Hydrocarbons are among the weakest acids known. Conversely, their conjugate bases are some of the strongest bases there are. Why is it so hard to remove a proton from a carbon atom? The answer is because the conjugate base is very unstable. When an excess of electron density develops on a carbon atom, the nucleus of that atom cannot offer much Coulombic attraction to stabilize it. Consider the hypothetical acid-base reactions shown in Figure 1.
Comparison of Four Acid-Base Reactions
In each case removal of a proton from the central atom generates an anion. The central atom of each anion has one more electron than it has protons in its nucleus. Figure 2 offers a schematic comparison of the central atom of an alkanide and an amide
Comparison of Charge Ratios in Alkanide and Amide Ions
The carbon has 2 electrons in its inner shell, and 5 electrons in its valence shell. The ratio of electrons to protons is 7/6 = 1.167. For the amide ion this ratio is 8/7 = 1.143. This is closer to 1/1 than in the case of carbon. Hence the amide ion is more stable than the alkanide ion. While the numerical difference in these ratios is small, the effect is huge: ammonia is 1012 more acidic than methane. Conversely, the amide ion is 1012 more stable than the methanide ion. You should extend this logic to hydroxide and fluoride ions.
Exercise 1 What is the electron/proton ratio in the hydroxide ion? What is the electron/proton ratio in the fluoride ion?
Now consider the acid-base reaction of acetone shown in Equation 1.
Experiments have shown that the pKa of acetone is 19. It is 1031 times more acidic than methane! This means that the conjugate base of acetone is 1031 times more stable than the conjugate base of methane. Why? What structural feature is there in the conjugate base of acetone that affords such stabilization? Clearly the answer must be the carbonyl group. As the proton is being removed from the carbon, the geometry about the carbon begins to change. As the negative charge increases, the geometry around the a-carbon changes in order to maximize the overlap between the orbital containing the lone pair of electrons and the pi orbitals of the carbonyl group.
Exercise 2 Draw orbital pictures of the enolate ions formed from the following compounds:
Exercise 3 If resonance stabilization of the conjugate base is responsible for the increase in the acidity of the C-H protons in acetone, then it stands to reason that the more resonance stabilization a conjugate base enjoys, the more acidic its conjugate acid will be. Select the appropriate order of acidity of the following carbon acids:
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Figure 3 presents several structures with annotations that illustrate important definitions associated with this area of chemistry.
Exercise 4 are geometric isomers diastereomers tautomers different compounds
Exercise 5 Select the tautomer(s) of
Exercise 6 Identify any enol ethers among structures A-D.
When acetone is treated with base in D2O, it is slowly converted into hexadeuteroacetone as shown in Equation 2.
Exercise 7 Indicate the number of hydrogen atoms that would be exchanged for deuterium atoms when the following compounds are subjected to the reaction conditions shown in Equation 2.
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Compounds that contain at least one hydrogen atom a to a carbonyl group quickly decolorize dilute aqueous solutions of dibromine and sodium hydroxide. Equation 3 illustrates the reaction for acetone.
The disappearance of the red-orange color indicates that the dibromine has been consumed. The products of the reaction, a-bromoacetone and sodium bromide are colorless. The reaction involves nucleophilic attack of C-a of an enolate ion on a molecule of dibromine. See Figure 5.
Given the chemistry described in Equation 3, it shouldn't be surprising that compounds which contain at least one hydrogen atom a to a carbonyl group also decolorize aqueous solutions of diiodine and sodium hydroxide. In the case of methyl ketones, the decoloration is accompanied by the formation of a molecule of iodoform, CHI3, a yellow solid. The decoloration of the solution and the formation of the yellow solid is taken as a positive result. Equation 4 illustrates the process for 2-pentanone. Note that the starting ketone is converted into a carboxylic acid that contains one fewer carbon atoms.
In reaction 4, each of the three hydrogens on the methyl group attached to the carbonyl carbon is replaced by an iodine atom. The hydrogen atom in the iodoform comes from the acid that is added in the second step of the process.
Exercise 8 Identify the compounds that would give a positive iodoform reaction:
If a strongly basic solution of a ketone or aldehyde containing at least one a-hydrogen is "quenched" with chlorotrimethylsilane, it is possible to isolate a silyl enolether as shown in Equation 5.
Equation 5 differs from Equations 3 and 4 in two important ways. First, the equilibrium constant for the first step of reaction 5, i.e. deprotonation of the a-hydrogen, is approximately 1019, while that for reactions 3 and 4 is about 10-3. This means that there is a very high concentration of the enolate ion in reaction 5, but not in reactions 3 and 4. The importance of this fact will become obvious when we discuss aldol condensations in general, and crossed aldol condensations in particular. Second, the oxygen atom of the enolate ion acts as the nucelophile. This reflects the fact that the Si-O bond is an extremely strong bond, much stronger than a Si-C bond.
In molecules which contain hydrogen atoms that are a to two carbonyl groups, the enol tautomer may actually be the predominant form of the compound even in the absence of base. Figure 4 shows a partial 1H-NMR spectrum of 2,4-pentanedione. Integration of the signals for the methyl groups of the keto and enol forms of this compond reveals the relative amounts of each tautomer.
A Partial NMR Spectrum of 2,4-Pentanedione
The enol/keto ratio depends upon the solvent. In the case of 2,4-pentanedione it varies between 3/1 and 4/1.
All of the lines of evidence described above may be rationalized by a process that involves the formation of an enolate ion. Figure 5 outlines the steps involved in the bromination of acetone.
Mechanistic Interpretation of the Bromination of Acetone
In the first step, the base abstracts a hydrogen atom from a methyl carbon. Electrons from the C-H bond move toward the carbonyl group, generating the enolate ion. In the second step, the negative charge on the oxygen atom moves back toward the carbon in order to regenerate the carbonyl group. As this happens, the electrons in the C-C double bond of the enolate ion form a bond to one of the bromine atoms in dibromine. This is a nucleophilic substitution reaction in which the enolate ion acts as a nucleophile and displaces a bromide ion from the dibromine.
In the isotope exchange experiment described in Equation 2, the enolate ion displaces -OD from D2O in a similar manner. The process is repeated until all of the hydrogen atoms have been replaced by deuteriums. Using D2O as the solvent insures that all of the hydrogen atoms will be exchanged.
In the iodoform reaction, the exchange of the three methyl hydrogens for iodine atoms produces a triiodomethylketone. The triiodomethyl group is then displaced by hydroxide ion as shown in Figure 6, again using 2-pentanone as an example.
The Last Stage of the Iodoform Reaction
Reaction of the triiodomethylketone with hydroxide ion as shown in Figure 6 is a nucleophilic acyl substitution reaction. The triiodomethanide ion is a reasonable leaving group since the negative charge on the carbon atom is stabilized by the three iodine atoms.
Consider the reactions shown in Equations 6 and 7.
Exercise 9 What is the equilibrium constant for reaction 6? Keq = What is it for reaction 7? Keq =
Given that the equilibrium constant for reaction 6 is approximately 103 times greater than that for reaction 7, how is it hydroxide ion prefers to act as a base rather than as a nucleophile? The answer is that it doesn't. Reaction 6 occurs in preference to reaction 7. In fact, for every time a hydroxide ion abstracts an a-hydrogen from one acetaldehyde molecule, 1000 hydroxide ions add to the carbonyl groups of other acetaldehyde molecules. However, reaction 6 is readily reversible. Reaction 7, while reversible, may also proceed forward if the enolate ion reacts with an electrophile such as water, dibromine, diiodine, or chlorotrimethylsilane. It's sort of like the tortoise and the hare; reaction 6 gets off to a speedy start, but reaction 7 wins out in the end. Reaction 7 is the first step in an important carbon-carbon bond making process called the aldol reaction.