LON-CAPA Sn2

Electrophilic Addition Reactions

Introduction

We have seen that treatment of an alkyl halide with a strong base can lead to the 1,2-elimination of hydrogen halide and the formation of an alkene as outlined in general terms in Scheme 1.

Scheme 1

Dehydrohalogenation: The Essentials

From one perspective this transformation may be considered an acid-base reaction in which the alkyl halide acts as an acid in the sense that it is a source of the elements of hydrogen halide, H and X. Viewed in this way, it is understandable that the reverse reaction, the conversion of an alkene to an alkyl halide, requires the addition of H and X to the alkene. In other words, the reaction of an alkene with an acid is an acid-base reaction in which the alkene is the base. This is the Lewis definition of a base as an electron pair donor; the alkene uses its pi electrons to form a bond with the hydrogen atom of the hydrogen halide. Scheme 2 outlines the process.

Scheme 2

Hydrohalogenation: The Essentials

Experimentally the reaction is often accomplished simply by mixing the alkene with the hydrogen halide in an inert solvent. Kinetic measurements indicate that this is a 2-step process. In the first step HX delivers a proton to the alkene to produce an intermediate carbocation. This intermediate then reacts rapidly with the halide ion to produce the final product. The first step involves addition of the electophilic proton to one of the doubly bonded carbon atoms. The second step involves addition of the nucleophilic halide ion to the other carbon atom. Since the first step is the rate determining step, the overall reaction is classified as an electrophilic addition reaction.

Exercise 1 What is the valence shell electron configuration of H⁺? Is this a filled shell, i.e. does H⁺ obey the filled shell rule ? Yes No

What is the valence shell electron configuration of Na⁺? Is this a filled shell, i.e. does Na⁺ obey the filled shell rule ? Yes No

Does Cl^- obey the filled shell rule ? Yes No

Exercise 2 When ethene, H₂C=CH₂, is treated with HCl, chloroethane, CH₃CH₂Cl, is formed rapidly. However, no reaction occurs when ethene is treated with NaCl. How does this support the idea that the reaction between ethene and HCl is an electrophilic addition?

In Scheme 2 the four substituents attached to the double bond are not specified. This is a reflection of the fact that the sigma-bonded framework of an alkene may be treated independently of the pi bond to which it is perpendicular. Having said that, we will now look at situations where the substituents obviously influence the outcome of the addition of electrophilic reagents to alkenes.

The Regiochemistry of Electrophilic Additions

Figure 1 presents the results of several reactions involving the electrophilic addition of HX to unsymmetrically substituted alkenes.

Figure 1

There's a Pattern Here

Exercise 3 Following the reaction sequence outlined in Scheme 2, draw the structures of the intermediate carbocations that are formed in reactions 1-4. Classify each carbocation as 1^o, 2^o, or 3^o. (Enter a 1, 2, or 3 in the approriate text box.)

Reaction 1 Reaction 2 Reaction 3 Reaction 4

Exercise 4 Following the reaction sequence outlined in Scheme 2, draw the structures of the intermediate carbocations that would be formed in reactions 1-4 if the proton were to add to the other carbon of the double bond. Classify each carbocation as 1^o, 2^o, or 3^o. (Enter a 1, 2, or 3 in the approriate text box.)

Reaction 1A Reaction 2A Reaction 3A Reaction 4 A

Exercise 5 Compare the carbocations formed in reactions 1-4 with their hypothetical counterparts in the alternative reactions 1A-4A. Select the reaction that produces the more stable intermediate. (Enter 1 or 1A, 2 or 2A, etc. in the text box.)

Reaction 1/1A Reaction 2/2A Reaction 3/3A Reaction 4/4A

The two possible products in reactions 1and 1A, 2 and 2A, etc. are called regioisomers. The word reflects the fact the proton adds to a different "region", i.e. carbon atom, of the double bond in reaction 1 than it does in reaction 1A, etc.

Exercise 6 Rationalize the result that reaction 4 is preferred over 4A despite the fact that a 2^o carbocation is involved in both reactions. Draw a structural diagram of the intermediate formed in reaction 4 to support your rationalization.

Exercise 7 State the correlation between the experimentally observed regiochemistry and the relative stabilities of the alternative carbocationic intermediates in the addition of hydrogen halides to unsymmetrically substituted alkenes.

The regiochemical outcomes summarized in Figure 1 suggest that addition of HX to an unsymmetricaly substituted alkene only produces one regioisomer. In fact, that's not always true. The product distribution depends on the relative stabilities of the alternative carbocationic intermediates. For example, addition of HCl to 2-pentene produces 2-chloropentane and 3-chloropentane in nearly equal amounts:

The carbocations leading to these two alkyl halides are both 2^o. They have about the same stability. They yield the regioisomeric products in about the same amounts. The greater the difference in stability between the two carbocations that can be produced by protonation of an unsymmetrically substituted alkene, the greater the difference in the amounts of the regioisomeric alkyl halides that are formed when those carbocations react with halide ions.

In our discussion of the Sn1 mechanism we saw that solvolysis of 2-bromo-2-methylpropane in aqueous ethanol yielded two substitution products:

Ethanol and water both act as nucleophiles toward the intermediate carbocation. This result suggests the possibility outlined in Scheme 3.

Scheme 3

Extending Basic Principles

In other words, treatment of an alkene with aqueous HCl should lead to the addition of H and OH as well as H and Cl to the double bond. The addition of "the elements of" water, H and OH, to an alkene is called hydration. It is one of the most useful methods for preparing alcohols.

Hydration of Alkenes

The goal of organic synthesis is always the preparation of a target molecule in the highest yield and the highest purity possible. To that end, Scheme 3 is flawed as a method of preparation of 2-methyl-2-propanol. The 2-chloro-2-methylpropane is an undesired side product. Every molecule of 2-methylpropene that is converted into 2-chloro-2-methylpropane is one less molecule that's available to form the desired product.

One way around this difficulty is to use an acid whose conjugate base is less nucleophilic than water. The two most common alternatives are sulfuric acid, H₂SO₄, and phosphoric acid, H₃PO₄. Their conjugate bases, HSO₄^-, and H₂PO₄^- are much less nucleophilic than water. Figure 2 details the hydration of propene with aqueous sulfuric acid. Compare the net result of this reaction, which adds H and OH to the double bond, to reaction 1 in Figure 1, which adds H and Br to the double bond.

Figure 2

Are You Dizzy Yet?

Notice that the sulfuric acid is catalytic in this reaction. In principle only one H₂SO₄ molecule is required to initiate the cycle wherein hydronium ions protonate the alkene and water molecules deprotonate the oxoniumion to produce the product.

Exercise 8 Draw Lewis structures for phosphoric acid and its conjugate base.

Exercise 9 The conjugate bases of sulfuric acid and phosphoric acid are relatively non-nucleophilic because They are large ions. Sulfuric and phosphoric acid are strong acids. These ions are stabilized by resonance. Sulur and phosphorous both have expanded valence shells.

Exercise 10 Create a scheme similar to that in Figure 2 for reactions 2-4 in Figure 1.

Exercise 11 What is the difference between hydration and hydrolysis? Give an example of each type of reaction.

Figure 2 uses propene as a specific example of the acid catalysed hydration of alkenes. Not all alkenes react at the same rate. In fact, as the data in Table 1 shows, more highly substituted alkenes are more reactive than propene toward electrophilic additon.

Table 1

Relative Reactivities of Alkenes

Name	Structure	k₂(M^-1sec^-1)	k_rel
propene	CH₃CH=CH₂	4.95 x 10^-8	1
Z-2-butene	CH₃CH=CH CH₃	8.32 x 10^-8	1.68
E-2-butene	CH₃CH=CH CH₃	3.51 x 10^-8	0.71
2-methyl-2-butene	CH₃CH=C (CH₃)₂	2.15 x 10^-4	4651
2,3-dimethyl-2-butene	(CH₃)₂C=C (CH₃)₂	3.42 x 10^-4	6909

Exercise 12 Draw the structures of the carbocations formed by protonation of each of the alkenes in Table 1. Then classify each carbocation as primary, secondary, or tertiary:

The carbocation formed from propene is

The carbocation formed from Z-2-butene is

The carbocation formed from E-2-butene is

The carbocation formed from 2-methyl-2-butene is

The carbocation formed from 2,3-dimethyl-2-butene is

While the extra CH₃ group lowers the energy of 2-methylpropene relative to propene, it lowers the energy of the carbocation that is produced by protonation of the double bond even more, i.e. a 3^o carbocation is more stable than a 2^o carbocation. The energy of the transition state leading to the 3^o carbocation is reduced similarly, and the activation energy for protonation of 2-methylpropene is less than that for propene.

The final aspect of the electrophilic addition reactions of alkenes that we'll consider is stereochemistry.