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

Elimination Reactions of Organic Chemistry
12.4 Elimination of Water (Dehydration)

In the presence of an acid, alcohols are protonated, and the very poor leaving ─OH group is converted to an extremely good leaving group, H2O. Once water leaves with its bonding electrons, a carbocation is formed, hence the likely mechanism is an E1 mechanism as shown in the example given in Reaction (12-30).

(12-30)Image

12.4.1 Dehydration Products

Now that we have established that the most likely mechanism is E1, a close examination of the E1 reaction mechanism will give a rationale for the formation of the different dehydration products shown in Reaction (12-30). In the first step of the mechanism, the alcohol OH group is protonated, and it is converted to a good leaving group, H2O. In the next step, a cation is formed by the elimination of water as shown in Reaction (12-31).

(12-31)Image

In the next step of the E1 mechanism, water abstracts a proton from the adjacent carbon of the carbocation to form the alkene products as shown in Reactions (12-32a) and (12-32b).

(12-32a)Image

(12-32b)Image

Note that the alkenes formed from Reactions (12-32a) and (12-32b) are the same. Abstraction however from the proton from the methyl group results in a different alkene product, as shown in Reaction (12-33).

(12-33)Image

As pointed out earlier, the more stable alkene product is the one that has the most alkyl substitution about the carbon—carbon double bond; Thus, the alkene product shown in Reactions (12-32a) and (12-32b) is the major product and Reaction (12-34) gives a summary of the elimination product distribution.

(12-34)Image

Let us examine a slightly more complex alcohol, in which it is possible to produce even more possible elimination products. Shown in Reaction (12-35) is the first step in the dehydration reaction of 1,2-dimethylcyclohexanol to form a carbocation.

(12-35)Image

From this carbocation, the elimination of either of the hydrogens of the three adjacent carbons results in three different products as shown in Reactions.

(12-36)Image

(12-37)Image

(12-38)Image

Note that there are three alkenes possible from the initially formed carbocation, the disubstituted, trisubstituted, and tetrasubstituted alkenes. We have seen from the study of the stability of alkenes that the more substituted alkenes are the more stable alkenes. Hence, the more stable alkene will be the major organic product for the above reaction (Reaction 12-38). Another example is shown in Reaction (12-31).

(12-39)Image

At this point, students should be able to predict the major product, based on knowledge of the type of mechanism and a rationale for predicting the major product based on the mechanism. Problem 12.5 is designed to get students to think analytically through this process.

Problem 12.5

Give the major organic product for the following reactions.

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12.4.2 Carbocation Rearrangement

Carbocations once formed will rearrange to form a more stable carbocation if possible. Since carbocations are formed for the E1 mechanism, unexpected elimination products are sometimes observed. An example is shown in Reaction (12-40), where an unexpected product is obtained due to rearrangement of a cation intermediate.

(12-40)Image

Since this is an E1 mechanism and involves the formation of a carbocation, the expected product is formed by an abstraction of a proton from the adjacent methyl group as shown in Reaction (12-41).

(12-41)Image

As we have seen before, it is possible for a secondary carbocation to become a more stable tertiary carbocation by migration of an adjacent methide group to form a more stable tertiary carbocation as shown in Reaction (12-42).

(12-42)Image

In the last step of the mechanism, deprotonation of this tertiary carbocation occurs to form the alkene as shown in Reaction (12-43).

(12-43)Image

Problem 12.6

Provide a step-by-step mechanism to explain the formation of the unexpected products shown for the following reactions.

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12.4.3 Pinacol Rearrangement

Let us now consider a diol, in which the alcohol functionalities are adjacent to each other. In the presence of an acid, the protonation of one OH functionality will occur and then water leaves to form a carbocation, as shown in Reaction (12-44).

(12-44)Image

The secondary carbocation formed in Reaction (12-44) is fairly stable since it is a secondary carbocation, but if there were an alkyl migration from the adjacent carbon, a carbocation with greater stability would result since the new carbocation would be in resonance with the electrons of the adjacent oxygen atom as shown in Reaction (12-45). Deprotonation of the carbocation intermediate results in a final product, which is an aldehyde.

(12-45)Image

Dehydration of diols of this type is known as the pinacol rearrangement. Note that the pinacol rearrangement occurs for 1,2-diols. As you can imagine, there are many possible products depending on the type of diol and which OH group is protonated. The protonation of the other ─OH group for the reaction given in Reaction (12-46) results in the same product as shown in Reactions (12-46) and (12-47).

(12-46)Image

Even though this carbocation is different from the one generated in Reaction (12-26), the migration of the methyl group generates the same resonance-stabilized carbocation, which gives the same aldehyde.

(12-47)Image

Thus, for this pinacol rearrangement, the overall reaction is shown in Reaction (12-48).

(12-48)Image

Problem 12.7

Predict the pinacol rearranged products of the following reactions.

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