(i)...and unsaturation, hybridization, classification of organic compounds and baeyer’s strain...

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All rights reserved.No part of this publication may be reproduced, stored in a database or retrieval system, ortransmitted in any form or by any means, electronic, mechanical, photocopying, recording,or otherwise, without the prior written Permission of the writter.

First Indian print, 2016

This edition is manufactured in Indian and is authorized for sale only in India, Bangladesh,Pakistan, Nepal, Sri Lanka & Maldeves.

Printed & Distributed by:Udaan Classes Pvt. Ltd.Rainbow building, Patna &Madhur SatyapushpaShubhash Nagar, Kota9122057123

Price: Rs. 499/-

Copyright © 2016 by Ajnish Gupta

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PrefaceThe guiding principle in writing this book was to create a textbook for students- a textbookthat presents the material in a way that they learn to solve all the questions along withthe strategy to approach the problems.In this book we mixed all our teaching experience of 12 years along with theoratical andexperimental knowledge to generate a hand book for all students to reason their way to asolution rather than memorize a multitude of facts, hoping they don’t run out of memory.This book covers mainly 4 units with 61 sections which are real tools of Organicchemistry, which a students must know before dealing any chemical reactions.Organic chemistry is very easy and conceptual subject and need proper understanding ofthe basics and stretegy to solve the questions in corret manner.This book will prepare your right mindset for learning Organic Chemistry. This mindset isessentially the one that focuses you on a small number of straight forward, repeated, fun-damental concepts and helps you to apply them in different ways to solve the variety ofproblems you face in organic chemistry.This book is complete as it not only covers theory in proper sequence but also providevarieties of questions along with 10 test papers to judge your knowledge before going tostart chemical reactions.This book also covers all the questions of AIEEE, IIT-Mains, IIT-Advanced, AIPMT & othermedical exams from 2000 to 2016.In this book balance has to be achieved between the number of questions and the quality ofthe questions, especially because it is relatively easy to frame a very large number ofmultiple-choice questions and theory of the subject.The questions in this book have been selected keeping three things in mind.First- the questions are such that they really test the understanding of the subject.Second- the questions cover all concepts.Third- the number of questions has been kept large enough to offer meaningful practice tothe students.We would like to thank the editors and production staff for their efforts in bringing out thebook. Suggestions from readers for the improvement of the book are welcome.

Ajnish Kumar Gupta & Bharti [email protected]

9122057123

mailto:[email protected]

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AcknowledgementWe are thankful to all the teachers who taught us during the concept building session ofour life specially Dr. Nizamuddin sir, senior Chandra Vijay Rao and Dr. Vijay PratimaMittal madam.We have written this book to remove the fever of organic chemistry from mind of students.We particularly want to thank many wonderful and talented students whom we havetaught over the years who in turn taught us how to be a good teacher and how we canhelp others.We want to make this book as user friendly as possible, and we will appreciate anycomments that will help us to achieve this goal in future editions.Finally, this edition has been presented before you with efforts to make it error free. Anythat remain are our responsibility; if you find any, please let us know so they can becorrected in future printing.

Ajnish Gupta & Bharti [email protected]

[email protected]



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Table of ContentsPage No.

Unit 1 : Basic understanding of organic chemistry 01Section 01 : Introduction of organic chemistry 02Section 02 : What is organic chemistry 04Section 03 : Representation of organic compounds 06Section 04 : Reason for the formation of large number of organic compounds 10Section 05 : Functional groups 12Section 06 : Homologue & Homologous series 16Section 07 : Nature of C, H & functional groups 19Section 08 : Saturated & Unsaturated molecules with important points 22Section 09 : Hybridization 25Section 10 : Classification of organic compounds 28Section 11 : Baeyer’s strain theory 31Exercise 1 : Subjective approach 33Answer : Subjective approach 39Exercise 2 : Objective approach 48Exercise 3 : World of competition 56Answer : Objective approach 59Answer : World of competition 60Approach : Objective approach (Exercise-2) 61Approach : World of competition (Exercise-3) 64

Unit 2 : Nomenclature of organic compounds 66Section 01 : Alkyl groups 67Section 02 : Basic concept of IUPAC nomenclature 69Section 03 : Points to write IUPAC nomenclature 72Section 04 : Points to write IUPAC nomenclature for multiple bonds & rings 76Section 05 : Points to write IUPAC nomenclature of compounds with functional groups 80Section 06 : Nomenclature of Bicyclic & Spiro compounds 85Section 07 : Some common names commonly used in organic chemistry 87Section 08 : Degree of unsaturation 103Section 09 : Chemically different hydrogen 106Exercise 1 : Subjective approach 109Answer : Subjective approach 123Exercise 2 : Objective approach 134Exercise 3 : World of competition 144

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Answer : Objective approach 149Answer : World of competition 150Approach : Objective approach (Exercise-2) 151Approach : World of competition (Exercise-3) 154Test 1 : Basic understanding of organic chemistry & nomenclature 156Test 2 : Basic understanding of organic chemistry & nomenclature 161Test 3 : Basic understanding of organic chemistry & nomenclature 166Answer : Test- 1, 2, 3 172Approach : Test paper 1 173Approach : Test paper 2 175Approach : Test paper 3 177

Unit 3 : Study of isomerism 179Section 01 : Basic concept of isomerism 180Section 02 : Chain isomerism 182Section 03 : Positional isomerism 184Section 04 : Functional isomerism 186Section 05 : Metamerism 188Section 06 : Basic concept of tautomerism 193Section 07 : % enol content 197Section 08 : Types of tautomerism 200Section 09 : Basic concept of geometrical isomerism 201Section 10 : Nomenclature used in geometrical isomerism 204Section 11 : Basic concept of optical isomerism 207Section 12 : Nomenclature used in optical isomerism 211Section 13 : Terms used in optical isomerism 215Section 14 : Compound showing optical isomerism along with calculation of number

of stereoisomers 218Section 15 : Conformational isomerism 221Section 16 : Conformation of cyclohexane 226Section 17 : Interconversion of one form of molecule into another 230Section 18 : Finding relationship between pair of compounds 232Section 19 : Physical properties of organic compounds 236Exercise 1 : Subjective approach 240Answer : Subjective approach 249Exercise 2 : Objective approach 259Exercise 3 : World of competition 277Answer : Objective approach 285Answer : World of competition 286Approach : Objective approach (Exercise-2) 288

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Approach : World of competition (Exercise-3) 293Test 4 : Concept of isomerism 297Test 5 : Concept of isomerism 303Test 6 : Concept of isomerism 308Answer : Test- 4, 5, 6 314Approach : Test paper 4 315Approach : Test paper 5 317Approach : Test paper 6 319

Unit 4 : Electronic effect & it’s application 321Section 01 : General concept of organic reactions 322Section 02 : Inductive effect 326Section 03 : Basic concept of resonance 328Section 04 : Points to draw resonating structures 331Section 05 : points to check stability of resonating structures 334Section 06 : Mesomeric effect 337Section 07 : Hyperconjugation 339Section 08 : Electromeric effect 341Section 09 : Organic reaction intermediates 343Section 10 : Carbocations 345Section 11 : Stability of carbocation 347Section 12 : Rearrangement of carbocations 349Section 13 : Classification of carbocation 354Section 14 : Carbanions 356Section 15 : Free radical 359Section 16 : Carbenes 363Section 17 : Nitrene 365Section 18 : Benzyne 367Section 19 : Ajnish rule & It’s application in organic chemistry 369Section 20 : General concept of Acid & Base 375Section 21 : Concept to decide relative acidic strength 378Section 22 : Concept to decide relative basic strength 381Exercise 1 : Subjective approach 384Answer : Subjective approach 401Exercise 2 : Objective approach 420Exercise 3 : World of competition 459Answer : Objective approach 469Answer : World of competition 471Approach : Objective approach (Exercise-2) 472Approach : World of competition (Exercise-3) 482

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Test 7 : Electronic effect & it’s application 485Test 8 : Electronic effect & it’s application 490Test 9 : Electronic effect & it’s application 496Answer : Test- 7, 8, 9 502Test 10 : Tools of organic chemistry 503Test 10 : Answer 508Approach : Test paper 7 509Approach : Test paper 8 511Approach : Test paper 9 513Approach : Test paper 10 515

Basic understanding of organic chemistry

Learning is easy at www.CoachMe.co.in 1

This unit is designed to lay the foundation for your organic chemistry. Here you will learn what isorganic chemistry, how organic compounds are represented, what is the reason for formation of largenumber of organic compounds, functional groups, homologues, nature of carbon & hydrogen, saturationand unsaturation, hybridization, classification of organic compounds and Baeyer’s strain theory.

1Unit

Basic understanding of Organic Chemistry

We wrote this book for one simple reason : We love writing. We get greatpleasure and satisfaction from taking a complicated subject, turning it arounduntil we see it clearly from a new angle, and then explaining it in simplewords. We write to explain chemistry to students today the way we wish ithad been explained to us years ago.

http://www.CoachMe.co.in


Learning is easy at www.CoachMe.co.in2

Have you ever thought that when you read anything from “any book” your eyes use an organiccompound (retinal) to convert visible light into nerve impulses. When you pick up this book from anyplace, your muscles do chemical reactions on sugars to give you energy. As you read the words andsentences of this book, gaps between your brain cells are being bridged by simple organic molecules(neurotransmitter amines) so that nerve impulse can be passed around your brain and you can understandall matters correctly and you did all that without consciously thinking about it.

You do not yet understand these processes in your mind but you are regularly carrying them outin your brain & body.

3CH3H C

O3CH

11-Cis retinal3H C

H

3CH

(Absorbs light when we see)

HO

NH

Serotonin2NH

(Human neurotransmitter)

You are not alone there.........No organic chemist, is so brilliant who can understand the detailed chemical working of human

mind or body very well.So don’t be sad and give a smile to yourself so that your journey should be cheerful....You do at least one thing after reading this page “Think the importance of organic chemistry and try

to search at least one thing which makes your life easy but not made of CARBON”.??

Are you thinking of this book, which you are reading presently - Made up of cellulose andindeed carbon

Notice board, Black or white board of schools - Made up of wood and indeed carbon. Cloths which you are wearing - Polyester, tericot, silk etc. - Polymer of organic compounds Perfumes, which you are using for good smell - Again organic molecule such as cis-jasmone

from jasmine flower and damascenone in smell of rose.

1Section

Introduction of Organic Chemistry

http://www.CoachMe.co.in2



Medicines at the time of your sickness - Made up of organic compounds such as paracetamoland aspirin

Sugars to make your life sweet - Made up of glucose, sucrose and indeed carbon. Are you thinking for opposite sex attraction- Pheromones & hormones are responsible for it, which

are again made up of carbon. Are you now thinking for drugs - Then they are again organic and made up of carbon such as

cocaine, nicotine, etc. Polythene which are used to carry things - made up of ethene which indeed is made up of

carbon Blue jeans, which you are wearing - made up of indigo dyes, made up of carbon. Petrol or diesel- Again made up of carbon Candles for your birthdays and parties for light - Again made up of carbon Are you thinking for bigger molecules such as Vitamins, Proteins, DNA, RNA – they are again

organic and made up of carbon. Trees, plants, flowers - They too are made up of carbon You, I or any human being along with all animals & plants- has some sort of carbon, in one

form or another.

3H C

O

OCocaine

MeOOCN

3H C 3CH

3CHDamascenone-The smell of rose

O

3CH

HOO

HO OH

OHO

Vitamin C

3CH

O 3CH

Cis Jasmone of jasmine

OHHO O

HO OHOH

Glucose

3CHOCOOH

O

Aspirin

3CH 3CH

OH

Vitamin A

3H C 3CH

3CH

N

2NH

HNN

NAdenine

N

2NHHNNH

N

O

Guanine

O

HN

NH

O

Indigo dye for blue colour O

NN

2NH

HCytocine

ONH

NH

Uracil

O

ONH

NH

Thymine

O3H C

Anything more which you want to suggest....Examples may varies from yours to our’s but from these example we want to give a clear cut picture

of importance of organic chemistry and its applications in daily life i.e. everything around us are organicin nature so

Love organic, feel organic and give a smile for organic....




What is Organic chemistry?Why do so many people study it?And why should I study it?The answer to these questions is all around you. Every living organism is made up of organic

compounds. The proteins that make your hair, skin and muscles; the DNA that controls your geneticheritage; the food that nourish you; the clothes that keep you warm and the medicines that heal you, areall organic chemicals. Anyone with a curiosity about life and living things must have a basic understandingof organic chemistry.

The foundation of organic chemistry dates from the mid 1700s, when chemistry was evolving fromalchemist’s art into a modern science. At that time unexplainable differences were noted betweensubstances obtained from living sources and those obtained from mineral sources. Compounds obtainedfrom plants and animals were often difficult to isolate, purify and even difficult to work. The Swedishchemist Torbern Bergman in 1770 was the first to differentiate “organic” and “inorganic” substancesand the term organic chemistry soon came to mean the chemistry of compounds found in living organism.

To many chemists at that time the only explanation for the differences in behavior of organic andinorganic compound was that the organic compounds must contain a peculiar “vital force” as a result oftheir origin in living substance. Due to this vital force theory chemist believed that organic compoundscould not be prepared and manipulated in the laboratory, as could inorganic compounds.

In 1816, however this vital force theory received a heavy blow when Michel Chevreul found thatsoap, prepared by the reaction of alkali with animal fatcould be separated into several pure organiccompounds,which he termed “fatty acid”. For the first time, one organic substance (fat) was convertedinto other organic substances (fatty acids and glycerin).

Animal fat Soap + Glycerin2H O

NaOH

soap Fatty acid3H O+

In 1828, vital force theory again got a heavy blow when Friedrich Wöhler prepared organic compoundurea, of human urine from inorganic compound ammonium cyanate.

–4 2 2

HeatNH OCN NH CONHAmmonium Cyanate Urea

2Section

What is Organic Chemistry




By the mid of 1800s, the weight of evidence was clearly against the vital force theory which ledWilliam Brande in 1848 to write “No definite line can be drawn between organic and inorganicchemistry......... Any distinction........ Must for the present be merely considered as matters of practicalconvenience calculated to further the progress of students”.

Chemistry today is unified. The same principles that explain the simplest inorganic compoundsalso explain the most complex organic ones. The only distinguishing characteristic of organic chemistryis that all contains the element carbon. Nevertheless, the division between organic and inorganic chemistry,which began for historical reasons, maintains its practical convenience... to further the progress of students.So, in general

“Organic chemistry is a branch of chemistry which deals with carbon and its compounds”

C6

12

Atomic number

Atomic massSymbol

–number of electrons–number of protons

From the periodic table we can easily find that it comprises of seven periods and eighteen groups.First and second groups elements are called as s block elements, thirteen to eighteen group elements arecalled p block elements, third to twelve group elements are d block while La to Lr which are downsideare called as f block elements.

When we carefully analyze the periodic table, it is clear that position of carbon is in second periodand fourteenth group and carbon is a non-metal.

Carbon is the basic element for life but it forms its compounds by combination with other elementssuch as H, N, O, F, Cl, Br, I, S, P etc.

In Organic chemistry we mainly deal with Structural determination – How to find out the structures of new compounds, even if they are

available only in invisibly small amount. Theoretical organic chemistry – How to understand those structures in terms of atoms and the

electrons that binds them together. Reaction mechanism – How to find out how those molecules reacts with each other and to

predict their reactions. Biological chemistry – How to find out what nature does and how the structures of biologically

active molecules are related to what they do.




Organic chemistry is the study of compounds that contain carbon but there are some other elementstoo associated with it such as H, O, and N. We know that carbon is tetravalent in nature and formcovalent compounds by sharing of electrons by generally H, F, Cl, Br, I, N, O, S and P. For sake ofsimplicity all halogens are generally represented by X. In the study of molecules, you will notice thatoxygen has two covalent bonds, nitrogen has three covalent bonds and carbon has four covalent bondswhen they are neutral. Neutral hydrogen & halogens each have one covalent bond. If atoms have morebonds or fewer bonds than the number required for neutral molecule then they would have either aformal charge or an unpaired electron. These numbers are very much important to remember when youare first drawing structures of organic compounds because they provide a quick way to recognize whenyou have made a mistake.

C N O H F Cl Br I X

This covalent bond may be of sigma ( )or pi ( ). Sigma bond is formed by lateral overlapping oforbitals. It may be of s-s, s-p or p-p but pi bonds are formed by sidewise overlapping of p-p orbitals interms of organic chemistry.

+s s s-s overlap

s p s-p overlap

y x+

y x

p orbital p orbitalzz

p-p overlap alongthe orbital axis

Py

y

Py

y

+

p-p sideways overlap

y y

3Section

Representation of Organic Compounds




As sigma bonds are strong as compared to pi bonds so generally pi bonds are involved in chemicalreaction with reagent, if present in molecule. When we compare the bond strength of bonds formed bys-p & p-p overlapping; p-p bond is stronger so generally s-p bond break on treatment with suitablereagent, such as in case of alkanes.

Organic chemistry concerns itself with the way in which these atoms are bonded together into stablemolecular structures and the way in which these structures change during chemical reactions.

For this organic molecules are represented by several ways1. Lewis structure or dot structure –Here valence electrons in a molecule are represented as dots.

When you draw Lewis structure you must make sure that hydrogen atom is surrounded by nomore than two electrons, and C, O, N and halogen (F, Cl, Br, I) atoms are surrounded by no morethan eight electrons. In other words, they must obey the octet rule. The valence electrons notused in bonding are called as non-bonding electrons or lone pair of electrons. Lewis structuresare useful because they show us which atoms are bonded together and tell us whether any atompossess lone pair of electrons or have a formal charge.

H C HH

H

.. .. H C OH

H

.. .. H.2. Dash structure or Kekule structure–In this representation, dash represents the bonds. Single

dash represents a single bond, double dash is used for double bond of any type such as of C=C,C=N, C=S etc. and triple dash represent triple bond of any type such as C C or C N, etc. In thisrepresentation lone pair of electrons of hetero atoms (O, N, S, and P etc.) may or may not beshown. This representation gives complete structural formula of any molecule.

H–C–C–HHH

HH C=C

H

H H

H H–C C–H

3. Condensed structure –In condensed structural representation, dash structures can furtherabbreviated by omitting some or all of the dashes representing covalent bonds and by indicatingthe number of identical groups attached to an atom by a subscript. The resulting expression ofthe compound is now called a condensed structural formula. For example, CH3CH3, CH2=CH2,CH CH, CH3OH. Similarly, CH3CH2CH2CH2CH2CH2CH2CH3 can be further condensed toCH3(CH2)6CH3.

4. Bond line structure – In this representation of organic molecules, carbon and hydrogen atomsbonded to carbon are not shown and the lines representing carbon-carbon bonds are drawn inzigzag fashion.

Even without writing the hydrogen atoms we know that they are there and we assume that anycarbon atom that does not appear to have its potential for four bonds satisfied is also attached to theappropriate number of hydrogen atoms.

We also rub out all the C representing carbon atoms and write only zig-zag line assuming that everykink in the line represents carbon atom.

Pentane Cyclohexane

Benzene

As the functional groups are the key to the organic chemistry so they are specifically shown if presentso the only atoms specifically written are heteroatoms such as oxygen, nitrogen, halogens, which makesthe functional groups, which we will discuss later.




O

OH

This H is shown becauseit is attached to an atom

other than C

all 4 bonds are shown to this c atom so no Hatoms are implied

these C atoms must also carry1 H atom because only 3 bonds

are shown for each atom

end C have 3 H atoms

these C atoms must also carry2 H atoms because only 2 bonds

are shown for each atom

every kink in thechain represents

a C atom

Organic molecules should be drawn to be realistic, economical and clear so here are 3 guidelines,which help you make structure more realistic. Guideline 1- Draw the chain of atoms as zigzag. Guideline 2 – Miss out the hydrogen attach to carbon atoms, along with the C-H bonds. Guideline 3 – Miss out the capital C representing carbon atoms.5. 3D structure – Of course, all the structures we have drawn give only an idea of the real structure

of the molecules. For example, the central carbon atom of CH3CH(NH2)COOH has tetrahedralarrangement around it but so far we have completely ignored it. Now we want to emphasizethis fact, so for this 3D structure of molecules are written using wedge-dashed projection andFischer projection.

COOH

3CH

H

2H N

A. Wedge-dash projection or perspective formula:-In this wedge-dash projection, the solid wedge ( ) is used to indicate a bond projectingabove the plane of paper, towards the observer and dashed wedge ( ) is used to depict thebond projecting below the plane of the paper and away from the observer. The bonds lying inthe plane of the paper are depicted by using a normal line ( ).

CH

H

H

H

bonds are awayfrom the observer(Dashed wedge)

bonds are towards the observer(Solid wedge)

bonds are inthe plane of paper

(Normal line)

Sometime we could miss out the hydrogen atom and draw something bit neater though slightly lessrealistic. Here we assume the missing hydrogen atom is below or above the plane by looking the 3attachments. If two bonds are in the plane and one is above the plane then hydrogen must be certainlybelow the plane and similarly if two bonds are in the plane and one is below the plane then hydrogenmust be above the plane.




COOH

H is assumed to be below the plane3CH

2NH2NHHCOOH

3CHor

2NHHCOOH

3CHH is assumed to be above the plane

2NH

COOH3CH

or

B. Fischer projection:–In this projection molecule is represented by horizontal and vertical lines. Groups on horizontalline are above the plane (towards the observer) and groups on vertical line are behind the plane(away from observer) and point of intersection of horizontal and vertical line is carbon.

C

H

Perspective formula of methane

H H

H 109.5°




What makes the carbon so special?What is it that sets carbon apart from all other elements in the periodic table?Why are there so many organic compounds?The answer lies in carbon’s position in the periodic table. Carbon is in the centre of second row

elements

Li

Elements of the second row of the periodic table

Be B C N O F

First think why molecules are formed from atoms? It is because of the reason that atoms combineswith same or with other atoms to form molecule so as to complete its octet and attain lower energy stateand hence become stable. That is the reason why noble gases are considered as inert gases, they generallydo not combine with itself or with other atoms because they have complete octet. But what about otheratoms? They have incomplete octet, so they must combines with same or other atoms to form moleculefor better stability.

Elements on the left hand side of carbon have less than 4 electrons in the valence shell (Li-1, Be-2, B-3) so they have more tendencies to loose electron to attain noble gas configuration for stability. That’swhy they generally forms compounds with Li+, Be2+, B3+ by losing 1, 2, 3 electrons respectively. Elementspresent downside in the same group too have similar tendency as that of Li, Be and B, hence formcompounds in the following states; Na+, K+, Rb+, Cs+, Mg2+, Ca2+, Sr2+, Ba2+, Al3+, Ga3+,etc.

Elements on the right hand side of carbon have more than 4 electrons in the valence shell (N-5, O-6,F-7). To complete their octet, valence electrons must be subtracted from 8 that’s why the valency of N is(8-5) i.e. 3, O is (8-6) i.e. 2 and that of F is (8-7) i.e. 1. It is much easier to gain 3, 2, 1 electrons to completetheir octet as compared to loosing 5,6,7 electrons to complete their octet. So these elements have moretendency to gain electrons and form compounds in the following states; N3-, P3-, As3-, Sb3-, Bi3-, O2-, S2-, Se2-

, Te2-, Po2-, F-, Cl-, Br-, I- .As elements present on the left hand side of carbon loose electrons to form compounds and elements

of right hand side gain electrons to form compounds so compounds formed are ionic in nature.But think about carbon and the elements present down side, which are present in the middle of each

period and have equal tendency to loose or gain electrons as they have 4 electrons in their octet. This led

4Section

Reason for the formation of large number oforganic compounds




carbon and other elements of this group (Si, Ge, Sn & Pb) to share electrons with itself and other elementsof periodic table to complete octet. As these compounds are formed by sharing of electrons so they areconsidered to be covalently bonded.

Carbon by sharing its electrons with other carbon atoms leads to formation of long chain carboncompounds which may be single, double or triple bonded, cyclic or acyclic, linear or branched. This self-linking property of carbon is called catenation. All the atoms of 14th group show the property of catenationbut it decreases down the group because of weak overlapping due to large size and follows order:

C > Si >> Ge > Sn > PbCarbon may also form multiple bonds with N, P, O, S etc. forming large number of functional group,

which we will discuss later.This is not the end of compound formation. Carbon forms many abnormal compounds with elements

of s, p & d blocks. So for sake of simplicity we are constructing an organic chemist’s periodic table withthe most important elements emphasized.

Elements, which are in dark box, are generally involved in making organic compounds along withdeuterium (D), which is an isotope of hydrogen (H).

18

K Ca Sc Ti Cr Mn Fe Co Ni Cu Zn Ga Ge As Se BrRb Sr Pd Ag Cd In Sn Sb Te I

Cs Ba Os Ir Pt Au HgFr Ra Hs Mt Ds Rg

3 4 5 6 7 8 9 10 11 12Na Mg

Li

H1

2

Al Si P S Cl

B C N O F13 14 15 16 17

D

T

Be

Kr

Xe

Tl Pb Bi Po At Rn

Ar

NeHe

La Ce Pr Nd Pm Sm Eu Gd Dy Ho Er Tm Yb Lu

Ac Th Pa U Np Pu Am Cm Cf Es Fm Md No Lr

Tb

Bk

Y Zr Nb Mo Tc

Hf Ta W Re

Rf Db Sg Bh

VRu Rh

At. mass = 12.01Valency = 4EN = 2.5

CarbonAt. radius = 67 pm

E. Con. = [He]2s pStructure = Hexagonal

2 2

As there are large number of atoms in periodic table which have valence electrons, atomic orbital ofcarbon may overlap with them and share its electron to form large number of compounds. But for thatmany other factors such as size, activation energy, electronegativity, electron affinity, catenation etc. areresponsible which all come under one word “Position” i.e. position of carbon in the periodic table. Thisword “position” include everything related with molecule formation therefore the main reason behindlarge number of organic compound is the position of carbon in the periodic table.




A functional group is an atom such as halogen (-X) or group of atoms such as carboxylic acid(-COOH) in a molecule that gives the molecule its characteristic chemical properties. They are the actiongroup or reactive site in a chemical reaction and the remaining hydrocarbon part remains inert.Eachfunctional group undergoes characteristic chemical reaction so by recognizing them it is possible topredict the reaction, which that molecule undergoes.

O

3CHHO 3CH

Carboxylic acidInert

hydrocarbon part

The concept of functional group is important for organic chemistry due to 3 reasons: Each functional group shows its characteristic chemical reaction i.e. a particular functional group

shows same nature of chemical reaction when present in any compound. Functional group helps in nomenclature of organic compounds. Each functional group have a

secondary suffix for it such as oic acid for carboxylic acid, ol for alcohol, etc. Functional group serves to classify organic compounds into different classes or families i.e.

Compounds with same functional group belong to same class.A molecule may have more than one functional group and are called as poly functional group

compounds or simply poly functional compounds and properties of each functional group may bemodified by the presence of other.

O

OH

O

HO

Carboxylic acid

Ketone

Alcohol

(Polyfunctional compound)As carbon has 4 valencies therefore compounds with all single bond, one double bond or with one

triple bond may be formed with any atoms, which satisfy octet rule.

5Section

Functional Groups




Functional group with bond only-Carbon may combine with those atoms which needs 1 electron for completing their octet (suchas hydrogen or halogen) or with some groups which require one electron for completing octetsuch as alkyl group (–CH3, –CH2CH3, etc.), hydroxyl group (–OH), alkoxy group (–OCH3 ,–OCH2CH3), amino group ( –NH2), thiol group ( –SH) or thioether (–SCH3) to form correspondingfunctional groups.

| |

| |– – –C H/C Alkane

|

|– –C X Alkyl halide

|

|– –C OH Alcohol

|

|– –C SH Thiol –C O – C Ether C – S –C Thioether

|C – N–H 1 amine

H

|C – N– C 2 amine

H

|C – N– C 3 amine

C

Functional group with one bond-Carbon forms 1 bond with many atoms such as O, N, S, P and even itself leading to formation

of different functional groups. Along C=O bond

OH/C – C–H Aldehyde

O

H/C – C– X Acid halide

O Carboxylic

acidH/C–C –OH

O O AcidanhydrideH/C–C –O–C –H/C

|

OH/C – C– N– H 1 amide

H

|

OH/C – C– N– C 2 amide

H

|

OH/C – C– N– C 3 amide

C

O

C – C–C Ketone

Along C=N bond

C=N–HC/H

C/HImine C=N–R

C/H

C/H

Alkyl imine(Schiff base)

C=N–OHR/H

R/HOxime 2C=N–NH

R/H

R/HHydrazone

C=N–N–PhC/H

C/HPhenylhydrazone

H

C=N–NC/H

C/H2,4-Dinitrophenyl

hydrazoneH2NO

2NO

2C=N–N–C–NHC/H

C/HSemicarbazone

H

O




Along C=C, C=S and C=P bond

Carbon disulphideS=C=S C=C Alkene Methylene triphenylphospheneH C=P Ph2

Ph

Ph Functional group with two bond-

AlkynesH/C C/H C/H C N Nitrile (Cyanide)

H/C N C Isonitriles (Isocyanide)

Carbon do not form 3 bonds with any atomAs the size of carbon is so small it do not form compounds which have 3 bonds. This does not

mean that no compound with 3– bonds exists. There are some d block elements, which form 3 bondsand total four bonds. In that case there are 1 , 2 & 1 bonds. This bond indicates the 3rd bond.

Stable quadruple bonds are most common among the transition metals such as rhenium, tungsten,molybdenum & chromium. Some common examples are

Chromium (II) acetateCr2 ( -O2CMe) 4(H2O) 2,

Potassium octachlorodirhenate (III), K2 [Re2Cl8].2H2O &Potassium octachlorodimolybdate K4 [Mo2Cl8]

Re Re

Cl Cl

ClClCl

Cl2-

Cl

Cl

Along with these functional groups there are some other functional groups which are alsoencountered in organic chemistry. So learn these too with very carefully so that your understandingabout functional group should be crystal clear. We have used R for alkyl group, which means simply thehydrocarbon part of the molecule. It may be acyclic or cyclic but attachments are same as given above orbelow. For more simplicity treat R as C so that you should get a better observation in molecules. Theseadditional functional groups are

Benzene

O

OORR

Carbonate ester Carboxylate

O

OR –

Hydroperoxy

OOR

HPeroxy

OOR

R

CarbonothionylH

S

R

Imide

O O

NH

R R

Hemiacetal

O–RHO

HRAcetal

O–R

HR

R–O

Hemiketal

O–R

R

HO

R

Ketal

O–R

R R

R–O

Orthoester

O–RR

R–O

O–ROrthocarbonate ester

O–RR–O

O–RR–OR–N

H

H1° Amine

R–NR

H2° Amine

R–NR

3° AmineR

R–NR

4° Ammonium saltRR

+ –HN=N =NAzide

N=NR

RAzo

R O NCyanate




R–N=C=OIsocyanate

O–NO

RNitrate

O

–

+ R N CIsonitrile

+ –O

Nitroso oxy

R NO R–N

O

NitroO

–+

NRNitroso

OR–S–OH

O

OSulphonic acid

SDisulfide

R SR R

Sulfinile

SR

OR–S–R

O

OSulfonyl

R–S–OHO

Sulfino

R S NThiocyanate

While identifying a functional group, look at its attachments very carefully. Same group with differentattachements may become different functional group. You will be surprised to note that same groupsuch as –OH when attached to single bonded carbon is alcohol, with benzene ring it is phenol, with C=Nit is oxime and with C=O it is called as carboxylic acid.

3H C

3H C

OH

3CHAlcohol

OH

Phenol

3H CN OH

3H COxime

O

Carboxylic acidOH3H C

Similarly many other differences are observed just on the basis of attachment such as

3H C

3H C

Cl

3CHAlkyl halide

2H CCl

Vinyl halide2H C Cl

Allyl halide

Cl

Aryl halide

O

Acid halideCl3H C

O 3CH3H CEther

O

O3CH

Ester3H C

OO

O3CH

Acid anhydride3H C

Amine2NHR

O

Amide2NHR

N

HydrazoneR R

2NH

As whole organic chemistry is based on functional groups and their interconversion so revise thesefunctional groups regularly for best picture of organic chemistry in mind.




A series of compounds in which members have same functional group but differ by one or more

2–CH – units (molecular weight 14 or multiple of 14) is called Homologous series (homos is Greek wordwhich means the same as). Members of homologous series are called homologues. Homologues are thecompounds with same general formula and possess similar chemical properties due to presence of samefunctional group (if any) i.e. all carboxylic acids are homologues to each other starting from formic acid(HCOOH) to any carboxylic acid with formula RCOOH where R is any acyclic saturated alkyl group. Allalkanes starting from one carbon to infinite carbon are homologues.

Example: 14

3 3

3 2 3

3 2 2 3

(A) CH(B) CH – CH Homologous sereis(C) CH – CH – CH of alkanes(D) CH – CH – CH –CH

Example: 2

(A)3CH –C–OH

O (B)

3 2CH –CH –C–OHO

(C) 3 2 2CH –CH –CH –C–OH

O (D)

3 2 3CH –CH –C–O–CHO

Here (a), (b), (c) are homologues but (d) is not homologue of (a), (b), (c) as it has different functionalgroup. One point should always kept in mind that homologues have same functional groups or in

other word have same type of chemical reactions so never insert CH2 unit to any bond whichcreate a compound with different functional group.

O

O

HH

3H C O

OH

Homologues2Insert CH unit

O

OH H 3H C

Homologues2Insert CH unit

2CH O

OH

6Section

Homologue & Homologous series




O

OH

3H C

Not Homologues

2Insert CH unit

O

O

3H C 3CH

Carboxylic acid Ester

As the compounds within a homologous series have the same general molecular formulaand the same functional group (amines, alcohol, carboxyl acid, ester, alkane, alkene, alkyneetc.), so they can be prepared using similar methods. Such as

O

3CHRCu/Heat

OH

3CHR

We can prepare any methyl ketone from any alcohol having CH3CH(OH)– (by changing thenumber of carbon in R) as Cu/heat is used to convert secondary alcohol into ketone, which wewill study later. Similarly we can prepare any carboxylic acid from primary alcohol (by changingthe number of carbon in R) using KMnO4/heat

O4KMnO /Heat

RR OHOH

Compounds within a homologous series show gradual change in physical properties due toincreased molecular size and mass, caused by the longer carbon chains. For example, ethane(CH3CH3), has a higher boiling point than methane (CH4). It is because of the reason that ethanemolecule experiences intermolecular attraction force than methane, as in a large molecule, theelectron cloud tends to be distorted at random to a greater extent. Thus, the London DispersionForces between ethane molecules are higher than that between methane molecules, resulting instronger forces of intermolecular attraction, raising the boiling point.

By observing the relative number of carbon and hydrogen atoms in acyclic alkanes it hasgeneral formula CnH2n+2 where n is any integer. So if alkane has 1 carbon atom it must have 4hydrogen atoms; if it has 2 carbon atoms then it must have 6 hydrogen atoms and so on. As weknow that carbon form s fou r covalent bond s and hyd rogen form s only one so there i s only onepossible structure for an alkane with molecular formula CH4 (methane) and only one structurefor an alkane with formula C2H6 (ethane). As the number of carbon increases the number ofpossible structures also increases. For example C4H10 have two, C5H12 have three, C6H14 have 5,and C7H16 have 9 possible structures. This number increases very rapidly as the number of carbonincreases such as C10H22 have 75 and C15H32 have 4347 possible structures which we will call lateras constitutional isomers.

As C15H32 have 4347 possible structures but all are having same molecular formula so they aresimply constitutional isomers which we will study later but starting from CH4, C2H6, C3H8 toC15H32 they differ (CH2)n unit, so they all are considered as homologues and have same nature ofchemical reactions and method of preparation. We use only the letter R for general alkyl group(alkane –1H) starting from 1 to infinite.




Here is a table, which helps you to understand homologue better with general formulaHomologous series General formula (Functional group if any) Example

Alkane CnH2n+2 (n=1) CH4, n=1Alkene CnH2n (n=2) C2H4, n=2Alkyne CnH2n-2 (n=2) C2H2, n=2Alcohol CnH2n+1OH (n=1) CH3OH, n=1

Carboxylic acid CnH2n+1COOH (n=0) CH2O2, n=0Carbohydrate Cn(H2O)y (n=3) C6H12O6 n=6

There are some direct chemical reactions from which we can convert one member of a homologousseries to the next member and such reactions are called as homologation reactions that we will study inchemical reactions

O

R OHCarboxylic acid

2SOClO

R Cl2 2CH N

OR

OHNext higher homologue of

carboxylic acid

2moist Ag O




When you carefully look over the organic compounds then you will find that carbon atoms are bondedwith other carbon atom or some other atoms such as H, N, S, P, O, X etc. to form molecule. This differencein attachment leads to difference in rate of chemical reactions. So it is important to analyze carefully theatoms directly attached to carbon in any organic molecule. On the basis of attachment of carbon withother carbon atoms, the carbon atoms in organic compounds are classified into four types Primary carbon (10C) – Carbon attached to none or one carbon is called as primary carbon. In

case when no other carbon is attached then such carbon is sometime called as super primarycarbon such as CH4.

Secondary carbon (20C) –Carbon attached to two carbons is called as secondary carbon. Tertiary carbon (30C) – Carbon attached to three carbon is called as tertiary carbon. Quaternary carbon (40C) – Carbon attached to four carbon is called as quaternary carbon.These carbon atoms may be in open chain or in ring; they may be neutral, charged or in radical form.

But the important part is the attachment with other atoms and for simplicity they are generally writtenwith single bonded forms. These forms may be derived from alkane, alkyl halide, alcohol, ether, amines,thioethers, thiols, etc.

3CH3H C

3CH

1°2°

3°1°

1°

3H C 3CH

3CH2°

2°

2°

2°

1° 1°

1°

3°4°

3H C

3CH

3CH

2°

1°

3CH

3H C3CH

1°

1°

1°1°

1°

3°4° 3°

Hydrogen’s attached at primary, secondary, tertiary carbons are called primary, secondary &tertiary hydrogen respectively. It is important to note that carbons are of four types buthydrogen’s are only of three types, as quaternary carbon does not have any hydrogen.

Only in case of methane primary carbon have 4 H but in all other alkanes and cycloalkanes each10, 20, 30 carbon have 3, 2, 1 H respectively. To find out total number of 10, 20, 30,40 carbon and 10,20, 30 hydrogen’s you have to count all the carbons or hydrogen of similar types.

In organic compounds other than alkane or cycloalkanes you must be more careful in reportingthe number H as individual carbon may have any number of H ranging from 3 to 0 as thesehydrogen are already replaced with some other atoms or groups.

7Section

Nature of C, H & Functional groups




O3CH

3CH3CHN3H C

3CH

1°

1°

1°

1°

1°2° 2°

1°H=152°H=23°H=0

OS

Cl

Br

Cl

1°H=52°H=83°H=0

1°

1°

2°2°

2°2°2°

3°

3H C1°

NHNH

HS1°

2°

3H C

2NO2°

1°

1°H=42°H=23°H=0

In organic chemistry some functional groups are formed just by replacement of one hydrogen,with an atom or groups such as alkyl halide, alcohol, primary amine, thiol, nitro compounds,sulphonic acid, etc.

3 3H C–CHOH

3H C–H+OH

Alkane Alcohol3 3H C–CH 3H C–H

Alkane Primary amine

2NH

2+NH

3 3H C–CH 3H C–H

Alkane Thioalcohol

SH

+SH 3 3H C–CH 3H C–H

Alkane Nitro alkane

2NO

2+NO

3 3H C–CH 3H C–H

Alkane Alkane sulphonic acid

3SO H

3+SO H

So alkyl halide formed from replacement of primary H, secondary H and tertiary H are called asprimary alkyl halide, secondary alkyl halide and tertiary alkyl halide respectively.

3H C Cl1° Alkyl halide

3H C2° Alkyl halide

Cl3CH

3H C3° Alkyl halide

3CH

Cl3CH

Similarly alcohol formed from replacement of primary H, secondary H and tertiary H is called asprimary alcohol, secondary alcohol and tertiary alcohol respectively.

3H C1° Alcohol

OH 3H C2° Alcohol

3CHOH

3H C3° Alcohol

3CHOH3CH

OH

1° Alcohol OH

2° Alcohol

3CH

OH

3° Alcohol

3CH

But some functional groups are formed by chemical reactions of two same or different functionalgroups such as ether, secondary amine, tertiary amine, thioether, ester, amide, acid anhydride, etc.

3CH3H CAlcohol

AlcoholO H + H O

3H C

EtherO

3CH

2–H O




O

Acid anhydride3H C

3CH3H C

Carboxylic acid

O H + H O2–H O O

O

3CH

O O

Carboxylic acid

O

Ester3H C

3CH3H C

Alcohol

O H + H O2–H O O 3CH

O

Carboxylic acid

O

Primary amide3H C

3CH3H C

Primary amine

O H + H NH2–H O NH 3CH

O

Carboxylic acid

3CH3H C

Primary amineNH

3–NHPrimary amine

2NH + H3H C

NH3CH

Secondary amine

3CH3H C

Secondary amine

N3–NH

Primary amine2NH + H

3H CN

3CH

Tertiary amine3CH

3H C

So be careful in reporting your answer for amines or amides. Secondary alcohol is that in which–OH is attached to secondary carbon but secondary amine is that in which nitrogen is directlyattached to two carbon and one hydrogen. Tertiary amine is one in which N is directly bondedto three carbon. So secondary or tertiary amine may be present over primary, secondary ortertiary carbon.

3H C 2NH1° Amine

3H C1° Amine

3CH2NH

3H C1° Amine

3CH

3CH2NH

2° Amine

NH

3° Amine

N3CH

2° Amine

NH

NH

2° Amine

NH

2° Amine

N

3° Amine

N

3° Amine

N

3° Amine




In organic chemistry when we look over the molecules we will find that some molecules are onlysingle bonded but some molecules are multiple bonded and that multiple bond may be of many typessuch as double bonds of C=C, C=N, C=O, C=S or triple bonds of C N or C C. To differentiate thesecompounds word saturated and unsaturated is used.

To understand these terms, take a glass full of water and add a spoon of sugar in it and stir it withspoon. You will find that whole sugar dissolves in it. But when you continuously go on adding spoons ofsugar in it and stirring it then after some time you will find that dissolution stop and sugar settles at thebottom of it. At that time you will say that no more sugar is dissolving in it i.e. it is saturated. Similarlywhen no further bond can be added to a molecule at room temperature then such compounds are calledas saturated compounds and if further bonds can be added in it then they are called as unsaturatedcompounds. As we know that single bond has no tendency to add anything at room temperature andhave only tendency to substitute some atom or group so they are considered as saturated compoundsand multiple bonded compounds have tendency to add bond so they are considered as unsaturatedcompounds.

| |Substitution reaction Addition reactionA – B C A –C B A B C –D A– B

DC

Saturated compounds may be cyclic or acyclic; may be only of carbon and hydrogen or may havesome hetero atoms such as halogens, nitrogen, oxygen or sulphur. The multiple bond may be of anyform such as C=C, C=N, C=O, C=S, C N or C C with or without rings.

Example of saturated compounds are-

3H C 3CH 3H C 3CH

Cl

3H C 3CH

OH

3H C 3CH

2NH

3H C 3CHO

3H C 3CH

S

3H C 3CH

SH

NH O

OH

NH

S

8Section

Saturated & unsaturated molecules withimportant positions




Example of unsaturated compounds are-

3H C 2CH CH3H C O

H3H C O

3H C 3CH

O

3H C OH O

3H C 2NH O

3H C OMe

NH O

N

S OH

If compounds are saturated with presence of only carbon & hydrogen then such compounds arecalled as saturated hydrocarbon, such as alkane and cycloalkane but if any hydrogen in it is replacedwith any other atom or group such as –Cl, –OH, –NH2, –SH then they are considered as derivatives ofsaturated hydrocarbons. Similarly if compounds are unsaturated and have only carbon & hydrogen,they are called as unsaturated hydrocarbon such as alkene, alkyne, cycloalkenes & cycloalkynes but ifhydrogen in it is replaced with any atom or group then they are considered as derivatives of unsaturatedhydrocarbon.

3 2 3 3 2 2

Alkane Derivative of alkaneCH CH CH CH CH CH Cl

2 3 3

Alkene Derivative of alkeneCH =CHCH Cl–CH=CHCH

When you again carefully analyze the functional groups you will find that some functional groupshave presence of carbon within it such as nitrile (–CN), aldehyde (-CHO), ketone (-CO-), carboxylic acid(-COOH), acid halide (-COX), acid anhydride (-COOCO-), ester (-COOC), amide (-CONH2) etc but somefunctional groups do not have carbons such as amine (–NH2), alcohol (-OH), thiol (-SH), nitro (-NO2),Sulphonic acid (-SO3H), ether (-O-), thioether (-S-), etc.

In organic chemistry some common positions such as zero (0), alpha ( ) , beta ( ) , gamma ( ) , delta

( ) , and epsilon ( ) have been commonly encountered so we must have a clear cut picture of it too.

If any functional group do not have carbon in it then the carbon directly attach to it is given theposition alpha followed by beta, gamma, delta and epsilon to next carbons in sequence such as

3H C αβδε Cl

3H C αβδε OH

3H C αβδε 2NH

3CH3H C α

βδ

β

2NO

3CH3H C α

βδ

β

3SO H

3CH3H C α

βδ

β

O3CH

α β

N+3H C 3CH

3CH

3CH

α

α

α

αβ

N

+

3CH3H C

βα α

β

αα

S+

3CH

βα α

β

α 3CHβ




But if any functional groups have carbon in it then that carbon is give position zero followed byalpha, beta, gamma, delta and epsilon to next carbons in sequence such as

3H C αβδε

O

H0 3H C α

βδεO

OH0 3H C α

βδεO

Cl0

3CH3H C α

ββ

3COOCH0

δ

3CH

3H C αββ

3COOCOCH0

δ

α

3CHO

ββα

β0

δ

O

Oαβ

δ0

ε

This zero (0), alpha ( ) , beta ( ) , gamma ( ) , delta ( ) , and epsilon ( ) have very much importancein organic chemistry as many named reactions such as Aldol reaction, Cannizzaro reaction, Claisenrearrangement, HVZ reaction etc involve these terms, which we will study later. Aldol condensation- When aldehyde having alpha hydrogen is treated with base, it first give

beta hydroxy aldehyde which later looses a water molecule to form alpha,beta-unsaturatedaldehyde as a product.

3CH –C–HO

3 2CH –CH–CH –C–H 3CH –CH=CH–C–HΔBaseOOHO

α ,β-unsaturatedaldehyde

β-Hydroxyaldehydeα

Aldehydehaving –H

000α α αββ




When we carefully look over the organic molecules we will find that some carbon atom is singlebonded while some others are multiple bonded. These single or multiple bonds create some fixed shapeor geometry in molecule such as linear, planner, tetrahedral, etc. Molecular geometry or molecularstructure is the 3-dimensional arrangement of the atoms that constitute a molecule. It determines severalproperties of a substance including its reactivity, polarity, phase of matter, color, magnetic properties,and even biological activities. The molecular geometry can be determined by various spectroscopicmethods & diffraction methods such as Infrared (IR), microwave & Raman spectroscopy can giveinformation about the molecule geometry. X-ray crystallography, Neutron diffraction and Electrondiffraction can give molecular structure for crystalline solids based on the distance between nuclei andconcentration of electron density. Gas electron diffraction can be used for small molecules in the gasphase. Nuclear magnetic resonance (NMR) & Förster (Fluorescence) resonance energy transfer (FRET)methods can be used to determine complementary information including relative distances, dihedralangles, angles, and connectivity. Molecular geometries are best determined at low temperature and canbe different as a solid, in solution, and as a gas.

As you learned earlier that these single or multiple bonds are formed by overlapping of atomicorbitals either by axial overlapping or by lateral overlapping, molecular geometries can be specified interms of bond lengths, bond angles and torsional angles.The bond length is defined to be the averagedistance between the centers of two atoms bonded together in any given molecule. A bond angle is theangle formed between three atoms across at least two bonds. For four atoms bonded together in a chain,the torsional angle is the angle between the plane formed by the first three atoms and the plane formedby the last three atoms.

To explain the shape of any molecule new concept of hybridization was proposed which is mixing ofatomic orbitals of same or nearly same energy and redistribution of energy to form new hybrid orbitalsof equal energy and same shape. So hybridization is simply a mathematical approach to explain theshape of any molecule i.e. if a molecule is tetrahedral then its tetrahedral shape is explained by sp3

hybridization and similarly if a molecule is planner then to explain it we have to use the concept of sp2

hybridization and for linear molecules sp hybridization. In organic chemistry when we have to find outhybridization of any atom always count the number of sigma bond and lone pair of electrons present onit (if any).

9Section

Hybridization




ATetrahedral shape

109.5°

A

Planner shape

120°

180°

Linear shape

A

As the shape of atomic orbitals is derived from Schrodinger wave equation so it is purely amathematical term. Some shapes commonly encounted in chemistry are

Trigonal plannerLinear shapeA.Tetrahedral shape

Trigonal bipyramidal

T shaped

Square pyramidal

Pentagonal bipyramidal

Planner pentagonal

Tricapped trigonal prismatic

Bent shape

Trigonal pyramidal

Seesaw

Octahedral

Square planar

Pentagonal pyramidal

Square antiprismatic

C.

E.

G.

I.

K.

M.

O.

B.D.

F.

H.

J.

L.

N.

P.

As organic compounds are carbon containing compounds and carbon has only s and p orbitals so thehybridization commonly observed in organic chemistry are sp, sp² & sp³. To calculate hybridization ofany atom always counts the number of sigma bonds and lone pair of electrons present on it. If the sum of number of sigma bonds and lone pair of electrons = 2, atom have sp hybridization. If the sum of number of sigma bonds and lone pair of electrons = 3, atom have sp² hybridization. If the sum of number of sigma bond and lone pair of electrons = 4, atom have sp³ hybridization.There is no role of positive charge in calculating the hybridization of any atom as that contains no

electrons or bonds. Negative charge is also considered as 2 electrons system similar to lone pair.

H

HHH

3sp

3sp3H C – OH

2sp

2H C2CH2sp

2sp

2sp

2H CCH2sp

2sp

sp

sp

sp2H C 2CHC

2sp 2sp 3CH

O3sp

3sp3sp

3sp 2sp

3sp

2sp2sp

2sp 2sp2sp

2sp

2sp2sp

2sp

As we told you earlier that it is simply a mathematical approach to explain the shape of molecule sothere may be some violation of it. These violations are commonly of three types: If system have A=B-C, where C have lone pair of electrons then one of the lone pairs present is

not counted for hybridization and geometry along C is planner & its hybridization will be eithersp or sp²

2H CO–H....

2sp 2H C

2NH2sp.. 2H C

S–H....

2sp




If system have A=B-C, where C have negative charge then –ve charge is not counted forhybridization and geometry along C is planner & its hybridization will be either sp or sp²

2H C–2CH

2sp 2H C

O.... –2sp 2H C

–NH

2sp.. You must remember these two structures where all atoms are sp² but molecule is not planner at

all.

Cyclooctatetraene = COT

or (Non Planner)

Overall Tub shaped

10 Annulene

or

Hydrogens present in the central carbon disturbe the plane

HH (Non Planner)




Up to this point you are familiar with the common functional groups, saturated and unsaturatedcompounds with their hybridization and shape so now we can classify the organic compounds intodifferent categories on the basis of their shape and stability. Organic compounds are divided into twoclasses- Acyclic or aliphatic & cyclic compound. Acyclic compounds or aliphatic compounds are thosewhich do not have any ring. The word aliphatic comes from aleiphar which mean fat or oil. Here carbonatoms can be joined together in straight chains or branched chains. They can be saturated, joined bysingle bonds (alkanes), or unsaturated, with double bonds (alkenes) or triple bonds (alkynes). Besideshydrogen, other elements can be bound to the carbon chain, the most common being oxygen, nitrogen,sulfur and halogen. The simplest aliphatic compound is methane (CH4). If aliphatic compounds arecyclic in nature then such compounds are called as alicyclic (Aliphatic + cyclic) such as cycloalkanes,cycloalkenes, cycloalkynes, cyclic esters, cyclic ketones etc. Aromatic compounds are the most important class of organic molecules due to their extra stability.

Many natural products such as DNA, RNA, hemoglobin, chlorophyll, medicines, and polymersare aromatic in nature. For a compound to show aromatic nature it must follow these fourconditions.

(I) Compounds should have at least one ring.(II) Each atom of the ring should have sp² or sp hybridization, no matter it is made up of C, N, S, O

or any other atom.(III) The ring should be planner or nearly planner, so that effective overlapping of pi bond is possible.(IV) The ring should follow Huckel’s rule i.e. the ring must have (4n + 2) electrons, where n is any

number starting from 0 to . If the value of n=0, ring must have 2 electrons; for n=1, 6 electrons must be present and so on.

If any compound violates any of the above conditions then it will not be considered as aromatic andnot get extra stability. For example, Hexa-1,3,5-triene (CH2=CH–CH=CH–CH=CH2) follows last threeconditions but violates the first one so is not an aromatic compounds. Aromatic compounds are further of two types – one with attachment of benzene nucleus and

other with absence of benzene nucleus. Those which contain benzene ring are called benzenoidwhile those which do not have benzene ring are considered as non-benzenoid.

Cyclic system may be homocyclic (carbocyclic) where each atom of the ring is made up of carbonor may be heterocyclic where at least one atom of the ring is hetero atom such as N, S or O.

10Section

Classification of organic compounds




Examples of 2 pi electrons aromatic system:

+

+

O

HO

HO

O

O

Ph

Ph

Ph

Ph

++

O

OO

O

–

–

NC CN

NC CN

3CH3H C

O

Ph Ph

++

O–

O–

O–

O–


3H C 3CH

Dimethyl fulvene

Calicene

N

N+

–

Fe

Ferrocene

–

+

Tropylium ion

2CH

+

+

2CH+

Group

SubstitutedBenzene

Examples of heterocyclic 6 pi electron aromatic system:

NH

Pyrole

Furan

O

Thiphene

S

NH

NImidazole

NPyridine

Pyrylium ion

O+

NN

3CH

Nicotine

PyrimidineN

N

NNiacin

COOH

N

2NHN

OH

N

2NHN

HO

Cytosine

NOH

NN

HO

Uracil

HNO OH

NOH

NN

HO

Thymine

HNO OH

3CH 3CH


Naphthalene N

Quinoline

Iso quinoleneN

Indole

NH

Benzofuran

O

Benzothiophene

S

Indenyl anion

Purine

NHNN N

Indole

NH

Azulene

Adenine

NHNN N

2NH

NHN

NNH +




Guanine

NHNN N

NHN

NOH

2H N 2H N

O

Examples of 14 pi electron aromatic system:

Anthracene

Phenanthrene

Phenalenyl anion

–

(14) Annulene

+

Anti- Aromatic compounds are very unstable compounds at room temperature so they eitherdimerize or trimerise to become alicyclic one.

Theoretically Antiaromatic compounds follow these 4 conditions: Compounds should have at least one ring. All atoms of ring should have sp² or sp hybridization, no matter it is made up of C, N, S, O or any

other atom. The ring should be planner or nearly planner, but it is important to note that there no effective

overlapping of pi bond is observed. Compound will not follow Huckel’s rule and ring must have (4n) electrons, where n = 1 to

HSome common examples of anti aromatic compounds are:

–

Cycloprop-2-enide ion

Cyclobuta-1, 3-diene

Cyclopenta-2,4-dienylium ion

+




When we carefully look over the cyclic saturated compounds we find that each atom is sp3 hybridizedso it must have bond angle 109028´ but in cycloalkanes this angle is mathematically 180-(360/n) where nis the number of atoms making ring i.e. in Cyclopropane this angle is 600; in Cyclobutane it is 900and soon. This difference in desired bond angle and real bond angle causes strain in bond which affects thereactivity as well as stability of molecule. Greater is the deviation from the theoretical angle greater is thestrain. To calculate the distortion or strain in ring we assume the atoms of ring in a plane, such as incyclopropane, all the 3 carbon atoms occupy one corner of an equilateral triangle with bond angle 600. Astwo corners bent themselves to form bond so strain too is divided equally. So strain in cyclopropane willbe ½ (109028´ – 600) = +24044´.

109°28'24°44'

60°24°44'

Deviation of bond angle in cyclopropane from normal tetrahedral angle

Distortion or strain = ½ (109028´ – bond angle of ring). So angle strains in some cycloalkanes arelisted in the table below.

Compound

CyclopropaneCyclobutaneCyclopentaneCyclohexane

CyclooctaneCycloheptane

No. of C in the ring

3456

87

Angle between the C atoms

60°90°108°120°

135°128°34'

Distortion or strain

24°44'9°44'0°44'–5°16'

–12°62'–9°33'

From the table it is clear that cyclopropane has the maximum distortion, so it is highly strainedmolecule and consequently more reactive than any of the monocyclic alkanes, which is clear from thereaction that ring can be opened very easily to relieve strain on reaction with Br2, HBr or H2/Ni at hightemperature. In contrast, cyclopentane & cyclohexane have least strain so they are found more readilyand are very stable as compared to cyclopropane.

Baeyer strain theory satisfactorily explains the typical reactivity and stability of smaller rings (fromC3 to C5) i.e. Stability order follows : Cyclopropane< Cyclobutane < Cyclopentane

11Section

Baeyer’s strain theory




But not valid for cyclohexane onwards because the strain again increases with the increase in numberof carbon atom but actually large rings are more stable. So molecular orbital theory is also consideredaccording to which covalent bond is formed by coaxial overlapping of atomic orbitals. The greater is theextent of overlap the stronger is the bond formed. In case of sp3 carbon, C-C bond will have maximumstrength if the C-C-C bonds have th e angle 109028´. If cyclopropane is an equilateral triangle then thebond angle of each C-C-C bond would be 600. Therefore it was proposed by Couson that in cyclopropanethe sp3 hybridized orbitals are not present exactly in one straight line due to mutual repulsion of orbitalof these bonds resulting thereby loss of overlap. This loss of overlap weakens the bond and is responsiblefor its instability and strain in molecule. Similarly, in case of cyclobutane, there is also loss of overlap butthe loss is less than in cyclopropane, so cyclobutane is more stable than cyclopropane. Overlapping oforbitals in large ring compound (5 or more carbon atoms) is however much better which accounts for thegreater stability of such compounds.

HH

HHH H

CC C

It is natural that when a molecule has strain within it, it will affect the stability of molecule. Thestability of molecule can be calculated easily by measuring heat of combustion which will give the measureof total strain and thermochemical stability which can be calculated mathematically.

Total strain = (No of C atom in the ring × observed heat of combustion/CH2) - observed heat ofcombustion/CH2 for n alkane.

Experimental data of total strain for different cycloalkanes

3456

8 - 1112-onwards

7

Heat of combustion KJ/CH2

697685664659

661 - 665657 - 661

662

Total strain in KJ

1201123512

32 - 880 - 48

35

No. of C in the ring

From the data above it is clear that strain decreases from C3 to C6 i.e. stability increases, but stabilityagain deteriorates from C7 to C11 ring system but interestingly increases from C12and attain the samestability of six membered ring. According to this theory, the carbon atoms in 5 membered and smallerrings can lie in one plane as explained by Baeyer but Sachse suggested that in six membered and higherrings the carbon atoms are present in different plane i.e. the ring is puckered. In this way the normalvalency angle 109028´ is retained and the ring produced is completely strainless. Thus he proposed thatcyclohexane exist in two puckered forms as boat and chair form in which chair form is more stable.These forms are readily interconvertible through half chair and twist boat forms simply by rotationabout the single bonds which we will study later in conformation.

Chair Half chair Twist boat Boat

These forms of cyclohexane are to give a real picture of it but on paper we commonly make itsplanner form.




You can’t mug-up organic chemistry because there’s too much of it. You can remember trivial thingslike name of compounds but that doesn’t help you to understand the principle behind the subject. Youhave to understand the principle because the only way to tackle organic chemistry is to learn to work itout. That is why we are first providing you some questions so that your understanding about the topicshould be increased. These problems will set you on your way but they are not the end of the journey, ashere you are beginning your journey to understand organic chemistry.

The problem would be of little use to you when you could not check your answers. For the maximumbenefit, you need to solve all the problems without looking the answers. Then you should compare youranswers or suggestions with ours before going to next chapter. If any answer is mismatch then darkenthose question numbers with red pen and again have a microscopic look over it and its theory andcorrect it before proceeding further. Here starts the first set of 60 questions.01. What is organic chemistry? Write at least 4 characters which distinguish organic compounds from

inorganic compounds.02. What is the reason behind the formation of large number of organic compounds?03. Which organic compound was first prepared in laboratory?04. Which atom is the centre of attraction in whole organic chemistry?05. What is the position of carbon in the periodic table?06. What is Octet Rule?07. What is electronegativity and electron affinity?08. What is the electronegativity value of C, H, N, O and F in Pauling scale?09. What is the valency of C, H, N, O and F?10. What is the symbol for the representation of all halogens?11. Write C-F, C-Cl, C-Br and C-I bonds in decreasing order of their bond strength.12. Draw structures of single bonded hydrocarbons with 6 carbon atoms having linear, branched and

cyclic frameworks.13. What is wrong in given structure?

H

HH

ONH

HHN

3H C H

OH

2NHSuggest better way to represent the molecules?

01Exercise

Subjective Approach




14. What is the carbon carbon bond length in ethane, ethene, ethyne & benzene?15. Which bond is weaker in C-H & C-C bond of alkane?16. Write the decreasing order of bond strength of carbon- carbon single, double and triple bond?17. How many electrons are involved in formation of single, double and triple bond?18. What are isotopes? Comment on the isotopes of carbon & hydrogen?19. Oxygen has atomic number 8 and have 3 isotopes with molecular mass 16, 17 & 18 respectively. How

many protons, electrons and neutrons does each of these isotopes have?20. What is Aufbau Principle?21. What is Pauli Exclusion Principle?22. What is Hund’s Rule?23. What is the basic difference between Principle and Rule?24. Potassium has atomic number 19 and atomic mass 39, with one unpaired electron in its valence shell.

Which orbital does the unpaired electron occupy?25. Write electronic configuration of F (Atomic no. 9), Cl (Atomic no. 17), Br (Atomic no. 35) & I (Atomic

no. 53)?26. What are ionic, covalent, co-ordinate & polar covalent bonds?27. Which of the following among given compound have most and least polar bond? NaI, LiBr, Cl2‚ ,

KCl28. What is dipole moment?29. What is formal charge and how it is calculated in molecules such as H3O+?

30. Write the Lewis structures of formic acid, formaldehyde & methanol?

31. Determine the partial positive charge on oxygen atom in a C=O bond if it has bond length 1.22Å andbond dipole moment 2.30 D?

32. Predict the relative length of HF, HCl, HBr & HI bonds?

33. Use the symbol and – to show the direction of polarity of the bonds shown in given compounds?

CH3-Cl, CH3-NH2, HO-Br, I-Cl, CH3-OH, CH3-MgBr & NH2-OH34. Name at least 10 organic compounds which make your life easy?35. What is functional groups and why they are important in organic chemistry?36. Name the compound in which -OH group is attach to tetrahedral carbon, C=C, aromatic ring and

C=O group?37. Convert these condensed form of molecules into bond line structures.

(A) C6H5CH(OH).(CH2)4COC2H5 (B) O(CH2CH2)2O(C) (CH3O) 2CHCH=CHCH(OMe)2

38. Draw bond line structures for these compounds showing the hydrocarbon framework clearly andshowing all the bonds present in functional groups(A) AcO(CH2)3NO2 (B) MeO2.CH2.OCOEt(C) CH2=CH.CO.NH(CH2) 2CN

39. What are the probable formulas for the following compoundsGeCl?, AlH?, CH?Cl2, SiF?, CH3NH?, AlCl?, CF2Cl?, NI? & PH?.




40. Why can’t organic molecule having formula C2H7, C2H7N & C3H5Br2 exist at room temperature?41. Fill the non bonding valence electrons if present that are missing from the following bond line

structure.

3H C 3CHO

O

3H C 3CH O

3H C Cl 3H C N O

3H C O 3CH

42. Convert the following molecular formulas into bond line structures leaving lone pair electrons asthey are assumed to be present there(A) C3H8 (B) CH5N (C) C2H6O (2 possibilities)(D) C2H4O (3 possibilities) (E) C3H9N (4 possibilities)

43. Sodium methoxide (NaOCH3) contains both covalent and ionic bonds. Indicate the linkage whichgives a clear cut picture of it.

44. What are homologues? Write at least 4 homologues of methane, formaldehyde, acetone and formicacid each.

45. How are molecules with a single functional group represented? Write a general formula of alkylchloride, alkyl alcohol, and alkyl carboxylic acid?

46. How do the compounds in homologous series differ in molecular formula, physical & chemicalproperties?

47. Name the word used for sulphur analogs of alcohol & ether?48. Find out the functional groups present in following compounds

3H C 3CH

SH

SHCompound with worst smell

(I)

3H C

3CH

SH

Compound with worst smell(II)

O

3CH

OO

Olean sex pheromone of olive fly(III)

2H C

Vinyl chlorideCl

(IV)

N

N

3CH

3CHCompound from coffee

(V)

Compound from cakes& biscuits

(VI)

O

O

OHO

3H C

Coryloneceramel roasted protein

(VII)

O NHOH

3CH

3CHZeneca's tenormin

For treatment & prevention of heart deasese

(VIII)

CHOMeO

HOVanillin of vanilla

(IX)

Pyridine(X)

N Phenol.

(XI)

OH

Aniline(XII)

2NH

Diazonium salt(XIII)

–2N Cl+

2CH

Styrene(XIV)

Cummene

(XV)

3H C 3CH

Cummene peroxide

(XVI)

3H C 3CHO–O–H

N 3CH

3CHS

O

HNH

R

OCOOH

Penicilin(XVII)




N3CHN

Nicotine(XVIII)

COOH

O 3CH

O

Aspirin(XIX)

HO

N

O

3CHH

Paracetamole(XX)

S

NN

3CH

EtOO

OO

NN NHN

3H C

3H C

Pfizer's sildenafil (Viagra)(XXI)

OTestosteron (Hormone)

(XXII)

3CH3CH OH

H H

H

3H C

O

OCocaine

N O 3CHO

(XXIII)

NQuinine(XXIV)

2H CN

HO3H C

O

3CHHO

NHN

OI

OH O

Fialuridine(Anti viral compound)

(XXV)

3CH

OH

Cl3H C

Chloroxylenol(XXVI)

ONH

HNO

Indigo dye(XXVII)

O

SO

2NH

ClO

NH

OOH

Furosemide (Sulpha drug)(XXVIII)

49. Give the structural formula for a 3 carbon compound containing each of the following functionalgroups(A) C=C (B) C C (C) Cl (D) OH (E) CHO (F) C=O (G) C N

50. Draw one possible structure for each of these molecules selecting any group of your choice for the“wild card” substituents for

1R2R

A

3Ar

O

1Ar

2Ar

51. Find the total number of primary, secondary, tertiary and quaternary carbon in given molecule.

3CH

3H C

3H CI

3H C

3H C

3CH

3CH

3CH3CH

3H CII

3H C

3CH

3CH3H CIII

OH Cl2H N

3CH3CH

3CHIV

3CH

3CHVO

3CH

HN

3CH

3CH3OCH

VI

3CH

VII

Br N OH

N

3CH

NO

O O

3H CVIII




52. Find the total number of primary, secondary and tertiary hydrogen present in given molecule.

3CH

3H C

3H CI

3CH

3H C

3H CII

3CH

3CH

3H C

3H CIII

Cl3CH

OH 3CH

3CH

3CHIV

3CH

3CHV

ClO

53. What are alicyclic compounds? Write at list 10 compounds with different chemical nature.54. Write the structure of smallest cyclic

(A) Alkane (B) Alkene (C) Alkyne(D) Ether (E) sec-Amine (F) t-Amine(G) Thioether (H) Ester (I) Acid anhydride(J) sec-Amide (K) t-Amide (L) Aromatic compound(M) Anti-aromatic compound.

55. Write the structure of open chain molecule that meet the following descriptions(A) Contains 2 sp² hybridized carbon and 2 sp³ hybridized C.(B) Contains only 4 carbons, all sp² hybridized.(C) Contains 2 sp hybridized carbons and 2 sp² hybridized C.

56. What bond angles do you expect for each of the following and what kind of hybridization do youexpect for the central atom in each case?(A) The C-O-C angle in CH3-O-CH3

(B) C-N-C angle in CH3-NH-CH3

(C) The C-N-H angle in CH3-NH-CH3

(D) The O=C-O angle in acetic acid57. Write the hybridization of each carbon in the given molecule?

3H C2CH

I

II

CH

H

H

H

IIIOH

3H C

2H C IV

2H N

Cl

COOH

3CH V

VI

VII

VIII

IX

X

2NH

XI

O

2CH.

XII

2CH

SHN

–

2CH+

XIII

N

3CH3CH3CH

3CH

3H C3H C

XIV

3CH

3CH HHH

HXV

58. State whether the given molecules are planner or non-planner.

I

II

III

IV

VC

H

H

H

H VI

3H C2CH

VII

2H C+

2CH

VIII

–2H C

2CH IX

3H C2CH

.

X

2H N2CH

XI

XII

H

H H XIII

O

XIV

S

NH

NXV




59. Classify the following molecules into saturated and unsaturated compounds?

I

3H C3CH

3CH3H C

II

Cl

3CH

3H C

IIIHO

IVBr

3H C

V

VI

VII

3H C

O3CH

VIII

3H C

O3CH

IX

3H C

O3CH

X

3H C

CN

3H C

3H CXICOOH

CN

XII

NH

XIII

XIV

XV

Cl

60. Classify the following molecules into aromatic, anti aromatic and non aromatic compounds.

I

II

III

IV

V

VI

VII

VIII

IX

X

XI

XII

.

XIII

XIV

XV

XVI

XVII

XVIII

XIX

XX

XXI

XXII

XXIII

XXIV

XXV

O

XXVI

S

NH

XXVII

NH

XXVIII

N

XXIX

XXX

XXXI

XXXII

XXXIII

XXXIV

2N Cl

HO OH

OO

Squaric acidXXXV

OO

+

–

SydnoneXXXVI

N

N

Ph

O

TroponeXXXVII

O

TropoloneXXXVIII

OH

O

+

Pyrylium ionXXXIX

Phenalenyl aniorXL

PentaleneXLI

HeptaleneXLII

XLIII

N+N–

XLIV

3CH3H C

XLV

XLVI

XLVII

O

Ph PhXLVIII

XLIX

L




01. Organic chemistry is the chemistry of carbon and its compounds with some exception of CO32– ,

CO, CO2 etc whose properties resembles more with inorganic compounds. In general organiccompounds(A) React more slowly and required higher temperature for reaction.(B) Undergoes more complex reactions and produce side products.(C) Has lower melting and boiling points with low solubility in water.(D) Are classified into families of compounds such as alcohols, ethers, carboxylic acid etc which

have different reactive groups so have different chemical reactions.02. Position of carbon in the periodic table.03. Urea04. Carbon05. 2nd period & 14th group.06. In trying to explain why atoms form bonds, Lewis proposed that an atom is most stable if it has

a filled outermost shell or an outer shell of eight electrons, which is called as Octet rule.According to it, an atom will give up, accept, or share electrons in order to achieve a filled shellor an outer shell that contains 8 electrons.

07. Electronegativity is a chemical property that describes the tendency of an atom or a functionalgroup to attract the bond pair electrons towards itself. An atom’s electronegativity is affected byboth its atomic number and the distance that its valence electrons reside from the charged nucleus.The higher the associated electronegativity number, the more an element attracts electronstowards it. There are many scales to measure it but Pauling scale is most common among these.Electron affinity of an atom or molecule is the amount of energy released when an electron isadded to a neutral atom or molecule to form a negative ion.

– –X e X

This property is measured for atoms and molecules in the gaseous state only, since in the solid orliquid states their energy levels would be changed by contact with other atoms or molecules.

08. 2.5 , 2.1, 3.0, 3.5, 4 .0 respectively09. 4, 1, 3, 2 & 1 respectively10. X11. C-F > C-Cl > C-Br > C-I12. Linear, branched and cyclic frameworks of 6 carbon hydrocarbons are

AnswersSubjective Approach

01




3CH3H C

3CH3H C

3CH

3CH3H C

3CH 3CH

3H C

3CH 3CH 3CH

3CH

3CH

3H C

3CH

3CH 3H C

3H C 3H C 3CH

3H C

3CH 3CH

3CH3H C

3CH

3H C

3CH

3CH

3CH

3CH3H C

3CH

3CH

3H C

13. Those structures are not according to geometry and may create confusions so betterrepresentations are

N

N 3CH

O

H

2NH

OH

3H C

14. 1.54Å, 1.34Å, 1.20Å and 1.39Å respectively15. C-H bond is weak16. Bond strength follows C C > C=C > C-C17. 2, 4 & 6 respectively18. Isotopes are variants of a particular element such as C or H. All isotopes of a given element share

the same number of protons but differs in its number of neutrons. The term isotope is formedfrom the Greek words “isos” means “equal” and “topos” means “place”, hence: “the same place,”meaning that different isotopes of a single element occupy the same position on the periodictable. The number of protons within the atom’s nucleus uniquely identifies an element, but agiven element may in principle have any number of neutrons. The number of nucleons (protonsand neutrons) in the nucleus is the mass number, and each isotope of a given element has adifferent mass number. For example, C-12, C-13 & C-14 are three isotopes of the element carbonwith mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6 which meansthat every carbon atom has 6 protons, so that the neutron number of these isotopes are 6,7 & 8respectively.H have three naturally occurring isotopes, sometimes denoted as 1H (Hydrogen), 2H (Deuterium,D), and 3H(Tritium, T) with 1 proton and 1, 2 , 3 neutron respectively.

19. Each isotopes of oxygen have 8 protons & 8 electrons but have 8,9,10 neutrons respectively.20. Aufbau principle (German word, Aufbau meaning “building up, construction”) is used to

determine the electronic configuration of an atom, molecule or ion. The principle postulates ahypothetical process in which an atom is “built up” by progressively adding electrons. As theyare added, they assume their most stable conditions (electron orbital) with respect to the nucleusand those electrons already there. According to the principle, electrons fill orbitals starting atthe lowest available (possible) energy levels before filling higher levels (e.g. 1s before 2s). Thenumber of electrons that can occupy each orbital is limited by the Pauli Exclusion Principle&Hund’s rule.




21. Pauli Exclusion Principle is the quantum mechanical principle that states that no two identicalfermions (particles with half-integer spin) may occupy the same quantum state simultaneously.A more rigorous statement is that the total wave function for two identical fermions is antisymmetric with respect to exchange of the particles. For example, no two electrons in a singleatom can have the same four quantum numbers; if n,l, and ml are the same, ms must be differentsuch that the electrons have opposite spins, and so on.

22. Hund’s rules refer to a set of rules which are used to determine the term symbol that correspondsto the ground state of a multi electrons atom. In chemistry, the first rule is especially importantand is often referred to as simply Hund’s rule. For a given electronic configuration, the term with maximum multiplicity has the lowest

energy. The multiplicity is equal to2S + 1, where S is the total spin angular momentum forall electrons. The term with lowest energy is also the term with maximum S.

For a given multiplicity, the term with the largest value of the orbital angular momentumnumber L has the lowest energy.

For a given term, in an atom with outermost sub shell half-filled or less, the level with thelowest value of the total angular momentum quantum number J (for the operator J = L + S)lies lowest in energy. If the outermost shell is more than half-filled, the level with the highestvalue of is lowest in energy.

These rules specify in a simple way how the usual energy interactions dictate the ground stateterm. The rules assume that the repulsion between the outer electrons is very much greater thanthe spin–orbit interaction which is in turn stronger than any other remaining interactions. Thisis referred to as the LS coupling regime. Full shells and sub shells do not contribute to the quantumnumbers for total S, the total spin angular momentum and for L, the total orbital angularmomentum. It can be shown that for full orbitals and sub orbitals both the residual electrostaticterm (repulsion between electrons) and the spin–orbit interaction can only shift all the energylevels together. Thus when determining the ordering of energy levels in general only the outervalence electrons need to be considered.

23. Rules can be violated but Principle can’t be violated and have no exceptions at all.24. 4S1

25. F 1s²2s²2p5 Cl [Ne]3s²,3p5 Br [Ar]4s²3d104p5 I [Kr]5s²4d105p526. Ionic bonds are formed by either loss or gain of electrons such as NaCl which is originally Na+Cl–

. Covalent bonds are formed by sharing of electrons between two atoms. This sharing may befrom same atoms or from different atoms such as H or O. If atoms are same then bonds arepurely covalent but if atoms are different then such bonds are considered as polar covalent.Coordinate bond is also a type of covalent bond but here sharing of electrons takes place from asingle atom such as compound of NH3 BF3 in which N shares its electron with B.

27. Most polar is KCl & least polar is Cl2.28. The negative and positive ends of a polar bond makes it a dipole. The polarity of the dipole is

indicated by the dipole moment ( ). The dipole moment of a bond is the product of magnitudeof the charge (e) on the atom (either the partial positive charge or the partial negative chargebecause they have the same magnitude) and the distance between the two charges (d). Dipolemoment is measured in a unit called Debye (D).

= e × d29. Formal charge is the difference between the number of valence electrons an atom has when it is

not bonded to any atom and the number of electrons it actually “owns” when it is bonded. Anatom “owns” all of its nonbonding electrons and half of its bonding electrons.

Formal charge = No of valence electrons – (No of non bonding electrons + ½ No of bondingelectrons)




Oxygen has 6 valence electrons but “own” electrons of oxygen in H3O+ is 5 (2 non bonding plus3, half of six bonding). As the number of number of “own” electron is 1 less than its valenceelectron so its formal charge is 1.

30. Lewis structures of formic acid, formaldehyde and methanol are respectively

O

CO

HH

Formic acid

O

CHH

Formaldehyde

C

Methanol

OHH

H H

31. If there was a full negative charge on the oxygen atom, the dipole moment would be(4.80 × 10-10 esu)×(1.22 × 10-8 cm) = 5.86 × 10-1 8 esu.cm = 5.86 DKnowing that the dipole moment is 2.30 D, partial positive charge on oxygen atom will be 2.30/5.86 = 0.39. From this we can conclude that oxygen has an excess of about 0.4 electron and carbonatom has a deficiency of 0 .4 electrons.

32. Bond length of HX followsHI (160 pm) > HBr (141 pm) > HCl (127 pm) > HF (92 pm)

33. –3CH Cl , –

3 2CH NH ‚ –HO Br , –I Cl , –3CH OH , –

3CH Mg Br & –2NH OH

34. Ten organic compounds areSN Compound Use

01

3CH

3CH3H COH

Menthol

Flavoring compound from the essentialoil of spearmint

023CH

O 3CH

Cis Jasmone of jasmine

Perfume distilled from jasmine

03Aniline

2NH

Basis for the Dyestuffs industry

04Phenol

OH

Antiseptic in surgery

053CH

3CH3CH

3CH3H C

Iso-octane

Major constituent of petrol




06

3CH 3CH

OH

Vitamin A

3H C 3CH

3CH

For treatment ofblindness

07

OHHO O

HO OHOH

Glucose

Energy source for living organism

08 OH

H

H

OH

HO H

2CH OH

HO O

(Sucrose)

H O

H

2HOH CHO

H

2CH OHH

OH Ordinary sugar isolated from sugarcaneor sugar beet

09N

Quinine

2H CN

HO3H C

Medicine used to treat malaria

10 3H C

SH

3CHPropanedithiol

HS

Worst smell in the world

35. Functional group is an atom or group of atoms in a molecule that gives the molecule itscharacteristic chemical properties. The importance of functional groups lies in the fact that it isthe site of attack of reagent, makes families for organic compounds and helps in nomenclature.

36. Alcohol, enol, phenol and carboxylic acid respectively37. Bond line structures are

(A)

OHO

(B)O

O(C)

3OCH

3OCH

3H CO

3H CO

38. Bond line structures are

(A)

O

OON+

O–(B) O

O O

O(C)

NH

N

O

39. Probable formulas are GeCl4 , AlH3 , CH2Cl2 , SiF4 & CH3NH2, AlCl3 , CF2Cl2 , NI3 , PH3 .




40. Carbon has only four valencies and donot have any vacant orbital so cannot expand its octet.That’s why molecules with such structures are not possible at room temperature.

41. Non bonding electrons are

3H C3CH

O××

×× 3H C

O××××

3CH 3H C

O

Cl××

××××

××××

3H C××N

3H C

O

O××

××

×××× 3CH

42. Bond line structures are(A) (B) 3 2H C – NH

(C) OH Oand (D) andO

H OHO

,

(E) and,2NH

2NH, NN

H

43. 3H C –O Na+

Ionic bond

Covalent bond

44. Homologues are the compounds with same general formula and possess similar chemicalproperties due to presence of same functional group (if any) i.e. all carboxylic acids arehomologues of each other starting from formic acid (HCOOH) to any carboxylic acid with formulaRCOOH where R is any acyclic saturated alkyl group. All alkanes starting from 1 carbon toinfinite carbon are homologues. Homologues differ by (CH2)n units. Homologues of Methane are Ethane, Propane, Butane & Pentane; Homologues of Formaldehyde (Methanal) are Ethanal, Propanal, Butanal and Pentanal; Homologues of Acetone are Butanone, Pentanone, Hexanone & Heptanone; Homologues of Formic acid are Acetic acid, Propanoic acid, Butanoic acid and Pentanoic

acid.45. Molecules with single functional groups are considerd to be derived from a hydrocarbon by replacing

one of its H atoms with a functional group. If the hydrocarbon is aliphatic then then the alkylgroup is represented by R, and the molecule containing functional group may be represented asRG where G is functional group such as RCl for alkyl chloride, ROH for alcohol and RCOOH forcarboxylic acid.

46. Compounds in homologous series differ by (CH2)n units which increases the size of alkyl group,so physical properties changes gradually such as general increase in melting point, boiling pointis observed with decrease in solubility in water. They undergo similar chemical reactions whoserate depends on the size of alkyl group and shape of molecule.

47. The word thio is usually used to indicate the presence of S. So sulphur analog of alcohol is thiol& ether is thioether.

48. The functional groups present are(I) Thiol (II) Thiol & Ketone (III) Ether(IV) Enyl halide (V) Imine (VI) Ether & Ketone(VII) Ketone & Enol (VIII) Ether, Alcohol & 2° Amine (IX) Ether, Phenol & Aldehyde




(X) Imine (XI) Phenol(XII) 1° Amine (XIII) Azo(XIV) Alkene & Benzene (XV) Alkyl Benzene(XVI) Peroxide (XVII) Amide, Thioether & Carboxylic acid(XVIII) Amine & Imine (XIX) Carboxylic acid & Ester(XX) Phenol & 2° Amide (XI) Amine, Amide, Ether & Sulphonamide(XXII) Alcohol, Alkene & Ketone (XXIII) Amine & Ester(XXIV) Amine & Alcohol, Imine (XXV) Alcohol, Ether, Enyl halide & Imine(XXVI) Phenol & Aryl halide (XXVII) Amine, Ketone & Alkene(XXVIII) Ether, Amine, Carboxylic acid & Sulphonamide

49. Structural formulas for compound with such skeletons are

(A)2CH

3H C (B) CH3H C (C) 3H CCl

Cl

3H C3CHor

(D) 3H C 3H C3CHor

HOOH (E)

3H C

OH (F)

3H C

O3CH (G)

3H CN

50. There are of course many possible structures. A could be heteroatom or a structural fragmentwhile Ar could be any of a very large number of substituted benzene rings or even other types ofaromatic rings. Four membered rings could have A = O, NH, CO, SO2 or even alkene while theR1 & R2 could be same or different. To make such structure always be careful that molecule mustfollow octet rule.

O3CH

3CH

3CHHN

Ph

2NH

COOH

2H C

3OCHPh

O

3OCH3OCH

SO O

The three aryl groups in second example, all might be different or some might be same.

O

OH

COOHOH

CHO

O

CHO OH

Cl

ON

O

O

O

51. Total number of primary, secondary, tertiary and quaternary carbons areCom. Primary C Secondary C Tertiary C Quaternary C

I 3 1 1 0II 7 0 3 1III 4 2 2 0IV 3 4 1 1V 4 2 2 0VI 4 3 1 0VII 3 7 1 0VIII 5 10 2 0




52. Total number of primary, secondary, tertiary hydrogens areCompound Primary H Secondary H Tertiary H

I 9 2 1II 15 0 1III 9 7 0IV 9 8 1V 8 4 1

53. Alicyclic compounds are simply the cyclic aliphatic compounds (Cyclic + Aliphatic), which arecommonly cycloalkane, cycloalkene, cycloalkyne, cyclic ether, cyclic secondary amines, cyclictertiary amine, cyclic ester, cyclic ketones, cyclic amide, cyclic acid anhydride.

54. Structures are

(A)

(B)

(C)

(D)

O

(E)

NH

(F)

N3CH

(G)

S

O(H)

O

O(I)O

O

O(J)

NH

O(K)

N3CH

(L)

+

–

(M)

55. (A) 3H C 3CH (B) 2CH2H C (C) 2CH

2H C

56. A. 109° 28´, sp3 B. 109° 28´, sp3 C. 109° 28´, sp3 D. 120° , sp2

57.3H C

2CH2sp

2sp

3sp

I

2sp

II

C

spH

H

H

H2sp

2sp

III

3sp

OH

3H C

2H C

3sp

2sp3sp

2sp

IV

2H N

3sp

Cl

COOH

3CH

2sp2sp 2All sp

V 2All sp

VI 2All sp

VII

2All spVIII

2All spIX

2All sp(X)

2NH

XI

O

2CH.

2sp 2sp3sp

3sp

2sp

3sp

XII

2CH2sp 2sp

3sp

2sp

S3sp

HN

–

2CH+

XIII2All sp

N

3CH3CH3CH

3CH

3H C3H C

XIV

3All sp 3CH

3CH HHH

H

XIV

3All sp

58. Overall shape of molecules areCom. Shape Com. Shape Com. ShapeI Planner II Non-Planner III Non-PlannerIV Planner V Non-Planner VI Non-PlannerVII Planner VIII Planner IX PlannerX Planner XI Planner XII PlannerXIII Planner XIV Planner XV Planner

59. Saturated and Unsaturated compound areComp. Nature Comp. Nature Comp. NatureI Saturated II Saturated III SaturatedIV Unsaturated V Saturated VI Unsaturated




VII Unsaturated VIII Unsaturated IX UnsaturatedX Unsaturated XI Unsaturated XII UnsaturatedXIII Unsaturated XIV Saturated XV Saturated

60. Classification of compounds areCom. Classification Com. Classification Com. ClassificationI Non-aromatic II Aromatic III Anti-aromaticIV Non-aromatic V Anti-aromatic VI AromaticVII Anti-aromatic VIII Non-aromatic IX Non-aromaticX Anti-aromatic XI Aromatic XII Non-aromaticXIII Aromatic XIV Aromatic XV AromaticXVI Non-aromatic XVII Aromatic XVIII Anti-aromaticXIX Non-aromatic XX Aromatic XXI Non-aromaticXXII Aromatic XXIII Aromatic XXIV AromaticXXV Aromatic XXVI Aromatic XXVII AromaticXXVIII Non-aromatic XXIX Aromatic XXX AromaticXXXI Aromatic XXXII Aromatic XXXIII AromaticXXXIV Aromatic XXXV Aromatic XXXVI AromaticXXXVII Aromatic XXXVIII Aromatic XXXIX AromaticXL Aromatic XLI Anti-aromatic XLII Anti-aromaticXLIII Aromatic XLIV Aromatic XLV AromaticXLVI Aromatic XLVII Non-aromatic XLVIII AromaticXLIX Aromatic L Aromatic




Single Correct Questions (SCQ) :1. First organic compound generated (1828) by Friedrich wohler is

(A) 2CO (B) + –

4N H OCN (C) 2 2H NCONH (D)2 2

SH N - C- NH

2. Formation of large number of organic compound is due to(A) size of carbon (B) Electronegativity of carbon(C) Catenation property of carbon (D) Position of carbon in the periodic table

3. How many e– are used to make a single, double & triple bond(A) 1,2,3 (B) 2,4,6 (C) 3,6,9 (D) 4,6,8

4. Electronegativity value on Pauling scale for C, H, N, O, F are(A) 2.1, 2.5, 3.5, 4.0 (B) 2.5, 2.1, 3.0, 3.5, 4.0 (C) 2.5, 2.1, 3.5, 3.0, 4.0 (D) 2.1, 2.5, 3.5, 3.0, 4.0

5. Total number of bonds in the given compound 2 2H C = C = CH - CH = C = CH is :

(A) 9 (B) 15 (C) 14 (D) 16

6. Molecular formula of Phenantharene is :-

(A) 14 14C H (B) 14 10C H (C) 14 12C H (D) 12 12C H

7. Number of sigma bonds in the given compound 3 2CH - CH - CH = CH - CN is :-

(A) 11 (B) 15 (C) 12 (D) 13

8. Find the functional group which is absent in penicillinN

SR – C – N

H

O

PenicillinO COOH

(A) Amine (B) Amide (C) Thio ether (D) Carboxylic acid

9. Homologue of 3CH COOH is

(A) 3 3CH COOCH (B) 3 2CH COCH OH (C) 3 2CH CH COOH (D) 2 3HCOCH OCH

02Exercise

Objective Approach




10. Number of 1°, 2°, 3° & 4° carbons present in given compound respectively is

(A) 5, 2, 1, 0 (B) 5, 1, 1, 1 (C) 5, 1, 0, 1 (D) 5, 1, 1, 0

11. Number of 1°, 2°, 3° Hydrogens present in given compound respectively is

(A) 15, 2, 1 (B) 20, 2, 1 (C) 15, 4, 0 (D) 15, 4, 1

12. Number of 3° & 2° carbon atoms respectively in the following compound are -

(A) 5, 6 (B) 6, 6 (C) 5, 7 (D) 4, 7

13. Number of 2° H atoms in the following compound OH

is

(A) 7 (B) 5 (C) 6 (D) 414. Which molecule is considered as saturated one.

(A) Alkane (B) Alkene (C) Alkyne (D) Aromatic comp15. Which molecule is considered as unsaturated one.

(A) (B)O

(C) NH

(D)

16. Central carbon in 2 2CH = C = CH will have which type of hybridization.

(A) sp (B) 2sp (C) 3sp (D) None

17. The given compound

HN 3CH is

(A) Alicyclic heterocyclic (B) Unsaturated homocyclic(C) Aromatic heterocyclic (D) Saturated heterocyclic

18. Which of the following is an alicyclic compound ?

(A) (B) (C)O

(D) O

19. The saturated heterocyclic compound is –

(A) NH

(B) (C)O

(D)

20. Which of the following do not have bridge head carbon in bicyclo compound.

(A) (B) (C) (D)

21. Which of the following is an unsaturated hydrocarbon.

(A) 3 2CH - CH - C N (B) 23 3CH COCH CH (C) (D)




22. Which molecule is not planner in shape.

(A) (B) (C) (D)

23. Which molecule is heterocyclic in nature.

(A)2NH

(B)N

(C)

OH

(D)

O3O – C – CH

COOH

24. Which molecule is anti-aromatic in nature.

(A)+

(B) (C) (D)

25. Which molecule is aromatic in nature.

(A)+

(B)+

(C) + (D)

26. Which of the following alkane do not have any 2° carbon.

(A) (B) (C) (D)

27. Maximum angle strain is observed in(A) Cyclopropane (B) Cyclobutane (C) Cyclopentane (D) Cyclohexane

Multiple Correct Questions (MCQ) :28. Which statement is correct for element with single bond.

(A) Nitrogen have one lone pair of electron. (B) Oxygen have two lone pair of electron.(C) Fluorine have three lone pair of electron. (D) Carbon have four lone pair of electron.

29. Which statement is correct for bond.(A) Sigma bond is formed by axial overlapping of orbitals.(B) Pi bond is formed by sidewise overlapping of p–p orbitals.(C) Sigma & Pi bonds are a type of ionic bond.(D) Co-ordinate bond is a type of covalent bond.

30. Which statement is correct for functional group.(A) Functional group shows its characteristic chemical reaction.(B) Functional groups helps in nomenclature of organic compounds.(C) Functional groups serves to classify organic compound into different classes or families.(D) All functional groups have same physical properites.

31. Functional group is present in vitamin C

O

HO

O

OHVitamin C

HOOH

is:

(A) Alcohol (B) Enol (C) Ester (D) Ether




32. Which functional group have absence bond ?(A) Alcohol (B) Aldehyde (C) Amine (D) Alkyl halide

33. Which functional group have bond ?(A) Ketone (B) Carboxylic acid (C) 3° amine (D) 2° amide

34. Which compound have C=N linkage ?(A) Imine (B) Amine (C) Oxime (D) Hydrazone

35. Which functional group is present is Viagra N

NHO

NSO

ON 3CHEtO

NN (Viagra)

3CH

(A) Amide (B) Sulphonamide (C) Amine (D) Ester

36. Which functional group is present in sulpha drug SO

O

2NH

Cl

NH

COOH

O

(A) Sulphonamide (B) Carboxylic acid (C) Amine (D) Alcohol37. Which compound(s) is/are homologue of CH3CH2CH2CHO ?

(A) CH2O (B) CH3CHO (C) CH3CH2CHO (D) CH3CH2COCH2CH3

38. Which statement is correct for homologues ?

(A) They have same type of chemical reaction. (B) They differ in 2 n(CH ) unit.

(C) They have same physical properties. (D) They are prepared by same general method.

39. Esters are formed by chemical reaction of(A) Carboxylic acid & alcohol (B) Acid halide & alcohol(C) Acid anhydride & alcohol (D) Aldehyde & alcohol

40. Which compound is unsaturated in nature ?

(A) (B) (C) (D)

41. Which compound is saturated in nature.

(A)OH

(B)NH2 (C)

O(D)

Cl

42. Which molecule have correct number of 2°C present.

(A) (3) (B) O

O O (0) (C) (3)

OH

OHHO(D)

NN N (3)

H

HH




43. Which molecule have correct number of 2 H present ?

(A)OH

(10) (B)2NH

(9) (C) (8)Cl

(D) (8)

44. Which molecule will have 3Sp hybridized carbon ?

(A) (B)Cl+

(C)O+

(D)

45. Which molecule is/are planner in shape ?

(A) (B) (C) (D)

46. Which molecules will exist at room temperature ?

(A)O

(B)O

(C)

O

(D)

47. Which molecule is/are polar in nature ?

(A) (B) (C) (D)

48. Which molecule is/are aromatic in nature ?

(A) O

(B) O

O

HO

HO(C)

CNNC(D)

O

49. Which molecule is/are aromatic in nature ?

(A) (B) (C) (D)

50. Which molecule have ( ) gamma position ?

(A) N+ (B) N (C) N+ (D) S+

51. Which molecule with correct shape is mentioned ?

(A) Tub shaped (B) non-planner

(C) Planner (D) HC CH Planner

52. The hybridizations of carbon atoms present in cumene is/are

(A) sp (B) 2sp (C) 3sp (D) 2dsp




Assertion / Reason type Questions (A/R) :Each question has 5 choice (A), (B), (C), (D) & (E) out of which only one is correct.(A) Statement-1 is true, Statement-2 is true and statement-2 is a correct explanation for statement 1.(B) Statement-1 is true, Statement-2 is true and statement-2 is not correct explanation for

statement 1.(C) Statement-1 is true and Statement-2 is false.(D) Statement-1 is false, Statement-2 is true(E) Both Statement-1 and Statement-2 is false.

53. Statement-1: The concept of ‘vital force theory’ was objected by Wohler by preparing urea fromNH4OCN.

Statement-2: ‘Vital force theory’ state that we can not prepare organic molecule in laboratory.54. Statement-1: CH3CH2CH(NH2)CH3 is a secondary amine.

Statement-2: In 2° amine, 2–NH group attached with 2° C

55. Statement-1: (CH3)3COH is a tertiary alcohol.Statement-2: In tertiary alcohol, –OH group is attached with tertiary carbon.

56. Statement-1: Oxalic acid & formic acid are homologue.Statement-2: Homologues are those compounds which have same nature of chemical reaction and

differ in 2 n(CH ) unit.

57. Statement-1: have 1°, 2°, 3° & 4° carbon.

Statement-2: 1°, 2°, 3° & 4° carbon is that carbon which is attached with 1,2,3 & 4 other carbonrespectively.

58. Statement-1: Cyclohexane is saturated while cyclohexene is unsaturated.Statement-2: Saturated compound is that which have only single bond while unsaturated compound

is that which have multiple bond of any form.59. Statement-1: Hybridization of oxygen atom in furan is sp2.

Statement-2: Hybridization is a mathematical approach to explain the shape of any atom in molecule.

60. Statement-1:+

is a aromatic compound.

Statement-2: Cyclic planner molecule with (4n + 2) electrons are called aromatic compounds.

61. Statement-1: Toluene is an alicyclic compound.Statement-2: Alicyclic compounds are simply the cyclic aliphatic compounds.

62. Statement-1: Angle strain in decreasing order follows-cyclopropane > Cyclobutane > Cyclopentane > Cyclohexane.

Statement-2: Angle strain is the half of angle difference between desired angle of sp3 atom and realangle of the atom in cycloalkane.

63. Statement-1: Loss of one H from alkane leads to alkyl group.Statement-2: Butane have four types of alkyl groups.




Match the Column type Questions (MTC) :64. Match the compounds written in Column-I with its nature in Column-II :-

Column - I Column - II

(A) (P) Aliphatic compound

(B)O

(Q) Aromatic compound

(C)3OCH

(R) Saturated compound

(D)N

(S) Unsaturated compound

(T) Heterocyclic compound65. Match the compounds written in Column-I with its nature in Column-II :-

Column - I Column - II

(A) (P) Aliphatic

(B) (Q) Aromatic

(C) (R) Alicyclic

(D)O

(S) Heterocyclic

(T) Homocyclic

Comprehension type Questions :Comprehension-01 :

Paracetamol is widely used as analgesic (pain reliever) & antipyretic (fever reducer) medicine. Basedon its structure, select the correct answer among this

3CH

O

N

H

HOParacetamol

66. Functional group present in paracetamol is(A) Alcohol (B) Amine (C) Ketone (D) 2° amide

67. Which statement is incorrect for paracetamol is(A) Paracetamol is unsaturated compound. (B) Paracetamol is derivative of hydrocarbon.(C) Paracetamol is heterocyclic compound. (D) Paracetamol is carbocyclic compound.




Comprehension-02 :

OH

I

2H /NiΔ

OH

II

Cu

O

III

2NH –OH H+

N–OH

IV

2H OH+ Polymerization Nylon-6

ON

HV

Δ

Based on this, answer the following questions.68. Incorrect statement about above compound is

(A) A functional group present in II is alcohol. (B) A functional group present in III is ketone.(C) A functional group present in IV is oxime. (D) A functional group present in V is lactone.

69. Incorrect statement about above compound is(A) Compound I is saturated (B) Compound II is saturated(C) Compound III is unsaturated (D) Compound IV is unsaturated

Integer type Questions :

70. Total number of primary carbon atoms present in is

71. Total number of aromatic compounds among the given molecule is

+

++

72. Total number of sigma bond ( ) present in benzene is

73. Total number of homocyclic compounds among the given molecule is

O

N

S

O

OH

OHHO

OO




1. In which of the following species is the underlined carbon having 3sp hybridisation ? [AIEEE-2002](A) 3CH COOH (B) 3 2CH CH OH (C) 3 3CH COCH (D) 2 3CH = CH - CH

2. The general formula n 2n 2C H O could be for open chain [AIEEE-2003]

(A) carboxylic acids (B) diols (C) dialdehydes (D) diketones

3. Which one of the following does not have 2sp hybridized carbon ? [AIEEE-2004](A) Acetonitrile (B) Acetic acid (C) Acetone (D) Acetamide

4. Which of the following represents the given mode of hybridisation 2 2sp - sp -sp-sp from left to right ?(A) 2H C = CH - C N (B) HC C - C CH [IIT-2003]

(C) 2 2H C = C = C = CH (D) 2 2H C = CH - CH = CH5. Which of the following molecules, in pure form, is (are) unstable at room temperature ? [IIT-2012]

(A) (B) (C)O

(D)O

6. The number of loan pair of electrons in melamine is [IIT-2013 (Advanced)](A) 4 (B) 6 (C) 6 (D) 8

7. In allene (C3H4), the type(s) of hybridization of the carbon atoms is (are): [IIT-2014 (mains)](A) sp and sp3 (B) sp2 and sp (C) only sp2 (D) sp2 and sp3

8. The number and type of bonds in 2–2C ion in CaC2 are: [IIT-2014 (mains)]

(A) One bond and one bond (B) One bond and two bond(C) Two bond and two bond (D) Two bond and one bond

9. For the compounds 3 3 3 3CH Cl, CH Br, CH I and CH F , the correct order of increasing C-halogen bond

length is : [IIT-2014 (mains)]

(A) 3 3 3 3CH F CH Cl CH Br CH I (B) 3 3 3 3CH F CH Br CH Cl CH I

(C) 3 3 3 3CH F CH I CH Br CH Cl (D) 3 3 3 3CH Cl CH Br CH F CH I

03Exercise

World of competitions




10. Match the orbital overlap figures shown in List-I with the description given in List-II and select thecorrect answer using the code given below the lists. [IIT-2014 (Advanced)]

List-I List-II

(P) 1. p-d antibonding Codes :-

P Q R S(A) 2 1 3 4(B) 4 3 1 2(C) 2 3 1 4(D) 4 1 3 2

(Q) 2. d-d bonding

(R) 3. p-d bonding

(S) 4. d-d antibonding

11. In the compound 2 3H C = C = CH – CH1 2 3 4

, the hybridization of 1st and 2nd carbon atom is: [BHU 2003]

(A) sp3-sp (B) sp3-sp3 (C) sp2-sp (D) sp2-sp2

12. Allyl cyanide contains and -bonds: [PMT (MP) 2004](A) 5 , 7 (B) 9 , 3 (C) 3 , 4 (D) 9 , 9

13. The C–H bond distance is longest in: [UGET (Med.) 2006]

(A) 2 2C H (B) 2 4C H (C) 2 6C H (D) 2 2 2C H Br

14. The correct order regarding the electronegativity of hybrid orbitals of carbon is: [AIPMT 2006]

(A) 2 3> spsp < sp (B) 2 3< spsp < sp (C) 2 3< spsp > sp (D) 2 3> spsp > sp

15. Number of and π bonds in 6 5H COOHC is: [DPMT 2007]

(A) 13 , 4 (B) 14 , 4 (C) 15 , 4 (D) 16 , 4

16. What is the percentage of p-character of hybrid orbits of carbon in methane, ethene and ethynerespectively? [SCRA 2007](A) 75, 66, 50 (B) 50, 66, 75 (C) 25, 33, 50 (D) 50, 33, 25

17. Compound which do not have sp2-hybridized C ? [JCECE 2008](A) Acetone (B) Acetic acid (C) Acetonitrile (D) Acetamide

18. The state of hybridization of Carbon in 1, 3, 5 are 3 2H C – CH = CH – CH – C CH123456

[CPMT 2008]

(A) 2 3, spsp, sp (B) 3 2, sp , spsp (C) 2 3, sp, spsp (D) 3 2, spsp, sp

19. Hybridization of N atom in pyridine N

is: [BHU 2008]

(A) 3sp (B) 2sp (C) sp (D) 3sp d20. The number of π bonds in the following compound 2 2O N – C C – NO is : [DPMT 2008]

(A) 2 (B) 3 (C) 4 (D) 121. Compound where underlined carbon use 3sp hybrid orbitals for bond formation is: [BCECE 2008]

(A) 3CH COOH (B) 3 2CH CONH (C) 3 2CH CH OH (D) 3 2CH CH = CH




22. The state of hybridization of 2 3 5 6C , C , C and C of 3CH – C – CH = CH – CH – C CH3CH

6 1234573CH

3CH is in the

following sequence [AIPMT 2009]

(A) 3 2 3sp, sp , sp and sp (B) 2 2 3sp, sp , sp and sp (C) 2 3 2sp, sp , sp and sp (D) 3 2 2sp , sp , sp and sp

23. Increasing order of carbon-carbn bond length for the following [CET (karnataka) 2011]

2 4 2 2 6 6 2 6C H C H C H C H(A) (B) (C) (D) is :

(A) B<C<A<D (B) C<B<A<D (C) D<C<A<B (D) B<A<C<D24. Considering state of hybridization of carbon atoms, find out the molecule among the following which

is linear? [AIPMT 2011](A) 3 3CH - CH = CH - CH (B) 3 3CH - C C - CH

(C) 2 2H C = CH - CH - C CH (D) 3 2 2 3CH - CH - CH - CH

25. Which one of the following is a non-benzenoid aromatic compound? [PMT (Kerala) 2011]

(A) Anthracene (B) Tropolone (C) Aniline (D) Benzoic acid

26. Only sp and sp2 hybrid orbitals are involved in the formation of [PMT 2012](A) 3 2CH - CH = CH (B) 3 3CH - CH (C) 3CH - C CH (D) 2 2H C = C = CH

27. The C–H bond and C–C bond in ethane are formed by which types of overlap? [CET 2012](A) sp3-s and sp3-sp3 (B) sp2-s and sp2-sp2 (C) sp–s and sp–sp (D) p–s and p–p

28. The radical 2CH.

is aromatic because it has [NEET 2013](A) 6 p orbitals & 6 unpaired electrons. (B) 7 p orbitals & 6 unpaired electrons.

(C) 7 p orbitals & 7 unpaired electrons. (D) 6 p orbitals & 7 unpaired electrons.

29. Which compound has same hybridization as its combusion product CO2 ? [AIPMT 2014]

(A) Ethane (B) Ethyne (C) Ethene (D) Ethanol

30. Among the following aromatic compound is [AIIMS 2015]

(A) (B) (C) (D)




01. (C) 02. (D) 03. (B) 04. (B) 05. (B)06. (B) 07. (C) 08. (A) 09. (C) 10. (B)11. (D) 12. (A) 13. (B) 14. (A) 15. (D)16. (A) 17. (A) 18. (D) 19. (C) 20. (C)21. (C) 22. (C) 23. (B) 24. (B) 25. (B)26. (D) 27. (A) 28. (A,B,C) 29. (A,B,D) 30. (A,B,C)31. (A,B,C) 32. (A,C,D) 33. (A,B,D) 34. (A,C,D) 35. (A,B,C)36. (A,B,C) 37. (A,B,C) 38. (A,B,D) 39. (A,B,C) 40. (B,C,D)41. (A,B,D) 42. (A,B) 43. (A,B,D) 44. (A,B,D) 45. (A,B,D)46. (A,C,D) 47. (B,C,D) 48. (A,B,C) 49. (A,B,C,D) 50. (B,C,D)51. (A,B,C,D) 52. (B,C) 53. (A) 54. (E) 55. (A)56. (B) 57. (A) 58. (A) 59. (B) 60. (A)61. (D) 62. (A) 63. (B) 64. (A-P,S ; B-P,R,T ; C-P,R ; D-Q,S,T)65. (A-Q,T ; B-P,R,T ; C-P ; D-Q,S) 66. (D) 67. (C) 68. (D)69. (A) 70. (4) 71. (6) 72. (12) 73. (6)

AnswersObjective Approach

02




1. (B) 2. (A) 3. (A) 4. (A) 5. (B,C)6. (B) 7. (B) 8. (B) 9. (A) 10. (C)11. (C) 12. (B) 13. (C) 14. (D) 15. (C)16. (A) 17. (C) 18. (D) 19. (B) 20. (C)21. (C) 22. (A) 23. (D) 24. (B) 25. (B)26. (D) 27. (A) 28. (A) 29. (B) 30. (A)

AnswersWorld of Competition

03




1. Urea (H2NCONH2)

2. Position of carbon in periodic table as it includes small size, electronegativity, catanation etc.

3. 2e– for single bond, 4e– for double bond & 6e– for triple bond.

4. Electronegativity of C, H, N, O, F are 2.5, 2.1, 3.0, 3.5 & 4.0 respectively.

5. Total no. of bond includes both & .

6. C4H10

7. 12 Sigma bond

8. Amine is absent.

9. Homologue of carboxylic is another carboxylic acid which differ in (CH2)n unit.

10. 5, 1, 1, 1

11. For hydrocarbons 1°C, 2°C, 3°H have 3, 2, 1 1°H, 2°H & 3°H respectively.

12. 5 tertiary & 6 secondary C.

13. 2°H is that H which is present on 2°C.

14. Compound with only single bond is called as saturated compound.

15. Compound with any type of multiple bonds are called as unsaturated compounds.

16. sp

17. Alicyclic with heterocyclic nature.

18. Aliphatic cyclic compounds are called as alicyclic compound.

19. Saturated heterocyclic compounds are single bonded heterocyclic compounds.

20. Spiro compound do not have bridze head C.

21. Unsaturated hydrocarbons are having either C = C or C C bond.

22. Cyclooctatetraene is non planner molecule.

23. Cyclic molecule having atleast 1 heteroatom (N, O, S) in ring are heterocyclic compound.

24. Cyclobutadiene is anti-aromatic.

ApproachObjective Approach

02




25. Tropylium ion is aromatic in nature.

26. 2° Carbon is that carbon which is attached with 2 C atoms.

27. Cyclopropane has maximum angle strain.

28. N, O, F have 1, 2, 3 lone pair of e– but carbon do not have any lone pair of e–.

29. Sigma & pi bonds are an example of covalent bond.

30. Functional groups have different physical properties.

31. Alcohol, Enol & Ester.

32. Alcohol, amine & alkyl halide do not have bond.

33. Ketone, Carboxylic acid & 2° amide have bond.

34. Imine, Oxime & hydrazone have C = N bond.

35. Amide, Sulphonamide & amine are present in Viagra.

36. Sulphonamide, amine are present in sulpha drug.

37. Homologue of aldehyde is another aldehyde differ in (CH2)n unit.

38. Homologue have different physical properties.

39. Esters are not formed by reaction of alcohol & aldehydes.

40. Compound with multiple bond is unsaturated.

41. Compound with only single bond is saturated in nature.

42. 2°C is that C which is attached with 2 C atoms.

43. 2°H is that H which is present on 2°C.

44. Sp3 hybridized carbon have 4 bonds.

45. Cyclooctatetraene is non-planner molecule.

46. Aliphatic & aromatic compound exist at room temperature but anti-aromatic compound do not existat room temperature.

47. Those molecules in which both ring become aromatic by breaking exocyclic double bond havepolarity.

48. A, B, C are aromatic by breaking exocyclic multiple bonds.

49. In case of polycyclic compound, e– on peripheri is counted for (4n + 2) e–.

50. Third position after functional group is considered as position.

51. Cyclooctatetraene is tub-shaped.

52. (Cumene)

53. No organic compound can be synthesized in lab is vital force theory.

54. Secondary amine are those compound in which 2 carbon atoms are attached with N.




55. Tertiary alcohol is that alcohol in which –OH group is attached with 3°C.

56. Oxalic acid & formic acid differ in CO2 unit.

57. 1°, 2°, 3°, 4°C is that carbon which is attached with 1, 2, 3 & 4 other carbon atoms.

58. Saturated compounds are single bonded while unsaturated compounds are multiple bonded.

59. Furan is planner so hybridization of O in it is sp2.

60. Cyclopropenyl cation is aromatic.

61. Totuene is aromatic compound.

62. Angle strain for lower rings follows : Cyclopropane > Cyclobutane > Cyclopentane > Cyclohexane.

63. Loss of H from alkane give alkyl group so butane have 4 alkyl group- n-butyl, sec-butyl, iso-butyl &tert-butyl.

64. Saturated compounds have only single bond while unsaturated compound have multiple bond.

Alicyclic compounds are also aliphatic in nature.

65. Homocyclic compounds are those in which each atom of ring is made up of C.

66. Functional group present in paracetamol is 2° amide.

67. Paracetamol have homocyclic ring as each atom of ring is made up of C.

68. Lactones are cyclic ester while Lactum is cyclic amide.

69. Saturated compounds are single bonded while unsaturated compounds are multiple bonded.

70. All terminal C is 1°C.

71. II, V, VI, VII, VIII & IX are aromatic.

72. bond includes both C–H & C–C bond.

73. I, III, VI, VII, VIII & IX are homocyclic compounds.




ApproachWorld of Competition

031. Sp2 carbon have total 3 bond.

2. Carboxylic acid can have general formula CnH2nO2 as it have degree of unsaturation = 2.

3.3CH – C N

Acetonitrile 3CH COOH

Acetic acid3 3CH COCH

Acetone3 2CH CONH

Acetamide

4. H2C=CH–C Nspspsp2 sp2

5. Anti-aromatic compounds are unstable at room temperature.

6.

NH2

NH2NH2

NN

N(Melamine)

7. H2C=C=CH2

spsp2 sp2

8.C C

Ca2+

9. Increasing order of C–X bond follows C–F < C–Cl < C–Br < C–I.

10. This question is based on overlapping of orbitals i.e. p & d orbitals. bonds are formed by axialoverlapping while bonds are formed by side wise overlapping. Bonding means overlapping ofsimilar orbital while anti-bonding means overlapping of dis-similar atomic orbital.

11. H2C=C=CH–CH3spsp2 sp2 sp3

3 421

12. CH2=CH–CH2–CN is allyl cyanide.

13. C–H bond is longest when carbon is sp3 hybridized.

14. Electronegativity of hybrid orbital follows sp > sp2 > sp3.

15. C6H5COOH is benzene carboxylic acid.




16. HC CHCH4sp3

CH2 2=CHsp2 sp

sp3, sp2, sp carbon have 75, 66, 50% p-character.

17.3 3CH COCH

Acetone3CH COOH

Acetic acid3CH CN

Acetonitrile3 2CH CONH

Acetamide

18. CH3 2–CH=CH–CH –C CH123456

sp3 sp2 sp2 sp3 sp sp

19. N atom have 2 & 1 lone pair of e– so it has sp2 hybridization.

20. C–C C–NO

O–

O

O–

21. sp3 hybridized orbital have 4 bonds.

22. CH3–C–CH=CH–CH–C CH1234567

CH3

CH3

sp3 sp3 sp2 sp2 sp3CH3

sp sp

23. H C31.54A° CH3 CH2

1.34A° CH2 HC1.20A°

CH 1.39A°

24. In compound having triple bond, 4 atoms are present in a line.

25.Anthracene

Tropolone

OHO

Aniline

NH2

Benzoic acid

COOH

26.sp2

H2 2C = C = CHsp sp2

27.

H

C

H

H H

H

C

H

sp3sp3

s–sp3sp3–sp3

28. Benzyl free radical is aromatic due to presence of benzene i.e. 6 p-orbital & 6 unpaired electrons.

29. HC CHCO2sp

H3 3C–CHsp sp2

H2 2C–CHsp3

CH3 2–CH –OHsp3

30. Cyclopropenyl cation is aromatic in nature.


(i)...and unsaturation, hybridization, classification of organic compounds and baeyer’s strain...

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