of NCERT Organic Chemistry - Coach Me€¦ · Haloalkane & Haloarenes 79 - 124 4. Alcohol, Phenol &...

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By

AJNISH GUPTA & BHARTI GUPTAProfessor of Organic Chemistry

Organic Chemistry

Problem Solving Strategyof NCERT

Questions

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All rights reserved. No part of this publication may bereproduced, 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 Reprint, 2014

This edition is manufactured in Indian and is autho-rized 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. 250/-

Copyright © 2014 by Ajnish Gupta

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PrefaceThe guiding principle in writing this book was to createa textbook for students- a textbook that presents thematerial in a way that they learn to solve all the ques-tions of NCERT along with the strategy to approachthe problems.In this book we mixed all our teaching experience of 10years along with theoratical and experimentalknowledge to generate a hand book for all students toreason their way to a solution rather than memorize amultitude of facts, hoping they don’t run out of memory.If you ask your teacher, senior or friend about NCERT,then they will surely say that NCERT questions are veryimportant to solve before giving board exam or any com-petitive exam as it is the basic theme for any board orcompetitive exam and nearly all the questions arederieved from NCERT only.But the problem in NCERT organic chemistry is thatthere are a lot of intermixing of concept involve in samechapter so many students get fear of it and generallyleave it by thinking that they can score good marks orrank without it, but they are fooling themself.Organic chemistry is very easy and conceptual subjectand need proper understanding of the basics andstretegy to solve the questions in corret manner.This book will prepare your right mindset for learningOrganic Chemistry. This mindset is essentially the onethat focuses you on a small number of straight forward,repeated, fundamental concepts and helps you to ap-ply them in different ways to solve the variety ofproblems you face in NCERT or other organic problems.

Ajnish Kumar Gupta & Bharti Gupta

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AcknowledgementWe are thankful to all the teachers who taught usduring the concept building session of our life speciallyDr. Nizamuddin sir, senior Chandra Vijay Rao and Dr.Vijay Pratima Mittal madam.We have written this book to remove the fever oforganic chemistry from mind of students.We particularly want to thank many wonderful andtalented students whom we have taught over the yearswho in turn taught us how to be a good teacher andhow we can help others.We want to make this book as user friendly as possible,and we will appreciate any comments that will help usto achieve this goal in future editions.Finally, this edition has been presented before you withefforts to make it errors free. Any that remain are ourresponsibility; if you find any, please let us know so theycan be corrected in future printing.

Ajnish Gupta & Bharti GuptaProfessors of Organic [email protected]

09122057123

mailto:[email protected]

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Table of ContentsUnit Topic Page. No.

1. Organic Chemistry- Some basicprinciple & techniques 01 - 46

2. Hydrocarbon 47 - 783. Haloalkane & Haloarenes 79 - 1244. Alcohol, Phenol & Ether 125 - 1725. Aldehyde, Ketone & Carboxylic acid 173 - 2246. Amines 225 - 2607. Biomolecules 261 - 2828. Polymers 283 - 3009. Chemistry in everyday life 301 - 313

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Dedicated to all those students whoare in fever of organic chemistry.

Dedication

Organic Chemistry-Some basic principles and Techniques

1Organic Chemistry is easy bywww.CoachMe.co.in

1Unit


http://www.CoachMe.co.in


Organic Chemistry is easy bywww.CoachMe.co.in

2

This unit give you an understanding of “Organic chem-istry- Some basic principles and Techniques” and cov-ers following topics:

What is Organic chemistry, Representation of Or-ganic compounds, Reason for the formation of largenumber of organic compounds, Functional groups,Homologue & Homologous series, Nature of C &H, functional groups, Saturated & unsaturated mol-ecules, Hybridization, Classification of organiccompounds, Alkyl groups, IUPAC nomenclature& Common name of Organic molecules.

Basic understanding of isomerism in organic chem-istry. Structural isomerism– Chain, Position,Function, Metamer, Tautomer. Stereoisomerism–Configurational & Conformational. Configura-tional– Geometrical & Optical isomerism.

Electronic effects- Inductive effect, Basic conceptof resonance, General cases of resonance, Electronflows in bond, How to draw resonating structures,Stabiity of resonating structrue, Mesomeric ef-fect, Hyperconjugation, Electromeric effect;Reaction intermediate-Carbocation, Carbanion,Free radicals, Carbene, Nitrene, Benzyne;Concept of Acid and Base- How to find out rela-tive Acidic & Basic strength, Scale to measureAcidic & Basic strength

Practical organic chemistry- Methods of purifica-tion of organic compounds, qualitative & quanti-tative analysis of organic compounds.

Objective

http://www.CoachMe.co.in2



Solved Example:Example 1.

How many and bonds are present in each of the followingmolecules?

(A) 3HC CCH = CHCH (B) 2 3CH = C = CHCH

Strategy.To solve such questions, always expand the structure and countthe number of & bond keeping given basic in mind.

A B A B A B

(A)H H H

Hσ = 10π = 3

(B)H H

Hσ =π =

92

H

HH

Example 2.What is the type of hybridisation of each carbon in the followingcompounds?

(A) 3CH Cl (B) 3 2(CH ) CO

(C) 3CH CN (D) 2HCONH

(E) 3CH CH = CHCN

Strategy.To find the hybridization of any atom, always count the numberof bonds & lone pairs of electrons. If




4

Sum of bond + lone pair of –e 2 sp hybridization

Sum of bond + lone pair of – 2e 3 sp hybridization

Sum of bond + lone pair of – 3e 4 sp hybridization

As carbon have no lone pair of electrons so always count thenumber of bonds in deciding the hybridization of carbonatom. There is no role of bond in deciding the hybridization of

any atom.

(A) H – C – Cl

H

H

34σ sp (B) 3 3CH –C–CH2sp 3sp3sp

O

(C) 3CH –C N3sp

(D)2H–C–NH

O

2sp

(E) 3CH –CH=CH–C N3sp 2sp 2sp

Example 3.Write the state of hybridisation of carbon in the followingcompounds and shapes of each of the molecules.

(A) 2H C = O (B) 3CH F (C) HC N

Strategy.Similar to above question, first expand the molecule & thencount the number of bond for deciding the hybridization ofcarbon.

(A) HC = O

H

2sp (B) H – C – F

H

H

3sp(C)

H–C N




As shape is the real structure of molecule which can beexplained with the help of hybridization.Generally following hybridization give following shape.

sp Linear

2sp Planner

3sp tetrahedral

So shape of A is planner, B is tetrahedral & C is linear.Example 4.

Expand each of the following condensed formulas into theircomplete structural formulas.

(A) 3 2 2 3CH CH COCH CH (B) 3 2 3 3CH CH = CH(CH ) CH

Strategy.To expand the molecule in correct form, always keep in mindthe valencies of some common atoms such as i.e. carbon alwaysform 4 bonds in neutral case. Always keep this concept for making a neutral structure.

C , N , O , S , H , F , Cl, Br , I

(A) H–C–C–C–C–C–H

H

H

H

H

H

H

H

H

O(B) H–C–C=C–C–C–C–C–H

H

H

H

H

H

H

H

H HH

H

H

Example 5.

For each of the following compounds, write a condensedformula and also their bond-line formula.

(A) 2 2 2 3 3 3HOCH CH CH CH(CH )CH(CH )CH

(B) N C – CH – C NOH




6

Strategy.

To write the condensed formula, always give 2 n(CH ) time for

more than one 2 3 n 3CH unit and (CH ) for CH unit.

For bond line formula, write the molecule in zig-zag form.

(A) 2 3 3 3 2HO(CH ) CH(CH )CH(CH ) (B) 2CH(OH)(CN)

zig-zag form.

HOOH

NC CNExample 6.

Expand each of the following bond-line formulas to show allthe atoms including carbon and hydrogen.

(A) (B)

(C) OH (D)

Strategy.In bond line formula, the number of H at each carbon = 4 – no.of visible bond.

(A)C

H

HC

C

C

C

C

HH

H H

CC

C

H H

HH

H

H

HH

HH

HH

(B) H–C–C–C–C–C–C–C–C–H

H

H

HH

HH

H

H

HH

HH

HH

HH




(C) H–C C–C–C C–H

H

H

OH

H

H

H

(D) H–C C

H

H H–C–H

H

H–C–H

H

C C–H

H

H

H H

Example 7.Structures and IUPAC names of some hydrocarbons are givenbelow. Explain why the names given in the parentheses areincorrect.

(A) 3 2 2 2 3CH – CH – CH – CH – CH – CH – CH – CH

3CH2,5,6-Trimethyloctane

[and not 3,4,7-Trimethyloctane]

3CH 3CH

(B) 3 2 2 2 3CH – CH – CH – CH – CH – CH – CH2 3CH CH 3CH

3-Ethyl-5-methylheptane[and not 5-Ethyl-3-methylheptane]

Strategy.(A) In writing the IUPAC name of organic compound, always

give minimum positions to substituents i.e. use lowestlocant for substituent.

(B) In writing the IUPAC name of organic compound onidentical position, numbering in chain starts according toalphabets of substituent. The alphabet which comes firstare given lower position.




8

Example 8.Write the IUPAC names of the compounds i-iv from their givenstructures.

(i) 3 2 2 2 2 3CH – CH – CH – CH – CH – CH – CH – CH

3CH

1 2 3 4 5 6 7 8

OH

Strategy.To write IUPAC name of any compound always use followingkeys(A) Use the concept of 2°prefix + 1°prefix + Word root

+ 1°suffix + 2°suffix in sequence.(B) Use the concept of position then alphabet then position if

functional group, multiple bond or substituents arepresent.

(C) For giving minimum position functional group thenmultiple bond then substituent.

(D) If more than one functional groups are present, then usethe concept of priority order.

(i)3 2 2 2 2 3CH –CH –CH–CH –CH –CH–CH –CH

3CH

1 2 3 4 5 6 87

OH6-Methyloctan-3-ol

(ii) 3 2 2 3CH –CH –C–CH –C–CH4

Hexane-2,4- dione

OO

2356 1

(iii) 3 2 2 2CH –C–CH –CH –CH –C–OH

5-Oxohexanoic acid1

O

2456 3

O

(iv)2HC C–CH=CH–CH=CH

Hexa-1,3-dien-5-yne123456




Example 9.Derive the structure of (i) 2-chlorohexane, (ii) Pent-4-en-2-ol,(iii) 3-Nitrocyclohexene, (iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxy-heptanal.

Strategy.To make the structure from given name, always take the helpof word root. Make the chain according to word root. Numberthem as 1, 2, 3 .... then place the substituent, multiple bond orfunctional group over it.

(i) 2

Cl

1 3 4 56 (ii)

OH

1 2 3 4 5

(iii)2NO2

3 (iv) 23

OH

(v) 6

OH

7 5 O

1234

H

Example 10.

Write the structural formula of :(A) o-Ethylanisole (B) p-Nitroaniline(C) 2,3-Dibromo-1-phenylpentane (D) 4-Eth yl-1-fluoro-2-nitrobenzene

Strategy.For disubstituted benzene, 2nd position is considered as ortho,3rd position as meta & 4th position as para with respect to firstsubstituent.

(A)

3OCH

(B)

2NO

2NH




10

(C) 1 2 3 4 5Br

BrPh

(D)2NO

F1 2

3456

Example 11.Using curved-arrow notation, show the formation of reactiveintermediates when the following covalent bonds undergoheterolytic cleavage.

(A) 3 3CH - SCH (B) 3CH - CN

(C) 3CH - Cu

Strategy.In heterolytic bond cleavage, bond breaks to acquire negativecharge over more electronegative atom & positive charge overless electronegative atom.

(A) 3 3 3 3CH – S – CH CH S – CH+

(B) 3 3CH – CN CH CN+

(C) 3 3CH – Cu CH Cu+

Example 12.Giving justification, categorise the following molecules/ions asnucleophile or electrophile:

+ ++ –- - • •• •3 2 5 3 3 3 2 2HS , BF , C H O , (CH ) N , Cl, CH - C = O , H N , NO

Strategy.To solve this question, remember the basic definition of nucleophile &electrophile.Nucleophiles- They are e– rich species with complete octet with eithernegative charge or with lone pair of electrons.




Electrophiles- They are the e– deficient species with either incompleteoctet, vacant orbital or +ve charge over it.

So nucleophiles are – –2 5 3 3 2HS , C H O , (CH ) N, H N while

electrophiles are +3 3 2BF , Cl , CH –C=O, NO

Example 13.Identify electrophilic centre in the following:

3 3 3CH CH = O, CH CN, CH I.

Strategy.Electrophilic centre is the electron deficient centre in themolecule. They arises due to difference in electronegativitybetween two atoms so electrophilic centre in above moleculesare

3 3 3CH HC = O H C C N H C – I

Example 14.Which bond is more polar in the following pairs of molecules:

(A) 3 3H C - H, H C - Br (B) 3 2 3H C - NH , H C - OH

(C) 3 3H C - OH, H C - SH

Strategy.Polarity of bond is judged on the basis of electronegativitydifference. Greater is the electronegativity difference, greateris the polarity. So(A) C–Br bond is more polar than C–H(B) C–O bond is more polar than C–N(C) C–O bond is more polar than C–S

Example 15.

In which C–C bond of 3 2 2CH CH CH Br, the inductive effect isexpected to be the least?




12

Strategy.As inductive effect is distance dependent so magnitude ofinductive effect diminishes as the number of intervening bondincreases.Hence, between 3rd carbon & hydrogen bond exert leastinductive effect.

2 2 2CH –CH –CH –BrH

least

Example 16.

Write resonance structures of -3CH COO and show the

movement of electrons by curved arrows.Strategy.

To write resonating structure of any molecule, first write thestructure of it and then put unshared electron or –ve charge onappropriate atoms, then draw arrow one at a time moving theelectron to get the other structure.

–3 3CH –C–O CH –C=O

O O–

Example 17.

Write resonance structures of 2CH = CH - CHO. Indicaterelative stability of the contributing structures.

Strategy.

Resonating structure of 2CH =CH–CHO will be

2 2 2CH =CH–C–H CH –CH=CH–C–H CH –CH=CH–C–H

O

I II III

O– O+

Relative stability of resonating structures are judged bygenerally these following points.




(A) Neutral molecules are generally more stable than charged.(B) Greater the number of bond, greater will be the stability.(C) Structure with complete octet for each atom is generally

more stable than those in which atleast one atom haveincomplete octet.

(D) Structures with negative charge on electronegative atom& positive charge on electropositive atom is more stablethan those in which electronegative atom have positivecharge & electropositive atom have negative charge.

(E) Two similar charge present neares to each other createselectronic repulsion which destabilize the molecule but twodissimilar charge present nearer to each other createselectronic attraction and stabilize the molecule.

(F) Resonating structure with linear conjugation is more stablethan structure with cross conjugation.So relative stability of resonating structures of

2CH =CH–CHO will be I > II > III.

Example 18.Explain why the following two structures, I and II cannot bethe major contributors to the real structure of 3 3CH COOCH .

3 3CH –C–OCH

I

O

3 3CH –C=OCH

II

O

Strategy.Ist structure is less contributor as it has atom with incompleteoctet while IInd structure is less contributor because of charge separationof positive & negative charge.

Example 19.

Explain why +

3 3(CH ) C is more stable than + +

3 2 3CH CH and C H isthe least stable cation.




14

Strategy.Stability of carbocation is generally decided by electronic effectsuch as resonance, hyperconjugation & inductive effect. Greateris the positive electronic effects, greater is the stability. So,

23 3 3 3(CH ) C CH CH CH9HC 3HC 0 HC

Example 20.On complete combustion, 0.246g of an organic compound gave0.198g of carbon dioxide and 0.1014g of water. Determine thepercentage composition of carbon and hydrogen in thecompound.

Strategy.% composition of C & H in compound will be

212 mass of CO 100% of C =

44 mass of compound

2 mass of water 100% of H =

18 mass of compound

12 0.198 100% of C = 21.95%

44 0.246

2 0.1014 100% of H =18 0.246

= 4.58%

Example 21.

In Dumas’ method for estimation of nitrogen, 0.3g of an organiccompound gave 50mL of nitrogen collected at 300K temperatureand 715mm pressure.

Strategy.Volume of nitrogen collected at 300K and 715 mm pressure is50 ml.Actual pressure = 715 – 15 = 700 mm




Volume of nitrogen at STP 273 700 50=300 760

= 41.9 mL

22,400 mL of N2 at STP weighs = 28g

41.9 mL of nitrogen weighs 28 41.9= g22400

% of nitrogen 28 41.9 100= 17.46%22400 0.3

Example 22.

During estimation of nitrogen present in an organic compoundby Kjeldahl’s method, the ammonia evolved from 0.5g of thecompound in Kjeldahl’s estimation of nitrogen, neutralized10mL of 1 M 2 4H SO . Find out the percentage of nitrogen inthe compound.

Strategy.

1M of 10mL 2 4 3H SO 1M of 20mL NH

1000mL of 1M 3NH contains 14g N

20mL of 1M 3NH contain 14 20 g N1000

So % of N 14 20 100 56.0%1000 0.5

Example 23.

In Carius method of estimation of halogen, 0.15g of an organiccompound gave 0.12 g of AgBr. Find out the percentage ofbromine in the compound.

Strategy.Molar mass of AgBr = 108 + 80 = 188 g/mol188 g AgBr contains 80 g Br




16

So 0.12 g AgBr contains 80 0.12 g Br188

Hence, % of Br 80 0.12 100188 0.15

= 34.04%

Example 24.

In sulphur estimation, 0.157 g of an organic compound gave0.4813 g of barium sulphate. What is the percentage of sulphurin the compound?

Strategy.Molecular mass of BaSO4 = 137 + 32 + 64 = 233 g233 g BaSO4 contains 32 g S

So, 0.4813 g BaSO4 contains 32 0.4813 g S233

% of S 32 0.4813 100233 0.157

= 42.10%




Exercise Problems:Exercise Problem 1.

What are hybridisation states of each carbon atom in thefollowing compounds?

2 3 2 3 2 2 6 6CH = C = O, CH CH = CH , (CH ) CO, CH = CHCN, C H

Strategy.To write the hybridization of any atom, always count thenumber of bond & lp of e–. IfSum of bond + lp = 2 sp hybridizationSum of bond + lp = 3 sp2 hybridizationSum of bond + lp = 4 sp3 hybridizationSo hybridization state of each carbon atom will be

2CH =C=O2sp

3 2CH –CH=CH3sp 2sp 2sp 3 3CH –C–CH

3sp 2sp

O

3sp

2CH =CH–CN2sp 2sp

2sp

2sp2sp

2sp

2sp2sp

Exercise Problem 2.

Indicate the and bonds in the following molecules :

6 6 6 12 2 2 2 2 3 2 3C H , C H , CH Cl , CH = C = CH , CH NO , HCONHCH

Strategy.

To find total number of & bonds present in any molecule,first expand the molecule and then look for single, double &triple bond. Single bond are only bond, double bond have




18

1 & 1 bond while triple bond have 1 & 1 bonds.

H

HH

HH

H HH HH

HH

HH

HHH

H

H

H

C ClClH

H

H

H

H

H

C NHO

O–

H

H

H

O

Exercise Problem 3.Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Strategy.To draw bond line formulas of any compound, first draw theskelelon of carbon taking the help of word root of IUPACnomenclature & then add substituent or functional group onappropriate position in it.

OH

Iso-propyl alcohol

O

H123

2,3-Dimethylbutanal

O

4 123567Heptan-4-one

Exercise Problem 4.Give the IUPAC names of the following compounds :

(A) (B) CN




(C) (D) BrCl

(E)O

Cl H(F) 2 2Cl CHCH OH

Strategy.To write the IUPAC name of any compound, always use(A) 2°prefix + 1°prefix + word root + 1°suffix + 2°suffix in

sequence.(B) Use the concept of position then alphabet then position if

functional group, multiple bond or substituents arepresent.

(C) For giving minimum position- Functional group- multiplebond- Substituent.

(D) If more than one functional groups are present, then usethe concept of priority order.

(A)Propylbenzene

(B)1

233-Methylpentane nitrile

45

CN

(C)

47

65321

2,5-Dimethylheptane (D)

47

65321

3-Bromo-3-chloroheptaneCl Br

(E)

O

H1233-Chloropropanal

Cl (F)2Cl–CH–CH –OH

Cl

12

2,2-DichloroethanolExercise Problem 5.

Which of the following represents the correct IUPAC name ofthe compounds concerned ?(A) 2,2-Dimethylpentane or 2-Dimethylpentane(B) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane




20

(C) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane(D) But-3-yn-1-ol or But-4-ol-1-yne.

Strategy.To solve such questions, first draw the incorrect structures fromgiven IUPAC name taking the help of word root of IUPAC &the correct it using main above concept.

(A)543

2,2-Dimethylpentane (correct)2-Dimethylpentane (incorrect)

21

If more than one substituents are present in molecule, thenindicate the position of each substituent.

(B)543

2,4,7-Trimethyloctane (correct)2,5,7-Trimethyloctane (incorrect)

21 6 78

Always place the substituents at minimum position.

(C)543

2-Chloro-4-methylpentane (correct)4-Chloro-2-methylpentane (incorrect)

21

Cl

If position of substituents are same from both end, thensubstituent with lower alphabet are given lower position.

(D)1 2 3

4HO

But-3-yn-1-ol (correct)But-4-ol-1-yne (incorrect)

If compound have functional group as well as multiple bondthen functional group is given lower position.




Exercise Problem 6.Draw formulas for the first five members of each homologousseries beginning with the following compounds.

(A) H - COOH (B) 3 3CH COCH (C) 2H - CH = CH

Strategy.(A) First five member of H–COOH will be

H–C–OHO

I 3CH –C–OH

O

II 3 2CH –CH –C–OH

O

III

3 2 2CH –CH –CH –C–OHO

IV 3 2 2 2CH –CH –CH –CH –C–OH

O

V

(B) First five member of 3 3CH COCH will be

3 3CH –C–CHO

I 3 2 3CH –C–CH –CH

O

II 3 2 2 3CH –C–CH –CH –CH

O

III

3 2 2 2 3CH –C–CH –CH –CH –CHO

IV 3 2 2 2 2 3CH –C–CH CH CH CH CH

O

V

(C) First five member of 2H –CH CH will be

2 2CH =CH 3 2CH –CH=CH 3 2 2CH –CH –CH=CHI II III

3 2 2 2CH –CH –CH –CH=CH 3 2 2 2 2CH –CH –CH –CH –CH=CHIV V




22

Exercise Problem 7.Give condensed and bond line structural formulas and identifythe functional group(s) present, if any, for :(A) 2,2,4-Trimethylpentane(B) 2-Hydroxy-1,2,3-propanetricarboxylic acid(C) Hexanedial

Strategy.The condensed & bond line formula of any compound can bedrawn by taking the help of word root.

(A)3 2 3CH –C–CH –CH–CH &

3CH

3CH

3CH

condensed form bond line formula

2,2,4-Trimethylpentane

(B)

2 2HOOC–CH –C–CH –COOH &


2-Hydroxy-1,2,3-propanetricarboxylic acid

OH

COOH

OH OO

OHOOHHO

(C)

2 2 2 2H–C–CH –CH –CH –CH –C–H


Hexanedial

O

O

OOH

H

Functional group present in (B) are alcohol & carboxylic acid& (C) is aldehyde while (A) do not have any functional group.




Exercise Problem 8.Identify the functional groups in the following compounds

(A)

CHO

OMeOH

(B)

2NH

O 2 2 2 5 2OCH CH N(C H )

(C)2CH=CH – NO

Strategy.Functional groups present in following compounds are

(A)

CHO

OMeOH Phenol

Ether

Aldehyde

(B)

2NH

Ether

Aromatic 1°amine

O 2 2 2 5 2OCH CH N(C H )

3°amine

(C) Nitro2CH=CH – NO

Ethylenic double bond




24

Exercise Problem 9.

Which of the two: - -2 2 2 3 2O NCH CH O or CH CH O is expected to

be more stable and why ?Strategy.

Stability of any molecule depends on the dispersion of electrondensity. Greater is the dispersion of charge, greater is thestability. So,

–2 2 2O N–CH –CH –O is more stable than –

3 2CH –CH –O because

2–NO group withdraws electron density by –I effect in firstwhich disperses –ve charge while –CH3 group donates electrondensity by +I effect which intensity the negative charge.

–2 2 2O N CH CH O –

3 2CH CH O(more stable) (less stable)

Exercise Problem 10.

Explain why alkyl groups act as electron donors when attachedto a system.

Strategy.Alkyl groups acts as electron donors when attached to a bonded system of C=C bond because of hyperconjugation. Inthis electronic effect C–H sigma bond overlap with adjacent bond.

2H – C – CH CH= 2H – C CH – CH=

H

H

H

H

+

(Propene)


Draw the resonance structures for the following compounds.Show the electron shift using curved-arrow notation.

(A) 6 5C H OH (B) 6 5 2C H NO




(C) 3CH CH = CHCHO (D) 6 5C H - CHO

(E) 26 5C H - CH (F)3 2CH CH = CH CH

Strategy.To write resonating structure of any molecule, first write thestructure of it and then put unshared electron or charge ifpresent on appropriate atom & then draw arrow one at a timeby moving the electrons to get other structures.

(A)

OH.. OH OH OH OH

(B)N

O– ON

O–

NO– OO–

NO– O–

NO– O–

(C)3 3CH – CH CH – C – H CH – CH – CH C – H= =

O O–

(D)

O

C–H C–H

O–

C–H

O–

C–H

O– O

C–H

(E)2CH 2CH 2CH 2CH 2CH

(F)3 2 3 2CH – CH CH – CH CH – CH – CH CH= =




26

Exercise Problem 12.What are electrophiles and nucleophiles ? Explain withexamples.

Strategy.Electrophile- They are the electron deficient species which havecapacity to accept a pair of e–. This electron deficiency may beof seen as(A) Positive charge such as H +, Cl+, Br+, NO2

+ , NO+, R+, RCO+

etc.

(B) Incomplete octet such as 3 3BF , AlCl .

(C) Vacant orbital- Such as 3 4 5FeCl , SiCl , SbCl etc.

Nucleophile- They are electron rich species which havecapability to donate a pair of e–. They have either –ve charge orhave lone pair of electrons with complete octet.

(A) Negative charge- Such as 2X , OH , NH , O R etc.

(B) Lone pair of e–- Such as 3 2 3NH , H O, CH OH etc.

Exercise Problem 13.Identify the reagents shown in bold in the following equationsas nucleophiles or electrophiles:

(A) – -3 3 2CH COOH +HO CH COO + H O

(B) –

3 3 3 2CH COCH +CN (CH ) C(CN)(OH)

(C) +

6 5 3 6 5 3C H + CH CO C H COCH

Strategy.As stated above nucleophiles are electron rich species witheither –ve charge or with lone pair of electrons with completeoctct, while electrophiles are electron deficient species witheither positive charge, vacant orbital or incomplete octct. So,




(A) –3CH COOH OH+ –

3 2CH COO H O+(Nucleophile but acts as a base)

(B) 3 3CH –C–CH + CN 3 3CH –C–CH

(Nucleophile)

3 3CH –C–CH

CN

OH

CN

OH+

O –

(C) 3CH –CO 3C–CH

+

(Electrophile)

O

3C–CH

O

H

Exercise Problem 14.Classify the following reactions in one of the reaction typestudied in this unit.

(A) –-3 2 3 2CH CH Br + HS CH CH SH+ Br

(B) 33 2 2 3 2(CH ) C = CH + HCl (CH ) ClC – CH

(C) –-3 2 2 2 2CH CH Br + HO CH = CH + H O + Br

(D) 2 3 2CH H O3 3 2 3 2 2(CH ) C - CH OH + HBr (CH ) CBrCH CH

Strategy.Organic reactions are mainly of 4 types-(A) Substitution reaction- Here generally sigma bonds are

broken & corresponding new sigma bonds are formed.

+ +

(B) Addition reaction- Here generally pi bonds are broken &corresponding 2 new sigma bonds are formed.




28

D

+

C

(C) Elimination reaction- Here generally sigma bonds arebroken & corresponding new pi bonds are formed.

DC

+

(D) Rearrangement reaction- Here generally attachment ofatom changes in carbon skeleton of substrate.

A–B–C A–C–BOn the basis of first attack of reagent, substitution reaction isof further 3 types.(A) Nucleophilic substitution reaction(B) Electrophilic substitution reaction(C) Free radical substitution reactionOn the basis of first attack of reagent, addition reaction is alsoof further 3 types.(A) Nucleophilic addition(B) Electrophilic addition(C) Free radical addition

So,

(A) 3 2 3 2CH – CH – Br HS CH – CH – SH Br

Here bond is broken & new bond is formed by attack of anucleophile so reaction is nucleophilic substitution reaction.

(B) 3 2 3 2

3

3

| | |

|

CH Cl HCH – C CH HCl CH – C – C H

CH

Here bonds are broken & corresponding 2 new sigma bonds




are formed by attack of an electrophile so reaction is electrophilicaddition reaction.

(C) 3 2 2 2 2CH – CH – Br O H CH CH H O Br

In this reaction 2 sigma bonds are broken & corresponding a pibond is formed, so reaction is elimination reaction.

(D) 3 2 3 2 3 2

3 3

3

| |

| |

CH CH

CH – C – CH OH HBr CH – C –CH CH H OBrCH

Here again bond is broken and corresponding new bondis formed by attack of a nucleophile & reaction followsnucleophilic substitution reaction, but as position of methylgroup change in substrate to product, so rearrangementreaction is also seen along with nucleophilic substitutionreaction.


What is the relationship between the members of followingpairs of structures ? Are they structural or geometrical isomersor resonance contributors?

(A)O O

(B) C C=D

D H

HC C=

DD

HH

(C) H–C–OH

O

H–C–OH

OH

Strategy.To find out structural & stereo isomers, always think for theIUPAC name. If IUPAC of isomers are different they arestreuctural isomers but if they have same IUPAC name, thenthey are stereoisomers.To find out resonating structure, always look for the positionof bonding electron & non-bonding electron as position ofatoms remains same in all resonating structures.




30

(A)O

Pentan-3-one

O

Pentan-2-one

As they have different IUPAC name, so are structural isomers.

(B)C C=

D

D H

HTrans-1,2-Deuteroethene

C C=DD

HHCis-1,2-Deuteroethene

As they have same IUPAC name but the spacial arrangementof deuterium along C=C double bond is different, so aregeometrical isomers.

(C)H–C–OH

OH–C–OH

OH

As position of atoms are same but have difference in positionof electrons, so are resonating structures.


For the following bond cleavages, use curved-arrows to showthe electron flow and classify each as homolysis or heterolysis.Identify reactive intermediate produced as free radical,carbocation and carbanion.

(A) 3 3 3 3CH O – OCH CH O+ OCH

(B) –+ OHO 2+ H OO–

(C)–+ Br

Br+

(D)E

++ E+




Strategy.The following reaction follows

(A) 3 3 3 3CH – O O – CH CH – O O – CH+... .

Here sigma bond is broken homolytically to form free radicals.

(B)

O

2H O+O

+ OH

H

Here base abstract proton heterolytically to form carbanion asa reaction intermediate.

(C)+ Br

Br

Here C–Br bond is broken heterolytically to form carbocationas a reaction intermediate.

(D)E

+ E+

Here benzene attacks over an electrophile heterolytically to formcarbocation as a reaction intermediate.

Exercise Problem 17.Explain the terms inductive and Electromeric effects. Whichelectron displacement effect explains the following correctorders of acidity of the carboxylic acids?

(A) 3 2 2Cl CCOOH > Cl CHCOOH > ClCH COOH

(B) 3 2 3 2 3 3CH CH COOH > (CH ) CHCOOH > (CH ) C.COOH

Strategy.Inductive effect- It is a permanent displacement of electrondensity along sigma bond and transmit along the carbon chain




32

when two atom have electronegativity difference. Thistransmission of polarity along bond is called as Inductiveeffect.

C C C Fδδδ+ δ–δδ+ δ+

C C C Oδδδ–

–δδ– δ–

Electromeric effect- It is a temporary displacement of electron density in carbon carbon multiple bond when a reagentattacks over the substrate.

2 2CH =CH + Br Brδ–δ–δ–δ–

Acidic character of any molecule depends on the stability ofconjugate base, so in both the cases inductive effect explainsthe relative stability of conjugate base & acidic strength follows

(A) 3 2 2Cl C–COOH > Cl CH–COOH > ClCH –COOH

The stability of conjugate base follows

– – –

2Cl C C O > Cl CH C O > Cl CH C OOCl

Cl

OCl O

(3-I of Cl) (2-I of Cl) (1-I of Cl)

(B) 3 2 3 2 3 3CH CH COOH > (CH ) CHCOOH > (CH ) CCOOH

The stability of conjugate base follows

– – –3 2 3 3CH CH C O > CH CH C O > CH C C O

O O O

3CH

3CH

3CH(Repulsion with

2 CH group)3

(Repulsion with3 CH group)3

(Repulsion with1 CH group)3




Exercise Problem 18.Give a brief description of the principles of the followingtechniques taking an example in each case.(A) Crystallisation (B) Distillation (C) Chromatography

Strategy.(A) Crystallisation- In this process we convert an impure

compound into a pure crystals. This process is based onthe difference in the solubility of the compound and theimpurities in a suitable solvent. Here the impure compoundis dissolved in a solvent in which it is sparingly soluble atroom temperature but appreciably soluble at highertemperature.Now solution is concentrated to get a nearly saturatedsolution at higher temperature. On cooling the solution,pure compound in the form of crystals separates.The best example of crystallisation is iodoformcrystallisation with alcohol & benzoic acid mixed withnaphthalene be purified by not water.

(B) Distillation- Distillation involves the process of heating aliquid to convert it into the vapour and then condensingthe vapour to get back to the liquid.This process of separation is applied only for thepurification of liquid which boil without decompositionat atmospheric pressure and contain non-volatileimpurities. So mixture of two liquids having sufficientdifference in their boiling points can be separated andpurified by this process.The best example are the separation of chloroform (bp334K) & aniline (bp 457K) can be done by it.

(C) Chromatography- It is a technique for the separation,purification & identification of constituents of mixtures.Chromatography is based on the principle of selectiveadsorption of components of a mixture between two phasei.e. a stationary phase and a moving phase. Here the




34

stationary phase can be a solid or a liquid while movingphase is generally liquid or gas.

Exercise Problem 19.Describe the method, which can be used to separate twocompounds with different solubilities in a solvent S.

Strategy.Two compounds with different solubilities is a solvent S can beseparated by the method called as Fractional crystallisation.In this process, a hot saturated solution of these two compoundsis allowed to cool, the less soluble compound crystalises outearlier than the more soluble compound. Now the crystals areseparated from the mother liquor & the mother liquor is againconcentrated and allowed to cool, then the crystal of the secondcompound is obtained.

Exercise Problem 20.What is the difference between distillation, distillation underreduced pressure and steam distillation ?

Strategy.This question is based on separation of compounds havingdifference in boiling point. Distillation simply involves theprocess of heating liquid to convert it into the vapours and thecondensation of the vapours to get back the liquid. Simple distillation is applied only for purification of those

liquids which boil without decomposition and containsnon-volatile impurities.

Distillation under reduced pressure is applied for thoseorganic liquids which decomposes at a temperature belowtheir boiling point.

Steam distillation is simply co-distillation with water. Thistechnique is used to separate substance which are steamvolatile and are immiscible with water.

Exercise Problem 21.Discuss the chemistry of Lassaigne’s test.




Strategy.Chemistry of Lassaigne’s test- Organic compounds arecovalently bonded so they are first converted to ions by fusingwith sodium at high temperature, then only we can test thepresence of perticular element in compound.In this process we first prepare sodium extract & then test thepresence of elements.Preparation of sodium extract- Here organic compounds arefirst fused with sodium and heated upto red hot in ignitiontube. Then we broke the red hot ignition tube in water & filterthe solution. The filterate obtained on this is called as sodiumextract. Now this sodium extract can be used to detect thepresence of N, S, X in organic compound.Test of Nitrogen- Here sodium extract is treated with ferroussulphate. If prussian blue colour is obtained, then compoundhave presence of N.

Na C N NaCN

4 2 2 4FeSO 2NaOH Fe(OH) Na SOFerrous hydroxide

(green)

2 4 6Fe(OH) 6NaCN Na [Fe(CN) ] 2 NaOHSodium ferrocyanide

On heating some Fe2+ ions are oxidized to Fe3+ ion3

4 6 4 6 34Fe 3Na [Fe(CN) ] Fe [Fe(CN) ] 12 NaFerric ferrocyanide

(Prussian blue)

Test of Nitrogen & Sulphur if present together.

Na C N S NaSCNSodium thiocyanate

33Fe NaSCN Fe(SCN) 3Na

Blook red colour




36

Test of sulphur- Here sodium extract is treated with leadacetate. If black ppt is obtained, then compound have presenceof S.

2Na S Na S

2 3 2 3Na S (CH COO) Pb PbS 2 CH COONa(Black ppt)

There is one more test of sulphur. Here sodium extract is treatedwith sodium nitroprusside solution. If a violet colour ppt isobtained, then compound have presence of S.

2 2 5 4 5Na S Na [Fe(NO)(CN) ] Na [Fe(CN) NOS]Sodium thionitroprusside

(violet ppt)

Test of halogen- Before testing halogens, sodium extract isboiled with conc. HNO3 to decompose sodium cyanide orsodium sulphide in the sodium extract, otherwise these ionswould interfere with silver nitrate test for halogen.

3 3NaCN HNO NaNO HCN

2 3 3 2Na S HNO NaNO H S

Now add AgNO3 to sodium extract. If white precipitate isobtained which is soluble in ammonium hydroxide but insolublein HNO3, then it indicates the presence of –Cl.

3 3NaCl AgNO AgCl NaNOWhite ppt

If pole yellow ppt is obtained which is sparingly soluble inammonium hydroxide, then it indicate the presence of –Br.

3 3NaBr AgNO AgBr NaNOPole yellow ppt

If yellow ppt is obtained which is insoluble in ammoniumhydroxide, then it indicates the presence of iodine.

3 3NaI AgNO AgI NaNOyellow ppt




Exercise Problem 22.Differentiate between the principle of estimation of nitrogen inan organic compound by (i) Dumas method and (ii) Kjeldahl’smethod.

Strategy.Dumas method- Here a known mass of the nitrogen containingorganic compound is heated with excess of CuO in anatmosphere of CO2, then nitrogen of organic compound isconverted into N2 gas. So the volume of N2 gas thus obtained isconverted into STP and the % of nitrogen can be determined.

228 Volume of N at STP% N 10022400 Mass of the substance taken

Kjeldahl’s method- Here a known mass of the nitrogencontaining organic compound is heated with excess of conc.

2 4H SO in presence of 4CuSO in Kjeldahl’s flask, then nitrogenof organic compound is converted into 4 2 4(NH ) SO . Thisammonium sulphate is then boiled with excess of NaOHsolution to liberate NH3 gas which is absorbed in a known excessof standard solution of 2 4H SO or HCl.

The volume of acid left after absorption of ammonia is estimatedby titration against a standard alkaline solution. From thevolume of the acid used, the percentage of nitrogen isdetermined by applying the mathermatical equation.

1.4 × Molarity of acid × basicity of acid × vol. of acid% N

Mass of substance taken

Exercise Problem 23.Discuss the principle of estimation of halogens, sulphur andphosphorus present in an organic compound.

Strategy.Principle of estimation of halogens- Here a known mass ofan organic compound is heated with fuming HNO3 in the




38

presence of AgNO3 in Carius tube in a furnace. By doing so,carbon & hydrogen present in the organic compounds areoxidised to CO2 and H2O respectively and halogen forms theprecipitate of AgX. Now this AgX is filtered, washed, dried &weighed. So,

Atomic mass of X x mass of AgX × 100% of halogen =

Molecular mass of AgX × mass of compound

Principle of estimation of sulphur- Here a known mass of anorganic compound is heated with Carius tube with sodiumperoxide on fuming HNO3. By doing so, sulphur is oxidised toH2SO4 and precipitated as BaSO4 by adding excess of BaCl2solution in water. Now this BaSO4 is filtered, washed, dried &weighed.

432 × Mass of BaSO 100% of S = 233 × Mass of compound

Principle of estimation of phosphorus- Here a known massof an organic compounds is heated with fuming HNO3. Bydoing so, phosphorus is oxidised to phosphoric acid andprecipitated as ammonium phosphomolybdate

4 3 4 3(NH ) PO .12 MoO by adding NH3 and ammonium

phosphomolybdate. Now 4 3 4 3(NH ) PO .12 MoO is filtered,washed, dried & weighed.

31 × Mass of ammonium molybdate 100% of P =

1877 × Mass of organic compound

Exercise Problem 24.Explain the principle of paper chromatography.

Strategy.Principle of paper chromatography- It is a type of partitionchromatography & based on principle of partition i.e. basedon continuous differential distribution of the variouscomponents of the mixture between the stationary and the




mobile phases.In this process, the solution of the mixture which is to beseparated is applied as a small spot at the base ofchromatography paper nearly 2cm above one end of the paperstrip. Now this paper is suspended in a suitable solvent. Thissolvent acts as mobile phase due to which solvent rises up thepaper.After some time the spots of the separated coloured compoundsare visible at different heights from the position of inital spoton the chromatogram.The coloured components of a mixture are identified byRetardation factor, Rf value, which is fix value for everycomponent.

.

.XY

solvent frontspot

base lineObservation on paper chromatograph

fDistance travelled by the compound (X)

R value = Distance travelled by the solvent (Y)


Why is nitric acid added to sodium extract before adding silvernitrate for testing halogens?

Strategy.Conc. HNO3 is added to sodium extract before adding silvernitrate before testing halogen is to remove NaCN or Na2S ifpresent in sodium extract, otherwise they give white ppt &black ppt with AgNO3 which creats confusion.

3 3NaCN HNO NaNO HCN

2 3 3 2Na S HNO NaNO H S




40

otherwise

3 3NaCN AgNO AgCN NaNOWhite ppt

2 3 2 3Na S AgNO Ag S NaNO(Black ppt)

Exercise Problem 26.Explain the reason for the fusion of an organic compound withmetallic sodium for testing nitrogen, sulphur and halogens.

Strategy.Organic compounds are covalently bonded so the detection ofelement is not easy so, they are first fused with sodium to convertthem into ion, which we get by easy chemical test.

Na C N NaCN

2Na S Na S

Na X NaX Exercise Problem 27.

Name a suitable technique of separation of the componentsfrom a mixture of calcium sulphate and camphor.

Strategy.Suitable technique used for separation of components frommixture of calcium sulphate & comphor is sublimation.As camphor is sublimable but CaSO4 is not so sublimation ofthe mixture gives camphor on the side of funnel while CaSO4is left in the china dish.

Exercise Problem 28.Explain, why an organic liquid vaporises at a temperature belowits boiling point in its steam distillation ?

Strategy.In steam distillation process, the mixture consisting of theorganic liquid and water boils when the sum of the vapour




pressure of the organic liquid and that of water becomes equalto the atmospheric pressure.Atmospheric pressure = Vapour pressure of the liquid + Vapour

pressure of the water.Since, vapour pressure of the liquid is lower than atmosphericpressure, the organic liquid vaporises at lower temperature thanits boiling point.


Will 4CCl give white precipitate of AgCl on heating it withsilver nitrate? Give reason for your answer.

Strategy.No, CCl4 will not give white precipitate of AgCl on heatingwith AgNO3 because CCl4 is covalently bonded compound andwill not ionize to give Cl– ion required for the formation ofAgCl as precipitate.

Exercise Problem 30.Why is a solution of potassium hydroxide used to absorb carbondioxide evolved during the estimation of carbon present in anorganic compound?

Strategy.As carbon dioxide is slightly acidic in nature, therefore it willreact with strong base like KOH to form K2CO3 & from theweight of CO2 obtained, % of carbon in the organic compoundcan be calculated.

2 2 3 2KOH CO K CO H O

212 × weight of CO formed% of C = 100

Weight of substance taken

Exercise Problem 31.Why is it necessary to use acetic acid and not sulphuric acidfor acidification of sodium extract for testing sulphur by leadacetate test?




42

Strategy.When H2SO4 is used for acidification of sodium extract fortesting sulphur by lead acetate test, then lead acetate itself reactwith H2SO4 to form white ppt of lead sulphate.

3 2 2 4 4 3(CH COO) Pb H SO PbSO CH COOHWhite ppt

Hence, this white ppt of PbSO4 will interfere with the followingtest of sulphur.

3 2 3(CH COO) Pb NaS PbS CH COONaBlack ppt

However, if acetic acid is used, it does not react with lead acetate, so itwill not interfere in the test.


An organic compound contains 69% carbon and 4.8%hydrogen, the remainder being oxygen. Calculate the massesof carbon dioxide and water produced when 0.20 g of thissubstance is subjected to complete combustion.

Strategy.Calculation of mass can be done by following formula.

212 × Mass of CO formed% of C = 10044 × mass of substance taken

212 × Mass of CO formed69 = 1000.2

So, mass of CO2 formed = 0.506 gram.

22 × Mass of H O formed% of H = 100Mass of substance taken

22 × Mass of H O formed4.8 = 10018 × 0.2

So, mass of H2O formed = 0.0864 g.




Exercise Problem 33.A sample of 0.50 g of an organic compound was treatedaccording to Kjeldahl’s method. The ammonia evolved wasabsorbed in 50ml of 0.5 M 2 4H SO . The residual acid required60 mL of 0.5 M solution of NaOH for neutralisation. Find thepercentage composition of nitrogen in the compound.

Strategy.Volume of H2SO4 taken = 50 mL of 0.5 M H2SO4

= 25 mL of 1.0 M H2SO4

Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH = 30 mL of 1.0 M NaOH

As, 2 4 2 4 2H SO 2NaOH Na SO 2H O

So, 1 mole of H2SO4 = 2 mole of NaOHHence, 30 mL of 1.0 M NaOH = 15 mL of 1.0 M H2SO4

Volume of acid used by ammonia = 25–15 = 10 mL

11.4 × N × Vol. of acid used% of Nitrogen =

Mass of organic compound

1.4 × 2 ×10% of Nitrogen = 56.00.5


0.3780 g of an organic chloro compound gave 0.5740 g of silverchloride in Carius estimation. Calculate the percentage ofchlorine present in the compound.

Strategy.Mass of organic compound taken = 0.3780 g.Mass of AgCl formed = 0.5740 g.




44

Mass of AgCl formed35.5% of Cl = × × 100143.5 Mass of substance taken

35.5 0.5740= × × 100 = 37.566%

143.5 0.3780


In the estimation of sulphur by Carius method, 0.468 g of anorganic sulphur compound afforded 0.668 g of bariumsulphate. Find out the percentage of sulphur in the givencompound.

Strategy.Mass of organic compound taken = 0.468 g.Mass of BaSO4 formed = 0.668 g.

432 Mass of BaSO formed% of S = × × 100233 Mass of substance taken

32 0.668= × × 100 = 19.60%

233 0.468


In the organic compound 2 2 2CH = CH - CH - CH - C CH, thepair of hydridised orbitals involved in the formation of :

2 3C - C bond is:

(A) 2sp - sp (B) 3sp - sp

(C) 2 3sp - sp (D) 3 3sp - sp

Strategy.To find out hybridised orbital first write hybridization of eachcarbon & then report your answer.So, hybridised orbitals involved in the formation of C2–C3 bondis sp2–sp3.




Exercise Problem 37.In the Lassaigne’s test of nitrogen in an organic compound,the Prussian blue colour is obtained due to the formation of :

(A) 4 6Na [Fe(CN) ] (B) 4 6 3Fe [Fe(CN) ]

(C) 2 6Fe [Fe(CN) ] (D) 3 6 4Fe [Fe(CN) ]

Strategy.In the Lassaigne’s test for nitrogen in an organic compound,the Prussian blue colour is obtained due to formation of

4 6 3Fe [Fe(CN) ] .

4 2 2 4FeSO 2NaOH Fe(OH) Na SO

2 4 6Fe(OH) 6NaCN Na [Fe(CN) ] 2 NaOH

34 6 4 6 34Fe 3Na [Fe(CN) ] Fe [Fe(CN) ] 12Na

Exercise Problem 38.Which of the following carbocation is most stable ?

(A) +

3 3 2(CH ) C.CH (B) 3 3(CH ) C

(C) +

3 2 2CH CH CH (D) +

3 2 3CH CH CH CH

Strategy.Greater is the number of hyperconjugating structure, greateris the stability of carbocation. So 3 3(CH ) C is most stablecarbocation.

3 2CH –C–CH

3CH

3CH

(0 HC)

3CH –C

3CH

3CH

(9 HC)




46

3 2 2CH –CH –CH(2 HC)

3 2 3CH –CH–CH –CH(5 HC)

Exercise Problem 39.The best and latest technique for isolation, purification andseparation of organic compounds is :(A) Crystallisation (B) Distillation(C) Sublimation (D) Chromatography

Strategy.The best & latest technique for isolation, purification &separation of organic compound is chromatography.

Exercise Problem 40.The reaction :

3 2 3 2CH CH I +KOH(aq) CH CH OH +KI

is classified as :(A) electrophilic substitution (B) nucleophilic substitution(C) elimination (D) addition

Strategy.The reaction is classified as nucleophilic substitution reactionas nucleophile (OH–) attacks over carbon directly attached to Ito replace it with itself.

3 2 3 2CH –CH –I CH –CH –OH + KI

HO


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