Preview Fundamentals of Organic Chemistry for the JEE Vol II (Main and Advanced) by Ananya Ganguly (2016)

Preview Fundamentals of Organic Chemistry for the JEE Vol II (Main and Advanced) by Ananya Ganguly (2016) Preview Fundamentals of Organic Chemistry for the JEE Vol II (Main and Advanced) by Ananya Ganguly (2016) Preview Fundamentals of Organic Chemistry for the JEE Vol II (Main and Advanced) by Ananya Ganguly (2016) Preview Fundamentals of Organic Chemistry for the JEE Vol II (Main and Advanced) by Ananya Ganguly (2016)

for the JEE (Main and Advanced)

ORGANIC CHEMISTRY

ORGANIC

CHEMISTRY

ANANYA GANGULY

Cover images: mathagraphics Shutterstock

This is the signature compilation of the class-tested notes of Ananya Ganguly, an iconic chemistry

coach This is different from other books due to the author’s unique teaching methodology and

authoritative and innovative approach in explaining the concepts The concepts are presented in a

structured way adopting student-friendly approach along with their applications Spread over two

volumes these books will act as a one-stop solution for the students to thoroughly revise organic

chemistry for JEE (Main and Advanced).

Salient Features

 Structured and systematic approach in explaining the concepts

 Class-tested notes with unique methodology and authoritative approach

Fundamentals of Organic Chemistry

Fundamentals of Organic Chemistry

for the JEE

(Main and Advanced) Volume II

Ananya Ganguly

Copyright © 2016 Pearson India Education Services Pvt Ltd

Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128, formerly known as Tutor­

Vista Global Pvt Ltd, licensee of Pearson Education in South Asia

No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent

This eBook may or may not include all assets that were part of the print version The publisher reserves the right to

remove any material in this eBook at any time

ISBN 978­93­325­4696-7

eISBN 978­93­325­8200­2

Head Office: A­8 (A), 7th Floor, Knowledge Boulevard, Sector 62, Noida 201 309, Uttar Pradesh, India

Registered Office: 4th Floor, Software Block, Elnet Software City, TS­140, Block 2 & 9, Rajiv Gandhi Salai, Taramani,

Chennai 600 113, Tamil Nadu, India

Fax: 080­30461003, Phone: 080­30461060

www.pearson.co.in, Email: companysecretary.india@pearson.com

Editor—Acquisitions: Jitendra Gaur

Editor—Production: G Sharmilee

Copyright © 2015 Pearson India Education Services Pvt Ltd

Copyright © 2012, 2013 Pearson India Education Services Pvt Ltd

This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or

otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in

which it is published and without a similar condition including this condition being imposed on the subsequent purchaser

and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in

or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying,

recording or otherwise), without the prior written permission of both the copyright owner and the publisher of this book

ISBN 978-93-325-4695-0

First Impression

Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128, formerly known as

TutorVista Global Pvt Ltd, licensee of Pearson Education in South Asia

Head Office: A-8(A), 7th Floor, Knowledge Boulevard, Sector 62, Noida 201 309, Uttar Pradesh, India

Registered Office: Module G4, Ground Floor, Elnet Software City, TS-140, Blocks 2 & 9, Rajiv Gandhi Salai,

Taramani, Chennai 600 113, Tamil Nadu, India

Fax: 080-30461003, Phone: 080-30461060

www.pearson.co.in, Email: companysecretary.india@pearson.com

Compositor: Mukesh Technologies Pvt Ltd.

Printed in India at

To my son Ayushman

Preparation of Alkyl Halide 1.6Physical Properties 1.18Aliphatic Nucleophilic Substitution Reaction 1.20The Variables in Nucleophilic Substitution 1.25The Nucleophile 1.28The Site of Substitution 1.38Solvent Effects 1.45Neighbouring Group Participation 1.64 Elimination Reactions 1.69Substrate Structure for E1 1.75Substitution and Elimination 1.85Organometallic Compounds 1.96Grignard Reagents 1.97Dihalogen Derivatives 1.106Trihalogen Derivatives 1.108Unsaturated Halides 1.111

C hapter 3 aLDehYDe aND KetONe 3.1–3.183

Nomenclature 3.1

Structure of a Carbonyl Group 3.5Methods of Preparation 3.5Physical Properties 3.20Reactions of Aldehyde and Ketones 3.23Carbon as Nucleophiles 3.27Oxygen as Nucleophile 3.30Sulphur as Nucleophile 3.35Nitrogen as Nucleophile 3.36Beckmann Rearrangement 3.39Reactions with Phosphorus and Sulphur Halide 3.47Aldol Condensations 3.53Cannizzaro Reaction 3.61Reaction of a, b-Unsaturated Aldehyde or Ketones 3.66Claisen Condensation 3.82Reformatsky Reaction 3.86Knoevenagel Reaction 3.88Benzilic Acid Rearrangement 3.92Darzen Condensation–Aromatic Aldehyde and Ketone 3.95Oxidation Reaction 3.96Baeyer-Villiger Oxidation 3.105Oppenauer Oxidation 3.112Reduction Reaction 3.113Clemmensen Reduction 3.113Wolff-Kishner Reduction 3.117Meerwein-Ponndorf-Verley Reduction 3.119Wittig Reaction 3.121Perkin Reaction 3.126Benzoin Condensation 3.128

C hapter 5 NItrOGeN CONtaINING COMpOUNDS 5.1–5.94

Diazonium Salts 5.33Nitro Compounds 5.45Cyanide or Nitriles (–C ≡ N) 5.49 Isocyanides 5.53

Urea (Carbamide) 5.55Aryl Nitro Compounds 5.58

C hapter 6 CarBOhYDrateS, aMINO aCIDS aND pOLYMerS 6.1–6.113

Classification 6.2D(+) Glucose, Dextrose (or) Grape Sugar 6.5Cyclic Structure of Glucose 6.6Reactions of Monosaccharides with Phenylhydrazine: Osazones 6.11Reducing and Nonreducing Sugars 6.12Glycoside Formation 6.13Periodate Oxidations: Oxidative Cleavage of Polyhydroxy Compounds 6.16Preparation of Sugar 6.19Action of Alkali 6.23Comparison of Glucose and Fructose 6.24The Anomeric Effect 6.25 Disaccharides 6.25

Polysaccharides 6.27

Amino Acids 6.30

Some Important Polymers 6.66

This page is intentionally left blankEditor—Acquisitions: Jitendra Gaur

Editor—Production: G Sharmilee

Copyright © 2015 Pearson India Education Services Pvt Ltd

Copyright © 2012, 2013 Pearson India Education Services Pvt Ltd

This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or

otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in

which it is published and without a similar condition including this condition being imposed on the subsequent purchaser

and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in

or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying,

recording or otherwise), without the prior written permission of both the copyright owner and the publisher of this book

ISBN 978-93-325-4695-0

First Impression

Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128, formerly known as

TutorVista Global Pvt Ltd, licensee of Pearson Education in South Asia

Head Office: A-8(A), 7th Floor, Knowledge Boulevard, Sector 62, Noida 201 309, Uttar Pradesh, India

Registered Office: Module G4, Ground Floor, Elnet Software City, TS-140, Blocks 2 & 9, Rajiv Gandhi Salai,

Taramani, Chennai 600 113, Tamil Nadu, India

Fax: 080-30461003, Phone: 080-30461060

www.pearson.co.in, Email: companysecretary.india@pearson.com

Compositor: Mukesh Technologies Pvt Ltd.

Printed in India at

To my son Ayushman

I am indeed very delighted to present the book Fundamentals of Organic Chemistry for the JEE (Main and Advanced)

Volume II to the readers with elaborated concepts along with more solved and unsolved problems.

In all competitive examinations, there are several vital areas that a candidate needs to master—organic chemistry being

one of them There are various books in the market on this topic having different approaches; however this book provides

simple, shortcut methods and time-saving tactics which are helpful during the examination to the students It will also help

you to gain confi dence through the right approach to a particular questions rather than attempting number of questions

This book would be extremely useful for the students who enroll for examinations like JEE (Main and Advanced) and for

other engineering entrance examinations

In order to bridge the gap between theory and practical, each concept is explained in detail, in an easy-to-understand

manner supported with numerous worked-out examples and practical These will stimulate thought and facilitate advanced

learning

One of the important factors that contribute to the success of a book is the way it has been developed This book refl ects

my experience and understanding of the requirements of the students The methods and approaches discussed in this book

are tried and tested modes of instruction in the class

Fundamentals of Organic Chemistry, Volume II cover topics like, Alkyl Halides and Aryl Halides, Alcohol, Phenol and

Ether, Aldehyde and Ketone, Carboxylic Acids and Derivatives, Nitrogen Containing Compounds, Carbohydrates, Amino

Acids and Polymers Volume I covers, IUPAC Nomenclature, General Organic Chemistry, Hydrocarbon, Electrophilic

Aromatic Substitution, and Analysis of Organic Compounds

I am sure that readers will appreciate this book and will fi nd this book very useful to prepare for various examinations

Your comments and suggestions would be very useful in improving the subsequent editions of this book

Although we have taken utmost care to prepare the manuscript and checking subsequent proofs, there may be a

possibility of some errors creeping inside the book We will welcome suggestions for further improvement of the book

Please mail us your suggestions on: chemistrycoach1@gmail.com

Ananya Ganguly

Fundamentals of Organic Chemistry, Volume II is the result of encouragement that I received from my students, who

insisted that, my knowledge and experience should benefi t a wider audience

I would like to thank my friends who have, over the years, been my support and strength Writing this book has been a

long but fulfi lling journey and I am fortunate to be assisted by a talented team of editors

I am indebted to my family for keeping me motivated during all stages of the project I always feel a divine power

sup-porting my efforts when my family is around They actually made me work harder on the project

I extend my sincere thanks to Pearson editorial team for their constant encouragement and support during the

publica-tion of this book

Acknowledgments

Alkyl Halides and

Aryl Halides

Introduction

The replacement of hydrogen atom(s) in a hydrocarbon, aliphatic or aromatic, by halogen atom(s) results in the formation

of an alkyl halide (haloalkane) and an aryl halide (haloarene), respectively

Haloalkanes contain halogen atom(s) attached to the sp3 hybridized carbon atom of an alkyl group, whereas haloarenes contain halogen atom(s) attached to sp2 hybridized carbon atom(s) of an aryl group

The general formula of saturated mono substituted alkyl halide is CnH2n+1X, where X is a halogen atom Alkyl halides are usually represented by R – X where R is an alkyl group



 CLASSIFICATION

Halides can be classified depending on the nature of carbon to which halogen is attached like alkyl or aryl or can be sified depending on the number of halogen, as di, tri, tetra halides

clas-According to the nature of halogen, they can be called as fluorides, chlorides, bromides or iodides

Alkyl halides can be classified as methyl halide, primary alkyl halide, secondary alkyl halide (2°) and tertiary alkyl halide (3°), according to the number of other carbon atoms attached to the carbon bearing the halogen atom

0HWK\OKDOLGH

;+

+

+ ;

++5

;+

5 ;

5

5

3ULPDU\DON\O KDOLGHƒ

6HFRQGDU\DON\O

R, R′ and R″ may be the same or different

HJ%HQ]\OLFKDOLGH

KDORJHQDWRPLVERQGHGWRDQ6SK\EULGLVHGFDUERQDWRPQH[WWRDQDURPDWLFULQJ

6LGHFKDLQDURPDWLFKDOLGHVRU$UKDORDONDQHV

+

& ;5

+

& ;5

*HPLQDOGLKDOLGH

>%RWKWKHKDORJHQDUHDWWDFKHGWRVDPHFDUERQ@

9LFLQDO'LKDOLGH

>%RWKWKHKDORJHQDWRPVDUHDWWDFKHGWRDGMDFHQWFDUERQDWRP@

(a) Halogen are always treated as substituents therefore preference will be given to longest chain and multiple bonds not

to the halide while selecting main chain or in numbering on main chain

(b) For different halogen preference will be given to alphabetical order

(c) Number the carbon atoms of the parent chain, beginning at the end nearer the first substituent, regardless of whether

it is an alkyl or halogen [NO2, NO, ethers are also treated as substituents like halogen] If both halogen and multiple bonds are present preference will be given to multiple bond

Alkyl halides can be given two kinds of names: common name and IUPAC name Here the compound is simply named

as an alkane with a halogen attached to side chain

Any halogen compound that contains a benzene ring is not classified as aryl halide e.g Benzyl chloride is not an aryl halide, but is a substituted alkyl halide.

2 (a) Classify alkyl halides on the basis of the bonding environment of the C of C—X Illustrate with general

formulas, (b) Use C2H4Br2 to classify dihalides

Ans (a) They are classified by the type of C to which X is bonded: primary (1°), secondary (2°), or tertiary (3°)

Their definitions and corresponding general formulas are (i) Primary: C is bonded to only one C; RCH2X (ii) Secondary: C is bonded to two C’s; R2CHX (iii) Tertiary: C is bonded to three C’s; R3CX

CH3X is unique because the C is bonded only to H’s It is simply called a methyl halide (b) The prefix gem (geminal) is used for two X’s on the same C and vie (vicinal) for two X’s on adjacent C’s; CH3CHBr2 is a gewi-dibromide and BrCH2CH2Br is a uic-dibromide

3 Supply (a) common and (b) IUPAC names for the C4H9I isomers and classify them as to whether they are 1°, 2°,

or 3°

Ans Common names have the alkyl group name followed by the halide name, i.e., ethyl bromide, C2H5Br In the IUPAC system alkyl halides are named as haloalkanes The longest alkane chain is numbered so that the smallest numbers are used to indicate the positions of substituents Several substituents are named alphabetically

&+&+&+&+,

Ans (a) 3-Bromo-1,1,2-trichloropropane; (b) 4-(1-bromoethyl)-3-chloro-2,2,7-trimethyloctane; (c)

2-chloro-methyl-1,1-dimethylcyclopentane; (d) cis-1,3-dibromocyclobutane, and (e) 1-chloro-3-methyl-2-butene

5 (a) Write structural formulas and IUPAC names for: (i) methylene bromide, (ii) chloroform, (iii) allyl bromide,

(iv) t-amyl chloride, and (v) neopentylbromide (b) Illustrate the use of the suffix “form” in naming trihalogen and trinitromethanes

Ans (a) (i) CH2Br2, dibromomethane; (ii) CHCl3, trichloromethane; (iii) H2C = CH—CH2Br, 3-bromo-l-pro-pene; (iv) (CH3)2CClCH2CH3, 2-chloro-2-methylbutane; and (v) (CH3)3CCH2Br, l-bromo-2,2-dimethyl-pro-pane (b) The “form” method is used for the HCX3 type of compounds: HCF3, fluoroform; HCCl3, chloroform; HCBr3, bromoform; HCl3, iodoform; HC(NO2)3, nitroform.

6 Give the structure of each of the following compounds: (a) 2,3-dibromo-3-ethylheptane, (b)

cis-2-bromo-chloromethylcyclohexane, (c) l-bromo-2-iodocyclobutene, (d) trans-9-chlorodecalin, and (e) 2-exo-3-endo- dichlorobicyclo[2.2.2]octane

,

%U (d)

+

&O(e) &O

&O

++

K



  o Q 

7 (a) Which alkyl halides can be made by free radical halogenation of alkanes? (b) Why is this method seldom used

for laboratory syntheses of alkyl halides? (c) How may polyhalogenation be minimized?

Ans (a) RCL and RBr, not RF or RI (b) Several isomeric monosubstituted halides are formed because most alkanes

have different types of H’s Furthermore, substitution of more than one H yields polyhalogenated compounds, (c) Polysubstitution is avoided by reacting the halogen with an excess of alkane For this reason, the hydrocarbon should be relatively inexpensive

8 Select the halides that can be made in good yields by free radical halogenation of the parent hydrocarbon Explain,

(a) CH3CH2Cl, (b) CH3CH2CH2CH2Cl, (c) Me3CCH2Cl, (d) Me3CCl, (e) Me3CBr, (f) chlorocyclo-propane, (g) H2C = CHCH2Cl, and (A) CsrH3CHBrCH = CHCH2CH3

Ans To get single monohalogenated products, all the reactive H’s of the parent hydrocarbon must be equivalent

This is true for (a) CH3CH3, (c) Me3CCH3, (f) cyclopropane, and (g) H2C = CHCH3 Although (g) has two types of H’s, the allylic H’s are much more reactive than the inert vinylic H’s The precursors for (b) and (d), which are CH3CH2CH2CH3 and Me3CH respectively, have two kinds of H’s and on chlorination give a mixture

of monochloro products However, (e) can be made in good yield even though there is more than one type of H, because bromination is much more selective than chlorination and the 3° H is replaced almost exclusively by Br Even though the parent hydrocarbon for (h), CH3CH2CH=CHCH2CH3, has only one kind of equivalent H and might be expected to give a single halogenated product, it actually gives two products The first propagation step produces an allylic radical whose resonance hybrid can react with Br2 in the second propagation step at either of two radical sites to form two monobromo substituted products

&+&+&+ &+&+&+

&+&+%U&+ &+&+&+

>&+&+±&+±&+&+&+@

4-Bromo-2-hexene is formed through an allylic rearrangement

2 From addition of halo acids on alkenes and alkynes Electrophilic addition of HX on alkenes and alkynes give the alkyl halides

5

&+

+;

;5(Where HX = HCl, HBr and HI)

Example

&++& +&

&+

+&O

&O

■ Reaction involves two steps:

First, formation of carbocation

■ As carbocation forms, that is planar, X can attack from both sides equally well, that’s why if the resultant centre is chiral reacemisation occur

■ Addition of HX to unsymmetrical alkene in general is governed by Markownikoff’s rule, but in actual and wide sense

by the stability of carbocation

&+&+±&+ &+±2+±&+

%U

+%U

■ The order of reactivity of HX: HI > HBr > HCl

■ The alkene which can form the most stable carbocation possesses the lowest activation energy and the fastest reaction

CH2 = CH2 < CH2 = CH – CH2 < CH2 – CH = CH – CH3 < (CH3)2C = C(CH3)2

■ Unsymmetrical alkene in the presence of peroxides or radical atmosphere, even coming in contact with atmospheric oxygen undergoes radical mechanism of addition, which is especially in the case of HBr (not HCI and HI) gives prod-uct opposite to Markownikoff’s rule (Anti-Markownikof’s)

3 2 HBr 3 2 2

peroxide

(Main product)

CH −CH =CH →CH −CH −CH −Br

&+ &+ &+±& &+ &+±&{&+ &+±& &+

&+±&+±&+ &+±&+±&+

+& +%U +&

&+ %U

3HUR[LGH

3 From addition of X2 on alkenes and alkynes: addition of halogens

Addition of X2 on alkenes give dihalo and that on alkyne give tetrahalo alkane

;5

&+

7HWUDKDORDONDQH+&

■ The reactivity order of X2 : F2 > Cl2 > Br2 > l2

■ The reactivity of alkene:

CH2 = CH2 < CH2 = CHCH3 < CH3CH = CH – CH3 < (CH3)2C = C(CH3)2

More substituted alkenes are more reactive

■ Halonium ion is the intermediate

&+±&+ &+±&+

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%U

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■ Stereochemically, Br2 gives anti-addition to alkene usually It implies that a suitably substrated alkene gives cific addition of alkene

stereospe-&+ &+ &+

+ %U

++

+

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WUDQVEXWHQH

1 Cis alkene + anti-addition → racemates

2 trans alkene-anti-addition →meso form

The above two facts are correct when IUPAC naming of that alkene can be done from either end of the carbon chain(structurally symmetrical alkene) If the alkene is structurally not symmetrical it will give racemic mixture provided product is chiral

Illustrations

9 Suggest alkene addition reactions to prepare the dihalogen compounds:

(a) 2,3-Dimethyl-2,3-dibromobutane (b) trans-1,2-Dibromocyclopentane (c) 1-Iodo-2-bromo-2-methylpentane (d) 2,2-Dibromobutane (e) 1,1-Dichloro-cw-2,3 dimethylcyclopropane

(c) +& &&+&+&+ ,%U ,&+&%U&+&+&+

0H0H

(d) H2C = CBrCH2CH3 → CHHBr 3CBr2CH2CH3

(e) FLV&+&+ &+&+&+&O&+ &2±. FLV&+&+±&+&+

&&O(f) H2C = CHCH2Br HBr

Alkyl Halides are best prepared from alcohols, which are easily accessible The hydroxyl group of an alcohol is replaced

by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride

Thionyl chloride is preferred because the other two products are escapable gases Hence the reaction gives pure

alkyl halides Phosphorus tribromide and triiodide are usually generated in situ (produced in solution) by the

reac-tion of red phosphorus with bromine and iodine respectively The preparareac-tion of alkyl chloride is carried out either by passing dry hydrogen chloride gas through a solution of alcohol or by heating a solution of alcohol in concentrated aqueous acid

(ii) Constant boiling with HBr (48%) is used for preparing alkyl bromide.

(iii) Good yields of R – I may be obtained by heating alcohols with sodium or potassium iodide in 95% phosphoric acid The order of reactivity of alcohols with a given haloacid is 3° > 2° > 1° The above method is not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than single bond

(d) Reaction of Alcohol with Lucas Reagent

1 (cone HCl + anhyd 2nCl2) Groves method This process is the replacement of “OH” group in primary and secondary alcohols with an “X’ atom by means of Hydrogen chloride or bromide in presence of Zinc chloride But for terti-ary alcohol, it readily reacts with concentrated hydrochloric acid in the absence of zinc chloride Zinc chloride is a lewis acid and consequently can co-ordinate with the alcohol

52+=Q&O 5²2²=Q&O† Ւ  61  †5>+2²=Q&OՒ @

+The R–O bond is weakened and so the complex readily forms a carbonium ion This test is applicable only for lower alcohols which are water soluble It is because alcohols having more than 6 carbon are almost insoluble in water hence two immiscible liquid is present before the addition of Lucas Reagent, so the presence of alkyl halide can not be detected properly

This reaction takes place via formation of carbocation so a rearranged halide will be formed

Anhydrous ZnCl2 acts as a Lewis acid and increases the leaving group ability of hydroxyl group by making it a weaker base

(SN1 mechanism) If the nature of R+ is such that it can undergo rearrangement the product will be mixture of isomeric alkyl chlorides The reaction also follows SN2 mechanism when the concentration of zinc chloride is low, the reaction is still catalysed, but no rearrangement occurs.

N

S 2

2 3 2 2 2

ZnCl + HCl + ROH → ZnCl− +ROH+→ZnCl +RCl+H O

Pyridine and dimethyl ammine also catalyse the reaction between alcohols and hydrochloric acid without rearrangement

2 Tosylation process can be used, –OTs is a weak nucleophile and better leaving group Due to its bulkyness it prefers

sn* reaction by backside attack of nucleophile

5

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+27V

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&+7V&OLV >3WROXHQHVXOSKRQ\OFKORULGH@

3. R – OH + NaBr →H SO4 2 R – Br + NaOH

For alkyl iodide a non-oxidising acid is used along with KI [H] is not preferred as it is a good reducing agent, and will convert alkyl iodide to alkane]

R – OH + KI H PO 3 4

→ RI + KOHDirect replacement of –OH group by halide ion is not possible as OH– ion is a strong nucleophile and can not be re-placed by weak nucleophile halide ion OH group can be made a weak nucleophile protonation tosylation of OH group so that it can be replaced by weak halide ion

+&O

+&O +&O

ULQJ H[SDQVLRQ

Ring expansion will takes place if ring size is (4 or 5 membered) and positive charge is present on carbon atom directly

attached to the ring (note: In &+†  ring expansion will not be preferred as this carbocation is stable due to bent bond overlapping)

Ring contraction will takes place if ring is larger than 6 membered and positive charge is present on the ring

4 Darzen Process Thionyl chloride reacts with straight-chain primary alcohols without rearrangement in the ence or absence of pyridine The reaction proceeds via a chloro sulphite Thionyl chloride is preferred because the other two products are escapable gases.

LQYHUWHGSURGXFW

3URWRQDWHGDON\O GLEURPRSKRVSKDWH YHU\JRRGOHDYLQJJURXS

10 Alkyl halides are most often prepared from alcohols, rarely by direct halogenation Supply structural formulas for

the alkyl halides synthesized in the reactions:

As NaCl and NaBr are less soluble in acetone, they precipitate and can be removed from solution by filtration

So that X– cannot displace back I– from R – I It facilitates the forward reaction according to Le-Chatelie’s principle NaI is soluble in acetone

Illustration

11 Supply reactions for the preparation from propene of (a) allyl iodide and (b) allyl fluoride:

Ans (a) CH3CH=CH2→Cl (hu) 2 ClCH2CH=CH2→NaI ICH2CH=CH2 + NaCl(s) Allyl chloride is readily converted

to the iodide with Nal in acetone because less soluble NaCl precipitates and is removed by filtration This drives the reaction to completion

(b) ClCH2CH=CH2 + AgF → FCH2CH = CH2 + AgCl(s)

(e) Hunsdiecker Reaction

The conversion of silver salts of carboxylic acids into organic halides by treating the silver carboxylates with halogens

in a refluxing inert solvent (such as carbon tetrachloride, chloroform or ether) under anhydrous conditions is known as

Hunsdiecker reaction1 The reaction offers a way of decreasing the length of carbon chain by one unit2 Although bromine

is the most often used halogen, chlorine and iodine have also been used

Step-I: In the first step the halogen and the silver carboxylate react to produce an acyl hypohalite as an intermediate.

5&22$J; 5

2

&

Step-II: This is the initiation step; the acyl hypohalite undergoes thermal decomposition into acyl and halogen free radicals

by means of homolytic fission

Important Points

(a) The free radical pathway for Hunsdiecker reaction is evidenced from the facts –

(i) isolation of the side product, R-R, which is formed presumably by radical recombination, is consistent with a

free-radical mechanism

(ii) except in a few cases, an optically active carboxylic acid leads to a racemic halide24 Besides, if ‘R’ is

neopen-tyl, there is no rearrangement, which would certainly happen with a carbocation These stereochemical results support the radical nature of the reaction

(b) Cristol-Firth modification: Cristol and Firth5 suggested an improved procedure for the reaction that consists in ing a solution of the carboxylic acid with mercuric oxide in carbon tetrachloride, followed by treatment with bromine The mechanism in this case also involves a mercury salt of the acid6

(c) A 1:1 ratio of the carboxylate to iodine (when is the reagent), the product becomes the usual halide ——— but a 2:1 ratio forms the ester, RCOOR This ester-forming reaction is termed as Simonini reaction1; the mechanism of this reaction is similar to that of the Hunsdiecker reaction

(iii) The reaction is applicable for preparing bridgehead halides –

2

1+

22

Mechanism

VXFFLQLPLGH

1±%U+%U2

1+%U2

If bromine radicals are present at high concentration, they prefer addition over substitution (Remember peroxide effect)

It they are present at low concentration substitution is preferred over addition (as it happens with N.B.S.) If is because the concentration of bromine radicals appears in rate equation of addition reaction but not in case of substitution reaction.NBS provide very low concentration of bromine by reacting with HBr, This selective bromination (at allylic or ben-zylic position) occurs because the intermediate leading to this product is stabilized by resonance, only one allylic hydrogen

Allylic halogenation can be used to convert alkene to diene

&+ &+±&+±&+ &+ &+±&+±&+

$OO\OLFN-PRO.FDOPRO

$ON\ON-PRO

.FDOPRO

9LQYOLFN-PRO

.FDOPROStereochemically, free radical halogenation gives racemic mixture if chirality appears in the product

&+±&+±&+±&+

Chlorination by sulphuryl chloride (SO2Cl2)

→ ++ - → +

+ → - +

Aryl Halide

The preparation and reactions of aryl halides have been discussed in the chapter Electrophilic Aromatic Substitution

& (OHFWURSKLOLFVLWH

G

G±;

Structure analysis shows that C–X bonds in an alkyl halide results from the overlap of a carbon Sp3 hybrid-orbital with

a halogen orbital Thus alkyl halide carbon atom have an approximately tetrahedral geometry With H – C – X bond angle near 109° Halogens increases in size as going down in the periodic table so the bond length of haloalkanes increases ac-cordingly and bond-strength decreases

Polarity of the bond decreases with increase in number of halogen

&+;!&+;!&+;!&;

0RVW

Aryl halides are less polar than alkyl halide due to resonance effect direction of dipole will be reversed

2 Lower halides (CH3Cl, CH3Br and C2H5Cl) with one and two carbon are gases All iodide and halide upto C18 are colorless, sweet – smelling liquids or solids Some chlorofluoro methane and chlorofluoro ethane are gases Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular force

of attraction are stronger in the halogen derivatives That’s why boiling points of halides are higher than those of the hydrocarbon of comparable molecular mass

Although Alkyl Halide is polar in nature, they are generally insoluble in water This is because of their inability to form hydrogen bonds while existing in water, or even break the hydrogen bond existing in water These are, however, soluble

in alcohol ether and benzene

3 Order of boiling point increases from fluoride to iodide and decreases from primary to tert Halide Iodine has larger surface area and outer electrons are loosely bound This makes iodine a highly polarizable atom A polarizable atom has increased London forces of attraction RF < RCl < RBr < RI with given halogen Boiling point increases with size

of carbon chain; straight chain alkene have higher B-pt than branched

5 Density increases from fluoride to iodide; RF and RCI are lighter than water but RBr and RI heavier than water

6 Stability decreases from fluoride to iodide due to decrease in Bond strength Alkyl iodides are sufficiently reactive to

be decomposed by light Liberation of iodine is responsible for the darkening of alkyl iodides on standing

2RI → R –– R + Ihν 2

7 The inflammability of organic halide is less than hydrocarbon It decreases with increase in halogen content, e.g., CCl4

is used as fire extinguisher

Illustrations

12 Compare R—X and R—H (alkyl halide and alkane) in following respects: (A) dipole moment (B) boiling point

(C) density and (D) solubility in H2O

Ans. (A) Dipole moment: X (halide) is more electronegative than carbon hence bonding pair

between carbon and X is attracted towards X

Hence R—X R Xd+-d- is more polar than R—H Thus dipole moment of R—X is greater than R—H

(B) Boiling point: Molecular weight of R—X is greater than R—H and also R—X is more

polar than R—H Hence b.p of R—X is greater than R—H

(D) Solubility in H2O: R—X is more polar than R—H hence R—X can have H-bonding with

H2O that makes it more soluble in it than R—H

R XH OH

d+ d- d+

d

- 13 (a) What two factors influence the boiling points of alkyl halides (as well as other compounds)? (b) Explain the

trends in the following boiling points (in °C): (i) Mel (42.4) > MeBr (3.56) > MeCl (–24.2) > MeF (–78.4) > CH4 (–161.7), (ii) PrBr (71.0) > i-PrBr (59.4), (iii) CCl4(76.8) > HCCl3 (61.3) > H2CCl2 (40.1) > H3CCl (–24.2), and (iv) C2F6(–79) > C2H6(–89)

Ans (a) (1) Van der Waal (London) forces of attraction depend on the overall shapes and sizes of molecules and on

their molecular weights Linear molecules have more surface contact and enhanced attraction than chain molecules, while spherically shaped molecules have very slight tangential contacts (2) Dipole-dipole attraction is a significant factor for RX’s, but not at all for alkanes (b) (i) All alkyl halides boil at a higher temperature than the parent alkane In this sequence the increases in molecular weight and size lead to increasing bp’s At the same time, dipole-dipole attractions decrease and this change should decrease the bp’s Clearly the effect of increasing van der Waal forces of attraction dwarfs the effect of decreasing dipole-dipole attraction (ii) As with alkanes ,branched C chains are more spherical-like and their smaller surface area results in lower boiling points (iii) Accumulation of Cl’s on CH4 increases the molecular weight and size, causing the bp to increase Notice that the Abp gets smaller as more Cl’s are introduced, a fact that may be due to an increase in the spherical nature of the molecule (iv) The perfluoroalkane, a fluorocarbon, is predicted to have the higher bp because it has the larger molecular weight What is surprising is the small difference in bp, a result due to F being only slightly larger than H and having a low polarizability

branched- 14 (a) List the densities of alkyl halides, water, and alkanes in decreasing order (b) List in decreasing order the

densities of the chloromethanes in (iii) of part (b) of the above question and H2O (c) Water is mixed with an immiscible liquid A in a separatory funnel to give two layers Describe what can be done to tell which layer is water and which is A (Do not solve the problem by taste or smell, it may not be healthy.)

Ans (a) RI > RBr > H2O > RCl > RF > RH (b) CCl4 > HCCl3 > H2CCl2 > H2O > H3CCl (c) Let a small amount of the lower (denser) layer drop into a test tube containing an equal volume of H2O If it dissolves (one layer is observed), it is H2O If two layers form, the lower layer is A

15 Discuss the ability of alkyl halides to dissolve (a) in H2O, (b) in organic solvents, and (c) salts

Ans (a) They are insoluble in H2O, probably because there is little (in the case of RF) to no H-bonding to H2O (b) They are soluble in alcohol, ether, and benzene, (c) Although they are polar, they do not dissolve salts



 ALIPHATIC NUCLEOPHILIC SUBSTITUTION REACTION

A large number of synthetically useful reactions fit into the classification of nucleophilic substitution at saturated carbon Every reaction has this com mon feature: a nucleophile replaces a substituent (the leaving group) on a saturated carbon atom The scope of the reaction is broad because many rea gents can function as nucleophiles toward a large variety of carbon com pounds

The SN2 pathway lies at the other extreme The nucleophile forms a bond to the saturated carbon atom at approximately

the same time that the bond to the leaving group breaks A concerted, one-step process occurs

1X±5²/ 5²1X/±

Reactions follow the SN2 mechanism when the reagent is a good nucleophile and when the carbon atom at which stitution occurs is relatively unhin dered Many reactions proceed by pathways which lie between the SN1 and SN2 extremes

sub-and can be considered to be mixtures of the two mechanisms

Mechanism of SN2 Reaction

The factors that affect the rate of the reaction are called the kinetics of the reaction The rate of a nucleophilic

substitu-tion reacsubstitu-tion, such as the reacsubstitu-tion of methyl bromide with hydroxide ion, depends on the concentrasubstitu-tions of both reagents

&++2±ĺ&+2+%U+2± &±%U²G±+2ÂÂÂÂ&ÂÂÂÂ%UG± +2±& %U±

The energy necessary to break the C–Br bond is supplied by simultaneous formation of the C–O bond When the transition state is reached, the central carbon atom has gone from its initial sp3 hybridization to a sp2 state, with an approxi-mately perpendicular p orbital One lobe of this orbital overlaps with the nucleophile and the other with the leaving group That is, the nucleophile, the central carbon and the leaving group are collinear, and remain so in the transition state for the reaction This is why a front-side SN2 mechanism has never been observed During the transition state, the three nonre-acting substituents and the central carbon are approximately coplanar The transition state involves trigonal bipyramidal geometry with a pentacoordinate carbon The concerted displacement mechanism implies both kinetic and stereochemical consequences

The rate law for the reaction may be expressed as:

rate ∝ [alkyl halide] [nucleophile]

The ‘proportional to’ sign (∝) can be replaced by an equal sign and proportionality constant (k) Because the rate of the reaction depends on the concentration of two reactants, it is a second-order reaction

rate = k [alkyl halide] [nucleophile]

The rate law tells us what molecules are involved in the transition state of the rate-determining step of the reaction The

transition state is bimolecular, that is, it involves two molecules This is not always the same as a second order mechanism

because, if an excess of nucleophile is present, that is, if it is solvent, the mechanism may still be bimolecular, though the experimentally determined kinetics will be of the first order Such kinetics is called pseudo-first order, as the concentration

of the solvent will not change significantly during the course of the reaction The rate constant describes how difficult it

is, to overcome the energy barrier of the reactions (how hard it is to reach the transition state) The lower the rate constant, the more difficult it is to reach the transition state

The reaction of methyl bromide with hydroxide ion is an example of an SN2 reaction In 1937, Hughes and Ingold

proposed a mechanism for SN2 reactions with the following experimental evidence:

1 The rate of the reaction depends on the concentration of the alkyl halide and on the concentration of the nucleophile

This means that both reactants are involved in the transition state of the rate-determining step

2 When the hydrogens of methyl bromide are successively replaced with methyl groups, the rate of the reaction with a

given nucleophile becomes progressively slower

3 The reaction of an alkyl halide in which the halogen is bonded to a chirality centre leads to the formation of

only one stereoisomer, and its configuration is inverted compared with the configuration of the reacting alkyl halide

The familiar SN2 and SN1 mechanisms describe only two of a possibly infinite number of pathways on a dimensional surface describing nucleophilic aliphatic substitution There is evidence that not all nucleophilic substitutions are heterolytic processes, meaning that both the electrons from the nucleophile are not always donated to the substrate at the same time Instead, some reactions have been found to occur by a process in which only one electron is transferred at

two-a time

A major mechanistic distinction between SN2 and SN1 reactions is the timing of the departure of the leaving group and the arrival of the nucleophile One tool to help visualize the relationship between these processes is a two-dimensional reaction-coordinate diagram shown in Figure overleaf

16 (a) Is the SN2 reaction exothermic or endothermic reaction (b) Draw an enthalpy diagram for an exothermic SN2.

Ans (a) The reaction is thermo chemically favourable only when a stronger base displaces a weaker base Hence,

successful displacements are exothermic (b) See Fig below This one-step reaction has one transition state and

17 (a) Use the general reaction between (S)-RCHDX and :Nu- to illustrate and explain with the aid of orbital

representations the stereochemical consequences of the SN2 reaction (b) Discuss the charge distribution in the transition state of this SN2 reaction

Ans (a) See Fig below In the TS the attacked C is sp2 hybridized with a p AO available for overlapping one of its lobes with an orbital of the incoming : Nu– while the other lobe overlaps with an orbital of the leaving group X– These overlaps account for the partial bonds drawn in the Fig.) The reaction is initiated by :Nu- beginning to overlap with the small lobe (tail) of the sp3 HO bonding with X In order to provide more bonding volume to give

a stronger bond, the tail becomes the larger lobe (head) and the head becomes the tail, inverting the configuration

of C The configuration of the OH compound is the opposite of that of the X compound (b) As = Nu– starts to bond to C, it loses some of its full charge and in the TS has a d- charge, as does X as it begins to leave as an anion There is little + charge on the C which becomes sp3 hydridized in the product

;

Mechanism of SN1 Reaction

Given our understanding of the SN2 reaction, if we were to measure the rate of reaction of tert-butyl bromide with water,

we would expect a relatively slow substitution reaction, as water is a poor nucleophile and tert-butyl bromide is sterically hindered to attack by a nucleophile However, we would actually discover that the reaction is surprisingly fast In fact,

it is over one million times faster than the reaction of methylbromide—a compound with no steric hindrance with water Clearly, the reaction must be taking place by a mechanism different from that of the SN2 reaction

As we have seen, a study of the kinetics of a reaction is one of the first steps undertaken when investigating the nism of a reaction If we were to investigate the kinetics of the reaction of tert-butyl bromide with water, we would find that doubling the concentration of the alkyl halide doubles the rate of the reaction We would also find that changing the concentration of the water has no effect on the rate of the reaction Knowing that the rate of this nucleophilic substitution reaction depends only on the concentration of the alkyl halide, we can write a rate law for the reaction

mecha-&

&+

%U+2+&

&+

UDWH N>DON\OKDOLGH@

&

&+2++%U+&

&+

Because the rate of the reaction depends on the concentration of only one reactant, it is a first-order reaction

Since the rate law for the reaction of tert-butyl bromide with water clearly differs from the rate law for the reaction of methyl bromide with hydroxide ion, the two reactions must follow different mechanisms We have seen that the reaction between methyl bromide and hydroxide ion is an SN2 reaction The reaction between tert-butyl bromide and water is an

S N1 reaction: S for substitution, N for nucleophilic and 1 for unimolecular The mechanism of an SN1 reaction is based

on the following experimental evidence

1 The rate law shows that the rate of the reaction depends only on the concentration of the alkyl halide This means that we must be observing a reaction whose rate-determining transition state involves only the alkyl halide The rate-

determining transition state is unimolecular, it involves only one molecule

2 When the methyl groups of tert-butyl bromide are successively replaced by hydrogens, the rate of the SN1 reaction decreases progressively This is opposite to the order of reactivity exhibited by alkyl halides in SN2 reaction

3 The reaction of an alkyl halide in which the halogen is bonded to a chirality centre leads to the formation of two stereoisomers: one with the same relative configuration as the reacting alkyl halide and the other with the inverted configu-ration Since the cation is planar, it is achiral Hence, starting with an optically active substrate, one gets racemic products The energy required to break the bond is supplied by the formation of many bonds between the ions (that are produced) and the solvent The dipole moment associated with the transition state is much more than the substrate Thus, the solvent stabilizes the transition state much more effectively than the substrate An SN1 reaction has two steps In the first step, the carbon-halogen bond breaks heterolytically, with the halogen retaining the previously shared pair of electrons In the second step, the nucleophile reacts rapidly with the carbocation that was formed in the first step

&+

&++& & 2++

&+

&++& & %U±

How does the reaction mechanism that we have just seen account for the three pieces of experimental evidence? First, because the alkyl halide is the only species that participates in the rate-limiting step, the mechanism agrees with the obser-vation that the rate of the reaction depends on the concentration of the alkyl halide and does not depend on the concentra-tion of the nucleophile

Second, a carbocation is formed in the slow step of an SNl reaction Because a tertiary carbocation is more stable and therefore easier to form than a secondary carbocation, which in turn is more stable and easier to form than a primary car-bocation, tertiary alkyl halides are more reactive than secondary alkyl halides, which are more reactive than primary alkyl halides in an SN1 reaction Thus, the reactivity order agrees with the observation that the rate of an SN1 reaction decreases

as the methyl groups of tert-butyl bromide are successively replaced by hydrogens

The relative reactivities of alkyl halides in an SN1 reaction are

3° alkyl halide > 2° alkyl halide > 1° alkyl halide

&) &±%U &)

The positively charged carbon of the carbocation intermediate is sp2 hybridized, and the three bonds connected to an sp2

hybridized carbon are in the same plane In the second step of the SN1 reaction, the nucleophile can approach the tion from either side of the plane

of experimental evidence for the mechanism of the SN1 reaction The SN1 reaction of an alkyl halide in which the leaving group is attached to a chirality centre leads to the formation of two stereoisomers: attack of the nucleophile on one side of the planar carbocation forms one stereoisomer, and attack on the other side produces the other stereoisomer



 THE VARIABLES IN NUCLEOPHILIC SUBSTITUTION

Many factors influence the course of nucleophilic substitution at a satu rated carbon atom The substrate and nucleophile have their individual reac tion characteristics Reaction medium (solvent) and conditions (temperature, concentration, etc.) can often determine the distribution of products as well as the mechanism In this section we will consider the variables and see how a balance of effects controls the overall reaction

The Leaving Group

In both the SN1 and the SN2 mechanisms, the leaving group departs with its bonding electrons at the rate determining step Thebetter leaving groups are better able to accommodate the original bonding electrons

The good leaving groups of the carboxylic acid family are good leaving groups when attached to a saturated carbon

atom also We again find that good leaving groups are the conjugate bases of strong acidsi.e, they should be weak bases

after departure

If an alkyl iodide, an alkyl bromide, at alkyl chloride and an alkyl fluoride (all having the same alkyl group) were lowec to react with the same nucleophile under the same conditions, we would fine that the alkyl iodide is the most reactive and the alkyl fluoride is the leas-reactive

al- Relative Rates of Reaction

HO– + RCH2l → RCH2OH + l– 30,000

HO– + RCH2Br → RCH2OH + Br– 10,000

HO– + RCH2CL → RCH2OH + Cl– 200

HO– + RCH2F → RCH2OH + F– 1

The only difference among these four reactions is the nature of the leaving group Apparently, the iodide ion is the best

leaving group and the fluoride ion is the worst This brings us to an important rule in organic chemistry that the weaker the

base, the better it is as a leaving group

The relative acidities of hydrogen halides are,

DON\OPHV\ODWH

25±26

or not a leaving group The lower the bond enthalpy of the bond between the carbon and the leaving group, the better the leaving group A poor leaving group can be made more reactive by coordination to an electrophilic species Hydroxide is

a very poor leaving group

The less basic the substituent, the more easily it is pulled off by solvent or pushed off by an attacking nucleophile The order of ease of displacement of groups is not fixed; it depends on the nature of the R group and on the conditions (nature

of solvent, that is, protic, aprotic, polar, nonpolar and so on) However, it appears that, in general, the order is:

OTf > OTs > OMs > I > Br > Cl > OH2+ > F > OAc > +NR3 > OR – OH > NR2

Remember that often a poor leaving group may be converted to a good leaving group by a simple reaction such as protonation For example, alkoxide and hydroxide are poor leaving groups, but on protonation they become a part of the good leaving groups alcohol and water

The stronger, and harder, as a base a leaving group is, the less readily can it be displaced; thus groups such as ΘOH,

ΘOR, ΘNH2 bonded to carbon by small, highly electronegative atoms of low polarisability (cf hard bases, above) cannot normally be displaced directly by other nucleophiles

Displacements that are otherwise difficult, or even impossible, to accomplish directly may sometimes be effected by modification of the potential leaving group—often through protonation—so as to make it weaker, and/or softer, as a base Thus ΘOH cannot be displaced directly by BrΘ, but is displaced readily if protonated first:

There are two main reasons for this: (a) BrΘ is now attacking a positively charged, as opposed to a neutral, species, and (b) the very weakly basic H2O is a very much better leaving group than the strongly basic ΘOH The well known use

of HI to cleave ethers results from IΘ being about the most nucleophilic species that can be generated in the strongly acid solution required for the initial protonation:

5±23K +† 5±23K ,Ւ 5,3K2+

+

†

The most important leaving groups are the conjugate bases of acids with pKa values below

Table 1.2 Relation of Leaving-Group Ability to the Acidity of the Conjugate Acid Leaving Group pKa of Conjugate Acid

CH3O– 15

NH2–

36 Poor

leaving groups

This is known as nucleophilic catalysis

 THE NUCLEOPHILE

A nucleophile uses its lone pair electrons to attack an electron-deficient atom other than a proton Nucleophilicity is a

measure of how readily the nucleophile is able to attack such an atom It is measured by a rate constant (k)

Addition of the nucleophile takes place after formation of the carbocation in reactions that proceed by the SN1

mecha-nism Therefore, the nucleophile does not influence the reaction rate of an S N 1 reaction By contrast, the SN2 nism requires participation of the nucleophile The nucleophile is often said to displace, or push off, the leaving group

mecha-Nucleophilicity of the reagent is extremely important in the SN2 reaction.

We would expect good nucleophiles to be good electron donors, i.e., good Lewis bases Nucleophilicity and basicity do correlate in many cases Their relation is most useful for comparison of a series of compounds in which the same atom is the nucleophile For instance, the oxygen nucleophile may be encountered as the reactive portion of an alcohol or a phenol The alcohol will be the better nucleophile because resonance delocalization of the oxygen electrons in the phenol reduces their availability This was the same rationale we used to account for acid-base characteristics of alcohols and phe nols

A similar type of relation is found for nucleophiles along a row of the periodic table The nitrogen atom is more philic than an oxygen atom in a series of similar compounds Again, this is a result of the greater availabil ity of an electron pair for bonding from the less electronegative nitrogen atom (Table)

nucleo-Table 1.3 Relation of Nucleophilicity to Basicity for Atoms in the Same Row of the Periodic nucleo-Table

N-atom Nucleophlles O-atom Nucleophiles N vs O Nucleophiles

&+2

&+&2±

+2

+1+2+1+2

Problem 3

Many nucleophiles are anions, but some anions are not nucleophiles Explain why BF 4 – is not a nucleophile.

Correlation of nucleophilicity with basicity is useful but not exact Two different kinds of reactions, normally in different

solvents, are being com pared Basicity is an equilibrium phenomenon which measures the position of equilibrium of a

reagent with a proton, usually in water Nucleophilicity involves the rate of reaction (kinetics) with a carbon atom, usually

in non-aqueous solvents

The basicity-nucleophilicity correlation is not useful for comparison of atoms down a family of the periodic table As atomic number increases, nucleophilicity increases and basicity decreases (Table)

Table 1.4 Relation of Nucleophilicity to Basicity Commonly Observed for Atoms in Same Family of the Periodic Table Group V Nucleophiles Group VI Nucleophiles Group VII Nucleophiles

'HFUHDVLQJ QXFOHRSKLOLFLW\

53

51

5652

,

%U

&O) 'HFUHDVLQJEDVLFLW\

The outer electrons of the larger atoms are diffused over greater volumes than are those of smaller atoms The localized electrons form weaker bonds to a proton, and the atoms are less basic However, the outer electrons of the larger atoms are also less tightly held by the nucleus They are more polarizable and are more available for forming bonds to

less-a cless-arbon less-atom; they less-are more nucleophilic Tless-able summless-arizes the reless-activities of less-a series of nucleophiles from different families of the periodic table

Bisulfite ion shows the rather interesting dependence of basicity and nucleophilicity on the atoms within a single ion Charge is delocalized over the oxygen and sulfur atoms The more basic oxygen atom typically reacts with a proton, and the more nucleophilic sulfur atom reacts at a carbon atom The bisulfite addition product from an aldehyde illustrates that

difference Ions in which charge is delocalized so that reaction can take place at two different atoms are known as

ambi-dent ions.

2+2²6²2

2+2²6 2

+2²6²&+52

+2²6 2

2

Table 1.5 Relative Rates of Nucleophilic Substitution (SN2)

Nucleophile k 2 (rel.) Nucleophile k 2 (rel.)

Nucleophile k 2 (rel.) Nucleophile k 2 (rel.)

 For the reaction Nu: + CH3I MeOH → Nu–CH3

 Method arbitrarily set as standard

pro-to a tetracoordinate carbon apro-tom can involve severe steric interactions

The important oxygen anions methoxide and tert-butoxide are illustra tive They are of similar basicity, being the

con-jugate bases of aliphatic alco hols Methoxide, however, is a small species which can easily approach a carbon atom during nucleophilic substitution In contrast, tert-butoxide is very bulky, so that steric restraints reduce its ability to function as a nucleophile

++

&

+²&²&²2

In comparing molecules with the same attacking atom, there is generally a direct relationship between basicity and

nu-cleophilicity Both describe a process involving the formation of a new bond to an electrophile by donation of an electron pair Stronger bases are better nucleophiles The hydroxide ion is a stronger base than a cyanide ion, but cyanide ion is a stronger nucleophile than hydroxide ion For example, a compound with negatively charged oxygen is a stronger base and

a better nucleophile than a compound with neutral oxygen (Table)

Table 1.6 Relation of Nucleophilicity to Basicity Stronger Base, Better Nucleophile Weaker Base Poorer Nucleophiles

In comparing molecules with attacking atoms of approximately the same size, the stronger bases are again the better

nucleophiles Within a group of nucleophiles that attack at the electrophile with the same atom, the nucleophilicity creases with decreasing basicity of the nucleophile Decreasing basicity is equivalent to decreasing affinity of an electron pair for a proton

de-RO– > HO– > C6H5O– > RCOO– >> ROH, H2O >>> RSO3–

This parallel relationship between nucleophilicity and basicity can be reversed by steric effects Therefore, less basic but sterically unhindered nucleophiles have a higher nucleophilicity than strongly basic but sterically hindered nucleo-

philes Thus, although t-butoxide ion is a stronger base than ethoxide ion, the bulky t-butoxide is a weaker nucleophile.

–CH3 > –NH2 > HO– > F– ; RS– >> Cl–; Et3N >> Et2O

In comparing molecules with attacking atoms that are very different in size, the direct relationship between

nucleophi-licity and basicity is maintained if the reaction occurs in the gas phase If, however, the reaction occurs in a solvent this relationship between nucleophilicity and basicity depends on the solvent In a comparison of the atomic centres from the same group of the periodic table, the trend is as follows:

Basic strength: RO– > RS–; ROH > RSH; F– > Cl– > Br– > I–

Nucleophilic ability: RS– > RO–; RSH > ROH; I– > Br– > Cl– >> F–

The divergence of nucleophilic ability from basic strength stems from the fact that as the atom-donating electron pair increases in size, the electrons in its outer shell will be further away from, and hence, held less tightly by the atomic nucleus These outer electrons are thus more polarizable They are more readily available to form a bond with the atom being attacked Polarizability appears to be much more important in nucleophilic ability than in the equilibrium situation involved in basicity; thus, species in which the relevant atom is large are commonly found to be better nucleophiles than their strength as bases might suggest

If the solvent is aprotic (it is not a hydrogen bond donor), the direct relationship between nucleophilicity and basicity holds For example, both the nucleophilicities and basicities of the halogens decrease with increasing size in an aprotic solvent such as dimethyl formamide

of experimental evidence for the mechanism of the SN1... volume to give

a stronger bond, the tail becomes the larger lobe (head) and the head becomes the tail, inverting the configuration

of C The configuration of the OH compound is the