Alkenes and Aromatics

By Peter Taylor

The Royal Society of Chemistry

Copyright © 2002 The Open University
All rights reserved.
ISBN: 978-0-85404-680-5

Contents

PART 1 ADDITION – PATHWAYS AND PRODUCTS,
1 ELECTROPHILIC ADDITION REACTIONS OF ALKENES, 11,
2 OTHER USEFUL ADDITION REACTIONS, 28,
LEARNING OUTCOMES, 32,
QUESTIONS: ANSWERS AND COMMENTS, 33,
FURTHER READING, 36,
ACKNOWLEDGEMENTS, 36,
PART 2 AROMATIC COMPOUNDS,
1 INTRODUCTION, 39,
2 THE STRUCTURE AND STABILITY OF BENZENE, 42,
3 ELECTRO PHI LIC AROMATIC SUBSTITUTION REACTIONS, 45,
4 THE EFFECTS OF SUBSTITUENTS, 61,
5 DIAZOIMIUM SALTS, 68,
APPENDIX SUMMARY OF REACTIONS USEFUL IN THE SYNTHESIS OF AROMATIC COMPOUNDS, 75,
LEARNING OUTCOMES, 77,
QUESTIONS: ANSWERS AND COMMENTS, 78,
FURTHER READING, 88,
ACKNOWLEDGEMENTS, 88,
PART 3 A FIRST LOOK AT SYNTHESIS,
1 STRATEGY FOR THE DISCOVERY OF NEW DRUGS, 91,
2 COMPOUNDS THAT MIMIC THE ACTION OF NOR ADRENALINE — AGONISTS, 96,
3 THE TARGET: β-AMINO ALCOHOLS, 101,
4 PLANNING THE SYNTHESIS OF PSEUDOEPHEDRINE, 102,
5 CARRYING OUT THE SYNTHESIS, 111,
6 THE SYNTHESIS OF β-AMINO ALCOHOLS, 122,
7 A DIFFERENT WAY OF LOOKING AT SYNTHETIC EFFICIENCY, 124,
LEARNING OUTCOMES, 130,
QUESTIONS: ANSWERS AND COMMENTS, 131,
FURTHER READING, 136,
ACKNOWLEDGEMENTS, 136,
CASE STUDY: INDUSTRIAL ORGANIC CHEMISTRY,
1 INTRODUCTION, 139,
2 PETROCHEMICALS, 144,
3 SPECIALITY AND FINE CHEMICALS, 162,
4 WHAT HAPPENS WHEN THE FOSSIL FUELS RUN OUT?, 170,
FURTHER READING, 172,
ACKNOWLEDGEMENTS, 172,
INDEX, 173,

CHAPTER 1

ELECTROPHILIC ADDITION REACTIONS OF ALKENES

1.1 Introduction

Addition reactions are those in which atoms or groups add to a molecule containing a double or triple bond, thereby reducing the degree of unsaturation; they are the reverse of elimination reactions. Some typical examples of addition reactions are shown below:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.1)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.2)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.3)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.4)

We shall concentrate on addition to alkenes. Although alkynes tend to undergo the same types of reaction, we shall not discuss their reactions in detail.

* Alkenes generally react by ionic mechanisms involving nucleophiles and electrophiles. Give definitions for each of these terms.

* nucleophile can be defined as a species possessing at least one non-bonded pair of electrons, which ultimately forms a new bond to carbon. An electruphile is a positively charged or positively polarized species that reacts with a nucleophile.

Alkenes generally provide the nucleophilic component of the addition. You may fiind it hard to picture how an alkene can act as a nucleophile. Figure 1.1 shows the bonding picture of a carbon-carbon double bond. Carbon-carbon double bonds are niatle up of a strong σ bond plus a weaker π bond. The two electrons in the π bond dominate the chemistry of alkenes. They can be thought of as providing a negatively charged cloud of electrons above and below the plane of the carbon atom framework. This electron-rich centre repels nucleophiles and attracts electrophiles.

So it is the pair of electrons in the π bond that acts as the nucleophile in the reactions of alkenes. Alkenes are certainly electron-rich, but they do not contain a non-bonded pair of electrons. However, although the π electrons are bonding electrons, they do react with electrophiles, as you will see. This is because the π electrons are polarizable; that is, they are far enough from the carbon nuclei to be susceptible to the influence of electrophiles.

One of the most characteristic reactions of alkenes is electrophilic addition, as exemplified by the addition of halogens (X2) and hydrogen halides (HX) across the double bond:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.5)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.6)

These reactions can be shown to proceed by a two-step mechanism, in which the first step involves reaction between the alkene and an electrophile. Reaction 1.7 shows the simplest form of this mechanism that is encountered. Notice that although this reaction is called an electrophilic addition reaction, the alkene is a nucleophile. This is because reactions are generally named after the nature of the reagent, and in this case the reagent is electrophilic.

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.7)

Look at the mechanism of the electrophilic addition reaction carefully, and try to understand the changes in the bonding.

Two of the electrons from the π system of the alkene form a new bond to the electrophile, which is given the symbol E+. The carbocation intermediate formed in this first step then reacts with a nucleophile, Nu-, to give the reaction product. So, in order for reactions such as this to occur, an alkene must be treated with a reagent that provides both an electrophile and a nucleophile.

1.2 Addition of HX

Reactions 1.8, 1.9 and 1.10 show that hydrogen iodide, hydrogen bromide and hydrogen chloride, respectively, all add to alkenes:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.8)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.9)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.10)

* Think back to the mechanism that we proposed for electrophilic addition. Which species do you think acts as the electrophile in Reactions 1.8, 1.9 and 110?

* All the hydrogen halides ionize as H+ and X-. The proton, H+, is a strong electrophile.

So the mechanisms of Reactions 1.8 and 1.9 are straightforward:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.11)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.12)

In principle, the addition of an electrophile to an alkene can lead to two different carbocation intermediates:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.13)

However, in Reactions 1.8 and 1.9 the two alkenes are symmetrically substituted, so the same carbocation is produced no matter which carbon-hydrogen bond is formed in the first step. For example:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.14)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.15)

* In principle, how many carbocation intermediates can be formed from the protonation of 2-methylpropene, (CH3C=CH2?

* This is an unsymmetrical alkene, so two distinct carbocation intermediates are possible, depending on whether the proton bonds to the central atom of the alkene or to the terminal carbon atom of the double bond. In theory, therefore, this reaction could lead to two products, 1.1 and 1.2:

[FORMULA NOT REPRODUCIBLE IN ASCII]

In practice, the only product is 2-chloro-2-methylpropane (1.1). The reason for this predominance is apparent when the relative stabilities of the two intermediate carbocations are considered.

* Which is more stable, a tertiary or a primary carbocation?

* The order of carbocation stabilities is

tertiary > secondary > primary > methyl (see Box 1.2)

A tertiary carbocation is more stable than a primary one because of the indluctive donating effect of the three alkyl groups attached to the charged carbon atom.

In electrophilic addition reactions the major product arises from the more-stable carbocation intermediate, because that reaction pathway has the lower energy of activation (see Figure 1.2). So the reactions are kinetically controlled; that is, the major product is the one that is formed faster. Other unsymmetrically substituted alkenes also give the major product by way of the more-stable carbocation intermediate. Thus, Reaction 1.16 proceeds via a secondary, rather than a primary, carbocation; and Reactions 1.17 and 1.18 proceeds via a tertiary, rather than a secondary, carbocation.

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.16)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.17)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.18)

As you can see from Reactions 1.16, 1.17 and I.18, in order to generate the more-stable carbocation, the proton adds to the less-substituted carbon atom of the double bond — that is, to the one that already has more hydrogen atoms attached.

The generalization that, in the addition of HX to an unsyrnnzetrical alkerze, the hydrogen always adds to the less-substituted carbon atom of the double bond, was made long before a mechanistic explanation was available. A Russian chemist, Vladimir Markovnikov (Box 1.3), put forward this empirical rule — now known as Markovnikov’s rule — after studying the products of a number of different HX addition reactions. Markovnikov’s rule is sometimes summarized as ‘to the one who has, will more be given’, because the hydrogen atom from HX goes to the alkene carbon atom with the greater number of hydrogens already attached.

So far, the alkenes that we have considered have been unsymmetrically substituted (for example, CH3CH=CH2) or symmetrically disubstituted (for example, CH3CH=CHCH3). A third possible category consists of alkenes, such as pent-2-ene, in which both carbon atoms of the double bond have the same number of alkyl substituents, but the substituents are different.

* How many products would you expect from the reaction of pent-2-ene, CH3CH=CHCH2CH3, with HBr?

* You might expect two products, because both possible carbocations formed by the addition of a proton are secondary.

In reactions such as this, a mixture of both possible products usually does result:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.19)

Note that Markovnikov’s rule applies only to the addition of HX. The earlier mechanistic discussion, however, applies to any electrophilic addition reaction: the major product will always result from the more-stable carbocation. We shall call this the mechanistic Markovnikov rule (see also Section 1.3).

* So, what will be the predominant product when HBr is added to C6H5CH=CHCH3?

* In both cases, a secondary carbocation is formed; however, 1.4 will be more stable than 1.3 because in 1.4 the positive charge is adjacent to a phenyl group.

This means that the charge can be spread further by resonance:

[FORMULA NOT REPRODUCIBLE IN ASCII]

Since the spreading of charge leads to a more-stable carbocation, 1.5 will be the predominant product.

The hydration of alkenes falls into the same category as the addition of HX. Although this reaction is rarely carried out in the laboratory, it is an important industrial process for preparing alcohols from alkene feedstocks, which in turn are obtained from crude oil. The alkene is usually passed into a 1:1 mixture of sulfuric acid and water. With 2-methylpropene, for example, 60–655% aqueous sulfuric acid is used to prepare 2-methylpropan-2-ol, presumably by way of the corresponding tertiary carbocation:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.20)

Hydrations are initiated by protons, so Markovnikov’s rule is again followed. as you can see from this reaction. The hydration reaction is the reverse of the acid-catalysed elimination of alcohols: low temperatures and a reaction in aqueous solution favour alcohol formation, whereas elimination is favoured by high temperatures and distillation of the alkene as it is formed.

This balance between addition and elimination can be exploited in order to bring about the migration of a double bond within a molecule. If an alkene is treated with a dilute acid, protonation followed by deprotonation can occur:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.21)

An equilibrium is established, which will favour the alkene that is more thermodynamically stable. This is usually the most-substituted alkene, so this method is sometimes used to convert a less-substituted alkene into a more-substituted one; hence, for example, in Reaction 1.22

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.22)

1.3 Addition of halogens and related compounds

Alkenes react readily with chlorine and bromine to produce 1,2-dihalides:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.23)

Let’s concentrate on the addition of bromine, although all the subsequent discussion applies equally well to the addition of chlorine. The mechanism of this reaction is not immediately obvious because bromine is a non-polar molecule. However, the addition does proceed by an ionic mechanism in which the halogen molecule provides the electrophile. In isolation, bromine is a covalent molecule with a symmetrical electron distribution. However, in the presence of the high electron density of the alkene double bond, polarization of bromine occurs, and one bromine atom becomes electrophilic (Figure 1.4).

By analogy with the addition of HX, the mechanism for the addition of bromine to an alkene can be written as shown in Reaction 1.24:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.24)

(The analytical applications of this type of reaction are discussed in Box 1.4.)

However, this is not the complete story. In the reaction of trans-but-2-ene with bromine, for example, the only product is the diastereomer that results from one bromine atom approaching from one face of the substrate and the other bromine atom approaching from the opposite face:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.25)

* Would this result be expected if a simple carbocation intermediate were involved?

* By analogy with the SN1 mechanism, you would expect the bromide ion to attack the carbocation indiscriminately. If this were the case, two products (diastereomers) would result:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.27)

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.28)

To explain these stereochemical observations, the postulated mechanism has to be slightly modified. One possibility is that, although a carbocation intermediate is still formed, a non-bonded electron pair on the bromine atom is donated to the carbocation to form a three-membered ring containing a positively charged bromine. This is called a cyclic bromonium ion. (In general, halogen additions proceed via a cyclic halonium ion; thus, if the electrophile were Cl2, the intermediate would be a chloronium ion.)

Reaction 1.29 shows the first two steps of the mechanism. The relative disposition of the substituents in the three-membered ring reflects that on the alkene, because the stxond step in Reaction 1.29 occurs very quickly, before rotation about the central1 carbon-carbon bond can occur. In fact, it occurs so fast that we usually represent Reaction 1.29 as a single step:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.30)

Once the bromonium ion intermediate has been formed, the bromide ion then attack!;, opening up the three-membered ring:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.31)

This is very much like an SN2 reaction, and attack occurs from the side opposite the ‘leaving group’. As you can see, the overall result is the anti-addition of bromine, and the product is rneso-2,3-dibroniobutane.

* Would attack of bromide ion at the other carbon atom of the alkene double bond (as shown in Structure 1.7) lead to a different product?

* This again leads to the overall anti-addition of bromine, but because of the internal symmetry of the molecule, the same product results, as shown in Scheme 1.4. (The two structures shown are simply two conformations of the same molecule, meso-2,3-dibromobutane.)

* What would be the final product of addition if, as is equally likely, the initial attack of the bromine molecule was from below the plane of the alkene (as in 1.8 rather than from above as in Reaction 1.30)?

* This time, the bromide ion subsequently attacks the bromonium ion so formed it either carbon atom from above, to give the same meso product, as shown in Scheme 1.5:

[FORMULA NOT REPRODUCIBLE IN ASCII]

So, symmetrically substituted trans alkenes lead only to meso products. However, symmetrically substituted cis alkenes react with bromine to produce enantiomeric products. This can be seen from the reaction of cis-but-2-ene:

[FORMULA NOT REPRODUCIBLE IN ASCII]

The notion of an intermediate bromonium ion to explain the observed anti-addition of bromine to alkenes was first suggested in 1938. As yet, bromonium ions have not been isolated, although they have been detected using spectroscopy. So. the two-step mechanism for electrophilic addition is substantiated by stereochemical and spectroscopic evidence.

Other evidence, which is also incompatible with a one-step addition mechanism, is obtained if the bromination reaction is carried out in the presence of other nucleophiles. Bromination reactions are normally carried out in an inert solvent such as tetrachloromethane. However, use of a solvent that is itself nucleophilic, such as methanol, results in a mixture of products:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.32)

* How would you account for the formation of the bromomethoxy compound in this reaction?

* Methanol competes with bromide ion as the nucleophile to capture the bromonium ion intermediate, and yields an ether as a byproduct:

[FORMULA NOT REPRODUCIBLE IN ASCII] (1.33)

(Continues…)Excerpted from Alkenes and Aromatics by Peter Taylor. Copyright © 2002 The Open University. Excerpted by permission of The Royal Society of Chemistry.
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