Chapter 7: Lecture PowerPoint

1. Chapter 7: Lecture PowerPoint An Overview of the Most Common Elementary Steps

2. 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.

3. 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

4. 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+).

5. 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).

6. 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.

7. Simplifying Grignard and Organolithium Compounds

8. 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⁻).

9. 7.2 Bimolecular NucleophilicSubstitution (SN2) Steps • In a bimolecular nucleophilic substitution (SN2) step, a molecular species (i.e., the substrate) undergoes substitution.

10. 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.

11. 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.

12. 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.

13. 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.

14. 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.

15. 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.

16. More Examples of Nucleophilic Addition Steps

17. More Examples of Nucleophile Elimination Steps

18. 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.

19. More E2 Examples

20. 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:⁻).

21. 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.

22. More Examples of Electrophilic Addition Steps

23. 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.

24. More Examplesof Electrophile Elimination Steps

25. 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.

26. 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.

27. 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.

28. 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.

29. Importance of Carbocation Rearrangements • Carbocation rearrangements are important to consider whenever carbocations are formed in an elementary step. • Heterolysis • Electrophilic addition

30. 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.

31. 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.

32. Charge Stability Favored over Bond Energy • Charge stability and bond energy can both differ. • Charge stability favors products. • Bond energy favors reactants.

33. 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).

34. 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.

35. 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

36. Mechanisms of Keto–Enol Tautomerization (Keto to Enol)

37. Mechanisms of Keto–Enol Tautomerization(Enol to Keto)

38. 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.

39. Relative Percentages of Ketoand Enol Forms

40. 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.

41. 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.

42. 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.

43. 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.

44. 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.