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

Elimination Reactions of Organic Chemistry
12.3 Elimination of Hydrogen and Halide (Dehydrohalogenation)

In the next sections of this chapter, we will determine how to apply the concepts of these two reaction mechanisms in order to achieve desired elimination products for different reactions. The first mechanism that we will examine is the E2 mechanism, in which a proton and a halogen anion are eliminated from a molecule in a basic medium to produce the elimination product. Reaction (12-19) gives a specific reaction that proceeds via the E2 mechanism.

(12-19)Image

We have seen from Reaction (12-5) that there are two possible products for these elimination reactions. Consider the reaction given in Reaction (12-19), since this reaction is carried out in a strong basic medium, an alkoxide anion, the mechanism is more than likely an E2. The base is relatively small and will abstract a proton to result in the more stable alkene as the major product. Note that since the carbon—carbon bond that is bonded to the leaving group can rotate, at some point making an acidic hydrogen trans to the leaving group, to give the E2 product. The E2 mechanism for the Reaction (12-19) is shown in Reaction (12-20).

(12-20)Image

The minor product will be obtained via the mechanism given in Reaction (12-21). In this mechanism, the base abstracts a proton from one of the methyl groups adjacent to the bromide leaving group.

(12-21)Image

If the same reaction were carried out in the absence of a strong base and just a solvent such as ethanol, the loss of the acidic hydrogen is much more difficult. As a result, the carbocation will be formed since the bromide is a good leaving group as shown in Reaction (12-22).

(12-22)Image

If the leaving group is an extremely good leaving group, such as the bromide or an iodide anion, this first step is very likely. In the presence of the solvent, which is only slightly basic, abstraction of the proton occurs to give the final alkene product, as shown in Reaction (12-23). Note that the solvent abstracts the proton to give the more stable alkene product, which is the alkene with more alkyl groups bonded to the alkene carbons.

(12-23)Image

You may be wondering why the Br does not abstract the proton to form the alkene. In this case, the Br is a much weaker base than that of C2H5OH. Remember that the pKa of HBr is −9.0; whereas, the pKa of protonated alcohols are around 0.0 (the pKa of H3O+ is −1.7). Thus, the major and minor elimination products of this E1 mechanism result from the intermediate carbocation that is formed in the RDS, and Reaction (12-24) shows the final steps to give both possible products.

(12-24)Image

As pointed out earlier, the alkene that has the most alkyl groups bonded to the carbons of the double bond is the most stable alkene, also known as the Zaitsev product, compared to the alkene with the least alkyl groups bonded to the carbons of the double bond, this product is known as the Hofmann product. The elimination reaction of another molecule, 1-bromo-1-methylcyclohexane, is shown in Reaction (12-25).

(12-25)Image

For this reaction, more than likely the mechanism is E1. The steps for the E1 mechanism are shown in Reactions (12-26) through (12-28). In the first step of the mechanism, the very good leaving group, Br, leaves to form a carbocation and a bromide anion.

(12-26)Image

Owing to the positive character of the carbocation, the electrons of the bonds adjacent to the charge will be attracted to that carbon. As a result of this shift in electron density, the hydrogens on the adjacent carbons are relatively acidic, and the pair of electrons of the C─H bond will serve to neutralize the carbocation by forming another bond, a pi (π) bond, and the result is the formation of a double bond as shown in Reaction (12-27).

(12-27)Image

In the final step of the mechanism, the bromide anion abstracts a proton from the protonated alcohol to form two very stable and neutral compounds as shown in Reaction (12-28).

(12-28)Image

By increasing the base strength of an elimination reaction, the percentage of elimination product by way of E2 mechanism is increased, compared to the E1 pathway. For example, when the reaction below is carried out in ethanol, compared to ethanol in the presence of potassium ethoxide, there is a higher percentage of the elimination product that proceeds via the E2 mechanism, compared to the E1 mechanism, as shown in Reaction (12-29).

(12-29)Image

Thus, by altering the reaction conditions, an elimination reaction can be made to proceed predominantly by either of the mechanisms described in this section, E2 or E1.

Problem 12.4

Give the major elimination products for each of the following reactions.

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