In our discussion of the aldol reaction we saw that competing reactions occur when aldehydes and ketones are treated with hydroxide ion. One the one hand, the hydroxide ion may add to the carbonyl carbon in a nucleophilic addition reaction, while on the other it may abstract an a-hydrogen. As the color coding in Figure 1 indicates that duality of reaction pathways stems from the fact that both the carbonyl carbon and the a-hydrogen of the aldehyde or ketone are electrophilic. For purposes of the following discussion it is important to note that addition of hydroxide ion to the carbonyl carbon generates a tetrahedral intermediate labeled T-1 in Figure 1.
Alternative Reaction Pathways for Aldehydes and Ketones
Aldehydes and Ketones vs. Esters
Esters are structurally related to aldehydes and ketones: all three classes of compounds contain a carbonyl group. It shouldn't be surprising then that esters display similar reactivity patterns to aldehydes and ketones. As we shall see, they also display some interesting and significant differences. Figure 2 presents an analogous scheme to that shown in Figure 1 for the reaction of simple esters with hydroxide ion. Nucleophilic addition of hydroxide to the carbonyl carbon generates a tetrahedral intermediate T-2.
Alternative Reaction Pathways for Esters
One of the most favorable pathways available to these tetrahedral intermediates involves regeneration of the carbonyl group. This is favorable because the carbonyl group is an especially stable entity. Figure 3 compares the alternative ways in which the tetrahedral intermediates T-1 and T-2 might regenerate the carbonyl group.
Alternative Fates of Tetrahedral Intermediates
In the case of aldehydes and ketones, regeneration of the carbonyl group is a reasonable alternative only when it results in expulsion of hydroxide ion from T-1. As we have seen, regeneration of the starting material by this path allows for the less likely alternative, abstraction of an a-hydrogen and the formation of aldol condensation products which follow that event.
Exercise 1 What is the approximate equilibrium constant for each of the alternatives shown in Figure 3?
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In the case of T-2, regeneration of the carbonyl group may be achieved by expulsion of hydroxide ion or alkoxide ion. In fact, the latter possibility is preferred. To understand why, it is necessary to consider the equilibria shown in Figure 4.
The equilibrium constant for expulsion of alkoxide ion from T-2 is approximately 1. Note that the product of this pathway is a carboxylic acid. Since this acid is formed in a strongly basic solution, it will be deprotonated rapidly. Given that the pKa of a carboxylic acid is about 5, the equilibrium constant for its deprotonation is approximately 1011. In other words, while the expulsion of hydroxide ion from T-2 (Figure 3) is about as likely as expulsion of alkoxide ion, the latter pathway is greatly preferred because a subsequent reaction has a much larger equilibrium constant. Consequently, treatment of an ester with aqueous sodium hydroxide results in the formation of the conjugate base of a carboxylic acid. More information is available.
In our discussion of the reactions of aldehydes and ketones with hydroxide ion, we saw that addition to the carbonyl carbon was more probable than abstraction of an a-hydrogen, but that the latter pathway was the one followed because the addition reaction was non-productive. In the case of structurally similar esters, i.e. esters containing at least one a-hydrogen, the more probable reaction is a productive reaction. It produces carboxylic acids. The process is called saponification.
The formation of carboxylic acids by treatment of esters with sodium hydroxide is known as saponification. Equation 1 summarizes the net transformation for the saponification of methyl salicylate, a fragrant component in oil of wintergreen.
The general procedure involves refluxing the ester in 6M NaOH until the mixture becomes homogeneous, indicating complete formation of the water-soluble carboxylate salt, RCO2-. Acidification of the mixture during the work-up produces the carboxylic acid.
Equation 2 provides another example of saponification of an even simpler ester, ethyl acetate.
Exercise 2 Draw the structures of the carboxylic acid and the alcohol that would be produced in each of the following reactions:
Exercise 3 Saturated fats are esters that may be represented by the general formula . Typically the values of x, y, and z range from 8-20. Saponifaction of fats produces glycerol (1,2,3-propanetriol) and three molecules of sodium carboxylates. The mixture of these compounds is used as soap. Acidification of the saponification mixture produces fatty acids. Draw the structures of glycerol and the three fatty acids that would be formed when x=8, y=10, and z=12.
Exercise 4 Saponification of esters is a specific example of a more general reaction type called nucleophilic acyl substitution. It is typical of derivatives of carboxylic acids- esters, acyl halides, amides, and anhydrides- and may be summarized in general terms by the following scheme:
Write equations for nucleophilic acyl substitution reactions where X = OH, and G = OCH3, Cl, N(CH3)2, and OC(=O)CH3. Estimate the equilibrium constant for each reaction. Hint- You need to know the pKa values of the conjugate acids of each G group.
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Exercise 5 Predict the relative reactivities of esters, acyl halides, amides, and anhydrides in nucleophilic acyl substitution reactions.
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NaOH vs NaOCH2CH3
Suppose that you were to treat ethyl acetate with sodium ethoxide rather than sodium hydroxide. What would happen? The answer; the Claisen condensation.
Exercise 6 Create a scheme similar to that shown in Figure 2 but using -:OR rather than -:OH as the nucleophile/base. Label the tetrahedral intermediate T-3.