Nucleophilic Carbon
Introduction
A nucleophile is Lewis base, i.e. an electron pair donor. Most often the electron pair is a lone pair on an atom in Group V, VI, or VII. As a Group IV element, a neutral carbon does not have any lone pairs. It is not nucleophilic. In order to generate a nucleophilic carbon atom it is necessary to form the conjugate base of a carbon acid. Figure 1 depicts this requirement in general terms.
Figure 1
Generating Nucleophilic Carbon
Organic chemists are interested in generating nucleophilic carbon because the formation of carbon-carbon bonds lies at the heart of organic synthesis, and the vast majority of carbon-carbon bond forming reactions involve the coupling of a nucleophilic carbon with an electrophilic carbon. Before we consider such reactions, however, we will survey briefly some of the most important methods of generating negatively charged carbon, beginning with deprotonation of carbon acids.
Carbon Acids
The term carbon acid refers to any species in which a hydrogen atom may be transferred to a base as shown in Figure 1. Generally carbon acids are extremely weak acids, although their pKa values vary by over 40 orders of magnitude! Representative examples of different types of carbon acids that span this range are listed in Table 1.
Table 1
pKa Values of Carbon Acids
Exercise 1 Following the example shown below, write equations for the deprotonation of the following carbon acids. The number in parentheses indicates the number of alternative deprotonation sites. In those situations where more than one deprotonation site is possible, indicate which site would be deprotonated preferentially, i.e. the most acidic site.
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The base that removes the proton from the carbon acid must be stronger than the conjugate base of the carbon acid if the equilibrium constant for the process is going to be greater than 1, i.e. if the reaction is going to generate a high concentration of the conjugate base of the carbon acid.
Exercise 2 Calculate the approximate value of Keq for each of the following reactions (Review the discussion of pKa values and equilibrium constants if you can't remember how to do this type of problem.). Enter your answer in the appropriate text box. A value of 1 x 10-8 would be entered as 10e-8.
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While it is not possible to deprotonate simple hydrocarbons like methane directly, there are compounds in which the pKa of a C-H bond is much lower than 50. The pKa of acetone, for example, is 19. If you treat acetone with a base whose conjugate acid has a pKa greater than 19, it should be possible to deprotonate one of the methyl groups. Equation 1 gives an example.
The base in Equation 1 is called diisopropylamide. It is the conjugate base of diisopropylamine, which has a pKa of approximately 38 (about the same as that of NH3).
Exercise 3 What is the equilibrium constant for reaction 1? Keq =
Exercise 4 What is the equilibrium constant for the following reaction? (Li + -:N[CH(CH3)2]2 is called lithium diisopropylamide, abbreviated LDA. It is widely used to deprotonate aldehydes and ketones.) 
Keq =
Enolate Ions
The fact that acetone is 30 orders of magnitude more acidic than an alkane must mean that the conjugate base of acetone is 30 orders of magnitude more stable than the conjugate base of propane. This huge increase in stability is attributed to the resonance interaction between the lone pair of electrons on the carbon and the pi system of the adjacent carbonyl group in the conjugate base. Figure 2 depicts this interaction from two different perspectives.
Figure 2
Resonance Stabilization
In the orbital representation red indicates areas of highest electron density and blue areas of lowest electron density. When a proton is removed from one of the methyl groups, the geometry (hybridization) at that carbon atom changes from tetrahedral (sp3) to trigonal planar (sp2). This allows for maximum overlap between the orbitals on the carbon and those of the adjacent pi bond; electron density is transferred from the carbon to the oxygen, thus stabilizing the anion. Such resonance stabilized carbanions are called enolate ions. They are versatile reagents that find wide application in organic synthesis.
Exercise 5 Draw structures for the resonance contributors for the enolate ions formed from the following compounds:
Exercise 6 Estimate the equilibrium constant when 1. Compound A is treated with a solution of KOH in Ethanol. Keq = 2. Compound B is treated with LDA in diisopropylamine. Keq = 3. Compound C is mixed with a concentrated aqueous solution of NaOH. Keq = Compound D is allowed to react with NaOCH3 in CH3OH. Keq =
There is one more approach to generating nucleophilic carbon by deprotonation of carbon acids that deserves comment. It involves deprotonation of carbon acids called phosphonium salts. The reaction constitutes the first step of a 3-step sequence for preparing alkenes known as the Wittig reaction. Phosphines, :PR3, are the phosphorous analogs of amines, :NR3. Like amines, phosphines are nucleophilic. They react with 1o and 2o alkyl halides by an Sn2 mechanism to produce phosphonium salts, +PR4. Equation 2 presents a simple example.
Triphenylphosphine, :P(C6H5)3 is the phosphine of choice for most Wittig reactions. The phosphonium salts are stable, white solids. Methyltriphenylphosphonium bromide melts from 230-234oC. It is available commercially as are many other alkyl triphenylphosphines. The positive charge on the phosphorous pulls electron density from the C-H bonds of the methyl group, thereby making those protons more acidic. The pKa of the methyl protons in methyl triphenylphosphonium bromide is approximately 15. Hence it is possible to deprotonate the methyl group using bases such as sodium amide or n-butyllithium. Equation 3 gives an example.
The phosphorous atom has empty 3-d orbitals it can use to accept the build up of electron density that occurs on the methyl carbon when it is deprotonated. As indicated by the double headed arrow in Equation 3, this is a form of resonance stabilization. The conjugate base of methyltriphenylphosphonium bromide is called an ylide (pronounced ill id). Ylides are nucleophilic reagents.
The examples we have discussed up to this point all involve deprotonation of carbon acids that are more acidic than simple alkanes. Since alkanes are the weakest acids, it is not possible to generate their conjugate bases by deprotonation. Consequently chemists have developed alternative methods to achieve this goal. One approach involves the formation of organometallic reagents, most notably organolithium and organomagnesium reagents.
Organometallic Reagents
When lithium metal is mixed with an alkyl or aryl halide in an inert solvent such as diethyl ether or tetrahydrofuran, the metal slowly dissolves producing a clear, colorless solution of a species that may be described simply by the formula RLi or ArLi. Equations 4 and 5 present representative examples. The products of these reactions are called methyl lithium and phenyl lithium, respectively.
Methyl lithium is the conjugate base of methane. As such it is an extremely strong base, one of the strongest known. It is also a very good nucleophile. The same is true of phenyl lithium, and of organolithium reagents in general. Organolithium reagents are too reactive to be isolated except under special conditions. Generally they are prepared when needed and used immediately. They may be stored at low temperatures in solution under an inert atmosphere for weeks or months, and solutions of various organolithium reagents of known concentration are available commercially.
Exercise 7 Complete the following equations:
Exercise 8 Referring to Table 1, what would you expect the approximate pKa of diethyl ether to be? Write equations as you did in Exercise 1 depicting the deprotonation of diethyl ether by methyl lithium. What would you expect the approximate value of the equilibrium constant to be? Keq =
Exercise 9 Is it possible to generate the conjugate base of pentane by deprotonation? Yes No
Exercise 10 Write an equation for the reaction of methyl lithium with water. What is the approximate equilibrium constant for this reaction? Keq =
Exercise 11 Why can't you use ethanol to prepare organolithium reagents?
Historically the first organometallic reagents of general synthetic utility were prepared by Victor Grignard in the early 1900s. Grignard discovered that alkyl and aryl halides react with magnesium metal in an inert solvent such as diethyl ether to produce organomagnesium compounds that may be described simply by the formula RMgX. These compounds are called Grignard regents. Equations 6 and 7 present representative examples.
Exercise 12 Write an equation showing how you would prepare each of the following Grignard reagents:
Like their organolithium relatives, Grignard reagents are highly reactive. They are generally prepared as needed, although they may be stored in solution for short periods provided they are kept cold and protected from moisture and oxygen. The carbon-metal bond in these organometallic reagents is highly polar covalent bond with the carbon atom bearing a partial negative charge and the metal atom a partial positive charge. The carbon-metal bond is not ionic in the way that an oxygen-metal bond or a nitrogn-metal bond is. Still, in terms of organic synthesis, it is useful to think of organometallic reagents as a source of free carbanions. Thus, CH3Li may be considered a source of -CH3 anions in the same way that NaOH is a source of -OH ions.
To summarize, we have seen two general methods for generating nucleophilic carbon, deprotonation of carbon acids and formation of organometallic reagents. We are now in a position to consider how to use these reagents to form carbon-carbon bonds.
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