particular species is present in more than one whole number oxidation states. Fe
3O4, Mn3O4,
Pb3O4 are some of the other examples of the compounds, which are mixed oxides, where we come
across with fractional oxidation states of the metal atom. However, the oxidation states may be in fraction as in O
2+ and O2– where it is +½ and –½ respectively.reaction (c), hydrogen of water has been
displaced by hydride ion into dihydrogen gas. Therefore, this may be called as displacement redox reaction. The reaction (d) involves disproportionation of NO
2
(+4 state) into NO2– (+3 state) and NO3–
(+5 state). Therefore reaction (d) is an example of disproportionation redox reaction. Unit 7.indd 245 10/10/2022 10:37:04 AM
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Problem 7.8
Write the net ionic equation for the
reaction of potassium dichromate(VI), K
2Cr2O7 with sodium sulphite, Na2SO3,
in an acid solution to give chromium(III) ion and the sulphate ion. Problem 7.7
Why do the following reactions proceed
differently ? Pb3O4 + 8HCl → 3PbCl2 + Cl2 + 4H2O and
Pb3O4 + 4HNO3 → 2Pb(NO3)2 + PbO2 +
2H2O
SolutionPb
3O4 is actually a stoichiometric
mixture of 2 mol of PbO and 1 mol of
PbO2. In PbO2, lead is present in +4
oxidation state, whereas the stable oxidation state of lead in PbO is +2.
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PbO
2
thus can act as an oxidant (oxidising agent) and, therefore, can oxidise Cl
– ion
of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction
Pb
3O4 + 8HCl → 3PbCl2 + Cl2 + 4H2O
can be splitted into two reactions namely:2PbO + 4HCl → 2PbCl
2 + 2H2O
(acid-base reaction)
+4 –1 +2 0
PbO2 + 4HCl → PbCl2 + Cl2 +2H2O
(redox reaction)
Since HNO3 itself is an oxidising agent
therefore, it is unlikely that the reaction
may occur between PbO2 and HNO3. However, the acid-base reaction occurs between PbO and HNO
3 as:
2PbO + 4HNO3 → 2Pb(NO3)2 + 2H2O
It is the passive nature of PbO2 against
HNO3 that makes the reaction different
from the one that follows with HCl. (a) Oxidation Number Method: In writing
equations for oxidation-reduction reactions,
just as for other reactions, the compositions and formulas must be known for the substances that react and for the products that are formed. The oxidation number method is now best illustrated in the following steps:
Step 1: Write the correct formula for each
reactant and product. Step 2: Identify atoms which undergo
change in oxidation number in the reaction by assigning the oxidation number to all elements in the reaction. Step 3: Calculate the increase or decrease
in the oxidation number per atom and for the entire molecule/ion in which it occurs. If these are not equal then multiply by suitable number so that these become equal. (If you realise that two substances are reduced and nothing is oxidised or vice-versa, something is wrong.
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Either the formulas of reactants or products are wrong or the oxidation numbers have not been assigned properly). Step 4: Ascertain the involvement of ions if
the reaction is taking place in water, add H
+ or
OH– ions to the expression on the appropriate
side so that the total ionic charges of reactants and products are equal. If the reaction is carried out in acidic solution, use H
+ ions in
the equation; if in basic solution, use OH– ions. Step 5 : Make the numbers of hydrogen
atoms in the expression on the two sides equal by adding water (H
2O) molecules to the
reactants or products. Now, also check the number of oxygen atoms. If there are the same number of oxygen atoms in the reactants and products, the equation then represents the balanced redox reaction. Let us now explain the steps involved in
the method with the help of a few problems
given below: 7.3.2 Balancing of Redox Reactions
Two methods are used to balance chemical
equations for redox processes. One of these methods is based on the change in the
oxidation number of reducing agent and the oxidising agent and the other method is based on splitting the redox reaction into two half reactions — one involving oxidation and the other involving reduction. Both these methods are in use and the choice of their use rests with the individual using them. Unit 7.indd 246 10/10/2022 10:37:04 AM
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redox reactions
Solution
Step 1: The skeletal ionic equation is: Cr
2O72–(aq) + SO32–(aq) → Cr3+(aq)
+ SO42–(aq)
Step 2: Assign oxidation numbers for
Cr and S
+6 –2 +4 –2 +3 +6 –2
Cr2O72–(aq) + SO32–(aq) → Cr(aq)+SO42–(aq)
This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.
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Step 3: Calculate the increase and
decrease of oxidation number, and make them equal: from step-2 we can notice that there is change in oxidation state of chromium and sulphur. Oxidation state of chromium changes form +6 to +3. There is decrease of +3 in oxidation state of chromium on right hand side of the equation. Oxidation state of sulphur changes from +4 to +6. There is an increase of +2 in the oxidation state of sulphur on right hand side. To make the increase and decrease of oxidation state equal, place numeral 2 before cromium ion on right hand side and numeral 3 before sulphate ion on right hand side and balance the chromium and sulphur atoms on both the sides of the equation. Thus we get
+6 –2 +4 –2 +3
Cr2O72–(aq) + 3SO32– (aq) → 2Cr3+ (aq) +
+6 –2
3SO42– (aq)
Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H
+ on the left to make ionic charges
equal
Cr2O72–(aq) + 3SO32–(aq)+ 8H+→ 2Cr3+(aq)
+ 3SO42– (aq)
Step 5: Finally, count the hydrogen
atoms, and add appropriate number of water molecules (i.e., 4H
2O) on the right
to achieve balanced redox change. Cr2O72– (aq) + 3SO32– (aq)+ 8H+ (aq) →
2Cr3+ (aq) + 3SO42– (aq) +4H2O (l)Problem 7.9
Permanganate ion reacts with bromide
ion in basic medium to give manganese dioxide and bromate ion. Write the
balanced ionic equation for the reaction. Solution
Step 1: The skeletal ionic equation is : MnO
4–(aq) + Br–(aq) → MnO2(s) + BrO3– (aq)
Step 2: Assign oxidation numbers for
Mn and Br
+7 –1 +4 +5
MnO4–(aq) + Br–(aq) →MnO2 (s) + BrO3– (aq)
this indicates that permanganate ion is the oxidant and bromide ion is the reductant.
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Step 3: Calculate the increase and
decrease of oxidation number, and make
the increase equal to the decrease. +7 –1 +4 +5
2MnO4–(aq)+Br –(aq) → 2MnO2(s)+BrO3–(aq)
Step 4: As the reaction occurs in the
basic medium, and the ionic charges are not equal on both sides, add 2 OH
– ions
on the right to make ionic charges equal. 2MnO4– (aq) + Br– (aq) → 2MnO2(s) +
BrO3–(aq) + 2OH–(aq)
Step 5: Finally, count the hydrogen atoms
and add appropriate number of water molecules (i.e. one H
2O molecule) on the
left side to achieve balanced redox change. 2MnO4–(aq) + Br–(aq) + H2O(l) → 2MnO2(s)
+ BrO3– (aq) + 2OH–(aq)
(b) Half Reaction Method: In this method,
the two half equations are balanced separately
and then added together to give balanced equation. Suppose we are to balance the equation
showing the oxidation of Fe
2+ ions to Fe3+ ions
by dichromate ions (Cr2O7)2– in acidic medium,
wherein, Cr2O72– ions are reduced to Cr3+ ions. The following steps are involved in this task. Step 1: Produce unbalanced equation for the
reaction in ionic for
m :
Fe2+(aq) + Cr2O72– (aq) → Fe3+ (aq) + Cr3+(aq)
(7.50)
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Step 2: Separate the equation into half-
reactions:
+2 +3
Oxidation half : Fe2+ (aq) → Fe3+(aq) (7.51)
+6 –2 +3
Reduction half : Cr2O72–(aq) → Cr3+(aq)
(7.52)
Step 3: Balance the atoms other than O and
H in each half reaction individually. Here the
oxidation half reaction is alr
eady balanced
with respect to Fe atoms. For the reduction half reaction, we multiply the Cr
3+ by 2 to
balance Cr atoms.
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Cr2O7 2–(aq) → 2 Cr3+(aq) (7.53)
Step 4: For reactions occurring in acidic
medium, add H2O to balance O atoms and H+
to balance H atoms. Thus, we get :
Cr2O7 2– (aq) + 14H+ (aq) → 2 Cr3+(aq) + 7H2O (l)
(7.54)
Step 5: Add electrons to one side of the half
reaction to balance the charges. If need be,
make
the number of electrons equal in the
two half reactions by multiplying one or both
half reactions by appropriate number. The oxidation half reaction is thus
rewritten to balance the charge:
Fe2+ (aq) → Fe3+ (aq) + e– (7.55)
Now in the reduction half reaction there
ar
e net twelve positive charges on the left hand
side and only six positive charges on the right
hand side. Therefore, we add six electrons on the left side. Cr
2O7 2– (aq) + 14H+ (aq) + 6e– → 2Cr3+(aq) +
7H2O (l) (7.56)
To equalise the number of electrons in both
the half reactions, we multiply the oxidation
half reaction by 6 and write as :
6Fe2+ (aq) → 6Fe3+(aq) + 6e– (7.57)
Step 6: We add the two half reactions to
achieve the overall r
eaction and cancel the
electrons on each side. This gives the net ionic
equation as :
6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq) → 6 Fe3+(aq) +
2Cr 3+(aq) + 7H2O(l) (7.58)Step 7: Verify that the equation contains the
same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number of atoms
and the charges. For the reaction in a basic medium, first
balance the atoms as is done in acidic medium. Then for each H
+ ion, add an equal number of
OH– ions to both sides of the equation.
|
Where
H+ and OH– appear on the same side of the
equation, combine these to give H2O. Problem 7.10
Permanganate(VII) ion, MnO4– in basic
solution oxidises iodide ion, I– to produce
molecular iodine (I2) and manganese
(IV) oxide (MnO2). Write a balanced ionic
equation to represent this redox reaction. SolutionStep 1:
First we write the skeletal
ionic equation, which is MnO
4– (aq) + I– (aq) → MnO2(s) + I2(s)
Step 2: The two half-reactions are:
–1 0
Oxidation half : I–(aq) → I2 (s)
+7 +4
Reduction half: MnO4–(aq) → MnO2(s)
Step 3: To balance the I atoms in the
oxidation half reaction, we rewrite it as:
2I– (aq) → I2 (s)
Step 4: To balance the O atoms in the reduction half reaction, we add two water molecules on the right:
MnO
4– (aq) → MnO2 (s) + 2 H2O (l)
To balance the H atoms, we add four H+
ions on the left: MnO
4 – (aq) + 4 H+ (aq) → MnO2(s) + 2H2O (l)
As the reaction takes place in a basic
solution, therefore, for four H+ ions, we
add four OH– ions to both sides of the
equation:
MnO4– (aq) + 4H+ (aq) + 4OH–(aq) →
MnO2 (s) + 2 H2O(l) + 4OH– (aq)
Replacing the H+ and OH– ions with water,
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redox reactions
the resultant equation is:
MnO4– (aq) + 2H2O (l) → MnO2 (s) + 4 OH– (aq)
Step 5 : In this step we balance the
charges of the two half-reactions in the
manner depicted as:
2I– (aq) → I2 (s) + 2e–
MnO4–(aq) + 2H2O(l) + 3e– → MnO2(s)
+ 4OH–(aq)
Now to equalise the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. 6I
–(aq) → 3I2 (s) + 6e–
2 MnO4– (aq) + 4H2O (l) +6e– → 2MnO2(s)
+ 8OH– (aq)
Step 6: Add two half-reactions to
obtain the net reactions after cancelling electrons on both sides. 6I
–(aq) + 2MnO4–(aq) + 4H2O(l) → 3I2(s) +
2MnO2(s) +8 OH–(aq)
Step 7: A final verification shows that
the equation is balanced in respect of the number of atoms and charges on both sides. 7.3.3
Redox Reactions as the Basis for Titrations
In acid-base systems we come across with a
titration method for finding out the strength of one solution against the other using a
pH sensitive indicator. Similarly, in redox systems, the titration method can be adopted to determine the strength of a reductant/oxidant using a redox sensitive indicator. The usage of indicators in redox titration is illustrated below:
(i)
In one situation, the reagent itself is
intensely coloured, e.g., per
manganate
ion, MnO4–. Here MnO4– acts as the self
indicator.
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The visible end point in this case is achieved after the last of the reductant (Fe
2+ or C2O42–) is oxidised and the first
lasting tinge of pink colour appears at MnO
4– concentration as low as 10–6 mol dm–3
(10–6 mol L–1). This ensures a minimal
‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry. (ii)
If there is no dramatic auto-colour change
(as with MnO4– titration), there are indicators
which are oxidised immediately after the
last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by Cr
2O72–, which
is not a self-indicator, but oxidises the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signalling the end point. (iii)
There is yet another method which is interesting and quite common. Its use is r
estricted to those reagents which are able
to oxidise I– ions, say, for example, Cu(II):
2Cu2+(aq) + 4I–(aq) → Cu2I2(s) + I2(aq) (7.59)
This method relies on the facts that
iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate ions (S
2O32–), which too is a redox
reaction:
I2(aq) + 2 S2O32–(aq)→2I–(aq) + S4O62–(aq) (7.60)
I2, though insoluble in water, remains in
solution containing KI as KI3. On addition of starch after the liberation
of iodine from the reaction of Cu2+ ions on
iodide ions, an intense blue colour appears. This colour disappears as soon as the iodine is consumed by the thiosulphate ions. Thus, the end-point can easily be tracked and the rest is the stoichiometric calculation only. 7.3.4
Limitations of Concept of Oxidation
Number
As you have observed in the above discussion,
the concept of redox pr
ocesses has been
evolving with time.
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This process of evolution is continuing. In fact, in recent past the oxidation process is visualised as a decrease in electron density and reduction process as an increase in electron density around the atom(s) involved in the reaction. 7.4
REDOX REACTIONS AND ELECTRODE
PROCESSES
The experiment corresponding to reaction
(7.15), can also be observed if zinc r
od is
dipped in copper sulphate solution. The redox reaction takes place and during the
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chemistry
reaction, zinc is oxidised to zinc ions and
copper ions are reduced to metallic copper due to direct transfer of electrons from zinc to copper ion. During this reaction heat is also evolved. Now we modify the experiment in such a manner that for the same redox reaction transfer of electrons takes place indirectly. This necessitates the separation of zinc metal from copper sulphate solution. We take copper sulphate solution in a beaker and put a copper strip or rod in it. We also take zinc sulphate solution in another beaker and put a zinc rod or strip in it. Now reaction takes place in either of the beakers and at the interface of the metal and its salt solution in each beaker both the reduced and oxidized forms of the same species are present.
|
These represent the species in the reduction and oxidation half reactions. A redox couple is defined as having together the oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction. This is represented by separating the
oxidised form from the reduced form by a vertical line or a slash representing an interface (e.g. solid/solution). For example in this experiment the two redox couples are represented as Zn
2+/Zn and Cu2+/Cu. In both cases, oxidised form is put before the reduced form. Now we put the beaker containing copper sulphate solution and the beaker containing zinc sulphate solution side by side (Fig. 7.3). We connect solutions in two beakers by a salt bridge (a U-tube containing a solution of potassium chloride or ammonium nitrate usually solidified by boiling with agar agar and later cooling to a jelly like substance). This provides an electric contact between the two solutions without allowing them to mix with each other.
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The zinc and copper rods are connected by a metallic wire with a provision for an ammeter and a switch. The set-up as shown in Fig.7.3 is known as Daniell cell. When the switch is in the off position, no reaction takes place in either of the beakers and no current flows through the metallic wire. As soon as the switch is in the on position, we make the following observations:
1. The transfer of electrons now does not
ta
ke place directly from Zn to Cu2+ but
through the metallic wire connecting the two rods as is apparent from the arrow which indicates the flow of current. 2. The e
lectricity from solution in one beaker
to solution in the other beaker flows by the migration of ions through the salt bridge. We know that the flow of current is possible only if there is a potential difference between the copper and zinc rods known as electrodes here. The potential associated with each
electrode is known as electrode potential. If the concentration of each species taking
part in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further the reaction is carried out at 298K, then the potential of each electrode is said to be the Standard Electrode Potential.
|
By convention, the standard electrode potential (E
) of hydrogen electrode is 0.00 volts. The
electrode potential value for each electrode process is a measure of the relative tendency of the active species in the process to remain in the oxidised/reduced form. A negative E
means that the redox couple is a stronger Fig.7.3 The set-up for Daniell cell. Electrons
pr
oduced at the anode due to oxidation
of Zn travel through the external circuit to the cathode where these reduce the copper ions. The circuit is completed inside the cell by the migration of ions through the salt bridge. It may be noted that the direction of current is opposite to the direction of electron flow. Unit 7.indd 250 10/10/2022 10:37:05 AM
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redox reactions
Reaction (Oxidised form + ne– → Reduced form) E / V
F2(g) + 2e– → 2F– 2.87
Co3+ + e– → Co2+ 1.81
H2O2 + 2H+ + 2e– → 2H2O 1.78
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O 1.51
Au3+ + 3e– → Au(s) 1.40
Cl2(g) + 2e– → 2Cl– 1.36
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O 1.33
O2(g) + 4H+ + 4e– → 2H2O 1.23
MnO2(s) + 4H+ + 2e– → Mn2+ + 2H2O 1.23
Br2 + 2e– → 2Br– 1.09
NO3– + 4H+ + 3e– → NO(g) + 2H2O 0.97
2Hg2+ + 2e– → Hg22+ 0.92
Ag+ + e– → Ag(s) 0.80
Fe3+ + e– → Fe2+ 0.77
O2(g) + 2H+ + 2e– → H2O2 0.68
I2(s) + 2e– → 2I– 0.54
Cu+ + e– → Cu(s) 0.52
Cu2+ + 2e– → Cu(s) 0.34
AgCl(s) + e– → Ag(s) + Cl– 0.22
AgBr(s) + e– → Ag(s) + Br– 0.10
2H+ + 2e– → H2(g) 0.00
Pb2+ + 2e– → Pb(s) –0.13
Sn2+ + 2e– → Sn(s) –0.14
Ni2+ + 2e– → Ni(s) –0.25
Fe2+ + 2e– → Fe(s) –0.44
Cr3+ + 3e– → Cr(s) –0.74
Zn2+ + 2e– → Zn(s) –0.76
2H2O + 2e– → H2(g) + 2OH– –0.83
Al3+ + 3e– → Al(s) –1.66
Mg2+ + 2e– → Mg(s) –2.36
Na+ + e– → Na(s) –2.71
Ca2+ + 2e– → Ca(s) –2.87
K+ + e– → K(s) –2.93
Li+ + e– → Li(s) –3.05 Increasing strength of oxidising agent
Increasing strength of reducing agent
1. A negative E means that the redox couple is a stronger reducing agent than the H+/H2 couple. 2. A positive E means that the redox couple is a weaker reducing agent than the H+/H2 couple.
|
Table 7.1 The Standard Electrode Potentials at 298 K
Ions are present as aqueous species and H2O as liquid; gases and
solids are shown by g and s respectively.reducing agent than the H+/H2 couple. A
positive E means that the redox couple is a
weaker reducing agent than the H+/H2 couple. The standard electrode potentials are very
important and we can get a lot of other useful information from them. The values of standard
electrode potentials for some selected electrode processes (reduction reactions) are given in Table 7.1. You will learn more about electrode reactions and cells in Class XII. Unit 7.indd 251 10/10/2022 10:37:05 AM
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chemistry
SUMMARY
Redox reactions form an important class of reactions in which oxidation and reduction
occur simultaneously. Three tier conceptualisation viz, classical, electronic and oxidation number, which is usually available in the texts, has been presented in detail. Oxidation, reduction, oxidising agent (oxidant) and reducing agent (reductant) have been viewed according to each conceptualisation. Oxidation numbers are assigned in accordance with a consistent set of rules. Oxidation number and ion-electron method both are useful means in writing equations for the redox reactions.
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Redox reactions are classified into four categories: combination, decomposition displacement and disproportionation reactions. The concept of redox couple and electrode processes is introduced here. The redox reactions find wide applications in the study of electrode processes and cells. EXERCISES
7.1 Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4
(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O
7.2 What are the
oxidation number of the underlined elements in each of the following
and how do you rationalise your results ? (a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
7.3 Justify that the following reactions are redox r
eactions:
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
(d) 2K(s) + F2(g) → 2K+F– (s)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
7.4 Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction. 7
.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72–
and NO3–. Suggest structure of these compounds. Count for the fallacy. 7.6 Write formulas for the following compounds:
(a) Mercury(II) chloride (b) Nickel(II) sulphate
(c) Tin(IV) oxide (d) Thallium(I) sulphate
(e) Iron(III) sulphate ( f ) Chromium(III) oxide
7.7 Suggest a list of
the substances where carbon can exhibit oxidation states from –4
to +4 and nitrogen from –3 to +5. 7.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing
agents in their reactions, ozone and nitric acid act only as oxidants.
|
Why ? 7.9 Consider the reactions:
(a) 6 CO2(g) + 6H2O(l) → C6 H12 O6(aq) + 6O2(g)
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redox reactions
(b) O3(g) + H2O2(l) → H2O(l) + 2O2(g)
Why it is more appropriate to write these reactions as :
(a) 6CO2(g) + 12H2O(l) → C6 H12 O6(aq) + 6H2O(l) + 6O2(g)
(b) O3(g) + H2O2 (l) → H2O(l) + O2(g) + O2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox
reactions. 7.10 The compound AgF2 is unstable compound. However, if formed, the compound acts
as a very strong oxidising agent. Why ? 7.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a
compound of lower oxidation state is formed if the reducing agent is in excess and
a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations. 7.12
How do you count for the following observations ? (a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufactur
e of benzoic acid from toluene
we use alcoholic potassium permanganate as an oxidant. Why ?
|
Write a balanced redox equation for the reaction. (b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains
bromide then we get red vapour of bromine. Why ? 7.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)
(b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq)
+ 2H2O(l)
(c) HCHO (l) + 2 Cu2+(aq) + 5 OH–(aq) → Cu2O(s) + HCOO–(aq) + 3H2O(l)
(d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
7.14 Consider the reactions :
2 S2O32– (aq) + I2(s) → S4 O6 2–(aq) + 2I–(aq)
S2O3 2–(aq) + 2Br2(l) + 5 H2O(l) → 2SO42–(aq) + 4Br–(aq) + 10H+(aq)
Why does the same reductant, thiosulphate react differ
ently with iodine and
bromine ? 7.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydr
oiodic acid is the best reductant. 7.16 Why does the following reaction occur ? XeO64– (aq) + 2F– (aq) + 6H+(aq) → XeO3(g)+ F2(g) + 3H2O(l)
What conclusion about the
compound Na4XeO6 (of which XeO64– is a part) can be
drawn from the reaction. 7.17 Consider the reactions:
(a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
(b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)
(c) C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) +
4NH3 (aq) + 2 H2O(l)
(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH–(aq) → No change observed. Unit 7.indd 253 10/10/2022 10:37:06 AM
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chemistry
What inference do you draw about the behaviour of Ag+ and Cu2+ from these
reactions ? 7.18 Balance the following redox reactions by ion – electron method :
(a) MnO4– (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium)
(b) MnO4– (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
7.19 Balance the following equations in basic medium by ion-electron method and
oxidation number methods and identify the oxidising agent and the reducing agent.
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(a) P4(s) + OH–(aq) → PH3(g) + HPO2– (aq)
(b) N2H4(l) + ClO3 –(aq) → NO(g) + Cl–(g)
(c) Cl2O7 (g) + H2O2(aq) → ClO2–(aq) + O2(g) + H+
7.20 What sorts of informations can you draw from the following reaction ? (CN)2(g) + 2OH–(aq) → CN–(aq) + CNO–(aq) + H2O(l)
7.21 The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+,
MnO2, and H+ ion. Write a balanced ionic equation for the reaction. 7.22 Consider the elements :
Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only postive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the
negative nor does the positive
oxidation state. 7.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess
of chlorine is r
emoved by treating with sulphur dioxide.
|
Present a balanced equation
for this redox change taking place in water. 7.24 Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction. 7.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight
of nitric oxide that can be obtained starting only with 10.00
g. of ammonia and 20.00 g of oxygen ? 7.26 Using the standard electrode potentials given in the Table 8.1, pr
edict if the reaction
between the following is feasible:
(a) Fe3+(aq) and I–(aq)
(b) Ag+(aq) and Cu(s)
(c) Fe3+ (aq) and Cu(s)
(d) Ag(s) and Fe3+(aq)
(e) Br2(aq) and Fe2+(aq). Unit 7.indd 254 10/10/2022 10:37:06 AM
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redox reactions
7.27 Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes
(ii) An aqueous solution AgNO3 with platinum electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes
(iv) An aqueous solution of CuCl2 with platinum electrodes. 7.28 Arrange the following metals in the order in which they displace each other from
the solution of their salts. Al, Cu, Fe, Mg and Zn. 7.29 Given the standard electrode potentials,
K+/K = –2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37V.
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Cr3+/Cr = –0.74V
arrange these metals in their increasing order of reducing power
. 7.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) +2Ag(s)
takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode. Unit 7.indd 255 10/10/2022 10:37:06 AM
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chemistry
Organic chemistry – sOme Basic
Princi Ples and Techniques
After studying this unit, you will
be able to
• under
stand reasons for
tetravalence of carbon and
shapes of organic molecules;
• write
structures of organic
molecules in various ways;
• classify the organic compounds;
• name the compounds according to IUPAC system of nomenclature
and also derive their structures from the given names;
•
understand the concept of organic reaction mechanism;
•
explain the influence of electronic displacements on structure and r
eactivity of organic compounds;
• recognise the types of organic reactions;
•
learn the techniques of purification of organic compounds;
•
write the chemical reactions involved in the qualitative analysis of organic compounds;
•
understand the principles involv
ed in quantitative analysis
of organic compounds.In the previous unit you have learnt that the element
carbon has the unique property called catenation due to
which it forms covalent bonds with other carbon atoms. It also forms covalent bonds with atoms of other elements like hydrogen, oxygen, nitrogen, sulphur, phosphorus and halogens. The resulting compounds are studied under a separate branch of chemistry called organic chemistry. This unit incorporates some basic principles and techniques of analysis required for understanding the formation and properties of organic compounds. 8.1 General
inTroduc Tion
Organic compounds are
vital for sustaining life on earth
and include complex molecules like genetic information
bearing deoxyribonucleic acid (DNA) and proteins that constitute essential compounds of our blood, muscles and
skin. Organic compounds appear in materials like clothing,
fuels, polymers, dyes and medicines. These are some of the important areas of application of these compounds. Science of organic chemistry is about two hundred
years old.
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Around the year 1780, chemists began to distinguish between organic compounds obtained from plants and animals and inorganic compounds prepared from mineral sources. Berzilius, a Swedish chemist proposed that a ‘vital force’ was responsible for the formation of organic compounds. However, this notion was rejected in 1828 when F. Wohler synthesised an organic compound, urea from an inorganic compound, ammonium cyanate. NH
4CNO NH2CONH2
Ammonium cyanate Urea
The pioneering synthesis of acetic acid by Kolbe
(1845) and that of methane by Berthelot (1856) showed
conclusively that organic compounds could be synthesised from inorganic sources in a laboratory. uniT 8
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organic chemistry – some basic principles and techniques
The development of electronic theory of
covalent bonding ushered organic chemistry
into its modern shape. 8.2 tetra Valence OF carBOn:
shaPes OF O rganic cOmPOunds
8.2.1 The Shapes of Carbon Compounds
The knowledge of fundamental concepts of
molecular structur
e helps in understanding
and predicting the properties of organic compounds. You have already learnt theories of valency and molecular structure in Unit 4. Also, you already know that tetravalence of carbon and the formation of covalent bonds by it are explained in terms of its electronic configuration and the hybridisation of s and p
orbitals. It may be recalled that formation and the shapes of molecules like methane (CH
4),
ethene (C2H4), ethyne (C2H2) are explained
in terms of the use of sp3, sp2 and sp hybrid
orbitals by carbon atoms in the respective molecules. Hybridisation influences the bond length
and bond enthalpy (strength) in compounds.
|
The sp hybrid orbital contains more s
character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp
3 hybrid orbital. The sp2 hybrid orbital
is intermediate in s character between sp
and sp3 and, hence, the length and enthalpy
of the bonds it forms, are also intermediate between them. The change in hybridisation affects the electronegativity of carbon. The greater the s character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an sp hybrid orbital with
50% s character is more electronegative than
that possessing sp
2 or sp3 hybridised orbitals. This relative electronegativity is reflected in several physical and chemical properties of the molecules concerned, about which you will learn in later units. 8.2.2
some characteristic Features
of π
Bonds
In a π (pi) bond formation, parallel orientation
of the two p orbitals on adjacent atoms is
necessary for a proper sideways overlap. Thus, in H2C=CH2 molecule all the atoms
must be in the same plane. The p orbitals are
mutually parallel and both the p orbitals are
perpendicular to the plane of the molecule. Rotation of one CH
2 fragment with respect
to other interferes with maximum overlap of p orbitals and, therefore, such rotation
about carbon-carbon double bond (C=C) is restricted.
|
The electron charge cloud of the π bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking
reagents. In general, π bonds provide the most
reactive centres in the molecules containing
multiple bonds. Problem 8.1
How many σ and π bonds are present in
each of the following molecules? (a) HC≡CCH=CHCH
3 (b) CH2=C=CHCH3
solution
(a) σC – C: 4; σC–H : 6; πC=C :1; π C≡C:2
(b) σC – C: 3; σC–H: 6; πC=C: 2. Problem 8 .2
What is the type of hybridisation of each
carbon in the following compounds? (a) CH3Cl, (b) (CH3)2CO, (c) CH3CN,
(d) HCONH2, (e) CH3CH=CHCN
solution
(a) sp3, (b) sp3, sp2, (c) sp3, sp, (d) sp2, (e)
sp3, sp2, sp2, sp
Problem 8.3Write the state of hybridisation of carbon
in the following compounds and shapes of each of the molecules. (a) H
2C=O, (b) CH3F, (c) HC≡N. solution
(a) sp2 hybridised carbon, trigonal planar;
(b) sp3 hybridised carbon, tetrahedral; (c)
sp hybridised carbon, linear. Unit 8.indd 257 10/10/2022 10:37:28 AM
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8.3 structural rePresen tatiOns
OF Organic cOmPOunds
8.3
.1 complete, condensed and Bond-line
structural Formulas
Structures of or
ganic compounds are
represented in several ways.
|
The Lewis
structure or dot structure, dash structure,
condensed structure and bond line structural formulas are some of the specific types. The Lewis structures, however, can be simplified by representing the two-electron covalent bond by a dash (–). Such a structural formula focuses on the electrons involved in bond formation. A single dash represents a single bond, double dash is used for double bond and a triple dash represents triple bond. Lone-pairs of electrons on heteroatoms (e.g., oxygen, nitrogen, sulphur, halogens etc.) may or may not be shown. Thus, ethane (C
2H6), ethene
(C2H4), ethyne (C2H2) and methanol (CH3OH)
can be represented by the following structural formulas. Such structural representations are called complete structural formulas.Similarly, CH
3CH2CH2CH2CH2CH2CH2CH3
can be further condensed to CH3(CH2)6CH3. For further simplification, organic chemists use another way of representing the structures, in which only lines are used. In this bond-line structural representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are drawn in a zig-zag fashion.
|
The only atoms specifically written are oxygen, chlorine, nitrogen etc. The terminals denote methyl
(–CH
3) groups (unless indicated otherwise by
a functional group), while the line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy the valency of the carbon atoms. Some of the examples are represented as follows:
(i) 3-Methyloctane can be represented in
various forms as:
(a) CH
3CH2CHCH2CH2CH2CH2CH3
|
CH3
These structural formulas can be further
abbreviated by omitting some or all of the
dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula. Thus, ethane, ethene, ethyne and methanol can be written as:
CH
3CH3 H2C=CH2 HC≡ CH CH3OH
Ethane Ethene Ethyne Methanol Ethane Ethene
Ethyne Methanol
(ii) Various ways of representing 2-bromo
butane are:
(a) CH3CHBrCH2CH3 (b)
(c) (b)
(c)
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In cyclic compounds, the bond-line formulas
may be given as follows:
Cyclopropane
Cyclopentane
chlorocyclohexane
Problem 8.4
Expand each of the following condensed
formulas into their complete structural formulas. (a) CH
3CH2COCH2CH3
(b) CH3CH=CH(CH2)3CH3
solution
(a)
(b) (b)
solution
Condensed formula:
(a) HO(CH2)3CH(CH3)CH(CH3)2
(b) HOCH(CN)2
Bond-line formula:
Problem 8.5
For each of the following compounds,
write a condensed formula and also their bond-line formula. (a)
HOCH2CH2CH2CH(CH3)CH(CH3)CH3(b)(a)
Problem 8.6
Expand each of the following bond-line
formulas to show all the atoms including carbon and hydrogen
(a)
(b)
(c)
(d)
solution
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Framework model Ball and stick model
Space filling model
Fig. 8.28.3.2 Three-Dimensional
Representation of Organic Molecules
The thr
ee-dimensional (3-D) structure of
organic molecules can be represented on paper by using certain conventions. For example, by using solid (
) and dashed
( ) wedge formula, the 3-D image of a
molecule from a two-dimensional picture
can be perceived. In these formulas the
solid-wedge is used to indicate a bond projecting out of the plane of paper, towards the observer.
|
The dashed-wedge is used to depict the bond projecting out of the plane of the paper and away from the observer. Wedges are shown in such a way that the broad end of the wedge is towards the observer. The bonds lying in plane of the paper are depicted by using a normal line (—). 3-D representation of methane molecule on paper has been shown in Fig. 8.1
Fig. 8.1 Wedge-and-dash representation of CH4Molecular Models
Molecular models are physical devices that are used for a better visualisation and perception of three-dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available. Commonly three types of molecular models are used: (1) Framework model, (2) Ball-and-stick model, and (3) Space filling model. In the framework model only the
bonds connecting the atoms of a molecule and not the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of atoms.
|
In the ball-and-stick model, both
the atoms and the bonds are shown. Balls represent atoms and the stick denotes a bond. Compounds containing C=C (e.g., ethene) can best be represented by using springs in place of sticks. These models are referred to as ball-and-spring model. The space-filling model emphasises the relative size of each atom based on its van der Waals radius. Bonds are not shown in this model. It conveys the volume occupied by each atom in the molecule. In addition to these models, computer graphics can also be used for molecular modelling. Unit 8.indd 260 11/10/2022 15:19:57
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8.4 Classifi Cation of organi C
Compounds
The existing large number of organic
compounds and their ever-incr
easing
numbers has made it necessary to classify them on the basis of their structures. Organic compounds are broadly classified as follows:
i. Acyclic or open chain compounds
These compounds are also called as aliphatic
compounds and consist of straight or branched chain compounds, for example:(homocyclic).
|
Cyclohexane Cyclohexene Cyclopropane
Sometimes atoms other than carbon are also present in the ring (heterocylic). Tetrahydrofuran given below is an example of this type of compound:
Tetrahydrofuran
These exhibit some of the properties similar to those of aliphatic compounds. (b)
aromatic compounds
Aromatic compounds are special types of
compounds. You will learn about these
compounds in detail in Unit 9. These include benzene and other related ring compounds (benzenoid). Like alicyclic compounds, aromatic comounds may also have hetero atom in the ring. Such compounds are called hetrocyclic aromatic compounds. Some of the examples of various types of aromatic compounds are:
Benzenoid aromatic compounds
Benzene Aniline Naphthalene
Non-benzenoid compound
TroponeIsobutane
Acetaldehyde Acetic acidCH3CH3
Ethane
ii Cyclic or closed chain or ring
compounds
(a) alicyclic compounds
Alicyc
lic (aliphatic cyclic) compounds contain
carbon atoms joined in the form of a ring
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Heterocyclic aromatic compounds
Furan Thiophene Pyridine
Organic compounds can also be classified
on the basis of functional groups, into families
or homologous series. 8.4.1 functional group
The functional group is an atom or a group
of atoms joined to the carbon chain which is r
esponsible for the characteristic chemical
properties of the organic compounds. The examples are hydroxyl group (–OH), aldehyde group (–CHO) and carboxylic acid group (–COOH) etc.
|
8.4.2 Homologous
series
A group or a series of organic compounds each
containing
a characteristic functional group
forms a homologous series and the members of the series are called homologues . The members
of a homologous series can be represented by general molecular formula and the successive members differ from each other in molecular formula by a –C
H2 unit. There are a number
of homologous series of organic compounds. Some of these are alkanes, alkenes, alkynes, haloalkanes, alkanols, alkanals, alkanones, alkanoic acids, amines etc. It is also possible that a compound contains
two or more identical or different functional gr
oups. This gives rise to polyfunctional
compounds. 8.5 nom EnClatur E of organ
iC
Compounds
Organic chemistry deals with millions of
compounds. In order to clearly identify them, a systematic method of naming has been developed and is known as the
iupaC
(international union of pure and applied
Chemistry) system of nomenclature. In this systematic nomenclature, the names ar
e
correlated with the structure such that the reader or listener can deduce the structure from the name. Before the IUPAC system of nomenclature,
however, or
ganic compounds were assigned
names based on their origin or certain
properties.
|
For instance, citric acid is named so because it is found in citrus fruits and the acid found in red ant is named formic acid since the Latin word for ant is formica. These names are traditional and are considered as trivial or common names. Some common
names are followed even today. For example, Buckminsterfullerene is a common name given to the newly discovered C
60 cluster (a
form of carbon) noting its structural similarity to the geodesic domes popularised by the famous architect R. Buckminster Fuller. Common names are useful and in many cases indispensable, particularly when the alternative systematic names are lengthy and complicated. Common names of some organic compounds are given in Table 8.1.
table 8.1 Common or trivial names of some
organic Compounds
8.5.1 the iupaC system of nomenclature
A systematic name of an
organic compound
is generally derived by identifying the parent
hydrocarbon and the functional group(s) attached to it. See the example given below. Unit 8.indd 262 11/10/2022 15:20:58
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By further using prefixes and suffixes, the
parent name can be modified to obtain the actual name. Compounds containing carbon and hydrogen only are called hydrocarbons. A hydrocarbon is termed saturated if it contains only carbon-carbon single bonds.
|
The IUPAC name for a homologous series of such compounds is alkane. Paraffin (Latin: little affinity) was the earlier name given to these compounds. Unsaturated hydrocarbons are those, which contain at least one carbon-carbon double or triple bond. 8.5.2
iu Pac nomenclature of alkanes
Straight chain hydrocarbons:
The names
of such compounds are based on their chain
structure, and end with suffix ‘-ane’ and carry a prefix indicating the number of carbon atoms present in the chain (except from CH
4
to C4H10, where the prefixes are derived from
trivial names). The IUPAC names of some straight chain saturated hydrocarbons are given in Table 8.2. The alkanes in Table 8.2 differ from each other by merely the number of -CH
2 groups in the chain. They are
homologues of alkane series.In order to name such compounds, the names
of alkyl groups are prefixed to the name of parent alkane. An alkyl group is derived from a saturated hydrocarbon by removing a hydrogen atom from carbon. Thus, CH
4
becomes -CH3 and is called methyl group. An
alkyl group is named by substituting ‘yl’ for ‘ane’ in the corresponding alkane.
|
Some alkyl
groups are listed in Table 8.3.
table 8.3 some alkyl groups
table
8.2 iu Pac names of some unbranched
Saturated Hydrocarbons
Branched chain hydrocarbons: In a
branched chain compound small chains of
carbon atoms are attached at one or more carbon atoms of the parent chain. The small carbon chains (branches) are called alkyl groups. For example:
CH
3–CH–CH2–CH3 CH3–CH–CH2–CH–CH3
CH3 CH2CH3 CH3
(a) (b) Abbreviations are used for some alkyl
gr
oups. For example, methyl is abbreviated as
Me, ethyl as Et, propyl as Pr and butyl as Bu. The alkyl groups can be branched also. Thus, propyl and butyl groups can have branched structures as shown below. CH
3-CH- CH3-CH2-CH- CH3-CH-CH2-
CH3CH3 CH3
Isopropyl- sec-Butyl- Isobutyl-
CH3 CH3
CH3-C- CH3-C-CH2-
CH3 CH3
tert-Butyl- Neopentyl-
Common branched groups have specific
trivial names. For example, the propyl gr
oups
can either be n -propyl group or isopropyl
group. The branched butyl groups are called
sec-butyl, isobutyl and tert-butyl group. We also encounter the structural unit,
–CH
2C(CH3)3, which is called neopentyl group.
|
Nomenclature of branched chain alkanes:
We encounter a number of branched chain alkanes. The rules for naming them are given
below. Unit 8.indd 263 10/10/2022 10:37:31 AM
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chemistry
separated from the groups by hyphens
and there is no break between methyl
and nonane.] 4. If two or more identical substituent groups are present then the numbers are separated by commas. The names of identical substituents are not repeated,
instead prefixes such as di (for 2), tri
(for 3), tetra (for 4), penta (for 5), hexa (for 6) etc. are used. While writing the name of the substituents in alphabetical order, these prefixes, however, are not considered. Thus, the following compounds are
named as:
CH3 CH3 CH3 CH3
CH3-CH-CH2-CH-CH3 CH3 C CH2CH CH3
1 2 3 4 5 1 2 3 4 5
2,4-Dimethylpentane 2,2,4-Trimethylpentane
H3C H2C CH3
CH3CH2CHCCH2CH2CH3
1 2 3 4 5 6 7
CH3
3-Ethyl-4,4-dimethylheptane
5. If the two substituents are found in equivalent positions, the lower
number
is given to the one coming first in the alphabetical listing.
|
Thus, the following compound is 3-ethyl-6-methyloctane and not 6-ethyl-3-methyloctane. 1 2 3 4 5 6 7 8
CH3 — CH2—CH—CH2—CH2—CH—CH2 —CH3
CH2CH3 CH3
6. The branched alkyl groups can be named by following the above mentioned procedur
es. However, the carbon atom
of the branch that attaches to the root alkane is numbered 1 as exemplified below. 4
3 2 1
CH3–CH–CH2–CH–
CH3 CH3
1,3-Dimethylbutyl-1. First of all, the longest carbon chain in the molecule is
identified. In the example
(I) given below, the longest chain has nine carbons and it is considered as the parent or root chain. Selection of parent chain as
shown in (II) is not correct because it has only eight carbons. 2. The
carbon atoms of the parent chain are
numbered to identify the parent alkane
and to locate the positions of the carbon atoms at which branching takes place due to the substitution of alkyl group in place of hydrogen atoms.
|
The numbering is done in such a way that the branched carbon atoms get the lowest possible numbers. Thus, the numbering in the
above example should be from left to right (branching at carbon atoms 2 and 6) and not from right to left (giving numbers 4 and 8 to the carbon atoms at which
branches are attached). 1 2 3 4 5 6 7 8 9
C C C C C C C C C
C CC
9 8 7 6 5 4 3 2 1
C C
C C C C C C C
C C C
3. The names of alkyl groups attached
a
s a branch are then prefixed to the
name of the parent alkane and position of the substituents is indicated by the appropriate numbers. If different alkyl groups are present, they are listed in alphabetical order. Thus, name for the compound shown above is: 6-ethyl-2-
methylnonane. [Note: the numbers are
Unit 8.indd 264 10/10/2022 10:37:31 AM
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The name of such branched chain alkyl group
is placed in parenthesis while naming the compound. While writing the trivial names of substituents’ in alphabetical order, the prefixes iso- and neo- are considered to be the part of the fundamental name of alkyl group. The prefixes sec- and tert- are not considered to be the part of the fundamental name. The use of iso and related common prefixes for naming alkyl groups is also allowed by the IUPAC nomenclature as long as these are not further substituted.
|
In multi-substituted compounds, the following rules may aso be remembered:
• If
there happens to be two chains of equal
size, then that chain is to be selected which
contains more number of side chains. • After selection of the chain, numbering
is to be done from the end closer to the substituent. 5-(2,2-Dimethylpropyl)nonane Problem 8.7
Structures and IUPAC names of some hydr
ocarbons are given below. Explain
why the names given in the parentheses are incorrect.3-Ethyl-1,1-dimethylcyclohexane
(not 1-ethyl-3,3-dimethylcyclohexane)Cyclic Compounds: A saturated monocyclic
compound is named by prefixing ‘cyclo’ to the corresponding straight chain alkane. If side chains are present, then the rules given above are applied. Names of some cyclic compounds are given below. 5-sec-Butyl-4-isopropyldecaneSolution
(a) Lowest locant number, 2,5,6 is
lower than 3,5,7, (b) substituents are in
equivalent position; lower number is
given to the one that comes first in the name according to alphabetical order.2,5,6- Trimethyloctane
[and not 3,4,7-Trimethyloctane]
3-Ethyl-5-methylheptane
[and not 5-Ethyl-3-methylheptane]
8.5.3 Nomenclature of Organic Compounds having Functional Gr
oup(s)
A functional group, as defined earlier, is an atom or a group of atoms bonded together in a
unique manner which is usually the site of 5-(2-Ethylbutyl)-3,3-dimethyldecane
[and not 5-(2,2-Dimethylbutyl)-3-ethyldecane]
Unit 8.indd 265 10/27/2022 2:13:15 PM
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chemistry
chemical reactivity in an organic molecule. Compounds having the same functional group undergo similar reactions. For example, CH
3OH, CH3CH2OH, and (CH3)2CHOH —
all having -OH functional group liberate hydrogen on reaction with sodium metal. The presence of functional groups enables systematisation of organic compounds into different classes.
|
Examples of some functional groups with their prefixes and suffixes along with some examples of organic compounds possessing these are given in Table 8.4. First of all, the functional group present
in the molecule is identified which determines the choice of appropriate suffix. The longest
chain of carbon atoms containing the functional group is numbered in such a way that the functional group is attached at the carbon atom possessing lowest possible number in the chain. By using the suffix as
given in Table 8.4, the name of the compound
is arrived at. In the case of polyfunctional compounds,
one of the functional groups is chosen as the
principal functional group and the compound is then named on that basis. The remaining functional groups, which are subordinate functional groups, are named as substituents using the appropriate prefixes. The choice of principal functional group is made on the basis of order of preference. The order of decreasing priority for some functional groups is:
-
cOOh, –sO3h, -cOOr (r=alkyl group), cOcl,
-CONH2, -cn,-hc=O, > c=O, -O h, -nh2, >
c=c<, -c≡C- . The –R,
C6H5-, halogens (F, Cl, Br, I), –NO2,
alkoxy (–OR) etc. are always prefix
substituents.
|
Thus, a compound containing both an alcohol and a keto group is named as hydroxyalkanone since the keto group is preferred to the hydroxyl group. For example, HOCH
2(CH2)3CH2COCH3 will
be named as 7-hydroxyheptan-2-one and not as 2-oxoheptan -7-ol. Similarly, BrCH
2CH=CH2
is named as 3-bromoprop-1-ene and not 1-bromoprop-2-ene. If more than one functional group of the
same type are present, their number is indicated by adding di, tri, etc. before the class
suffix. In such cases the full name of the
parent alkane is written before the class suffix. For example CH2(OH)CH2(OH) is named as
ethane–1,2–diol. However, the ending – ne of the parent alkane is dropped in the case of compounds having more than one double or triple bond; for example, CH
2=CH-CH=CH2 is
named as buta–1,3–diene. Problem 8.8
Write the IUPAC names of the compounds
i-iv from their given structures. solution
• The functional group present is an alcohol (OH).
|
Hence the suf
fix is ‘-ol’. • The longest chain containing -OH has eight carbon atoms. Hence the
corresponding saturated hydr
ocar-
bon is octane. • The -OH is on carbon atom 3. In
addition, a methyl group is attached at 6
th carbon. Hence, the systematic name of this compound is 6-Methyloctan-3-ol. solution
The functional group present is ketone (>C=O), hence suffix ‘-one’. Presence of two keto groups is indicated by ‘di’,
hence suffix becomes ‘dione’. The two keto groups are at carbons 2 and 4. The longest chain contains 6 carbon atoms, hence, parent hydrocarbon is hexane.
|
Thus, the systematic name is Hexane-2,4-dione. Unit 8.indd 266 10/10/2022 10:37:32 AM
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organic chemistry – some basic principles and techniques
table 8.4 some Functional groups and classes of Organic compounds
Unit 8.indd 267 10/10/2022 10:37:32 AM
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chemistry
solution
Here, two functional groups namely
ketone and carboxylic acid ar
e present. The principal functional group is the carboxylic acid group; hence the parent chain will be suffixed with ‘oic’ acid. Numbering of the chain starts from carbon of – COOH functional group. The keto group in the chain at carbon 5 is indicated by ‘oxo’. The longest chain including the principal functional group has 6 carbon atoms; hence the parent hydrocarbon is hexane. The compound is, therefore, named as
5-Oxohexanoic acid. solution
The tw
o C=C functional groups are
present at carbon atoms 1 and 3, while the C≡C functional group is present at carbon 5. These groups are indicated by suffixes ‘diene’ and ‘yne’ respectively. The longest chain containing the functional groups has 6 carbon atoms; hence the parent hydrocarbon is hexane.
|
The name of compound, therefore, is Hexa-1,3-dien-5-yne. Problem 8.9Derive the structure of (i) 2-Chlorohexane,
(ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene, (iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxy-heptanal. solution
(i) ‘hexane’ indicates the presence of
6 carbon atoms in the chain. The functional group chloro is present at carbon 2. Hence, the structure of the compound is CH
3CH2CH2CH2CH(Cl)CH3. (ii) ‘pent’ indicates that parent
hydrocarbon contains 5 carbon atoms in the chain. ‘en’ and ‘ol’ correspond to the functional groups C=C and -OH at carbon atoms 4 and 2 respectively. Thus, the structure is
CH
2=CHCH2CH (OH)CH3. (iii) Six membered ring containing a
carbon-carbon double bond is implied by cyclohexene, which is numbered as shown in (I). The prefix 3-nitro means that a nitro group is present on C-3.
|
Thus, complete structural formula of the compound is (II). Double bond is suffixed functional group whereas NO
2 is prefixed
functional group therefore double bond gets preference over –NO
2 group:
(iv) ‘1-ol’ means that a -OH group is present at C-1. OH is suffixed functional group and gets preference over C=C bond. Thus the structure is as shown
in (II):
(v) ‘heptanal’ indicates the compound
to be an aldehyde containing 7 carbon atoms in the parent chain. The
‘6-hydroxy’ indicates that -OH group is present at carbon 6. Thus, the structural formula of the compound is: CH
3CH(OH)
CH2CH2CH2CH2CHO. Carbon atom of –
CHO group is included while numbering the carbon chain. 8.5.4 nomenclature of substituted
Benzene compounds
For IUPAC nomenclature of substituted
benzene compounds, the substituent is
placed as prefix to the word benzene as shown in the following examples. However, common names (written in bracket below) of many substituted benzene compounds are also
universally used. Unit 8.indd 268 10/10/2022 10:37:32 AM
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If benzene ring is disubstituted,
the position of substituents is defined
by numbering the carbon atoms of the ring such that the substituents are located at the lowest numbers possible.For example, the compound(b) is named as 1,3-dibromobenzene and not as
1,5-dibromobenzene.
|
Substituent of the base compound is
ass
igned number1 and then the direction
of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in alphabetical order. Some examples are given below. Methylbenzene Methoxybenzene Aminobenzene (Toluene) (Anisole)
(Aniline)
Nitrobenzene Bromobenzene
1,2-Dibromo-
benzene1,3-Dibromo-
benzene1,4-Dibromo-
benzene(a) (b) (c)
In the trivial system of nomenclature
the terms ortho
(o), meta ( m) and para ( p)
are used as prefixes to indicate the relative positions 1,2;1,3 and 1,4 respectively. Thus, 1,3-dibromobenzene (b) is named as
m-dibromobenzene (meta is abbreviated as
m-) and the other isomers of dibromobenzene
1,2-(a) and 1,4-(c), are named as ortho (or
just o-) and para (or just p -)-dibromobenzene,
respectively. For tri - or higher substituted benzene
derivatives, these prefixes cannot be used and the compounds ar
e named by identifying
substituent positions on the ring by following the lowest locant rule. In some cases, common name of benzene derivatives is taken as the base compound.1-Chloro-2,4-dinitrobenzene
(not 4-chloro,1,3-dinitrobenzene)
2-Chloro-1-methyl-4-nitrobenzene
(not 4-methyl-5-chloro-nitrobenzene)
2-Chloro-4-methylanisole 4-Ethyl-2-methylaniline
3,4-Dimethylphenol
When a benzene ring is attached to an
alkane with a functional group, it is considered
as substituent, instead of a parent. The name for benzene as substituent is phenyl (C
6H5-,
also abbreviated as Ph). Unit 8.indd 269 10/10/2022 10:37:33 AM
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chemistry
different carbon skeletons, these are referred
to as chain isomers and the phenomenon is termed as chain isomerism. For example, C
5H12 represents three compounds:
Isomerism
Structural isomerism Stereoisomerism
Chain
isomerismPosition
isomerismFunctional
group
isomerismMetamerism Geometrical
isomerismOptical
isomerism8.6 isOmerism
The phenomenon of existence of two or more compounds possessing the same molecular formula but differ
ent properties is known
as isomerism.
|
Such compounds are called as isomers. The following flow chart shows different types of isomerism. 8.6.1
structural isomerism
Compounds having the sam
e molecular
formula but different structures (manners
in which atoms are linked) are classified as structural isomers. Some typical examples of different types of structural isomerism are given below:
(i) Chain isomerism: When two or more
compounds have similar molecular formula but (a) (b)
(c) (d)Problem 8.10
Write the structural formula of:
(a) o
-Ethylanisole, (b) p-Nitroaniline,
(c) 2,3 - Dibromo -1 - phenylpentane,
(d) 4-Ethyl-1-fluoro-2-nitrobenzene. solution
CH3
CH3 C CH3
CH3
Neopentane
(2,2-Dimethylpropane)
(ii) Position isomerism: When two or
more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers and this phenomenon is termed as position isomerism. For example, the molecular formula C
3H8O represents two
alcohols:
(iii) Functional group isomerism: Two or
more compounds having the same molecular formula but different functional groups are called functional isomers and this phenomenon is termed as functional group isomerism. For example, the molecular formula C
3H6O represents an aldehyde and
a ketone: CH3
CH3CH2CH2CH2CH3 CH3−CHCH2CH3
Pentane Isopentane
(2-Methylbutane)
OH
CH3CH2CH2OH CH3−CH-CH3
Propan-1-ol Propan-2-ol
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(iv) Metamerism: It a rises due to different
alkyl chains on either side of the functional
group in the molecule. For example, C4H10O
represents methoxypropane (CH3OC3H7) and
ethoxyethane (C2H5OC2H5). 8.6.2 Stereoisomerism
The compounds that have the same
constitution and sequence of covalent bonds but differ in relative positions of their atoms or groups in space are called stereoisomers. This special type of isomerism is called as stereoisomerism and can be classified as geometrical and optical isomerism.
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8.7
FUNDAMENTAL CONCEPTS IN
ORGANIC REACTION MECHANISM
In an organic reaction, the or
ganic molecule
(also referred as a substrate) reacts with an appropriate attacking reagent and leads to the formation of one or more intermediate(s) and finally product(s)
The general reaction is depicted as follows : in understanding the reactivity of organic
compounds and in planning strategy for their synthesis. In the following sections, we shall learn
some of the principles that explain how these reactions take place. 8.7.1 Fission of a Covalent BondA covalent bond can get cleaved either by : (i)
heterolytic cleavage, or by (ii) homolytic cleavage. In heterolytic cleavage, the bond breaks
in such a fashion that the shar
ed pair of
electrons remains with one of the fragments. After heterolysis, one atom has a sextet
electronic structure and a positive charge and the other, a valence octet with at least one lone pair and a negative charge. Thus, heterolytic
cleavage of bromomethane will give C
+H3 and
Br– as shown below. A species having a carbon atom possessing
sextext of electr
ons and a positive charge is
called a carbocation (earlier called carbonium
ion). The
H3 ion is known as a methyl cation
or methyl carbonium ion. Carbocations are classified as primary, secondary or tertiary
depending on whether one, two or three carbons are directly attached to the positively charged carbon. Some other examples of carbocations are: CH
3C+H2 (ethyl cation, a
primary carbocation), (CH3)2C+H (isopropyl
cation, a secondary carbocation), and (CH3)3C+
(tert-butyl cation, a tertiary carbocation).
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Carbocations are highly unstable and reactive species. Alkyl groups directly attached to the positively charged carbon stabilise the carbocations due to inductive and hyperconjugation effects, which you will be studying in the sections 8.7.5 and 8.7.9. The observed order of carbocation stability is: C
+H3 < CH3C+
H2 < (CH3)2C+
H < (CH3)3C+
. These
carbocations have trigonal planar shape with positively charged carbon being sp
2
hybridised. Thus, the shape of C+H3 may be
considered as being derived from the overlap of three equivalent C(sp
2) hybridised orbitals
with 1s orbital of each of the three hydrogen Organic
molecule(Substrate)[Intermediate] Product(s)Attacking
Reagent
Byproducts
Substrate is that reactant which supplies
carbon to the new bond and the other reactant is called reagent. If both the reactants supply carbon to the new bond then choice is arbitrary and in that case the molecule on which attention is focused is called substrate. In such a reaction a covalent bond
between two carbon atoms or a carbon and some other atom is broken and a new bond is formed. A sequential account of each step,
describing details of electron movement, energetics during bond cleavage and bond formation, and the rates of transformation of reactants into products (kinetics) is referred to as reaction mechanism. The knowledge of reaction mechanism helps O H
CH3−C-CH3 CH3−CH2—C= O
Propanone Propanal
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atoms. Each bond may be represented as
C(sp2)–H(1s ) sigma bond.
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The remaining
carbon orbital is perpendicular to the molecular plane and contains no electrons. [Fig. 8.3(a)]. Fig. 8.3 (a) Shape of methyl carbocation
The heterolytic cleavage can also give a
species in which carbon gets the shared pair of electrons. For example, when group Z attached to the carbon leaves without
electron pair, the methyl anion is
formed. Such a carbon species carrying a negative charge on carbon atom is called carbanion. Carbon in carbanion is generally sp
3 hybridised and its structure is distorted
tetrahedron as shown in Fig. 8.3(b). Fig.
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8.3 (b) Shape of methyl carbanion
Carbanions are also unstable and reactive species. The organic reactions which proceed through heterolytic bond cleavage are called ionic or heteropolar or just polar reactions. In homolytic cleavage, one of the electrons
of the shared pair in a covalent bond goes with each of the bonded atoms. Thus, in homolytic cleavage, the movement of a single electron takes place instead of an electron pair. The single electron movement is shown by ‘half-headed’ (fish hook:
) curved arrow. Such cleavage results in the formation of neutral species (atom or group) which contains an unpaired electron. These species are called free radicals. Like carbocations and carbanions, free radicals are also very reactive. A homolytic cleavage can be
shown as:
Alkyl
free radical
Alk
yl radicals are classified as primary,
secondary, or tertiary. Alkyl radical stability increases as we proceed from primary to tertiary:
,
Methyl Ethyl Isopropyl Tert-butyl
free free free free
radical radical radical radical
Organic reactions, which proceed by
homolytic fission are called free radical or homopolar or nonpolar reactions.
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8.7.2
substrate and reagent
Ions are generally not formed in the r
eactions
of organic compounds. Molecules as such
participate in the reaction. It is convenient to name one reagent as substrate and other as reagent. In general, a molecule whose carbon is involved in new bond formation is called substrate and the other one is called reagent. When carbon-carbon bond is formed, the choice of naming the reactants as substrate and reagent is arbitrary and depends on molecule under observation. Example:
(i)
CH2 = CH2 + Br2 → CH2 Br – CH2Br
Substrate Reagent Product
(ii)
Nucleophiles and Electrophiles
Reagents attack the reactive site of the
substrate. The reactive site may be electron
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deficient portion of the molecule (a positive
reactive site) e.g., an atom with incomplete electron shell or the positive end of the dipole in the molecule. If the attacking species is electron rich, it attacks these sites. If attacking species is electron deficient, the reactive site for it is that part of the substrate molecule which can supply electrons, e.g., π electrons in a double bond. A reagent that brings an electron pair to
the reactive site is called a nucleophile (Nu:) i.e., nucleus seeking and the reaction is then called nucleophilic .
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A reagent that takes away
an electron pair from reactive site is called electrophile (E
+) i.e., electron seeking and the
reaction is called electrophilic. During a polar organic reaction, a
nucleophile attacks an electrophilic centre of the substrate which is that specific atom or part of the substrate which is electron deficient. Similarly, the electrophiles attack at nucleophilic centre, which is the electron rich centre of the substrate. Thus, the electrophiles receive electron pair from the substrate when the two undergo bonding interaction. A curved-arrow notation is used to show the movement of an electron pair from the nucleophile to the electrophile. Some examples of nucleophiles are the negatively charged ions with lone pair of electrons such as hydroxide (HO
– ), cyanide (NC–) ions and
carbanions (R3C:–). Neutral molecules such
as etc., can also act as
nucleophiles due to the presence of lone pair of electrons. Examples of electrophiles
include carbocations (C+H3) and neutral
molecules having functional groups like carbonyl group (>C=O) or alkyl halides
(R
3C-X, where X is a halogen atom). The
carbon atom in carbocations has sextet configuration; hence, it is electron deficient
and can receive a pair of electrons from the nucleophiles. In neutral molecules such as alkyl halides, due to the polarity of the C-X bond a partial positive charge is generated
on the carbon atom and hence the carbon atom becomes an electrophilic centre at which a nucleophile can attack.Problem 8.11
Using curved-arrow notation, show the
formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
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(a) CH
3–SCH3, (b) CH3–CN, (c) CH3–Cu
Solution
Problem 8.12Giving justification, categorise the
following molecules/ions as nucleophile or electrophile:
Solution
Nucleophiles: HS–,C2H5O–,(CH3)3N:H2N:–
These species have unshared pair of
electrons, which can be donated and shared with an electrophile. Electrophiles: BF
3,C1+ H3–C+=O,N+
O2. Reactive sites have only six valence
electrons; can accept electron pair from a nucleophile. Problem 8.13
Identify electrophilic centre in the following: CH
3CH=O, CH3CN, CH3I. Solution
Among CH3HC*=O, H3CC*≡N, and
H3C*–I, the starred carbon atoms are
electrophilic centers as they will have
partial positive charge due to polarity of the bond. 8.7.3
Electron Movement in Organic
Reactions
The movem
ent of electrons in organic
reactions can be shown by curved-arrow
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notation. It shows how changes in bonding
occur due to electronic redistribution during the reaction. To show the change in position of a pair of electrons, curved arrow starts from the point from where an electron pair is shifted and it ends at a location to which the pair of electron may move. Presentation of shifting of electron pair is
given below :
(i)
from π bond to
adjacent bond position
(ii) from π bond to
adjacent atom
(iii) from atom to adjacent
bond position
Move
ment of single electron is indicated by
a single barbed ‘fish hooks’ (i.e. half headed
curved arrow).
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For example, in transfer of hydroxide ion giving ethanol and in the dissociation of chloromethane, the movement of electron using curved arrows can be depicted as follows:
8.7.4 electron displacement effects in
covalent Bonds
The electron displacement in an organic
molecule may take place either in the ground
state under the influence of an atom or a substituent group or in the presence of an appropriate attacking reagent. The electron displacements due to the influence of
an atom or a substituent group present in the molecule cause permanent polarlisation of the bond. Inductive effect and resonance effects are examples of this type of electron displacements. Temporary electron displacement effects are seen in a molecule
when a reagent approaches to attack it. This type of electron displacement is called electromeric effect or polarisability effect. In the following sections we will learn about these types of electronic displacements.8.7.5
inductive effect
When a covalent bond is formed between
atoms of different electronegativity, the
electron density is more towards the more electronegative atom of the bond. Such a shift of electron density results in a polar covalent bond. Bond polarity leads to various electronic effects in organic compounds. Let us consider cholorethane (CH
3CH2Cl)
in which the C–Cl bond is a polar covalent
bond. It is polarised in such a way that the carbon-1 gains some positive charge (δ
+)
and the chlorine some negative charge (δ–).
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The fractional electronic charges on the two atoms in a polar covalent bond are denoted by symbol δ (delta) and the shift of electron density is shown by an arrow that points from δ
+ to δ– end of the polar bond. δδ+ δ+ δ−
CH3→CH2→Cl
2 1
In turn carbon-1, which has developed
partial positive charge (
δ+) draws some
electron density towards it from the adjacent C-C bond. Consequently, some positive charge (δδ
+) develops on carbon-2 also, where δδ+
symbolises relatively smaller positive charge as compared to that on carbon – 1. In other words, the polar C – Cl bond induces polarity in the adjacent bonds. Such polarisation of σ-bond caused by the polarisation of adjacent
σ-bond is referred to as the inductive effect. This effect is passed on to the subsequent bonds also but the effect decreases rapidly as the number of intervening bonds increases and becomes vanishingly small after three bonds. The inductive effect is related to the ability of substituent(s) to either withdraw or donate electron density to the attached carbon atom. Based on this ability, the substitutents can be classified as electron-withdrawing or
electron donating groups relative to hydrogen. Halogens and many other groups such as nitro (- NO
2), cyano (- CN), carboxy (- COOH),
ester (COOR), aryloxy (-OAr, e.g. – OC6H5), etc.
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are electron-withdrawing groups. On the other hand, the alkyl groups like methyl (–CH
3) and
ethyl (–CH2–CH3) are usually considered as
electron donating groups. Unit 8.indd 274 10/10/2022 10:37:34 AM
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Problem 8.14
Which bond is more polar in the following
pairs of molecules: (a) H3C-H, H3C-Br
(b) H3C-NH2, H3C-OH (c) H3C-OH,
H3C-SH
solution
(a) C–Br,
since Br is more electronegative
than H, (b) C–O, (c) C–O
Problem 8.15
In which C–C bond of CH3CH2CH2Br, the
inductive effect is expected to be the least? solution
Magnitude of inductive effect diminishes
as the number of intervening bonds
increases. Hence, the effect is least in the bond between carbon-3 and hydrogen. 8.7.6
resonance structure
Ther
e are many organic molecules whose
behaviour cannot be explained by a single Lewis
structure. An example is that of
benzene. Its cyclic structure containing alternating C–C single and C=C double bonds shown is inadequate for explaining its characteristic properties. As per the above representation, benzene
s
hould exhibit two different bond lengths,
due to C–C single and C=C double bonds. However, as determined experimentally benzene has a uniform C–C bond distances of 139 pm, a value intermediate between the C–C single(154 pm) and C=C double (134 pm) bonds.
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Thus, the structure of benzene cannot be represented adequately by the above structure. Further, benzene can be represented equally well by the energetically identical structures I and II. benzene cannot be adequately represented
by any of these structures, rather it is a hybrid of the two structures (I and II) called resonance structures. The resonance
structures (canonical structures or contributing structures) are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability. Another example of resonance is provided
by nitromethane (CH3NO2) which can be
represented by two Lewis structures, (I and II). There are two types of N-O bonds in these structures. howev
er, it is known that the two n–O
bonds of nitr
omethane are of the same
length (intermediate between a n–O single
bond and
a n=O double bond). the actual
st
ructure of nitromethane is therefore
a resonance hybrid of the two canonical
forms I and II. The
energy of actual structure of the molecule
(the resonance hybrid) is lower than that of
any of the canonical structures.
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The difference
in energy between the actual structure and the lowest energy resonance structure is called the resonance stabilisation energy or simply the resonance energy . The more the number
of important contributing structures, the more is the resonance energy. Resonance is particularly important when the contributing structures are equivalent in energy. The following rules are applied while writing
resonance structur
es:
The
resonance structures have (i) the same
positions of nuclei and (ii) the same number of unpaired electrons. Among the resonance structures, the one which has more number of covalent bonds, all the atoms with octet of electrons (except hydrogen which has a duplet), less separation of opposite charges, (a negative charge if any on more electronegative atom, a positive charge if any on more electropositive atom) and more dispersal of charge, is more stable than others. Benzene
Therefore, according to the resonance
theory (Unit 4) the actual structure of
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Problem 8.16
Write resonance structures of CH3COO–
and show the movement of electrons by
curved arrows. solution
Fir
st, write the structure and put
unshared pairs of valence electrons on appropriate atoms. Then draw the arrows one at a time moving the electrons to get the other structures. Problem 8.17
Write resonance structures of
CH2=CH–CHO. Indicate relative stability
of the contributing structures.
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solutionsolution
The two structures are less important
contributors as they involve charge
separation. Additionally, structure I contains a carbon atom with an
incomplete octe
t.
8.7.7 resonance effect
The resonance effect is defined as ‘the polarity
pr
oduced in the molecule by the interaction
of two π -bonds or between a π -bond and
lone pair of electrons present on an adjacent
atom’. The effect is transmitted through the chain. There are two types of resonance or mesomeric effect designated as R or M effect. (i) Positive Resonance Effect (+R effect) In this effect, the transfer of electrons is away
from an atom or substituent group attached to the conjugated system. This electron displacement makes certain positions in the molecule of high electron densities. This effect in aniline is shown as :
Stability: I > II > III
[I: Most stable, more number of covalent bonds, each carbon and oxygen atom has an octet and no separation of opposite charge II: negative charge on more electronegative atom and positive charge on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable]. Problem 8.18
Explain why the following two structures, I and II cannot be the major contributors to the real structure of CH
3COOCH3. (ii) Negative Resonance Effect (- R effect)
This effect is observed when the transfer of
electrons is towards the atom or substituent group attached to the conjugated system. For example in nitrobenzene this electron displacement can be depicted as :
The atoms or substituent groups, which
represent +R or –R e
lectron displacement
effects are as follows :
+R effect: – halogen, –OH, –OR, –OCOR, –NH2,
–NHR, –NR2, –NHCOR,
– R effect: – COOH, –CHO, >C=O, – CN, –NO2
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The presence of alternate single and
double bonds in an open chain or cyclic system
is termed as a conjugated system.
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These systems often show abnormal behaviour. The examples are 1,3- butadiene, aniline and nitrobenzene etc. In such systems, the π-electrons are delocalised and the system
develops polarity. 8.7.8
electromeric effect ( e effect)
It is a temporary effect. The or
ganic compounds
having a multiple bond (a double or triple
bond) show this effect in the presence of an attacking reagent only. It is defined as the complete transfer of a shared pair of
π-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. The effect is annulled as soon as the attacking reagent is removed from the domain of the reaction. It is represented by E and the shifting of the electrons is shown by a curved arrow (
). There are two distinct types of
electromeric effect. (i) Positive Eelctromeric Effect (+E effect) In this effect the π−electrons of the multiple bond are transferred to that atom to which the reagent gets attached.
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For example: system or to an atom with an unshared
p orbital. The σ electrons of C—H bond of the
alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p orbital. Hyperconjugation is a
permanent effect. To understand hyperconjugation effect, let
us take an example of CH
3 C+H2 (ethyl cation) in
which the positively charged carbon atom has an empty p orbital. One of the C-H bonds of the methyl group can align in the plane of this empty p orbital and the electrons constituting
the C-H bond in plane with this p orbital can then be delocalised into the empty p orbital as depicted in Fig. 8.4 (a). (ii) Negative Electromeric Effect (–E effect) In this effect the π - electrons of the multiple bond are transferred to that atom to which the attacking reagent does not get attached. For example:
When inductive and electromeric effects
operate in opposite directions, the electomeric effect predominates. 8.7.9
hyperconjugation
Hy
perconjugation is a general stabilising
interaction. It involves delocalisation of
σ electrons of C—H bond of an alkyl group
directly attached to an atom of unsaturated Fig.
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8.4(a) Orbital diagram showing
hyperconjugation in ethyl cation
This type of overlap stabilises the
carbocation because electron density from the adjacent σ bond helps in dispersing the positive charge. In general, greater the number of alkyl
groups attached to a positively charged carbon
atom, the greater is the hyperconjugation interaction and stabilisation of the cation. Thus, we have the following relative stability of carbocations :
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Hyperconjugation is also possible in
alkenes and alkylarenes. Delocalisation of electrons by
hyperconjugation in the case of alkene can
be depicted as in Fig. 8.4(b).Problem 8.19 Explain why (CH
3)3C+ is more stable
than CH3C+H2 and C+H3 is the least stable
cation. solution
Hyperconjugation interaction in (CH3)3C+ is
greater than in CH3C+H2 as the (CH3)3C+
has nine C-H bonds. In C+H3, vacant p
orbital is perpendicular to the plane
in which C-H bonds lie; hence cannot
overlap with it. Thus, C+H3 lacks
hyperconjugative stability. 8.7.10 types of Or
ganic reactions and
Mechanisms
Organic r
eactions can be classified into the
following categories:
(i) Substitution reactions
(ii) Addition reactions
(iii) Elimination reactions
(iv) Rearrangement reactions
You will be studying these reactions in
Unit 9 and later in class XII. 8.8 methOds OF Puri Ficati On OF
Organic cOmPOunds
Once an organic compound is extracted
fr
om a natural source or synthesised in the
laboratory, it is essential to purify it.
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Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it. The common techniques used for
purification are as follows :
(i)
Sublimation
(ii) Crystallisation
(iii) Distillation
(iv) Differential extraction and
(v) Chromatography
Finally, the purity of a compound is
ascertained by deter
mining its melting or
boiling point. Most of the pure compounds
have sharp melting points and boiling points. New methods of checking the purity of an organic compound are based on different Fig. 8.4(b) Orbital diagram showing
hyperconjugation in propene
There are various ways of looking at the
hyperconjugative effect. One of the way is to regard C—H bond as possessing partial ionic character due to resonance. The hyperconjugation may also be
regarded as no bond resonance. Unit 8.indd 278 10/10/2022 10:37:35 AM
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types of chromatographic and spectroscopic
techniques. 8.8.1 sublimation
You have learnt earlier that on heating, some
solid substances change fr
om solid to vapour
state without passing through liquid state. The purification technique based on the above principle is known as sublimation and is used to separate sublimable compounds from non-sublimable impurities.
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8.8.2
crystallisation
This is one of the most commonly used
techniques for the purification of solid organic
compounds. It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. On cooling the solution, pure compound crystallises out and is removed by filtration. The filtrate (mother liquor) contains impurities and small quantity of the compound. If the compound is highly soluble in one solvent and very little soluble in another solvent, crystallisation can be satisfactorily carried out in a mixture of these solvents. Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. Repeated crystallisation becomes necessary for the purification of compounds containing impurities of comparable solubilities. 8.8.3
distillation
This important method is used to separate
(i) volatile liquids from nonvolatile impurities and (ii) the liquids having sufficient dif
ference
in their boiling points.
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Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Chloroform (b.p 334 K) and aniline (b.p. 457 K) are easily separated by the technique of distillation (Fig 8.5). The liquid mixture is taken in a round bottom flask and heated carefully. On boiling, the vapours of lower
boiling component are formed first. The vapours are condensed by using a condenser and the liquid is collected in a receiver. The vapours of higher boiling component form later and the liquid can be collected separately. Fractional Distillation: If the difference
in boiling points of two liquids is not much,
simple distillation cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously.
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The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask (Fig.8.6, page 280). Vapours of the liquid with higher boiling
point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become richer in more volatile component. By the
Fig.8.5 Simple distillation. The vapours of a
substance formed are condensed and
the liquid is collected in conical flask. Unit 8.indd 279 10/10/2022 10:37:35 AM
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time the vapours reach to the top of the
fractionating column, these are rich in the more volatile component. Fractionating columns are available in various sizes and designs as shown in Fig.8.7. A fractionating column provides many surfaces for heat exchange between the ascending vapours and the descending condensed liquid.
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Some of the condensing liquid in the fractionating column obtains heat from the ascending vapours and revaporises. The vapours thus become richer in low boiling component. The vapours of low boiling component ascend to the top of the column. On reaching the top, the vapours become pure in low boiling component and pass through the condenser and the pure liquid is collected in a receiver. After a series of successive distillations, the remaining liquid in the distillation flask gets enriched in high boiling component. Each successive condensation and vaporisation
Fig.8.7 Different types of fractionating columns.unit in the fractionating column is called a
theoretical plate. Commercially, columns
with hundreds of plates are available. One of the technological applications of
fractional distillation is to separate different fractions of crude oil in petroleum industry. Fig.8.6 Fractional distillation. The vapours of lower boiling fraction reach the top of the column first followed by vapours of higher boiling fractions.Distillation under reduced pressure: This method is
used to purify liquids having very high boiling points and those, which decompose at or below their boiling points.
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Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water pump or vacuum pump (Fig.8.8). Glycerol can be separated from spent-lye in soap industry by using this technique. Unit 8.indd 280 10/10/2022 10:37:35 AM
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Fig.8.8 Distillation under reduced pressure. A liquid boils at a temperatur e below its
vapour pressure by reducing the pressure. Steam Distillation: This technique is
applied to separate substances which are
steam volatile and are immiscible with water. In steam distillation, steam from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the volatile organic compound is condensed and collected. The compound is later separated from water using a separating funnel.
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In steam distillation, the liquid boils when the sum of vapour pressures due to the organic liquid (p
1)
and that due to water (p2) becomes equal to
the atmospheric pressure (p), i.e. p =p1+ p2. Since p1 is lower than p, the organic liquid
vaporises at lower temperature than its boiling point. Thus, if one of the substances in the
mixture is water and the other, a water
insoluble substance, then the mixture will boil close to but below, 373K. A mixture of water and the substance is obtained which can be separated by using a separating funnel. Aniline is separated by this technique from aniline – water mixture (Fig.8.9, Page 282).8.8.4
differential extraction
When an organic compound is pr
esent in an
aqueous medium, it is separated by shaking
it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by separatory funnel. The organic solvent is later removed by distillation or by evaporation to get back
the compound. Differential extraction is carried out in a separatory funnel as shown in
Fig. 8.10 (Page 282).
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If the organic compound is less soluble in the organic solvent, a very large quantity of solvent would be required to extract even a very small quantity of the compound. The technique of continuous extraction is employed in such cases. In this technique same solvent is repeatedly used for extraction of the compound. 8.8.5
chromatography
Chromatography is an important technique
extensively used to separate mixtur
es into
their components, purify compounds and also to test the purity of compounds. The
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name chromatography is based on the Greek
word chroma, for colour since the method
was first used for the separation of coloured substances found in plants. In this technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the
mixture get gradually separated from one another. The moving phase is called the mobile phase. Based on the principle involved,
chromatography is classified into different categories.
|
Two of these are:
(a)
Adsorption chromatography, and
(b) Partition chromatography. a)
Adsorption Chromatography : Adsor-
ption chromatography is based on the fact
that different compounds are adsorbed on an adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina. When a mobile phase is allowed to move
over a stationary phase (adsorbent), the components of the mixture move by varying distances over the stationary phase. Following are two main types of chromatographic techniques based on the principle of differential adsorption. (a)
Column chromatography, and
(b) Thin layer chromatography.Fig.8.10 Differential extraction. Extraction of com-
pound takes place based on difference
in solubility Fig.8.9 Steam distillation. Steam volatile component volatilizes, the vapours condense
in the condenser and the liquid collects in conical flask. Unit 8.indd 282 10/10/2022 10:37:36 AM
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organic chemistry – some basic principles and techniques
Column Chromatography: Column
chromatography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (Fig.
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8.11). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube. An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column (Fig.8.11).The glass plate is then placed in a closed jar containing the eluant (Fig. 8.12a). As the eluant rises up the plate, the components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and separation takes place. The relative adsorption of each component of the mixture is expressed in terms of its retardation factor i.e. R
f value (Fig.8.12 b). Rf = Distance moved by the substance from base line (x)
Distance moved by the solvent from base line (y)
Fig.8.11 Column chromatography.
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Differ ent
stages of separation of components of a mixture. Thin Layer Chromatography: Thin layer
chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin layer (about 0.2mm thick) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate or chromaplate. The solution of the mixture to be separated is applied as a small spot about 2 cm above one end of the TLC plate. The spots of coloured compounds are
visible on TLC plate due to their original colour. The spots of colourless compounds, which are invisible to the eye but fluoresce in ultraviolet light, can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine. Spots of compounds, which adsorb iodine, will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate.
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For example, amino acids may be detected by spraying the plate with ninhydrin solution (Fig.8.12b). Fig.8.12 (b) Developed chromatogram.Fig.8.12 (a) Thin layer chromatography. Chromatogram being developed. Unit 8.indd 283 11/10/2022 15:22:47
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chemistry
spot on the chromatogram. The spots of the
separated colourless compounds may be
observed either under ultraviolet light or by the use of an appropriate spray reagent as discussed under thin layer chromatography. 8.9
qualitati Ve analysis OF Organic
cOmPOunds
The elements present in organic compounds
ar
e carbon and hydrogen. In addition to
these, they may also contain oxygen, nitrogen, sulphur, halogens and phosphorus. 8.9.1
detection of carbon and hydrogen
Carbon and hydrogen ar
e detected by heating
the compound with copper(II) oxide. Carbon
present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue). C + 2CuO
2Cu + CO2
2H + CuO Cu + H2O
CO2 + Ca(OH)2 CaCO3↓ + H2O
5H2O + CuSO4 CuSO4.5H2O
White Blue
8.9.2 detection of Other elements
Nitrogen, sulphur, halogens and phosphorus
present in an organic compound are detected
by “lassaigne’s test”.
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The elements pr
esent
in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:
Na + C + N NaCN
2Na + S Na2S
Na + X Na X
(X = Cl, Br or I)
C, N,
S and X come from organic compound. Cyanide, sulphide and halide of sodium so
formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract. (A) Test for Nitrogen The sodium fusion extract is boiled with
iron(II) sulphate and then acidified with Partition Chromatography: Partition
chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted
at the base with the solution of the mixture is
suspended in a suitable solvent or a mixture of solvents (Fig. 8.13).
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This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial
Fig.8.13 Paper chromatography. Chromatography paper in two different
shapes. Unit 8.indd 284 10/10/2022 10:37:37 AM
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organic chemistry – some basic principles and techniques
concentrated sulphuric acid. The formation
of Prussian blue colour confirms the presence
of nitrogen. Sodium cyanide first reacts
with iron(II) sulphate and forms sodium hexacyanidoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanidoferrate(II) to produce iron(III) hexacyanidoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.
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