Organic Chemistry: Concepts and Applications - Headley Allan D. 2020

Nucleophilic Substitution Reactions at sp3 Carbons
15.6 Bimolecular Substitution Reaction Mechanism (SN2 Mechanism)

For a nucleophilic substitution reaction to take place via a SN2 mechanism, the electrons of the nucleophile must get close enough to the partial positive electrophilic carbon of the electrophile to actually “push” out the leaving group, and in the process a new organic product is formed as shown in Reaction (15-18).

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If this general substitution reaction were visualized in slow motion, one would observe that as the nucleophile approaches the electrophile, a partial bond is formed between the nucleophilic atom and the electrophilic atom; at same time, the bond between the atom of the leaving group and the electrophilic carbon is partially broken. As the reaction proceeds further, the bond to the electrophile and the nucleophile is fully formed; and also, the bond to the leaving group and the electrophile is fully broken to form the products. Reaction (15-19) gives an illustration of this process.

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This type of reaction is described as bimolecular because two molecules, the nucleophile and the electrophile, are involved in this very important step of the mechanism for the formation of the products; this is called the rate-determining step (RDS). As a result, this mechanism is described as a SN2 mechanism, substitution nucleophilic bimolecular, and Figure 15.1 gives the energy profile for the mechanism of this type of substitution reaction.

15.6.1 The Electrophile of SN2 Reactions

For a reaction to proceed via this pathway, or mechanism, it is obvious that there must be enough space for the nucleophile to approach the electrophilic carbon. A methyl halide, which has three very small hydrogen atoms bonded to the electrophilic carbon, has enough space to accommodate the nucleophile and, as a result, a substitution reaction involving a methyl halide proceeds very favorably by this mechanism. On the other hand, a tertiary halide, which has three much larger alkyl groups bonded to its electrophilic carbon, does not have enough space for the nucleophile to approach the electrophilic carbon of the electrophile. Hence, a substitution reaction involving a tertiary halide will proceed extremely slowly to the point that it is not considered to proceed by this mechanism (SN2 mechanism) as illustrated below.

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Figure 15.1 Energy profile of SN2 reaction of a nucleophile with an electrophile.

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Figure 15.2 Energy profile of SN2 reaction where Ea1 represents the reaction with an electrophile that is not crowded, such as methyl halide, and Ea2 represents the reaction with a more crowded electrophile.

An alternate way of understanding this concept is that the energy of activation for a substitution reaction involving a methyl halide is less than the energy of activation for the reaction involving a more crowded alkyl halide as illustrated in Figure 15.2.

Thus, the order of reactivity of electrophiles for substitution reactions that proceed by a SN2 mechanism is methyl > 1o > 2o > 3o (the reactivity of the 3° is so slow that it is considered unreactive by this mechanism).

Order of reactivity of electrophiles for substitution reactions that proceed by a SN2 mechanism: methyl > 1o> 2o>>> 3o

Problem 15.6

Which of the following pairs of molecules would react the fastest by a SN2 reaction mechanism?

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15.6.2 The Nucleophile of SN2 Reactions

For a substitution reaction to proceed by this SN2 mechanism, it should now be obvious that the nucleophile must be small enough in order to gain access to the electrophilic carbon. The term nucleophilicity is often used to describe the ability of a nucleophile to gain access to the electrophilic carbon of an electrophile. The nucleophilicity trend is shown in Figure 15.3. This trend is based on the effective size of the nucleophile in a particular solvent.

Thus, the most favorable substitution reactions are those in which there is a good nucleophile (small) and a methyl or primary halide. As pointed out earlier, nucleophiles are also Lewis bases, which contain at least one unshared pair of electrons. As a result, a nucleophile for an SN2 reaction cannot be a strong base since as you might expect, an acid—base reaction, which leads to an elimination reaction will take place instead of the desired substitution reaction. Large strong bases, such as lithium diisopropylamide (LDA) or potassium tert-butoxide, that were discussed in Chapter 7 and shown in Figure 15.4 are typically never used for substitution reactions (Figure 15.10).

You will recall from Chapter 12 that strong bases are required for elimination reactions, specifically E2 reactions, and the nucleophiles described in this chapter can also serve as bases for elimination reactions. Thus, there is always a competition between substitution reactions vs. elimination reactions when these reactions are carried out. Thus, it is very important that the appropriate reaction conditions are in place to ensure the desired type of reaction and eventually the desired organic product. To ensure substitution reactions, a small weak base is preferred. On the other hand, to ensure elimination reaction, strong bases are required, often at elevated temperatures. Strong bulky bases such as potassium tert-butoxide or LDA are often used to essentially guarantee elimination reactions. As you might have guessed, bulky electrophile in the presence of even a small strong base will undergo substantial elimination reaction. Problem 15.7 is designed to have students combine the knowledge gained from Chapter 12 on elimination with the information discussed thus far so to determine possible products.

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Figure 15.3 Nucleophilicity trend for nucleophiles of SN2 reactions.

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Figure 15.4 These bulky strong Lewis bases do not serve as good nucleophiles; instead, they serve primarily as strong bases for acid—base reactions.

Problem 15.7

Give the substitution and elimination products for each of the reactions shown below. Indicate which product (elimination or substitution) do you think would be the major product.

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15.6.3 The Solvents of SN2 Reactions

As you might expect, these reactions are carried out in a solvent and the nature of the solvent can influence the outcomes of substitution reactions. Nonpolar, aprotic solvents are typically the best solvents for substitution reactions that proceed via this mechanism pathway. The fluoride anion is a very small nucleophile owing to fluorine being the most electronegative atom of the periodic table. Hence, it is a good nucleophile for SN2 reactions, but its ability to act as a good nucleophile depends of the solvent. In polar protic solvents, such as alcohols, it is highly solvated, which increases its effective size, and does not make it a good nucleophile for SN2 reactions in these types of solvents, as illustrated in Figure 15.5.

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Figure 15.5 Solvated fluoride anion with a protic solvent, methanol.

On the other hand, in a nonpolar aprotic solvent, such as acetone, the fluoride is not highly solvated, and hence, it is effectively a small and effective nucleophile. Table 15.1 gives a description of commonly used solvents of organic chemistry.

Table 15.1 Common protic and aprotic organic solvents.

Solvent

Type of solvent

Structure

Water

Protic-polar

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Acetone

Nonprotic-polar

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Methanol

Protic-polar

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Ethanol

Protic-polar

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Dimethyl sulfoxide (DMSO)

Nonprotic-polar

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Diethyl ether

Nonprotic

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Tetrahydrofuran (THF)

Nonprotic-polar

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Carbon tetrachloride

Nonprotic-polar

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Problem 15.8

i. Of the following pairs of nucleophiles, which would be a better nucleophile in an aprotic polar solvent, such as dimethyl sulfoxide (DMSO)?Image

ii. Select different solvents that could be used to carry out the following SN2 reactions.Image

15.6.4 Stereochemistry of the Products of SN2 Reactions

One important outcome of substitution reactions that proceed via the SN2 mechanism is if the electrophilic carbon is stereogenic, the product has the opposite configuration compared to the starting material. A look at Reaction (15-20) explains why an inversion takes place. If the nucleophile approaches from the opposite side of the leaving group, then an inversion of configuration takes place to form the product.

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Problem 15.9

For the following reactions, give the major organic product and assign absolute configuration of the chiral carbon in both reactant and product.

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Thus, this process can be used to synthesize a particular enantiomer by taking advantage of the stereospecificity of SN2 reactions and a reaction that was covered in Section 15-3-5 in which an OH group was converted to a good leaving group using TsCl. This concept is shown in Reaction (15-21).

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Problem 15.10

Show how to carry out the following transformations, carefully note the stereochemistry in both the reactant and product.

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In analyzing SN2 reactions to predict the organic products, the following factors should be considered.

1. Electrophile: CH3 > 1o > 2o ≫̸ 3 o

2. Stereochemistry: stereospecific reaction in which inversion results. (Remember that a stereospecific reaction converts a reactant with a specific stereochemistry into another as a product.)

3. Nucleophile: small and weak base.

4. Leaving group: need leaving groups that are weak stable conjugate bases (good leaving group)

5. Solvent: must be nonionizing, which are typically nonprotic solvents. Good solvents include acetone and dimethyl sulfoxide (DMSO).

6. Mechanism: concerted in that the attack of the nucleophile and the leaving of the leaving group occur in a single step.

15.6.5 Intramolecular SN2 Reactions

Note that it is possible for the nucleophile and the electrophile to be on the same molecule. In this case, the reaction is described as an intramolecular substitution reaction as shown in Reaction (15-22), the mechanism is shown below.

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A similar intramolecular substitution SN2 reaction can be used to synthesize oxiranes as shown in Reaction (15-23).

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For these reactions, a weak base, such as sodium hydroxide, can be used to shift the equilibrium in the direction of the formation of an alkoxide ion for the reaction to occur as shown in Reaction (15-24). As expected, there are other organic products including the elimination products and substitution products.

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We have already discussed in Chapter 8 that halohydrin can be formed from the addition reaction of water and bromine to alkenes. Using an alkene as a starting compound for the synthesis of an epoxide by a combination of these reactions is illustrated in Reaction (15-25).

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