We have seen that electrophilic addition reactions of alkenes involve a 2-step process. In the first step, which is rate-determining, an electrophilic reagent adds to the pi bond. In the second step a nucleophilic reagent adds to the electron deficient intermediate that was formed in the first step. Now we are going to take a look at a situation where that sequence is reversed, i.e. nucleophilic addition precedes electrophilic addition. This chemistry is typical of aldehydes and ketones, which may be thought of as the oxygen analogs of alkenes. Therefore, it may be helpful if we begin our discussion of nucleophilic addition reactions with a comparison of alkenes and aldehydes and ketones.
Figure 1 compares the structures and boiling points of the alkenes and their carbonyl cousins.
Exercise 1 Draw pictures to represent the intermolecular interactions that are responsible for the fact that formaldehyde has a higher boiling point than ethene.
Exercise 2 Would you expect 2-methylpropene oracetone to be more soluble in water?
Because the pi electron distribution is unsymmetrical, aldehydes and ketones are more polar than their hydrocarbon analogs. Formaldehye, for example, is completely miscible with water. In fact, the dipole-dipole interactions between formaldehyde and water are strong enough that they lead to covalent bond formation as shown in Figure 3.
The process animated in Figure 3 is represented alternatively in Equation 1.
The process outlined in Equation 1 may be generalized as shown in Equation 2 where R and R' represent H, alkyl, or aryl groups. If the atom that is attached to the carbonyl carbon is a heteroatom, i.e. a halogen, nitrogen, oxygen, or sulfur, a different course of reaction is observed.
As the arrows in Equations 1 and 2 indicate, the addition of water to an aldehyde or ketone is an equilibrium. The value of the equilibrium constant depends upon the substituents attached to the carbonyl carbon. Table 1 lists the values of the equilibrium constants for formaldehyde, acetaldehyde, acetone, and chloral.
Exercise 4 Write the equilibrium expression for Equation 1. What is the value of Keq when [H2C=O] = 0?
Exercise 5 Write equations analagous to Equation 1 for the addition of water to acetaldehyde, acetone, and chloral. Write the corresponding equilibrium expressions.
Exercise 6 Is the decrease in Keq that occurs on going from formaldehyde to acetone more likely due to changes in electronegativity steric factors hyperconjugation ?
The chemistry expressed in Equation 2 may be generalized even more as shown in Equation 3.
In other words, it is possible to added the "elements of" an alcohol, H and OR, to an aldehyde or ketone in the same way you can add the "elements of" water, H and OH. If you start with an aldehyde, the product formed is called a hemiacetal. If you start with a ketone, it is called a hemiketal.
Figure 2 offers a mechanistic rationalization of Equation 3.
The reaction is initiated by the addition of a proton to the pi bond of the carbonyl group, generating a resonance stabilized carbocation. A molecule of alcohol bonds to the carbonyl carbon of this intermediate, forming a protonated alcohol. Deprotonation yields the observed product. Note that the hydrogen ion is catalytic in this reaction.
Exercise 9 Draw the structure of the ion that would be formed if the alcohol bonded to the oxygen atom rather than the carbon atom of the carbocation intermediate. Why isn't that pathway observed?
Imagine starting with 1000 molecules of the aldehyde and 1000 molecules of the alcohol. Once the reaction comes to equilibrium 969 molecules of product will have been formed and 31 molecules of starting material will remain: [969]/[31][31]= 1.008. A simple way to increase the amount of product would be to increase the amount of alcohol in the mixture. If you were to increase the number of alcohol molecules from 1000 to 1200, at equilibrium the would be 995 molecules of product formed and 5 molecules of aldehyde and 205 molecules of alcohol remaining: [995]/[5][205] = 0.97. In other words, using an additional 200 molecules of alcohol results in the formation of 26 more molecules of product. To force the conversion of all the aldehyde to product, you would use a large excess of the alcohol. Unfortunately, this complicates the situation since hemiacetals and hemiketals also react with alcohols to form acetals and ketals.
When an aldehyde or ketone is mixed with an alcohol in the presence of a catalytic amount of acid, several concurrent equilibria are established. The outcome of these coupled reactions depends upon the equilibrium constants of the individual reactions and upon the reaction conditions as mentioned above. Figure 3 summarizes several of the equilibria that are involved in the conversion of an aldehyde or ketone to an acetal or ketal, respectively.
Equation 4 reduces the transformation described in Figure 3 to its essentials.
Exercise 11 Two of the equilibria in Figure 3 involve proton transfers. Write equations to depict a proton transfer from each of the intermediate oxonium ions to a. a molecule of an alcohol b. a molecule of aldehyde or ketone.
Whenever a molecule contains an aldehyde or ketone and an alcohol functional group, the possibility exists for an intramolecular addition of the alcohol to the carbonyl group as shown in Equation 5. The product of reaction 5 is known as a cyclic hemiacetal or a cyclic hemiketal, depending on whether the carbonyl group was part of an aldehyde or a ketone, respectively.
Exercise 12 Write equations analogous to Equation 5 for compounds in which n = 1, R = H, CH3; n = 2, R = H, CH3; n = 3, R = H, CH3.
Exercise 13 Note that there is a chiral center generated in Equation 5. If the alcohol group adds to the top face of the pi bond, one enantiomer is formed. If it adds to the bottom face, the other enantiomer is formed. Draw the structure of the product that would be formed if the addition occurred as shown below. Would the addition yield the R or S configuration at the new chiral center?
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