Preview Organic Chemistry—A Modern Approach (VolumeII) by Nimai Tewari (2018)

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Organic chemistry

a modern approach

Volume-II

A bout the A uthor

Nimai Tewari is a retired Associate Professor in Department of Chemistry Katwa College (affiliated to University of Burdwan), West Bengal A Ph.D in Organic Chemistry from Calcutta University,

he has taught the subject for a period of more than three decades

He has published various research papers in national and international journals Apart from Organic Chemistry—A Modern Approach, Volume-II (including Volume-I of this title), Dr Tewari has authored four more books on Organic Chemistry for undergraduate and postgraduate students His research interest includes Organic Synthesis and Heterocyclic Chemistry

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2 Reactions of the Carbonyl Group and Reactions at the a-Carbon

2.1 Introduction 2.2

2.2 Nucleophilic Addition to Carbon-Oxygen Double Bond 2.5

2.2.1 Structure of the Carbonyl Group 2.5

2.2.2 Effect of Structure on Reactivity 2.5

2.2.3 Stereochemistry of Nucleophilic Addition 2.8

Solved Problems 2.10

2.3 Addition of Oxygen Nucleophiles: Water and Alcohols (ROH) 2.11

2.3.1 Addition of Water (Hydration) and Formation of gem-diols 2.11

2.3.2 Mechanism for Hydrate Formation 2.11

2.3.3 Kinetics of Hydrate Formation 2.11

2.3.4 Thermodynamics of Hydrate Formation 2.13

2.3.5 Formation of Stable and Isolable Hydrates 2.14

2.3.6 Addition of Alcohols and Formation of Hemiacetals and Acetals 2.162.3.7 Mechanism for Acid-catalyzed Hemiacetal and Acetal Formation 2.182.3.8 Mechanism for Base-catalyzed Hemiacetal Formation (Base-catalyzed Acetal Formation does not take place) 2.20

2.3.9 Hydrolysis of Acetals 2.20

2.3.10 Acetals as Protecting Groups 2.22

2.3.11 Tetrahydropyranyl Group as a Protecting Group 2.23

Solved Problems 2.24

Study Problems 2.33

2.4 Addition of Sulphur Nucleophiles: Thiols (RSH) and Sodium Bisulphite

(NaHSO3) 2.38

2.4.1 Addition of Thiols 2.38

2.4.2 Mechanism for Thioacetal Formation 2.38

2.4.3 Thermodynamics for Thioacetal Formation 2.39

2.4.4 Importance of Thioacetals in Organic Synthesis 2.39

2.4.5 Addition of Sodium Bisulphite (NaHSO3) 2.40

2.4.6 Mechanism for Bisulphite Additions 2.40

2.4.7 Applications of Bisulphite Addition Reaction 2.41

Solved Problems 2.42

Study Problems 2.45

2.5 Addition of Hydride Ion (:H①

) and addition of electrons 2.462.5.1 Reduction of carbonyl compounds by lithium aluminium hydride

(LiAlH4) 2.472.5.2 Mechanism for the reduction of carbonyl compounds by lithium aluminium hydride (LiAlH4) 2.47

2.5.3 Reduction of aldehydes and ketones by sodium borohydride (NaBH4) to form primary and secondary alcohols, respectively 2.50

2.5.4 Mechanism for the reduction of an aldehyde or a ketone by sodium

borohydride (NaBH4) 2.512.5.5 Reduction of carboxylic acids, amides and nitriles by borane (BH3) 2.512.5.6 Reduction of Acid Chlorides to the Corresponding Aldehydes by Lithium tri-tert-butoxyaluminium Hydride [LiAlH(O-t-Bu)3] and Reduction of Esters and Nitriles to the Corresponding Aldehydes by Diisobutylaluminum Hydride (i–Bu2AlH or DIBAL–H): 2.53

2.6 Addition of Nitrogen Nucleophiles 2.84

2.6.1 Addition of Primary Amines and Formation of Imines 2.84

2.6.2 Addition of Compounds like Hydroxylamine, Hydrazine and Substituted Hydrazines (Similar to Primary Amines and Formation of Imine

Derivatives) 2.862.6.3 Reductive Amination 2.88

2.6.4 Wolff-Kishner Reduction 2.89

2.6.5 Important uses of the Imine-Forming Reaction 2.91

2.6.6 Addition of Secondary Amines and Formation of Enamines 2.91

Solved Problems 2.96

Study Problems 2.104

2.7 Addition of Carbon Nucleophiles 2.106

2.7.1 Addition of Hydrogen Cyanide (HCN) 2.106

2.7.2 Addition of Grignard and Other Organometallic Compounds 2.1092.7.3 Addition of Acetylides 2.116

2.9.1 Halogenation at the a-carbon of Aldehydes and Ketones 2.208

2.9.2 Alkylation at the a-carbon of Aldehydes and Ketones 2.214

2.9.3 The Claisen Condensation (Addition of Acid Derivatives to the a-carbon of Carbonyl Compounds) 2.227

2.10.2 Reactions of Carboxylic Acids 2.290

2.10.3 Reactions of Acid Chlorides 2.311

3 Electrophilic and Nucleophilic Substitutions in Aromatic Systems 3.1–3.181Introduction 3.1

3.1 Electrophilic Aromatic Substitution 3.3

3.1.1 Characteristics of Electrophilic Aromatic Substitution 3.3

Solved Problems 3.24

Study Problems 3.39

3.1.2 Formation of Carbon–Nitrogen s-Bond (Nitration, Nitrosation

and Diazo coupling) 3.43Solved Problems 3.52

Study Problems 3.61

3.1.3 Formation of Carbon-Halogen s-Bond (Halogenation):

Bromination and Chlorination, Iodination, Fluorination 3.63Solved Problems 3.67

Study Problems 3.70

3.1.4 Formation of Carbon–Sulphur s-Bond: Sulphonation and

Chlorosulphonation 3.72Solved Problems 3.78

Study Problems 3.81

3.1.5 Formation of Carbon–Carbon s-Bond: Friedel–crafts Reaction,

Houben–Hoesch Reaction, Gattermann Formylation, Gattermann–

Koch Fromylation, Reimer-Tiemann Reaction, and Kolbe-Schmitt Reaction 3.83

Solved Problems 3.107

Study Problems 3.121

3.1.6 Use of electrophilic Aromatic Substitution Reactions in Organic

Synthesis 3.124Solved Problems 3.141

Study Problems 3.146

3.2 Nucleophilic Aromatic Substitution 3.147

3.2.1 The Addition-Elimination or SNAr Mechanism 3.148

3.2.2 The Elimination-Addition or Benzyne Mechanism 3.155

4.1.2 Nucleophilic Rearrangements to Electron-Deficient Nitrogen 4.60

P refACe

While teaching Organic Chemistry at the undergraduate level, I observed that students have to plough through several specific books within a short span of time and as a consequence, they often fail to assimilate the subject matter It is this difficulty of the students that motivated me to undertake this concise and systematic work on Organic Chemistry in the light of the modern structural theories The book has been designed primarily for students who have taken an honours course in Organic Chemistry at the undergraduate level

The book is organised into four chapters: (i) Addition to Carbon-Carbon Multiple Bond, (ii) Reactions of the Carbonyl Group and Reaction at the a-carbon to the Carbonyl Group, (iii) Electrophilic and Nucleophilic Substitutions in Aromatic Systems, and (iv) Rearrangement Reactions

I believe that Organic Chemistry can be best understood and learnt by solving problems Each article of every chapter concludes with a number of solved as well as study problems to develop the proficiency of students that can be acquired only by practice The book presents

a large number of reactions supported with mechanistic explanation and diagrams

I am pleased to note that Volume-I of this title has been received well by both teachers and students This volume will also be highly useful for students in advanced courses who have had a basic course in Organic Chemistry in B.Sc (Chemistry) Hons It also aims to help the students preparing for competitive examinations like NET, GATE and SLET

Acknowledgements

I would like to express my deep sense of gratitude to Mrs Vibha Mahajan, Director, Science and Engineering Portfolio, for successful publication of this book I also wish to thank Mr Suman Sen and Mr P L Pandita for taking keen interest in publishing this book I am grateful to all of them

I also owe a debt of gratitude to my friends and former colleagues for their constructive suggestions and to my students who encouraged me constantly I extend my heartfelt

gratitude to my wife, Mrs Dali Tewari, and my daughter, Aindrila Tewari (Mukherjee), for their support and encouragement during the long period of writing this book

Valuable suggestions from the readers for further improvement of the book are most welcome These can be shared at: n.tewari54@gmail.com

Nimai Tewari

1.1 Electrophilic Addition

1.1.1 Addition of hydrogen halides

1.1.2 Addition of Water and Alcohol

1.1.3 Oxymercuration-Demercuration

1.1.4 Addition of Halogens

1.1.5 Hydroboration-Oxidation

1.1.6 Ozonolysis of Alkenes and Alkynes

1.1.7 Addition of Carbenes and

a, b-Unsaturated Aldehydes and Ketones

1.2.3 Michael Addition 1.3 Addition of radicals and radical substitution of benzylic and allylic hydrogens

1.3.1 Radical addition of halogens and hydrogen halides and radical polymerization

1.3.2 Allylic chlorination and allylic bromination

The characteristic feature of the structure of an alkene is the carbon–carbon double bond

It is thus the functional group of alkenes and it determines the characteristic reactions that alkenes undergo These reactions are of two types:

(a) There are reactions that take place at the double bond itself and as a result, the

double bond is destroyed These reactions are called addition reactions and these are called so because in these reactions two molecules combine to yield a single molecule of the product The reagent is simply added to the substrate, in contrast

to a substitution reaction in which part of the reagent is substituted into a part of the substrate This type of reaction can be depicted by using E for an electrophilic portion of a reagent and Nu for a nucleophilic portion, as follows:

C==C + E—Nu addition

E

Nu When an electrophile (or a positive pole of a dipole) initiates the process, the reaction

is termed as electrophilic addition reaction and when a nucleophile (or a negative pole of a dipole) initiates the process, the reaction is said to be a nucleophilic addition reaction

Since free radicals also seek electrons, therefore, alkenes also undergo free-radical addition Besides these additions involving charged species and free radicals, alkenes undergo some molecular additions in which two unsaturated molecules

or two such parts of a single molecule add to form a cyclic product Therefore, depending upon the electronic nature of the addendum used, addition reactions may be classified as:

1 Electrophilic additions

2 Nucleophilic additions

3 Free-radical additions

4 Molecular additions

(b) There are some reactions that take place not at the carbon–carbon double bond,

but at certain positions having special relationships to the double bond The double bond, though not involved, plays an important role in the reaction These are free radical substitution reactions known as allylic substitutions For example:

to a reagent that is seeking electrons It thus follows that in many reactions, the carbon–carbon double bond acts as a source of electrons, i.e., it acts as a nucleophile The molecules or ions with which it reacts are those that are deficient in electrons

and can accept an electron pair, that is, are electrophiles Each reaction starts with the addition of an electrophile to one of the sp2 hybridized carbons of the alkene and ends with the addition of a nucleophile to the other sp2 hybridized carbon The result is that the p bond breaks and the two sp2 carbons form two new s bonds with the reagent.

2 In an addition reaction, one p bond and one s bond are converted into two s bonds The energy released in making two s bonds exceeds that required to break one s bond and one p bond and, therefore, addition reactions are usually exothermic and energetically favourable

Hydrohalogenation, for example, is the electrophilic addition of hydrogen halides HX(X = Cl, Br and I) to alkenes to form alkyl halides

d + d –

H——C——C——XAlkene (X = Cl, Br, I)

Hydrogen halide

Alkyl halide

Some other reactions of this type that we shall study in this chapter include addition

of sulphuric acid, water (in the presence of an acid catalyst), halogen, etc

Mechanism The elelctrophilic addition to C == C usually occurs in two steps:

Step 1: The incoming reagent with its electrophilic portion in the front or the electrophile with its vacant orbital approaches the p cloud of the alkene in a perpendicular direction from above or below the plane of the six atoms to form a p-complex which may or may not be

an actual intermediate The p-complex then collapses to carbocation through the formation

of a s bond between the electrophilic portion of the reagent or the electrophile and any one

of the doubly bonded carbon atoms The overall result in the formation of a carbocation and nucleophile, Nu , * from the alkene and the reagent, E — Nu Electrophiles containing

an unshared pair of electrons form a three-membered ring in which the electrophile bears

a positive charge This may be an intermediate when addition of halogens occurs or a true cyclic product when addition of neutral electrophiles such as carbenes, nitrenes, etc occurs

It is the rate-determining (slow) step of the reaction.

Step 2: The nucleophile combines with the intermediate carbocation or the membered cyclic cation to yield the addition product It is a fast step

three-The energy diagram for the reaction may be represented as follows:

T.S.I

T.S.II

Intermediate carbocation

or a three-membered cyclic cation

Progress of the reaction

C==C + E—Nu

E—C—C—Nu

In contrast to electrophilic additions, nucleophilic addition to carbon–carbon multiple bond is less common and this is because the negative charge cloud of the p orbital shields the molecule from attack of nucleophilic reagents Since the C == C moiety can readily offer its loosely held p electrons to an electron-deficient species (electrophile), the additions are mostly electrophilic in nature

Since the addition of the electrophile is the rate-determining step of the reaction, therefore, the rate of electrophilic addition to a carbon–carbon multiple bond depends on the relative electron density of a p bond The more the multiple bond is electron rich, the more it is reactive towards an electrophilic addition Electron-donating groups would, therefore, be expected to enhance the reactivity, and electron-withdrawing group would, therefore, be expected to decrease the reaction rate The following relative rates of addition of Br2 to different alkenes in CH2Cl2 solution at –78°C are actually observed in practice:

Kinetics: Since the step 1, i.e., the electrophilic addition to carbon–carbon double bond,

is the slow step and the T.S of this step involves both the substrate and the reagent (the electrophile), therefore, the reaction is expected to follow second-order kinetics

rate = k [substrate] [reagent]

Stereochemistry: The steroechemistry of addition is often important in delineating the mechanism of a reaction Because the carbon atoms of a double bond are both trigonal planar (sp2), the elements E and Nu can be added to them from the same side or from opposite sides

triple bonds are less susceptible to electrophilic attack than double bonds In general, the triple bonds are less susceptible to electrophilic attack than double bonds, even though the concentration of electrons in a triple bond is higher than that in a double bond The reasons are as follows:

1 Because of smaller distance between the two triple bonded carbon atoms and better p orbital overlap, the electrons in the triple bond are held more tightly and hence, they are poorly available to an electrophile, i.e., it is harder for an attacking

electrophile to pull out a pair There is evidence from far-UV spectra to support this conclusion.

2 An electrophile adds to a double bond to give a saturated carbocation A similar addition to a triple bond gives a vinylic cation A vinylic cation is less stable than the corresponding saturated carbocation because it possesses the linear (sp) arrangement instead of planar (sp2) arrangement which, as suggested by quantum machanics, is the stabler configuration of carbocation

3 If the reaction takes places via a bridged-ion intermediate, then those obtained from triple bond (I) will be highly strained and hence, less stable than those obtained from double bonds (II)

1.1.1 Addition of hydrogen halides

1.1.1.1 Reaction

The addition of hydrogen halides HX(X = F, Cl, Br and I) to alkenes to form alkyl halides

is known as hydrohalogenation These additions are usually carried out by dissolving the hydrogen halide in a solvent, such as acetic acid or methylene chloride, or by bubbling the gaseous HX directly into the alkene

a halide ion (X*) are formed This step is highly endothermic and its energy of activation

is very high As a consequence, it takes place slowly, and it is the rate-determining step

of the reaction In Step 2, the halide ion reacts with the highly reactive carbocation by donating an electron pair to form an alkyl halide This exothermic step has a very low energy of activation So, it takes place rapidly

1.1.1.3 The Order of Reactivity of Hydrogen Halides

The order of reactivity of hydrogen halides is HI > HBr > HCl > HF The reactivity of hydrogen halides depends on their acid strength The acid strength decreases in the order:

HI > HBr > HCl > HF (this follows from the fact that the H — F bond is by far the strongest and the H — I bond is the weakest) and because of this, the order of reactivity of HX is

HI > HBr > HCl > HF

1.1.1.4 Markownikoff’s Rule

The problem of constitutional orientation (ascertaining which one of the two doubly bonded carbons ultimately forms bond with the electrophile and which one combines with the nucleophile) does not arise in the cases of addition of symmetrical reagents like Br2, Cl2, etc (addendum) to both symmetrical (e.g., CH2 == CH2,CH3

CH == CH CH3, etc.) and unsymmetrical substrates e.g., CH3CH == CH2, CH3CH == C(CH3)2, etc However, this is very important for the addition of an unsymmetrical reagent (e.g.,

H2O, H2SO4, HBr, HCl, etc.) to an unsymmetrical alkene In most cases, the problem can

be solved by applying a rule known as Markowniff's rule

Definition: If an unsymmetrical alkene combines with a hydrogen halide (H — X), the halide ion adds to the carbon atom bearing the lesser number of hydrogen atoms

The rule is further extended to the cases of other polar addition to unsymmetrical alkenes

It is stated as: ‘the negative part of an addendum adds to that carbon atom of the double bond of an unsymmetrical alkene which contains the lesser number of hydrogen atoms’

d + d

-== 2 + ææææY — ZÆ — 2 == 2 2 2

RCH CH RCHZ CH Y (YZ H O, H SO , ICl,etc.)

Modern statement: The modern statement of the Markownikoff ’s rule is as follows: In the ionic addition of an unsymmetrical addendum to a double bond of an unsymmetrical alkene, the positive portion of the addendum attaches itself to a doubly bonded carbon so

as to yield the more stable carbocation as an intermediate

The addition of HI to 2-methyl-2-butene takes place according to the Markownikoff's rule:

(minor product)(major product)

Explanation: The Markownikoff addition of HI to 2-methyl-2-butene may be explained

in terms of stabilities of carbocation intermediates that could be formed by attachment

of a proton to one or the other doubly bonded carbon atom Addition of the electrophile,

H!, to 2-methyl-2-butene may yield two different carbocations: tert-pentyl cation (a 3° carbocation) and isopentyl cation (a 2° carbocation) In tert-pentyl cation, the positive carbon is stabilized by the inductive effect involving two methyl and one ethyl groups and hyperconjugative effect involving eight a–H atoms On the other hand, in isopentyl cation, the positive carbon is stabilized by +I effect of two alkyl groups (isopropyl and methyl groups) and hyperconjugative effect involving four a–H atoms So, the tert-pentyl cation is relatively more stable than the isopentyl cation and the reaction leading to tert-pentyl cation has a lower energy of activation than the reaction leading to isopentyl cation The tert-pentyl cation is, therefore, formed at a faster rate than isopentyl cation and consequently, 2-iodo-2-methylbutane is obtained as the major product in this kinetically controlled reaction Hence, the composition of the products obtained is controlled by the stability of the carbocations involved

Regioselective reaction: A reaction in which two or more constitutional isomers (they differ in how their atoms linked) could be obtained as products but one of them predominates or obtained as the only product, is called a regioselective reaction Since 2-methyl-2-butene reacts with HI to produce 2-iodo-2-methylbutane predominantly and propene reacts with HBr to form 2-bromopropane exclusively, therefore, these reactions are regioselective reactions.

Energy diagram: The energy diagram for the addition of HI to (CH3)2C == CHCH3 is as follows:

Exceptions of Markownikoff ’s rule: Due to special structural features of alkenes addition often takes place not according to the Markownikoff ’s rule For example:

(i) CH2==CH—CF3+HCl ææÆCl CH CH CF2 2 3 (anti-Markownikoff product)

(ii) CH2==CH—N(CH ) I≈ 3 3 *+HIææÆICH CH N(CH ) I2 2≈ 3 3 * (anti-Markownikoff product)When H! adds to CH2 == CH –– CF3, one of the carbocations that could be formed is primary (I) and the other is secondary (II) In the carbocation II, the powerful electron-attracting – CF3 group is directly attached to the positive carbon and so, it destabilizes the carbocation

by intensifying the positive charge more than I in which the – CF3 group is not directly attached to the positive carbon As a result, the reaction proceeds through the relatively more stable carbocation I, although a primary one, to give the anti-Markownikoff product ClCH2CH2CF3

Since a primary carbocation is too unstable to be formed, the anti-Markownikoff product

is probably obtained through a concerted addition process

Addition of H! to the doubly bonded carbons of the ammonium salt may produce a primary carbocation (III) and a secondary carbocation (IV) Because of repulsive destabilization, the carbocation IV (although a secondary one) with adjacent positive charges is relatively less stable than the carbocation III (a primary carbocation) in which the charges are separated

by a saturated carbon atom Thus, the proton becomes attached to the carbon adjacent to nitrogen to give the relatively stable carbocation III Subsequent reaction of III with I①gives the anti-Markownikoff product, I CH CH N (CH ) I 2 2 ≈ 3 3 *

Another exception to Markownikoff ’s rule concerns the addition of HBr to alkenes when the addition is carried out in the presence of peroxides (i.e., compounds with the general formula ROOR) This (peroxide effect or Kharasch effect) will be discussed in article 1.3.4.3

1.1.1.5 Addition of HX to Cumulated Dienes

Cumulated dienes undergo electrophilic addition of HX in accordance with the Markownikoff ’s rule, even though attack by the electrophile (H!) at the sp hybridized carbon would give a stable allylic cation For example:

2 2 3 2

Br

|

CH C CH HBr CH C CHAllene 2 Brom propeneo

or 1, 2 Propadiene

== == + ææÆ ==

-

-—

Explanation: Addition of H! to the central sp hybridized carbon of allene or 1,2- propadiene does not give immediately a resonance-stabilized allylic cation because the three p orbitals are not parallel and it requires a rotation about the C — C bond for delocalization of electrons to take place Therefore, attack by H! at the central carbon produces initially a relatively unstable 1° carbocation However, when H! adds to a terminal

sp2 carbon, a relatively stable secondary vinylic cation is produced So, the activation energy for the Markownikoff addition is lower than that for the anti-Markownikoff addition and consequently, the addition occurs in accordance with the Markownikoff's rule

d+

Br—H 2-attack

d–

1-attack H—Br

d+ d–

1 2

C—C—CH

H

C—CH

H

+

HH

∫∫

Br + CH ==CH—CH2 2

An unstable 1° carbocation

90° rotationØ

CH ==CH—CH Br3-Bromopropene(not obtained)

+

1.1.1.6 Reaction of hydrogen bromide with 1, 3-butadiene (kinetic control

versus thermodynamic control of a chemical reaction)

When 1,3-butadiene is treated with HBr in the absence of air, a reaction occurs by an ionic mechanism to form both the 1,2- and 1,4-addition products However, the proportions in

which these two products are obtained are markedly affected by the temperature at which the reaction is carried out.

When the reaction is carried out at low temperature (–80°C), a mixture containing

80 percent of the 1,2-product and 20 percent of the 1,4-product is obtained, i.e., the major product is formed by 1,2-addition

Mechanism: The reaction proceeds through the following two steps:

Step 1: In this step (which is the same whether the 1,2-product or the 1,4-product is being formed), the addition of H! takes place at C-1 (but not at C-2 which give a much less stable 1° carbocation) of 1,3-butadien because the resulting carbocation is a relatively more stable (resonance-stabilized) allylic cation It is the rate-determining step of the reaction

Step 2: This is the product-determining step The allylic cation is, in fact, a resonance hybrid of a secondary allylic cation (I) and a primary allylic cation (II) Because a secondary allylic cation is more stable than a primary allylic cation, the contribution of I to the hybrid is greater than that of II Because of this, in the allylic cation, the interior carbon becomes relatively more positive than the terminal carbon and, therefore, the activation energy for the formation of 1,2-product is less than that for the formation of 1,4-product

As a consequence, the bromide ion attacks C-2 (the more positive carbon) more readily than C-4 (the less positive carbon) to give the 1,2-addtion product, 3-bromobut-1-ene, predominantly Under these conditions, the reaction is said to be under kinetic (or rate) control because the product formed more rapidly predominates in the product mixture The predominant product is called the kinetic product The 1,4-addition product 1-bromobut-2-ene, however, is relatively more stable than the 1,2-addition product 3-bromobut-1-ene and it is called the thermodynamic product

Energy profile diagram: This behaviour of 1,3-butadiene and hydrogen bromide can be more fully understood if we examine the energy profile diagram for the reaction shown in the following figure.

The energy of activation leading to the 1,2-product is less than that leading to the product As a result, the 1,2-product is formed at a rate faster than the 1,4-product and predominates at lower temperature Since an internal double bond is more stable than

1,4-a termin1,4-al double bond bec1,4-ause of hyperconjug1,4-ative effect, therefore, the 1,4-product

is thermodynamically more stable than the 1,2-product and so, its valley is placed at a lower level than that of the 1,2-product The product formation is irreversible because lower temperature does not provide sufficient energy to lift either product out of its deep potential energy valley

At higher temperature (40°C), the intermediate ions have sufficient energy to cross both the barriers with relative case Sufficient energy is also available for ionization of the products Therefore, both the reactions are now reversible Since the 1,4-product is thermodynamically more stable, its activation energy for ionization is much greater than that of the 1,2-product and consequently, the 1,4-product ionizes more slowly than the 1,2-

product Under these conditions, the 1,2-product, which is still formed rapidly, rearranges

to the more stable 1,4-product through the allyl cation because the overall change is energetically favourable So, when an equilibrium is established between the products, the more stable 1,4-product predominates in the product mixture

Under these conditions, the reaction is said to be under thermodynamic (or equilibrium) control because by equilibration the more stable product predominates The predominant product is called thermodynamic (or equilibrium) product

1 Which out of 2-hexene and 3-hexene is expected to be more suitable for using as the substrate for synthesizing 3-bromohexane(CH3CH2CHBrCH2

CH2CH3)? Give your reasoning

Solution Being an unsymmetrical alkene 2-hexene (CH3CH==CHCH2CH2CH3) reacts with HBr to form a mixture of 2-bromohexane (50%) and 3-bromohexane (~50%) and 3-bromohexane (~50%) and this is because the reaction proceeds through the formation of two carbocations However, the symmetrical alkene 3-hexene (CH3CH2CH==CHCH2CH3) reacts with HBr to form 3-bromohexane exclusively because the reaction proceeds through the formation of a single carbocation Therefore, 3-hexene is more suitable as the substrate for synthesizing 3-bromohexane

2 The addition of HBr to which of the following alkenes is more regioselective and why?

Methylenecyclohexane 1-MethylcyclohexeneSolution When H! adds to methylenecyclohexane, one of the carbocations that could

be formed is tertiary and the other is primary A tertiary carbocation is very much stable than a primary carbocation, which, in fact, is too unstable to be formed Thus, 1-bromo-1-methylcyclohexane will be only product of this reaction

On the other hand, when H! adds to 1-methylcyclohexene, one carbocation that could be formed is tertiary and the other is secondary Since a tertiary carbocation and a secondary carbocation are closer in stability, therefore, both 1-bromo-1-methylcyclohexane and 1-bromo-2-methylcyclohexane will be formed as the major product and the minor product, respectively.

Hence, the addition of HBr to methylenecyclohexane is relatively more regioselective

3 The addition of HCl to 3,3-dimethylbut-1-ene leads to the formation of

an unexpected product, 2-chloro-2,3-dimethybutane, in somewhat greater yield than 3-chloro-2,2-dimethylbutane, the expected Markownikoff product

(CH ) C — CH CH (CH ) CCHCl CH (CH ) CClCH (CH )3,3-Dimethylbut-1-ene 3-Chloro-2,2-dimethylbu tane 2-Chloro-2,3-dimethylbutane

(minor product) (major product)

Explain this observation

Solution The reaction starts like a normal addition of HCl, that is, by protonation of the double bond to yield the more stable carbocation with the greater number of alkyl groups

at the electron-deficient carbon Reaction of this carbocation with Cl* occurs as expected

to yield the expected Markownikoff product as the minor product

This intermediate secondary (2∞) carbocation undergoes rearrangement (the methyl group moves with its pair of bonding electrons from the carbon adjacent to the positive carbon)

to give a relatively more stable tertiary carbocation The major product is formed by the attack of Cl* on the new carbocation

4 When HCl is allowed to react with 3-methylcyclohexene, methylcyclohexane and 1-chloro-1-methylcyclohexane are obtained Draw a mechanism to explain this result.

1-chloro-3-Solution The addition of H! to 3-methylcyclohexene forms two different secondary carbocations However, one of them undergoes rearrangement by a hydride shift to yield a more stable tertiary carbocation Nucleophilic attack of Cl* on the secondary and the tertiary carbocations produces 1-chloro-3-methylcyclohexane and 1-chloro-1-methylcyclohexane, respectively

5 Explain the mechanism of the following reaction:

Solution When H! adds to a-pinene, a stable 3∞ carbocation is formed predominantly But this 3∞ carbocation readily rearranges to a less stable 2∞ carbocation which is normally not favourable In fact, the rearrangement expands a strained four-membered ring to a much less-strained five-membered ring, and this relief of strain provides a driving force for the rearrangement Nucleophilic attack of Cl* on the 2∞ carbocation forms bornyl chloride

6 Methylenecyclobutane react with hydrogen chloride to give ylcyclobutane (the anti-Markownikoff product), while methylenecyclo-hexane reacts with hydrogen chloride to yield 1-chloro-1-methylcyclo-hexane (the Markownikoff product) Account for these observations.Solution When H! adds to methylenecyclobutane, one of the carbocations that could

chlorometh-be formed is primary (I) and the other is tertiary (II) The 3∞ carbocation II suffers from considerable angle strain (120∞ – 90∞ = 30∞), whereas the 1∞ carbocation has small angle strain (109.5∞ – 90∞ = 19.5∞) Because of greater angle strain, the carbocation II, although

a 3∞ one, is less stable than the 1∞ carbocation I and the addition of HCl proceeds through the carbocation I to give anti-Markownikoff product chloromethylcyclobutane

Since a primary carbocation is too unstable to be formed, the anti-Markownikoff product

is probably obtained through a concerted addition process

When H! adds to methylenecyclohexane, one of the carbocations that could be formed

is tertiary (III) and the other is primary (IV) The carbocation III suffers from a very little angle strain Then, from electronic point of view the 3∞ cabocation III is relatively more stable than the 1∞ carbocation IV and therefore, the reaction proceeds through the carbocation III to give the Markownikoff product 1-chloro-1-methylcyclohexane

7 Predict the major product in each of the following reactions and give your reasoning.

(b) The addition of H! to the two sp carbons of propyne may produce one secondary

vinylic cation (I) and one primary vinylic cation (II) Since the secondary vinylic cation is relatively more stable (stabilized by the +I and hyperconjugative effect of the methyl group) than the primary vinylic cation, therefore, the reaction actually proceeds through the secondary vinylic cation which subsequently reacts with Br*

to yield 3-bromopropene

2-Bromopropene adds another molecule of HBr to form 2,2-dibromopropane as follows:

Addition of H! to the C-1 carbon of 2-bromopropene leads to the formation of the carbocation III in which bromine withdraws electrons by its –I effect and donates electron by its +R effect Since –I 〉 + R, bromine is net electron-withdrawing and

it destabilizes the carbocation But the +I and hyperconjugative stabilizing effect

of the two methyl groups is large enough to compensate this destabilization and

in fact, it is more stable than the 1∞ carbocation IV (that could be formed by the addition of H! to C-2) which is further destabilized by the –I effect of bromine The reaction thus proceeds through the carbocation III to give 2,2-dibromopropane (c) When H! adds to styrene (PhCH == CH2), one of the carbocation that could be formed

is secondary benzylic and the other is primary The secondary benzylic carbocation

is much more stable (stabilized by resonance) than the primary carbocation and for this reason, the reaction proceeds through the secondary benzylic carbocation

to give PhCHBrCH3 nearly exclusively

(d) Since one mole of HCl is allowed to react with this unsymmetrical diene, it adds

preferentially to the more reactive or more electron-rich C-1—C-2 double bond to give 5-methylhex-1-ene (CH2==CHCH2CH2CClMe2) as the major product through the formation of a stable 3° carbocation HCl also adds to the C-5—C-6 double bond

to give 5-chloro-2-methylhex-1-ene as the minor product through the formation of

a less stable 2° carbocation

8 Explain the following observation:

CH3

CH3

CH3Solution When H! adds to 1-methyl-1-vinylcyclobutane, one of the carbocations that could be formed is secondary and the other is primary The secondary carbocation readily rearranges by ring expansion (a 1,2-bond shift) to give a tertiary carbocation, which is more stable and, because of the five-membered ring, has less angle strain Thus, the reaction proceeds not through the 1∞ carbocation but through the rearranged 3∞ carbocation to give 1-bromo-1,2-dimethylcyclopentane exclusively

9 Explain the stereochemical outcome of the following reaction:

2 Ethylpent 1 ene ( ) 3 Bromo 3 methylhexane

Is the reaction stereoselective?

Solution Initial addition of the electrophile H! (from HBr) occurs from either side of the planar double bond to form a carbocation Both modes of addition (from above or below the plane) generate the same achiral carbocation

The positively charged carbon is sp2 hybridized, so the three atoms to which it is bonded lie in a plane When the bromide ion attacks the carbocation from above the plane, one enantiomer of 3-bromo-3-methylhexane is formed, but when it attacks the carbocation from below the plane, the other enantiomer is formed Because attack from either direction occurs with equal facility or equal probability, equal amounts of the two enantiomers are obtained, i.e., a racemic mixture (optically inactive) is obtained.

An addition reaction that forms a compound with one chirality centre from a reactant without any chirality centre is not stereoselective because it does not select for a particular stereoisomer Thus, the reaction is not a stereoselective one

10 Is the temperature at which a reaction changes from being kinetically

controlled to being thermodynamically controlled same for all the reaction? Explain

Solution The temperature at which a reaction changes from being kinetically controlled

to being thermodynamically controlled actually depends on the particular reaction Each reaction that is irreversible under mild conditions, a reversible under vigorous conditions has a particular temperature at which the changeover occurs In fact, the temperature required to shift the reaction of a conjugated diene with an electrophilic reagent from kinetic control to thermodynamic control depends on both the diene and the electrophilic reagent The reaction of 1,3-butadiene with HCl, for example, remains under kinetic control at 45∞C, even though the addition of HBr to the same diene is under thermodynamic control at that temperature Since a C — Cl bond is stronger than a C — Br bond, therefore, a higher temperature is required for the products to undergo the reverse reaction Thermodynamic control is achieved only when sufficient energy is applied to allow the reaction to be reversible

11 Give the major 1,2- and 1,4-addition products of the following reaction and

indicate which is the kinetic product and which is the thermodynamic product

+ HCl

CH2

H C3

Solution The addition of H! takes place at the methylene carbon because this leads to the formation of a carbocation in which the positive charge is shared by a tertiary allylic and a secondary allylic carbon If the proton were to add to the other end of the conjugated system, the carbocation would be less stable because in that case the positive charge would be shared by a primary allylic and a secondary allylic carbon Therefore, 3-chloro-3,4-dimethylcyclohexene is the 1,2-product and 3-chloro-1,6-dimethylcyclohexene is the 1,4-product.

3-Chloro-3,4-dimethylcyclohexene is the kinetic product because, in the transition state (relatively more stable) for its formation, the positive charge is concentrated on a tertiary carbon 3-chloro-1,6-dimethylcyclohexene is the thermodynamic product because it is relatively more stable due to its more highly substituted double bond (more stabilized by hyperconjugation)

1 What alkene should be used to synthesize 2-bromopentene and why?

2 C6H5CH == CHC6H5 reacts with HBr faster than CH3CH == CHCH3, even though both compounds are 1,2-disubstituted alkenes Explain

3 What would be the major product obtained from the addition of HBr to each of the following compounds?

5 Propose a mechanism for each of the following reactions:

[Hint: Me2CClCHMe2 (major) and Me2CCHClMe (minor)]

8 Compare the regioselectivity of the reactions of HBr with

(a) CF3CH == CH2, (b) BrCH == CH2 and (c) CH3OCH == CHCH3

9 Use the Hammond postulate to explain why (CH3)2C == CH2 reacts faster than

CH3CH == CH2 in electrophilic addition of HBr

10 Give the mechanism and the stereochemical course involved in the reaction that

take place when 1,2-dimethylcyclohexene is allowed to react with HB to yield 1-chloro-1,2-dimethylcyclohexane

11 Two stereoisomeric alkenes X and Y (molecular formula C6H12) react with HI

to give the same single product and undergo catalytic hydrogenation to produce hexane Identify X and Y

[Hint: X and Y are cis-and trans-3-hexene.]

12 Predict the products (stereoisomers) of the following reaction and give the

stereoselective?

13 Rank the following alkenes in order of their increasing rate of reaction with HBr

and explain the order:

14 Write down the structures of two different alkenes that would react with HBr to

yield each of the following compounds as the major (or only) addition product

Br

|(CH ) CCH(CH ) (b)

15 Predict the products of the following reaction and account for the formation of both

products:

(CH ) CH3 2 —CH==CH2+HBrææÆA (major) B (minor)+

[Hint: The major product is obtained through the rearrangement of the initial carbocation involving a hydride shift.]

16 Only one of the following three alkyl bromides can be prepared as the major product

of the addition of HBr to an alkene Identify the alkene and explain why the other two cannot be prepared as the major product

CH CH CH CH CH Br CH CHBr CH CH CH CH CHBr C(CH ) C H

CB

A

17 Predict the major product of each of the following reactions and show the

stereoisomers that could be obtained:

18 Suggest a mechanism for the following reaction:

19 Arrange the following compounds in order of increasing reactivity towards addition

20 Explain why the reaction (a) takes place according to the Markownikoff's rule,

while the reaction (b) does not:

(a) CH2==CHF+HFææÆCH CHF3 2

(b) CH2==CHCF3 +HFææÆFCH CH CF2 2 3

21 Predict the products expected to be formed when one mole of 1,3-pentadiene is

allowed to react with 1 mole of HBr What is the major product?

22 Identify the major 1,2- and 1,4-addition products expected to be obtained when

each of the following compounds is treated with HCl Indicate which is the kinetic product and which is the thermodynamic product

23 When acetylene is allowed to react with HCl, vinyl chloride is obtained through the

formation of vinyl cation However, when vinyl bromide is refluxed with alcoholic AgNO3, the same vinyl cation is not obtained

—

HC CH HCl CH C H Cl CH CH ClAcetylene vinyl cation Vinyl chloride

CH CH Br \/ CH C H AgBrVinyl bromide vinyl cation

≈

== æææææææÆ == + Ø

Explain these observations

[Hint: Vinyl bromide is thermodynamically more stable than acetylene because of resonance:

For this reason, the activation energy for the formation of vinyl cation from vinyl bromide is greater than that required for its formation from acetylene This explains why the same vinyl cation is generated from acetylene by addition of H! but not from vinyl bromide by C — Br bond cleavage.]

24 In HCl addition of some akenes (e.g., 2-methyl-1-butene, 2-methyl-2-butene,

1-methylcyclopentene, etc.), it is found that the anti-addition occurs following the rate law: rate = k [alkene] [HCl]2, while some alkenes (e.g., styrene, 1-phenyl propene, indene, etc.) that undergo syn-addition follow the rate law: rate = k [alkene] [HCl] Explain

[Hint: A third-order rate expression is observed in anti-addition of HBr because the T.S involves proton transfer to the alkene from one HCl molecule and capture

of Cl①

ion from the second The reaction occurs as follows:

The anti-stereochemistry is consistent with the expectation that the attack of Cl①occurs from the opposite side of the p bond to which the proton is delivered.

A second-order rate expression is observed in syn-addition of HCl because in addition of a proton and attack by Cl①

, only one HCl molecule is involved The stereochemistry is consistent with the expectation that the attack of Cl①

occurs from the same side of the p bond to which the proton is delivered In fact, an ion pair is formed by alkyne protonation and it collapses to the product faster than rotation takes places Syn -addition thus occurs because the proton and the halide ion are initially on the same face of the molecule

25 Predict the regiochemistry of the following addition reactions Explain your

as an acid catalyst These reactions, too, are also regioselective, and the addition of water

to the double bond occurs in accordance with the Markownikoff’s rule For example:

Mechanism: 2-Methylpropene undergoes acid-catalyzed hydration in accordance with the Markownikoff’s rule (both water and alkene are unsymmetrical) as follows: