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7.01 Introduction to Substitution and Elimination Reactions Alkyl halides commonly undergo two general types of reactions. When treated with a nucleophile, an alkyl halide can undergo a substitution reaction, in which the nucleophile replaces the halogen. When treated with a base, an alkyl halide can undergo an elimination reaction, in which a π bond (an alkene) is formed:

During our coverage of substitution and elimination reactions, we will use the term substrate to refer to the alkyl halide The halogen withdraws electron density via induction, rendering the adjacent carbon atom electrophilic, and therefore subject to attack by a nucleophile

The halogen can serve as a leaving group, and substitution/elimination processes can only occur when a leaving group is present. Good leaving groups are the conjugate bases of strong acids As a rule of thumb, a good leaving group is the conjugate base of an acid with a pKa < 0. For this reason, chloride, bromide, and iodide are all good leaving groups, while fluoride is not (the pKa of HF is 3.2).

7.02 Nomenclature and Uses of Alkyl Halides

Structure of Alkyl Halides In an alkyl halide, each carbon atom can be described in terms of its proximity to the halogen, using letters of the Greek alphabet. The alpha (α) position is the carbon atom connected directly to the halogen, while the beta (β) positions are the carbon atoms connected to the α position:

An alkyl halide will have only one α position, but there can be as many as three β positions. Alkyl halides are classified as primary (1°), secondary (2°), or tertiary (3°) based on the number of alkyl groups connected to the α position.

Naming Alkyl Halides Recall from Section 4.02 that systematic (IUPAC) names of alkanes are assigned using four discrete steps:  1.Identify and name the parent.  2.Identify and name the substituents.  3.Number the parent chain and assign a locant to each substituent.  4.Assemble the substituents alphabetically.

7.03 SN2 Reactions

Kinetics These observations are consistent with a second-order process that has the following rate equation: Rate=k [alkyl halide] [nucleophile]

Based on their observations, Ingold and Hughes concluded that the mechanism must exhibit a step in which the alkyl halide and the nucleophile collide with each other. Because that step involves two chemical entities, it is said to be bimolecular. Ingold and Hughes coined the term SN2 to refer to bimolecular nucleophilic substitution reactions:

stereospecificity of SN2 Reactions There is another crucial piece of evidence that led Ingold and Hughes to propose the concerted mechanism. When the α position is a chiral center, a change in configuration is generally observed, as illustrated in the following example:

The reactant exhibits the R configuration, while the product exhibits the S configuration. That is, this reaction is said to proceed with inversion of configuration The requirement for inversion of configuration means that the nucleophile can only attack from the back side (the side opposite the leaving group) and never from the front side (Figure 7.2). There are two ways to explain why the reaction proceeds through back-side attack:  1.The lone pairs of the leaving group create regions of high electron density that effectively block the front side of the substrate, so the nucleophile can only approach from the back side.  2.Molecular orbital (MO) theory provides a more sophisticated answer. Recall that molecular orbitals are associated with the entire molecule (as opposed to atomic orbitals, which are associated with individual atoms). According to MO theory, the electron density flows from the HOMO of the nucleophile into the LUMO of the electrophile. As an example let's focus our attention on the LUMO of methyl bromide (Figure 7.3).

The transition state (drawn in brackets) will soon be discussed in more detail. This reaction is said to be stereospecific, because the configuration of the product is dependent on the configuration of the starting material. The Effect of Substrate Structure on the Rate of an SN2 Process TABLE7.2EFFECT OF SUBSTITUENTS ON THE RATES OF SN2 REACTIONS R─Br→acetone25°CI⊝R─I+Br⊝ Effect of βSubstituents

Effect of α-Substituents Structure

Relative ratea

Structure

Relative ratea

H3Cα─Br

145

1

1

0.8

0.008

0.04

Negligible

0.00001

All of these observations are consistent with a concerted process in which the nucleophile encounters steric hindrance as it approaches the alkyl halide. To understand the nature of the steric effects that govern SN2 reactions, we must explore the transition state for a typical SN2 reaction, shown here in general form.

Notice that dotted lines are used to indicate the bonds that are in the process of being broken or formed, and the double-dagger symbol, just outside the brackets, indicates that the drawing represents a transition state rather than an intermediate

The highest point on the curve (highlighted) represents the transition state. The relative energy of this transition state determines the rate of the reaction. If the transition state is high in energy, then the energy of activation (Ea) will be large, and the rate will be slow. If the transition state is low in energy, then Ea will be small, and the rate will be fast. With this in mind, we can now explore the effect of steric hindrance on the reaction rates of SN2 processes, and we can justify why tertiary alkyl halides are unreactive toward SN2.

If the hydrogen atoms in Figure 7.5 are replaced with alkyl groups, steric interactions cause the transition state to be even higher in energy, raising Ea for the reaction. Compare the relative energy diagrams for reactions involving methyl, primary, and secondary alkyl halides (Figure 7.6).

For a methyl halide, Ea is relatively small, and the reaction is rapid. In contrast, Ea for the reaction of a secondary alkyl halide is relatively high, and the reaction is slower. With a tertiary alkyl halide, there are three substituents connected to the α position, and the SN2 transition state is so high in energy that the reaction occurs too slowly to be observed. For similar reasons, an SN2 process will also not occur when the starting alkyl halide has three substituents connected to

a β position. As an example, we revisit the structure of neopentyl bromide (first shown in Table 7.2):

This compound is a primary alkyl halide, but it has three methyl groups attached to the β position. These methyl groups provide steric interactions that raise the energy of the SN2 transition state (Figure 7.7), and as a result, the rate of SN2 is too slow to be useful.

7.04 Nucleophilic Strength and Solvent Effects in SN2 Reactions

Nucleophilicity The term “nucleophilicity” refers to the rate at which a nucleophile will attack a suitable electrophile. A strong nucleophile will give a relatively fast SN2 reaction, while a weak nucleophile will give a relatively slow SN2 reaction. For this reason, a strong nucleophile is generally required in order for an SN2 reaction to be efficient and practical.

Another factor that impacts nucleophilicity is polarizability, which is often even more important than charge. Recall that polarizability describes the ability of an atom to distribute its electron density unevenly as a result of external influences. Polarizability is directly related to the size of the atom and, more specifically, to the number of electrons that are distant from the nucleus.

For SN2 reactions, the nucleophile is generally ionic, as is the leaving group, so a polar solvent is required in order to solvate these ionic species. Furthermore, the transition state also has ionic character, so a polar solvent helps stabilize the transition state as well. For these reasons, SN2 reactions generally cannot be performed in nonpolar solvents, such as benzene, unless special

techniques are employed (we will see one such technique in Chapter 13). Polar solvents are broadly classified into two categories: protic and aprotic. Protic solvents contain a hydrogen atom connected directly to an electronegative atom, while polar aprotic solvents lack such a hydrogen atom. Several examples are shown for each category:

Notice that each of the protic solvents has at least one hydrogen atom (shown in red) connected to an electronegative atom (such as O or N). Aprotic solvents also have hydrogen atoms, but none of those hydrogen atoms are connected to an electronegative atom. It is observed that SN2 reactions are generally much faster when performed in polar aprotic solvents, rather than protic solvents. TABLE7.3EFFECT OF SOLVENTS ON THE RATES OF SN2 REACTIONS

Nucleoph ile

Relative rate in MeOHa (protic solvent)

Relative rate in DMFa (polar aprotic solvent)

Increase in rate (due to polar aprotic solvent)

I⊝

43b

6,500b

× 150

Br⊝

1

21,000

× 21,000

Cl⊝

0.04

41,000

× 1,030,000

F⊝

0.0006

49,000

× 82,000,000

A couple of trends become apparent when we compare the data in Table 7.3:  •For each of the nucleophiles shown, the rate of reaction is significantly enhanced when the solvent is changed from protic (methanol) to polar aprotic (DMF). The extent of the rate enhancement is different for each nucleophile, although fluoride is the most affected by the change in solvent—the rate is 82 million times faster in a polar aprotic solvent. These data indicate that the



use of a polar aprotic solvent will speed up the rate of an SN2 process by many orders of magnitude. •If we compare the relative rates of the halides in methanol (protic solvent), we find that iodide reacts most rapidly, and fluoride is the least reactive. However, if we compare the relative rates in DMF (aprotic solvent), we find the opposite trend—fluoride reacts most rapidly.

To understand these observations and trends, we must explore the interaction between the solvent and the nucleophile. When NaCl is dissolved in a protic solvent, such as water, both the Na+ and Cl− ions are well solvated.

Protic solvents have electronegative atoms with lone pairs (in this case oxygen) that can stabilize Na+ ions, and protic solvents also have the ability to form hydrogen bonds with the Cl− ions, thereby stabilizing them as well. In contrast, a polar aprotic solvent can only stabilize the cations, not the anions. For example, dimethylsulfoxide (DMSO) can stabilize the Na+ ions, but it lacks the ability to stabilize the Cl− ions via hydrogen bonds. Why? In DMSO, the electron-poor region (δ+) is located at the center of the compound, surrounded by lone pairs and methyl groups (highlighted below). For steric reasons, this δ+ region is relatively inaccessible to an anion dissolved in the solvent, so anions are not stabilized by the solvent.

The result is that nucleophiles are less stabilized (higher in energy) when placed in a polar aprotic solvent. Polar aprotic solvents enhance the rate of an SN2 process by raising the energy of the nucleophile, giving a smaller Ea (Figure 7.9).

In protic solvents, fluoride ions can form very strong hydrogen bonds with the solvent, so the fluoride ions are very tightly bound to their solvent shell, rendering them mostly unavailable to function as nucleophiles. In such an environment, fluoride ions are weak nucleophiles because they would have to shed part of their solvent shell, which would be a large increase in energy. However, when a polar aprotic solvent is used, the fluoride ions do not form hydrogen bonds and they are not tightly bound to the solvent shell, rendering them fully available to function as nucleophiles. The case of fluoride is a powerful illustration of the fact that nucleophilicity is strongly dependent on the identity of the solvent. Indeed, nucleophilicity can be sensitive to other factors as well, including the identity of the starting alkyl halide and the temperature. 7.06 Introduction to E2 Reactions At the beginning of this chapter, we noted that alkyl halides can undergo substitution as well as elimination reactions. Thus far, we have exclusively explored substitution (SN2) reactions of alkyl halides. In this section, we will begin to explore elimination reactions. These reactions involve the use of bases, rather than nucleophiles, so we must quickly review some differences between strong bases and weak bases. When treated with a strong base, an alkyl halide can undergo a type of elimination process, called beta elimination, also known as 1,2-elimination, in which a proton is removed from the β position, the halide is ejected as a leaving group (X−) from the α position, and a double bond is formed between the α and β positions:

When an alkyl halide undergoes a beta elimination reaction, the process is also called a dehydrohalogenation, because H and X are removed from the substrate.

Both mechanisms involve proton transfer and loss of a leaving group, but consider the timing of these events. In the first mechanism, both events occur simultaneously. That is, they occur in a concerted fashion. However, in the second mechanism, they occur separately, in a stepwise fashion

Kinetics These observations are consistent with a second-order process that has the following rate equation: Rate=k [alkyl halide] [base]

Based on these observations, we conclude that the mechanism must exhibit a step in which the alkyl halide and the base collide with each other. Because that step involves two chemical entities, it is said to be bimolecular. Bimolecular elimination reactions are called E2 reactions:

The Effect of Substrate Structure on the Rate of an E2 Process In Section 7.04, we saw that the rate of an SN2 process with a tertiary alkyl halide is generally so slow that it can be assumed that the alkyl halide is inert under such reaction conditions. It might therefore come as a surprise that tertiary alkyl halides undergo E2 reactions quite rapidly A substitution reaction occurs when the reagent functions as a nucleophile and attacks the α position, while an elimination reaction occurs when the reagent functions as a base and

abstracts a proton from a β position. With a tertiary substrate, steric hindrance prevents the reagent from functioning as a nucleophile at an appreciable rate, but the reagent can still function as a base without encountering much steric hindrance (Figure 7.10).

7.07 Nomenclature and Stability of Alkenes In the previous section, we saw that alkyl halides can undergo beta elimination reactions to give alkenes. Before we continue our coverage of beta elimination reactions, we must first familiarize ourselves with some features of alkenes, including rules for naming them as well as the relationship between structure and stability. Recall that naming an alkane, or an alkyl halide, requires four discrete steps:  1.Identify the parent.  2.Identify the substituents.  3.Assign a locant to each substituent.  4.Arrange the substituents alphabetically. Alkenes are named using the same four steps, with the following additional guidelines:  When naming the parent, replace the suffix “ane” with “ene” to indicate the presence of a C═C double bond:



When choosing the parent of an alkene, use the longest chain that includes the π bond:

When numbering the parent chain of an alkene, the π bond should receive the lowest number possible despite the presence of alkyl substituents:

IUPAC nomenclature also recognizes common names for many simple alkenes. Here are three such examples:

IUPAC nomenclature recognizes common names for the following groups when they appear as substituents in a compound:

In addition to their systematic and common names, alkenes are also classified by their degree of substitution.

Stability of Alkenes In general, a cis alkene will be less stable than its stereoisomeric trans alkene. The source of this instability is attributed to steric strain exhibited by the cis isomer. This steric strain can be visualized by comparing space-filling models of cis-2-butene and trans-2-butene

The difference in energy between stereoisomeric alkenes can be quantified by comparing their heats of combustion:

Both reactions yield the same products. Therefore, we can use heats of combustion to compare the relative energy levels of the starting materials (Figure 7.12). This analysis suggests that the trans isomer is 4 kJ/mol more stable than the cis isomer.

When comparing the stability of alkenes, another factor must be taken into account, in addition to steric effects. We must also consider the degree of substitution. By comparing heats of combustion for isomeric alkenes (all with the same molecular formula, C6H12), the trend in Figure 7.13 emerges. Alkenes are more stable when they are highly substituted. Tetrasubstituted alkenes are more stable than trisubstituted alkenes. This effect, called hyperconjugation, is a stabilizing effect because it enables the delocalization of electron density. In a similar way, alkyl groups can also stabilize the neighboring sp2hybridized carbon atoms of a π bond. Once again, the resulting delocalization of electron density is a stabilizing effect.

7.08 Regiochemical and Stereochemical Outcomes for E2 Reactions Regioselectivity of E2 Reactions In many cases, an elimination reaction can produce more than one possible product:

In this example, the β positions are not identical, so the double bond can form in two different regions of the molecule. This consideration is an example of regiochemistry, and the

reaction is said to produce two different regiochemical outcomes. Both products are formed, but the more substituted alkene is generally observed to be the major product. For example:

The reaction is said to be regioselective. This trend was first observed in 1875 by Russian chemist Alexander M. Zaitsev (University of Kazan), and as a result, the more substituted alkene is called the Zaitsev product. However, many exceptions have been observed in which the Zaitsev product is the minor product. For example, when both the substrate and the base are sterically hindered, the less substituted alkene is often the major product. The less substituted alkene is called the Hofmann product.

TABLE7.4PRODUCT DISTRIBUTION OF AN E2 REACTION AS A FUNCTION OF BASE

BASE

ZAITSEV

HOFMANN

71%

29%

28%

72%

8%

92%

Stereoselectivity of E2 Reactions The examples in the previous sections focused on regiochemistry. We will now focus our attention on stereochemistry. This substrate has two identical β positions so regiochemistry is not an issue in this case. Deprotonation of either β position produces the same result. But in this case, stereochemistry is relevant, because two possible stereoisomeric alkenes can be obtained:

Both stereoisomers (cis and trans) are produced, but the trans product predominates. Consider an energy diagram showing formation of the cis and trans products (Figure 7.14). Using the Hammond postulate (Section 6.06), we can show that the transition state for formation of the trans alkene must be more stable than the transition state for formation of the cis alkene. This reaction is said to be stereoselective because the substrate produces two stereoisomers in unequal amounts.

Stereospecificity of E2 Reactions In the previous example, the β position had two different protons. In such a case, b...