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 We must remember that… • 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. • 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 promotes the flow of electrons from O to H and results in the formation of the new O-H bond.

3. Lewis Acids and Lewis Bases • A Lewis base is an electron-pair donor, whereas a Lewis acid is an electron-pair acceptor. • In an elementary step, electrons tend to flow from an electron-rich site to an electron-poor site. • In an elementary step, a Lewis base tends to form a bond with a Lewis acid. • Here, H2N⁻ is the Lewis base and CH3OH is the Lewis acid.

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 (true generally of group IA cations (i.e., Li+, Na+, and K+).

5. Organometallic Compounds andGrignard 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; and lithium dialkylcuprate [Li+(R-Cu-R)-]. • These kinds of organometallic compounds are useful reagents for forming new carbon–carbon bonds.

6. Metal-Carbon Bond • The carbon–metal bond in an organometallic compound exhibits significant covalent character. • The carbon–metal bonds are significantly polar due to the difference in electronegativity between carbon (2.55) and the metal (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. 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.

8. 7.2 Bimolecular NucleophilicSubstitution (SN2) Steps • In a bimolecular nucleophilic substitution (SN2) step, a molecular species (i.e., the substrate) undergoes substitution.

9. 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 “bimolecular” because it contains two separate reacting species in an elementary step. • The step’s molecularity is 2. • It is named “nucleophilic” because a nucleophile is the species that reacts with the substrate through the use of a lone pair.

10. The Nucleophile and The Leaving Groupcontinued… • Leaving groups are typically conjugate bases of strong acids and are relatively stable with a negative charge. • A nucleophile tends to be attracted by and form a bond to an atom that bears a partial or full positive charge.

11. 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.

12. 7.3 Bond-Formation (Coordination) and Bond-Breaking (Heterolysis) Steps • The proton transfer and the SN2 steps have a bond that is formed and a separate bond that is broken simultaneously. • However, there is also bond formation and bond breaking that occur as independent steps. • In the following coordination step, a bond is formed.

13. Heterolytic Bond Dissociation/Heterolysis • A heterolytic bond dissociation steps, or heterolysis,occurs when a bond is broken and the two electrons are not distributed equally to the atoms initially involved in the bond. • Heterolysis steps are the reverse of coordination steps.

14. 7.4 Bimolecular Elimination (E2) Steps • A bimolecular elimination (E2) step can take 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 result of the E2 steps is the incorporation of a carbon–carbon multiple bond into a molecule at a particular site.

15. More E2 Examples

16. Electron-rich to electron poorand E2 steps • The base in an E2 step is the electron-rich species, but the hydrogen atom that the base attacks is not particularly electron poor; instead, the electron-poor atom is the carbon atom bonded to the leaving group. • Thus, the movement of electrons from the electron-rich site to the electron-poor site is depicted with two curved arrows originating from the strong base (B:-)

17. 7.5 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.

18. More Examples of Nucleophilic Addition and Nucleophile Elimination Steps

19. The Importance of Nucleophilic Addition and Nucleophile Elimination Nucleophilic addition and nucleophile elimination are helpful because: • They can cause the number of bonds between two atoms of significantly different polarity to change, thus leading to reduction and oxidation reactions. • They allow two relatively large organic species to be joined to make one even larger organic species. • When nucleophilic addition and nucleophile elimination steps occur sequentially, substitution can take place at a polar π bond. Much of these advantages will be seen in future chapters!

20. 7.6 Electrophilic Addition and Electrophile Elimination Steps • An electrophilic addition step occurs when a nonpolar π bond approaches a strongly electron-deficient species (the electrophile) and a bond forms between an atom of the π bond and the electrophile. • 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.

21. Electrophilic Addition Examples

22. 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 below, the positively charged C atom is electron poor, whereas the C-H single bond is electron rich (H+ is the electrophile that has been eliminated).

23. Electrophile Elimination Example • H+ cannot exist on its own in solution. • Any base that is present in solution will therefore assist in the removal of a proton in an electrophile elimination step. • If water is present it will remove the proton to generate the π bond.

24. 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-methyl 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.

25. Carbocation Rearrangements (cont’d) • 1,2-Methyl shifts are different from that of the hydride shift because the migrating methyl group is transferred to an adjacent atom. • Carbocation rearrangements are important to consider whenever carbocations are formed in a particular reaction. • A single bond to hydrogen or carbon on an adjacent atom is relatively electron rich because two electrons are localized in the bonding region. • The single curved arrow is used to depict a carbocation rearrangement.

26. 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, and that tendency increases with increasing stability of the products relative to the reactants.

27. Charge Stability • 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 since one covalent bond is formed, giving carbon an octet, and none are broken.

28. More on the Driving Forces of Reactions • The charge in the products is located on a different atom than it is in the reactants. • Additionally, one covalent bond is broken and another is formed. • The negative charge is better accommodated on Cl than on F. • The H-F bond that is formed (569 kJ/mol) is much stronger than the H-Cl bond that is broken (431 kJ/mol).

29. Charge Stability and Bond Energy • Charge stability and bond energy can both differ. • Charge stability favors products because the negative charge is better accommodated on Cl than on N. • Bond energy favors the reactants because the H-Cl bond that is broken (431 kJ/mol) is stronger than the H-N bond that is formed (389 kJ/mol).

30. Charge Stability Favored Over Bond Energy • Charge stability favors the products because the negative charge is better accommodated on Cl than on O.

31. 7.9 Keto–EnolTautomerization • Charge stability does not always dictate a chemical reaction’s outcome.

32. More On Keto–EnolTautomerization • The base, called an enolate anion, has two resonance structures that delocalize the negative charge onto a C atom and an O atom. • One of two constitutional isomers can be generated depending upon which atom is protonated. • One is an enol, and the other is a compound containing a carbonyl group (C=O)—either a ketone or an aldehyde—and is referred to as the keto form.

33. Relative Percentages of Ketoand Enol Forms • For most enolate anions, protonation leads predominantly to the ketoform. • The enol form is present only in trace amounts, suggesting that the keto form is significantly more stable.

34. Keto-EnolTautomerization • While in solution keto and enol forms equilibrate with one another. • When an enol is produced in a reaction, it will spontaneously rearrange to the more stable keto form.

35. Keto–EnolTautomerizationcontinued… • 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.

36. Keto-EnolTautomerization and Sugars • Glycolysis is the metabolic pathway breaks down simple carbohydrates for their energy. • This process involves back-to-back tautomerization reactions catalyzed by the enzyme phosphohexoseisomerase(shown in purple on the next slide).

37. Keto-EnolTautomerization and Sugarscontinued…

38. Summary and Conclusions • Curved arrow notation reflects the flow of electrons from an electron-rich site to an electron-poor site. • Metal cations from group IA typically behave as spectator ions. • Bimolecular nucleophilic substitution (SN2) involves a substrate that has a leaving group (L). • A nucleophile generally contains an atom that has a full or partial negative charge and possesses a lone pair of electrons. • Other reactions covered were Coordination steps and Heterolytic bond dissociation (heterolysis) steps, Bimolecular Elimination (E2), Nucleophilic addition/elimination, and finally 1,2-hydride and 1,2-methyl shifts. • In an 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.

39. Summary and Conclusionscontinued… • 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 electrophilic addition, 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.

40. Summary and Conclusionscontinued… • In a 1,2-hydride shift or 1,2-alkyl shift, a C-H orC-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 keto form is in equilibrium with its enol form via proton transfer steps.