7.1 Mechanisms as Predictive Tools: The Proton Transfer Step Revisited

A reaction mechanism can be very helpful as a predictive tool. In this section we begin to see how mechanisms help, by revisiting curved arrow notation for a proton transfer step and examining specifically how it can be used to make predictions about bond formation and bond breaking.

Even though our discussion here is in the context of proton transfer steps, we can apply the same logic to other elementary steps as well. Effectively, then, this section establishes a process by which to analyze an elementary step. The utility of the process becomes apparent when we discuss new elementary steps later in this chapter.

7.1a Curved Arrow Notation: Electron Rich to Electron Poor

Curved arrow notation was introduced in Section 6.1 as a means for keeping track of valence electrons in an elementary step. It can be far more powerful than that, though, if we recall two concepts:

1. Opposite charges attract; like charges repel.

2. Atoms in the first and second rows of the periodic table must obey the duet and octet rules, respectively.

With these ideas in mind, examine Equation 7-1, which shows the curved arrows for the proton transfer between HCl and HO–.

A two-part illustration shows the flow of electron from an electron-rich site, a hydroxyl group to an electron-poor site, a hydrogen atom of hydrogen chloride, through a curved arrow notation. The first part shows an electron-rich site, hydroxyl group, with oxygen of the hydroxide anion containing three lone pairs of electrons, representing a negative charge interacting with the hydrogen atom of hydrochloride. The hydrogen atom of hydrochloride group is marked delta plus, which is single bonded to the chloride group, marked with three lone pairs of electrons. A curved arrow beginning from the oxygen atom of a hydroxyl group is pointing toward the hydrogen atom of hydrochloride is labeled curved arrow drawn from electron-rich to electron-poor. Another curved arrow is marked from a single bond between a hydrogen atom and chloride ion of hydrogen chloride pointing toward chloride ion. The second part shows an oxygen atom of a hydroxyl group containing two lone pairs of electrons single bonded to a hydrogen atom and a chloride anion containing four lone pairs of electrons.

Notice that HO– bears a full negative charge. This excess electron density on O means the electrons on O are somewhat destabilized due to their mutual repulsion. The proton on HCl bears a partial positive charge, δ+, because Cl is more electronegative than H. As a result, the electrons on O are attracted to the proton on HCl. This simultaneous charge repulsion among the electrons on O and their attraction to H facilitates the flow of electrons from O to H and results in the formation of the new O—H bond.

This example illustrates one of the most important guidelines for drawing curved arrows:

In an elementary step, electrons tend to flow from an electron-rich site to an electron-poor site.

In Equation 7-1, the HO– anion is relatively electron rich and the H atom on HCl is relatively electron poor, denoted by the light red screen behind the negatively charged O atom and the light blue screen behind the H bearing a partial positive charge. (This is the same color scheme used in electrostatic potential maps to represent areas of more and less electron density, respectively.) To help you keep track of those atoms, we have kept the red and blue screens the same in the products.

The curved arrow in Equation 7-1 that goes from the H—Cl bond to the Cl atom is needed to maintain H’s duet of electrons. Without that curved arrow, and hence without breaking that bond, H would end up with two bonds, which is one too many.

YOUR TURN 7.1
SHOW ANSWERS

In this proton transfer step, identify the electron-rich and electron-poor sites, and label the curved arrow that connects the two as “electron rich to electron poor.”

A two-part illustration shows the transfer of protons between electron-rich and electron-poor sites. The first part shows sulfhydryl anion, with sulfide of the sulfhydryl group containing three lone pairs of electrons, representing negative charge interacting with hydrogen atom of hydrogen bromide. The hydrogen atom of hydrogen bromide is single bonded to the bromide group, marked with three lone pairs of electrons. A curved arrow beginning from sulfur atom of sulfhydryl group points toward the hydrogen atom of hydrogen bromide. Another curved arrow is marked from a single bond between a hydrogen atom and bromide ion of HBr pointing toward bromide ion. The second part shows sulfur atom of sulfhydryl group containing two lone pairs of electrons single bonded to a hydrogen atom and a bromide anion containing four lone pairs of electrons.

A two-part illustration shows the transfer of protons between electron-rich and electron-poor sites. The first part shows sulfhydryl anion, with sulfide of the sulfhydryl group containing three lone pairs of electrons, representing negative charge interacting with hydrogen atom of hydrogen bromide. The sulfur atom is labeled �Electron-rich site�. The hydrogen atom of hydrogen bromide is single bonded to the bromide group, marked with three lone pairs of electrons. The hydrogen atom is labeled �Electron-poor site�. A curved arrow beginning from sulfur atom of sulfhydryl group points toward the hydrogen atom of hydrogen bromide and labeled �Electron rich to electron poor�. Another curved arrow is marked from a single bond between a hydrogen atom and bromide ion of HBr pointing toward bromide ion. The second part shows sulfur atom of sulfhydryl group containing two lone pairs of electrons single bonded to a hydrogen atom and a bromide anion containing four lone pairs of electrons.

YOUR TURN 7.2
SHOW ANSWERS

The curved arrow notation for the following proton transfer is faulty. What is unacceptable about it? In the space provided here, make the necessary corrections to the curved arrow notation.

A two-part illustration shows a faulty interpretation of transfer of protons between electron-rich and electron-poor sites. The first part shows a hydroxyl group, with an oxygen atom carrying three lone pairs of electrons, representing negative charge interacting with a hydrogen atom of a hydrochloride. The hydrogen atom of the hydrochloride group is single bonded to the chloride group, marked with three lone pairs of electrons. The second part shows an oxygen atom of a hydroxyl group containing two lone pairs of electrons single bonded to a hydrogen atom and a chloride anion containing four lone pairs of electrons.

This is faulty because H cannot form two covalent bonds. A second arrow needs to be drawn to show the H—Cl bond breaking.

A two-part illustration shows a faulty interpretation of transfer of protons between electron-rich and electron-poor sites. The first part shows a hydroxyl group, with an oxygen atom carrying three lone pairs of electrons, representing negative charge interacting with a hydrogen atom of a hydrochloride. The hydrogen atom of the hydrochloride group is single bonded to the chloride group and the single bond is interacting with chloride group, marked with three lone pairs of electrons. The second part shows an oxygen atom of a hydroxyl group containing two lone pairs of electrons single bonded to a hydrogen atom and a chloride anion containing four lone pairs of electrons.

Connections Trimethylamine [(CH3)3N, Problem 7.2] is a gas that is often associated with the odor of rotting fish. Sensors have been developed to test for trimethylamine to assess the freshness of fish.

A girl holding a rotten fish in one hand while closing her nose to stop the odor with another hand.

Solved Problem 7.1

Identify the electron-poor H atom in methanol. Draw the mechanism by which methanol acts as an acid in a proton transfer reaction with H2N–.

Condensed structural formula of methanol. The condensed structural formula of methanol consists of a carbon atom single bonded to three hydrogen atoms and an oxygen atom carrying two lone pairs of electrons. The oxygen atom is further attached to a hydrogen atom by a single bond.

Think
SHOW SECTION

What kinds of charges characterize an electron-poor atom? Are there any formal charges present? Any strong partial charges? When H2N– and CH3OH are combined, what curved arrow can we draw to depict the flow of electrons from an electron-rich site to an electron-poor site?

Solve
SHOW SECTION

CH3OH does not bear any full charges, but the highly electronegative O places a strong partial positive charge on the H to which it is bonded. That H is therefore electron poor, and a blue screen is placed behind it as a reminder. H2N– bears a full negative charge and is therefore electron rich. As a reminder, a red screen is placed behind N. To represent the flow of electrons from the electron-rich site (H2N–) to the electron-poor site (CH3OH), draw a curved arrow from a lone pair on N to the H atom on O. A second curved arrow is needed to make sure H has only one bond, not two.

A chemical reaction shows an equation representing transfer of protons between electron-rich, amino group and electron-poor, methanol. The first part of the equation shows a condensed structural formula of methanol, consisting of a central carbon atom single bonded to three hydrogen atoms and an oxygen atom carrying two lone pairs of electrons. Further, the oxygen atom is single-bonded to a hydrogen atom, marked delta plus. The methanol is shown to react with an amino group, with nitrogen anion of an amino group containing two lone pairs of electrons, marked with a negative charge. A curved arrow from negatively charged nitrogen is shown pointing toward the positively charged hydrogen atom of methanol, labeled electron-rich to electron-poor. Another curved arrow marked on a single bond between and an oxygen atom and a positively charged hydrogen atom of methanol, point toward a lone pair of electrons present on the oxygen atom. The second part shows a condensed structural formula of a molecule, with a central atom single-bonded to three hydrogen atoms and an oxygen anion containing two lone pairs of electrons and a hydrogen atom single-bonded to the nitrogen atom, containing a lone pair of electrons of the amino group.

problem 7.2 Identify the electron-rich and electron-poor sites in each of the reactant molecules shown here. Draw the curved arrows and the products for the proton transfer between these two molecules and label the curved arrow that represents the flow of electrons from an electron-rich site to an electron-poor site.

An illustration shows an equation to check for electron-rich and electron-poor sites. The equation shows a condensed structural formula of trimethylamine, represented as a central nitrogen atom single-bonded to three methyl groups each, arranged in a trigonal planar geometry, shown reacting with a water molecule. It is followed by a rightward arrow to read a question mark.

7.1b Simplifying Assumptions Regarding Electron-Rich and Electron-Poor Species

Contrary to what is suggested in Equation 7-1, we cannot simply add hydroxide anion (HO–) to HCl to initiate a proton transfer reaction, because of the following restriction:

Anions do not exist in the solid or liquid phase without the presence of cations, and vice versa.

We can, however, add a source of HO–, such as NaOH; NaOH is an ionic compound, so in solution it dissolves as Na+ and HO–.

This initially may seem to complicate the picture, because Na+ is electron poor. We can thus envision Na+ reacting with an electron-rich site of another species. However, Na+ is relatively inert in solution and tends not to react. It behaves instead as a spectator ion.

In general, group 1A metal cations (i.e., Li+, Na+, and K+) tend to behave as spectator ions in solution.

Consequently, when we envision the flow of electrons from an electron-rich site to an electron-poor site, we can disregard such metal cations.

Solved Problem 7.3

Draw the necessary curved arrows for the proton transfer between KOCH3 and HCN in solution.

Think
SHOW SECTION

What are the electron-rich and electron-poor sites in each compound? Are there any simplifying assumptions we can make?

Solve
SHOW SECTION

KOCH3 is an ionic compound that dissolves in solution as K+ and CH3O–. We can therefore treat K+ as a spectator ion and consider just CH3O– as the reactive species it contributes. HCN has an electron-poor H atom due to the high effective electronegativity of the sp-hybridized C atom. Thus, a curved arrow is drawn from the electron-rich O to the electron-poor H to initiate the proton transfer. A second curved arrow is drawn to break the H—C bond to avoid two bonds to H.

A three-part illustration shows an equation representing proton transfer between electron-rich and electron-poor sites in a reaction between potassium methoxide and hydrogen cyanide in a solution. The first part shows a condensed molecular structure of potassium methoxide, with a central carbon atom single-bonded to three hydrogen atoms and an oxygen atom, containing three lone pairs of electrons carrying a negative charge, and a potassium ion carrying a positive charge. The potassium ion is labeled, treat K cation as a spectator ion. It is further followed by a rightward arrow marked simplify. The second part shows a condensed structural formula of methoxide, with a negatively charged oxygen atom labeled electron-rich site and a condensed structural formula of hydrogen cyanide showing a central carbon atom single-bonded to a hydrogen atom, marked delta plus and labeled electron-poor site which is further triple-bonded to a nitrogen atom. A curved arrow is drawn from a negatively charged oxygen atom with its head pointing toward a partially positively charged hydrogen atom. Another curved arrow is drawn from a single bond between hydrogen and carbon with its head pointing toward the carbon atom of hydrogen cyanide. The third part of the illustration shows a condensed structural formula of a molecule representing a central carbon atom single-bonded to three hydrogen atoms and an oxygen atom, containing two lone pairs of electrons. The oxygen atom is further single-bonded to a hydrogen atom. A negatively charged carbon atom, containing a lone pair of electrons is attached to a nitrogen atom by a triple bond.

problem 7.4 Use curved arrow notation to indicate the proton transfer between NaSH and CH3CO2H.

A two-part illustration shows a simplification of organometallic compound resulting in the formation of a carbanion. The first part shows a condensed structural formula showing a central carbon atom, marked delta minus, labeled �electron-rich� surrounded by three vacant single bonds with a metal attached to the fourth one. The metal ion is marked delta plus and labeled �electron-poor.� It is followed by a rightward arrow labeled �simplify� leading to the second part. The second part shows a condensed structural formula of carbanion with a central carbon atom, containing a lone pair of electrons and a negative charge surrounded by three vacant single bonds. The caption reads, �Figure 7-1 Simplifying assumptions in organometallic compounds (Left) Because of the high polarity in a carbon�metal bond, organometallic compounds contain an electron-rich site on C and an electron-poor site on the metal. (Right) We can usually ignore the reactivity of the metal-containing portion and treat the organometallic compound simply as a carbanion, which is electron rich. The quotation marks remind us that the carbanion does not actually exist in solution.�
FIGURE 7-1 Simplifying assumptions in organometallic compounds (Left) Because of the high polarity in a carbon–metal bond, organometallic compounds contain an electron-rich site on C and an electron-poor site on the metal. (Right) We can usually ignore the reactivity of the metal-containing portion and treat the organometallic compound simply as a carbanion, which is electron rich. The quotation marks remind us that the carbanion does not actually exist in solution.

Similar ideas allow us to make simplifying assumptions for organometallic compounds, which contain a metal atom bonded directly to a carbon atom. Examples of organometallic compounds include alkyllithium (R—Li); alkylmagnesium halide (R—MgX, where X = Cl, Br, or I), also called a Grignard reagent; and lithium dialkyl cuprate [Li+(R—Cu—R)–]. These kinds of organometallic compounds are useful reagents for forming new carbon–carbon bonds (see Chapter 10).

To simplify how we treat organometallic compounds in mechanisms, we must first recognize that the carbon–metal bond is a polar covalent bond. The carbon atom’s electronegativity (2.55) is significantly greater than that of the metal (Li = 0.98, Mg = 1.31, and Cu = 1.90), so there is a large partial negative charge on carbon (making it electron rich) and a large partial positive charge on the metal atom (making it electron poor), as shown in Figure 7-1.

When considering organometallic compounds in the flow of electrons from an electron-rich site to an electron-poor site, we can often treat the metal-containing portion as a spectator.

A three-part illustration shows a simplification of three compounds highlighting an electron-rich carbon atom. The first part shows a simplification of organometallic compound Li single-bonded to CH 2CH 3 into CH 2CH3, where a carbon atom linked to H2 contains a lone pair of electrons carrying a negative charge. The second part shows a simplification of phenyl-magnesium bromide. It shows a condensed structural formula represented as an aromatic benzene ring, with a carbon atom at ortho-position linked to Mg Br. It is simplified into a benzene anion, shown as a benzene aromatic ring, with a carbon atom at ortho-position containing a lone pair of electrons and a negative charge. The third part shows a condensed structural formula of a compound Li-Cu single-bonded to two methyl groups each, arranged in a bent. It is simplified into a methyl group, with carbon atom, containing a lone pair of electrons and marked negatively charged. The caption reads, �Simplifying some specific organometallic compounds (a) LiCH2CH3 is simplified to negative charge CH2CH3. (b) C6H5MgBr is simplified to C6H25 negative charge. (c) (CH3)2CuLi is simplified to H3C negative charge.�
FIGURE 7-2 Simplifying some specific organometallic compounds (a) LiCH2CH3 is simplified to –CH2CH3. (b) C6H5MgBr is simplified to C6H5-. (c) (CH3)2CuLi is simplified to H3C–.

In most reactions involving organometallic compounds, the product that we are interested in isolating contains the organic portion of the organometallic compound and not the metal-containing portion. Therefore, even though the carbon–metal bond is covalent, it often helps to simplify organometallic compounds in a way that allows us to disregard the metal-containing portion, in much the same way as we did with the group 1A metal cations:

Therefore, as shown on the right in Figure 7-1, we can often treat organometallic compounds simply as electron-rich carbanions—compounds in which a negative formal charge appears on C. Applying this simplifying assumption allows us, for example, to treat CH3CH2Li as a source of , C6H5MgBr as a source of , and (CH3)2CuLi as a source of , as shown in Figure 7-2.

Solved Problem 7.5

What are the products of the proton transfer step between C6H5MgBr and H2O?

A chemical reaction shows reaction between phenyl-magnesium bromide and water to represent proton transfer. The reaction shows a condensed structural formula consisting of an aromatic benzene ring with a carbon atom at its ortho-position, linked to magnesium bromide. It is shown to react with a water molecule followed by a rightward arrow to read a question mark.

Think
SHOW SECTION

What are the electron-rich sites? What are the electron-poor sites? What can be disregarded?

Solve
SHOW SECTION

In H2O, O is electron rich, and both H atoms are electron poor. In C6H5MgBr, we can disregard the MgBr portion because it contains the metal, and treat the reactive species as . The proton transfer is initiated by drawing a curved arrow from the electron-rich C on to an electron-poor H on H2O.

A chemical representation of proton transfer when phenyl-magnesium bromide reacts with water is shown. It shows a condensed structural formula consisting of an aromatic benzene ring, with a carbon atom at ortho-position containing a lone pair of electrons and a negative charge reacting with a molecule, shown as central oxygen atom single-bonded to two hydrogen atoms in bent position, with the oxygen atom, containing two lone pairs of electrons. A curved arrow is drawn from the carbon atom of the benzene ring with its head pointing to one of the hydrogen atoms bonded to the oxygen atom and labeled electron-rich to electron-poor. It is followed by a rightward arrow to show linkage of a hydrogen atom to the ortho-carbon atom of the benzene ring by a single bond and an oxygen atom, containing two lone pairs of electrons carrying a negative charge, single bonded to another hydrogen atom.

problem 7.6 Use curved arrow notation to show the proton transfer step that occurs between CH3Li and CH3OH. Predict the products of this reaction.

The strategy we applied to simplifying organometallic compounds can also be used to simplify hydride reagents, such as lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4), which are commonly used as reducing agents (see Chapter 17). LiAlH4 consists of Li+ ions and ions, as shown on the left in Figure 7-3a. We can treat Li+ as a spectator ion, leaving as the reactive species (Fig. 7-3a, middle). According to the Lewis structure of , the H atoms [electronegativity (EN) = 2.20] are covalently bonded to an Al metal atom (EN = 1.61). The excess negative charge built up on each H atom means that we can treat LiAlH4 as a source of hydride, :H–, as indicated on the right in Figure 7-3a. For similar reasons, we can treat NaBH4 as a source of :H–, too.

A two-part illustration shows a two-step simplification of hydride reagents, lithium aluminum hydride, and sodium borohydride. The first part shows a condensed structural formula of lithium aluminum hydride as a lithium cation, labeled �Group 1, A metal cation is a spectator ion� linked to a negatively charged unit of aluminum hydride shown as a central aluminum atom linked to four hydrogen atoms by a single bond each. Next step titled, �disregard the metal portion� shows a negatively charged unit of aluminum hydride as a central aluminum atom linked to four partially-negatively charged hydrogen atoms by a single bond each. It is further simplified to show an electron-rich hydrogen atom, containing a lone pair of electrons, marked with a negative charge. The second part shows a condensed structural formula of sodium borohydride as sodium cation, linked to a negatively charged unit of borohydride, shown as a central boron atom linked to four hydrogen atoms by a single bond each. Next step shows negatively charged unit of borohydride as a central boron atom linked to four partially-negatively charged hydrogen atoms by a single bond each. It is further simplified to show an electron-rich hydrogen atom, containing a lone pair of electrons, marked with a negative charge. The caption reads, �Simplifying assumptions in hydride reagents (a) Lithium aluminum hydride, LiAlH4, consists of Li positive and AlH4 negative ions. The reactive species is AlH4 negative (middle), which can be treated simply as a lone pair of electrons with H negative charge. The quotation marks indicate that a lone pair of electrons with H2 does not actually exist in solution. (b) Sodium borohydride, NaBH4, consists of Na positive and BH4 negative ions. The reactive species is BH4 negative (middle), which can be treated as a lone pair of electrons with H2, too.�
FIGURE 7-3 Simplifying assumptions in hydride reagents (a) Lithium aluminum hydride, LiAlH4, consists of Li+ and AIH4- ions. The reactive species is AIH4- (middle), which can be treated simply as : H–. The quotation marks indicate that : H– does not actually exist in solution. (b) Sodium borohydride, NaBH4, consists of Na+ and BH4- ions. The reactive species is BH4- (middle), which can be treated as : H–, too.

problem 7.7 Use curved arrow notation to show the proton transfer step that takes place between LiAlH4 and water. Predict the products of the reaction. Do the same for the proton transfer step between NaBH4 and phenol (C6H5OH).

Later, we will need to be careful when making these simplifying assumptions, particularly when treating organometallic compounds as carbanions, R–, and hydride reagents as hydride anions, H–. Experimentally, significant differences in reactivity are observed among the different organometallic compounds and among the different hydride reagents. These differences are discussed more extensively in the context of nucleophilic addition reactions in Chapters 17 and 18. For now, you should take the time to commit to memory the simplifications in Figures 7-2 and 7-3.