An Overview of the Most Common Elementary Steps 71 Mechanisms as Predictive Tools The Proton Transfer Step Revisited Curved Arrow Notation Electron Rich to Electron Poor Remember the following concepts ID: 935724
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Slide1
Chapter 7: Lecture PowerPoint
An Overview of the Most
Common Elementary Steps
Slide27.1 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.
Slide3Electron Rich to Electron Poor
In an elementary step, electrons tend to flow from an electron-rich site to an electron-poor site.
H
2N⁻ is electron rich and CH3OH is electron poor.Note that:Red highlight = Electron richBlue highlight = Electron poor
Same color scheme as electrostatic potential maps in Section 1.7
Slide4Simplifying 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+).
Slide5Organometallic 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 reagentLithium dialkylcuprate [Li+(R—Cu—R)⁻]
These kinds of organometallic compounds are useful reagents for forming new carbon–carbon bonds
(see Chapter 10).
Slide6Carbon−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.
Slide7Simplifying Grignard
and Organolithium Compounds
Slide8Hydride Reagents
Hydride reagents commonly function as reducing agents.
These include lithium aluminum hydride (LiAlH
4) and sodium borohydride (NaBH4).Li+ is a spectator ion and AlH4⁻ is the reactive species.Hydride reagents can
be simplified as a
source of hydride (:H
⁻
).
Slide97.2 Bimolecular Nucleophilic
Substitution (S
N
2) StepsIn a bimolecular nucleophilic substitution (SN2) step, a molecular species (i.e., the substrate) undergoes substitution.
Slide10The 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.
Slide11Characteristics 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.
Slide12Characteristics 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.
Slide137.3 Bond-Formation (Coordination) andBond-Breaking (Heterolysis) Steps
The proton transfer and the S
N
2 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
.
Slide14Heterolytic 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.
Slide157.4 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.
Slide16More Examples
of Nucleophilic Addition Steps
Slide17More Examples
of Nucleophile Elimination Steps
Slide187.5 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.
Slide19More E2 Examples
Slide20Electron-Rich to Electron-Poor Sites
and 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:⁻).
Slide217.6 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.
Slide22More Examples
of Electrophilic Addition Steps
Slide23Carbocations 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.
Slide24More Examples
of Electrophile Elimination Steps
Slide25Electrophile 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.
Slide267.7 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.
Slide27Carbocation Rearrangements
continued…
In a
1,2-alkyl shift, an alkyl group migrates, rather than a hydride anionA 1,2-methyl shift is a specific type of a 1,2-alkyl shiftThe 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.
Slide28Electron 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.
Slide29Importance of Carbocation Rearrangements
Carbocation rearrangements are important to consider whenever
carbocations
are formed in an elementary step.HeterolysisElectrophilic addition
Slide307.8 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.
Slide31Evaluating 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.
Slide32Charge Stability Favored over Bond Energy
Charge stability and bond energy can both differ.
Charge stability favors products.
Bond energy favors reactants.
Slide33Charge Stability Favored over Bond Energy
continued…
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).
Slide34Other 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.
Slide357.9 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 the
a
(alpha) carbon
In the enol form, the hydrogen atom appears on the oxygen atom instead
Slide36Mechanisms of Keto
–
Enol
Tautomerization (Keto to Enol)
Slide37Mechanisms of Keto
–
Enol
Tautomerization(Enol to Keto)
Slide38Equilibrium between
Keto
and Enol FormsFor 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.
Slide39Relative Percentages of Keto
and
Enol Forms
Slide40Keto 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.
Slide41Sugar 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.
Slide42Summary and Conclusions I
C
urved 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
.
Slide43Summary and Conclusions
II
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 on
the 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.
Slide44Summary and Conclusions
III
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 keto form is in equilibrium with its enol
form
via proton transfer steps.