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Chapter 7: Lecture PowerPoint. An Overview of the Most Common Elementary Steps. 7.1 Mechanisms as Predictive Tools: The Proton Transfer Step Revisited. Curved Arrow Notation: Electron Rich to Electron Poor Remember the following concepts: Opposite charges attract; like charges repel.
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Chapter 7: Lecture PowerPoint An Overview of the Most Common Elementary Steps
7.1 Mechanisms as Predictive Tools: The Proton Transfer Step Revisited Curved Arrow Notation: Electron Rich to Electron Poor Remember the following concepts: • Opposite charges attract; like charges repel. • Atoms in the first and second rows of the periodic table must obey the duet and octet rules, respectively. • Electrons on O are attracted to the proton on HCl. • Electrons simultaneously are repelled by O. • Electrons flow from O to H, forming a new O—H bond.
Electron Rich to Electron Poor • In an elementary step, electrons tend to flow from an electron-rich site to an electron-poor site. • H2N⁻ is electron rich and CH3OH is electron poor. • Note that: • Red highlight = Electron rich • Blue highlight = Electron poor • Same color scheme as electrostatic potential maps in Section 1.7
Simplifying Assumptions Regarding Electron-Rich and Electron-Poor Species • Anions do not exist in the solid or liquid phase without the presence of cations, and vice versa. • Metal cations behave as spectator ions, which is true generally of group 1A cations (i.e., Li+, Na+, and K+).
Organometallic Compounds and Grignard Reagents • Organometallic compounds have a metal atom bonded directly to a carbon atom. • Organometallic compounds include: • Alkyllithium (R—Li) • Alkylmagnesium halide (R—MgX, where X = Cl, Br, or I) • Also called a Grignard reagent • Lithium dialkylcuprate [Li+(R—Cu—R)⁻] • These kinds of organometallic compounds are useful reagents for forming new carbon–carbon bonds (see Chapter 10).
Carbon−Metal Bond • The carbon–metal bond acts as a polar covalent bond. • Carbon is more electronegative than the metal. • (C = 2.55, Li = 0.98, Mg = 1.31, and Cu = 1.90) • Organometallic compounds can simply be treated as electron-rich carbanions—compounds in which a negative formal charge appears on C.
Hydride Reagents • Hydride reagents commonly function as reducing agents. • These include lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). • Li+ is a spectator ion and AlH4⁻ is the reactive species. • Hydride reagents can be simplified as asource of hydride (:H⁻).
7.2 Bimolecular NucleophilicSubstitution (SN2) Steps • In a bimolecular nucleophilic substitution (SN2) step, a molecular species (i.e., the substrate) undergoes substitution.
The Nucleophile and the Leaving Group • A nucleophile of an SN2 reaction forms a bond to the substrate at the same time that a bond is broken between the substrate and a leaving group. • The step is said to be bimolecular because it contains two separate reacting species in an elementary step. • The step’s molecularity is 2.
Characteristics of a Substrate • The substrate is the molecule that contains a leaving group. • Leaving groups are relatively stable with a negative charge. • Leaving groups are typically conjugate bases of strong acids. • A nucleophile tends to be attracted by and form a bond to an atom that bears a partial or full positive charge.
Characteristics of a Nucleophile Species that act as nucleophiles generally have the following two characteristics: • The nucleophile has an atom that carries a full negative charge or a partial negative charge. • The atom with the negative charge on the nucleophile has a pair of electrons that can be used to form a bond to an atom in the substrate.
7.3 Bond-Formation (Coordination) andBond-Breaking (Heterolysis) Steps • The proton transfer and the SN2 steps have a bond that is formed and a separate bond that is broken simultaneously. • Bond formation and bond breaking can occur as independent steps, however. • In the following coordination step, a bond is formed. • Recall the general chemistry concepts of Lewis acid–base reactions: • A Lewis acid is an electron-pair acceptor. • A Lewis base is an electron-pair donor. • The product is called the Lewis adduct.
Heterolytic Bond Dissociation/Heterolysis • A heterolytic bond dissociation step, or heterolysis,occurs when a bond is broken and the two electrons end up on one of the atoms initially involved in the bond. • Heterolysis steps are the reverse of coordination steps. • Heterolysis andcoordination steps donot take place inisolation; they usually compose one step of a longer mechanism.
7.4 Nucleophilic Addition and Nucleophile Elimination Steps • A nucleophilic addition step occurs when a nucleophile adds to a polar π bond. • A nucleophile elimination step is the reverse of nucleophilic addition.
7.5 Bimolecular Elimination (E2) Steps • A bimolecular elimination (E2) step takes place when a strong base attacks a substrate in which a leaving group and a hydrogen atom are on adjacent carbon atoms. • Both the H atom and the leaving group (L) are eliminated from the substrate. • The E2 step results in the generation of a new π bond between the two carbon atoms.
Electron-Rich to Electron-Poor Sitesand E2 Steps • The base in an E2 step is the electron-rich species; the electron-poor atom is the carbon atom bonded to the leaving group. • The movement of electrons from the electron-rich site to the electron-poor site therefore is depicted with two curved arrows originating from the strong base (B:⁻).
7.6 Electrophilic Addition and Electrophile Elimination Steps • An electrophilic addition step occurs when a nonpolar π bond donates electrons to a strongly electron-deficient species (the electrophile or E+), forming a new bond between the two. • The product of the electrophilic addition step is a carbocation, which is highly unstable and will react further because it has a positive charge and lacks an octet.
Carbocations and Electrophile Elimination • Carbocations are typically unstable, so the reverse of electrophilic addition is also a common elementary step. • In the reverse step, called electrophile elimination, an electrophile is eliminated from the carbocation, generating a stable, uncharged, organic species. • In the electrophile elimination step shown, the positively charged C atom is electron poor, whereas the C—E single bond is electron rich.
Electrophile Elimination Explained • H+ cannot exist on its own in solution. • Any base that is present in solution, such as water, will therefore assist in the removal of a proton in an electrophile elimination step.
7.7 Carbocation Rearrangements:1,2-Hydride Shifts and 1,2-Alkyl Shifts • A carbocation can also undergo a rearrangement—the 1,2-hydride shift or the 1,2-alkyl shift. • A hydride anion (H⁻) is said to shift because a hydrogen atom migrates along with the pair of electrons initially making up the C—H bond. • A “1,2 shift” refers to the migration of an atom or group (in this case, a hydride) to an adjacent atom.
Carbocation Rearrangements continued… • In a 1,2-alkyl shift, an alkyl group migrates, rather than a hydride anion • A 1,2-methyl shift is a specific type of a 1,2-alkyl shift • The numbering system is no different from that of a 1,2-hydride shift because the migration group (the methyl group) is transferred to an adjacent atom. • Like the hydride shift, the methyl group migrates with a pair of electrons.
Electron Rich to Electron Poor in Carbocation Rearrangements • A single bond to hydrogen or carbon on an adjacent atom is relatively electron rich because two electrons are localized in the bonding region. • A single curved arrow is used to depict a carbocation rearrangement.
Importance of Carbocation Rearrangements • Carbocation rearrangements are important to consider whenever carbocations are formed in an elementary step. • Heterolysis • Electrophilic addition
7.8 The Driving Force for Chemical Reactions • The driving force for a reaction reflects the extent to which the reaction favors products over reactants. • Driving force increases with increasing stability of the products relative to the reactants.
Evaluating Charge Stability and Bond Energy • Charge stability heavily favors products because, although there are two formal charges in the reactants, there are no formal charges in the products. • Total bond energy favors products because one covalent bond is formed, giving carbon an octet, and none are broken.
Charge Stability Favored over Bond Energy • Charge stability and bond energy can both differ. • Charge stability favors products. • Bond energy favors reactants.
Charge Stability Favored over Bond Energycontinued… • Charge stability favors the products because the negative charge is better accommodated on Cl than on O. • Bond energy favors the reactants because a s and a p bond are formed, while two s bonds are broken. • A s bond is typically stronger than a p bond (see Chapter 3).
Other Important Factors • Sometimes you will need to consider other factors. • In the first step, charge stability favors products, but the reaction does not occur. • In the second step, charge stability and bond energy both favor reactants, but the step can still occur. • Learn more about this in Chapter 9.
7.9 Keto–Enol Tautomerization • In aqueous acidic or basic solutions, aldehydes and ketones exist in rapid equilibrium with a rearranged form, called an enol. • As a ketone or aldehyde, the species is called the keto form. • In the enol form, the species has a carbon that is simultaneously part of a C=C functional group and an –OH functional group. • Isomers in equilibrium are called tautomers. • This specific equilibrium is called keto–enol tautomerization. • In the keto form, a hydrogen atom is on thea (alpha) carbon • In the enol form, the hydrogen atom appears on the oxygen atom instead
Equilibrium between Ketoand Enol Forms • For most tautomerization equilibria, the keto form is in much greater abundance than the enol form. • This suggests that the keto form is significantly more stable.
Keto Form Is Generally Favored • The predominance of the keto form does not stem from a difference in charge stability, but rather is an outcome of a greater total bond energy in the keto form than in the enol form.
Sugar Transformers: Tautomerization in the Body • Glycolysis is the metabolic pathway that breaks down simple carbohydrates for their energy. • Isomerase enzymes are responsible for tautomerization of sugars in cells.
Summary and Conclusions I • Curved arrow notation reflects the flow of electrons from an electron-rich site to an electron-poor site. • Metal cations from group 1A typically behave as spectator ions. • Organometallic compounds can be simplified and thought of as electron-rich carbanions. • Bimolecular nucleophilic substitution (SN2) involves a substrate that has a leaving group (L),which is replaced by a nucleophile (Nu⁻). • A nucleophile generally contains an atom that has a full or partial negative charge and possesses a lone pair of electrons. • Other reactions covered: coordination steps and heterolytic bond dissociation (heterolysis) steps, bimolecular elimination (E2), nucleophilic addition/elimination, and 1,2-hydride and 1,2-methyl shifts.
Summary and ConclusionsII • In a nucleophilic addition step, a nucleophile forms a bond to the positive end of a polar C—X multiple bond, forcing a pair of electrons from a π bond onto X. • The nucleophile is relatively electron rich, and the atom at the positive end of the polar C—X multiple bond is relatively electron poor. • In a nucleophile elimination step, a new C—X π bond is formed at the same time that a leaving group is expelled. • In a bimolecular elimination (E2) step, a base deprotonates a hydrogen onthe substrate at the same time that a leaving group is expelled, leaving an additional bond between the atoms to which the hydrogen and the leaving group were initially bonded. • In an electrophilic addition step, a pair of electrons from a nonpolar π bond forms a bond to an electrophile, an electron-deficient species. • In an electrophile elimination step, an electrophile is eliminated from a carbocation species and a nonpolar π bond is formed simultaneously.
Summary and ConclusionsIII • In a 1,2-hydride shift or 1,2-alkyl shift, a C—H or C—C bond adjacent to a carbocation is broken, and the bond is reformed to the C atom initially with the positive charge. The positive charge moves to the C atom whose bond is broken. • Charge stability and total bond energy are two major factors that contribute to a reaction’s driving force. • In a keto–enol tautomerization, the ketoform is in equilibrium with its enolform via proton transfer steps.
