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Some Arrow-Pushing Guidelines (Section 1.14)
1. Arrows follow electron movement.
2. Some rules for the appearance of arrows
- The arrow must begin from the electron source. There are two sources:
a. An atom (which must have a lone pair to give)
b. A bond pair (an old bond that breaks)
- An arrow must always point directly to an atom, because when electrons move, they
always go to some new atom.
3. Ignore any Spectator Atoms. Any metal atom is always a “spectator”
- When you have a metal spectator atom, realize that the non-metal next to it must have
negative charge
4. Draw all H’s on any Atom Whose Bonding Changes
5. Draw all lone-pairs on any Atom whose bonding changes
6. KEY ON BOND CHANGES. Any two-electron bond that changes (either made or
broken) must have an arrow to illustrate:
- where it came from (new bond made) or
- an arrow showing where it goes to (old bond broken)
7. Watch for Formal Charges and Changes in Formal Charge
- If an atom’s charge gets more positive Þ it’s donating/losing an electron pair Þ arrow
must emanate from that atom or one of it’s associated bonds. There are two “more
positive” transactions:
- When an anion becomes neutral. In this case, an arrow will emanate from the
atom. The atom has donated a lone pair which becomes a bond pair.
- When a neutral atom becomes cationic. In this case, the atom will be losing a
bond pair, so the arrow should emanate from the bond rather than from the atom.
- If an atom’s charge gets more negative Þ it’s accepting an electron pair Þ an arrow must
point to that atom. Ordinarily the arrow will have started from a bond and will point to
the atom.
8. When bonds change, but Formal Charge Doesn’t Change, A “Substitution” is Involved
- Often an atom gives up an old bond and replaces it with a new bond. This is
“substitution”.
- In this case, there will be an incoming arrow pointing directly at the atom (to illustrate
formation of the new bond), and an outgoing arrow emanating from the old bond that
breaks
4.16 Reactive Intermediates: Stability Patterns
- Shortlived, unstable, highly reactive intermediates
- Normally lack normal bonding
These are tremendously important:
1. They will be the least stable intermediate in any multistep mechanism
2. When formed, they are products of the rate-determining step
3. Factors that stabilize them will speed up reaction rates
Thus it is very important to know their stability patterns!
Class Structure Stability Pattern
Carbocations Allylic > 3º > 2º > 1º > methyl > alkenyl
(vinyl, aryl)
Electron
Poor
Electrophilic/
Acidic
Carbon
Radicals
Allylic > 3º > 2º > 1º > methyl > alkenyl
(vinyl, aryl)
Electron
Poor
Electrophilic/
Acidic
Carbanions Allylic > alkenyl (vinyl, aryl) > methyl > 1º >
Electron
Rich
Nucleophilic/
Basic
Notes
1. Both carbocations and radicals have the same pattern. So you don’t need to memorize them
twice!
2. Carbanions are almost exactly the reverse, except that being allylic is ideal for both.
3. All benefit from resonance (allylic).
4. Cations and radicals both fall short of octet rule. As a result, they are both electron deficient.
Carbanions, by contrast, are electron rich.
5. Alkyl substituents are electron donors. As a result, they are good for electron deficient cations
and radicals (3º > 2º > 1º > methyl) but bad for carbanions.
6. Alkenyl (vinyl or aryl) carbons are inherently a bit electron poor. This is excellent for
carbanions, but terrible for cations or radicals.
C
C
C
2. Product Stability/Reactivity : The more stable the product, the more favorable its formation
will be. In terms of rates, this means that the more stable the product, the faster the reaction.
(The concept here is that the more stable the product, the more favorable it will be to make that
product.)
Key note: Often the “product” that’s relevant in this context will not be the final product of
the reaction, but will be the “product” of the rate determining step.
Why: Because as the stability of the anion products increases from A to D, the reactivity of
the parent acids increase
- Reactivity of alkanes toward radical halogenation
Why: Because as the stability of the radical produced during the rate-determining-step
increases, the reactivity of the parent alkane increases
1, E1 Reactivity
Why: Because as the stability of the cation produced in the rate-determining step increases,
the reactivity of the parent halide increases as well
CH
3
NH
2 OH
OH
O
CH
2
Na NHNa
ONa
ONa
O
A
B
C
D
H
3
C
CH
3 < <
3 ° plus resonance
Br
Br
Br
Br
3 ° plus resonance
3. Transition-State Stability/Reactivity : The more stable the transition state, the faster the
reaction will be. (The concept here is that the lower the transition state, the more easily it will be
crossed.)
2 Reactivity
Why: The pattern reflects the relative stability of the transition states. In the case of 3˚ versus 2˚ versus 1˚, the issue is
steric congestion in the transition state. The transition states for the more highly substituted halides are
destabilized. In the case of allylic halides, the transition state is stabilized for orbital reasons, not steric
reasons.
Br
Br
Br
Br
1° plus allylic
4. S
N
1 Reactions.
Recognition : A. Neutral, weak nucleophile. No anionic nucleophile/base, and
B. 3º or 2º alkyl halide. (Controlled by cation stability).
(1º alkyl halides undergo S
N
2 instead. For 2º alkyl halides, S
N
1 is often accompanied by
variable amounts of E1.)
Predicting product: Remove halide and replace it with the nucleophile (minus an H atom!)
Stereochemistry : Racemization. The achiral cation intermediate forgets any stereochem.
Mech : Stepwise, 3 steps, via carbocation. Be able to draw completely.
5. E1 Reactions. 3º > 2º > 1º (Controlled by cation stability)
Recognition : A. Neutral, weak nucleophile. No anionic nucleophile/base, and
B. 3º or 2º alkyl halide. (Controlled by cation stability).
(For 2º alkyl halides, E1 is often accompanied by variable amounts of S
N
Orientation : The most substituted alkene forms
Predicting the major product: Remove halide and a hydrogen from the neighboring carbon
that can give the most highly substituted alkene. The hydrogen on the neighboring carbon can
be cis or trans.
Stereochemistry : Not an issue. The eliminating hydrogen can be cis or trans..
Mech : Stepwise, 2 steps, via carbocation. Be able to draw completely.
Sorting among S
N
2, S
N
1, E2, E1: How do I predict?
Step 1: Check nucleophile/base.
N
1/E1 mixture of both
N
2/E2.
Step 2: If anionic , and in the S N
2/E2 , then Check the substrate.
o 1º S
N
2
o 2º S
N
2/E2 mixture. Often more S
N
2 , but not reliable…
o 3º E
OCH
3 Br
S
N
1: resonance >3º>2º>1º> alkenyl
HOCH
3
OCH
3
H Br
HOCH
3
Br
Br
slow
step
OCH
3
H
Br
E1: 3 º>2º>1º
HOCH
3
H
Br
H
H + H Br
Br
slow
step
6.16 Comparing S N
2 vs S
N
1
S
N
1 S
N
2
1 Nucleophile Neutral, weak Anionic, strong
2 Substrate 3º R-X > 2º R-X 1º R-X > 2º R-X
Allylic effect… Allylic Helps Allylic helps
3 Leaving Group I > Br > Cl I > Br > Cl
4 Solvent Polar needed Non-factor
5 Rate Law K[RX] k[RX][Anion]
6 Stereochemistry
(on chiral, normally 2º R-X)
Racemization Inversion
7 Ions Cationic Anionic
8 Rearrangements Problem at times Never
6.21 Comparing E2 vs E
E1 E
1 Nucleophile/Base Neutral, weak, acidic Anionic, strong, basic
2 Substrate 3º R-X > 2º R-X 3 º RX > 2º RX > 1º RX
Allylic effect… Allylic Helps Non-factor
3 Leaving Group I > Br > Cl I > Br > Cl
4 Solvent Polar needed Non-factor
5 Rate Law K[RX] k[RX][Anion]
6 Stereochemistry Non-selective Trans requirement
7 Ions Cationic Anionic
8 Rearrangements Problem at times Never
9 Orientation Zaitsev’s Rule: Prefer
more substituted alkene
Zaitsev’s Rule: Prefer more
Substituted alkene (assuming
trans requirement permits)
Comparing S
N
2 vs S
N
1 vs E2 vs E1: How Do I Predict Which Happens When?
Step 1: Check nucleophile/base.
N
1/E1 mixture of both
2/E2.
Step 2: If anionic , and in the S
N
2/E2 pool, then Check the substrate.
o 1º S
N
o 2º S
N
2/E2 mixture. Often more S
N
2 , but not reliable…
o 3º E
Notes:
1º R-X S
N
2 only No E2 or S
N
1/E1 (cation too
lousy for S
N
1/E1; S
N
2 too fast
for E2 to compete)
3º R-X E2 (anionic) or
S
N
1/E1 (neutral/acidic)
No S
N
2 (sterics too lousy)
2º R-X mixtures common
Reaction Mechanisms (see p. 310)
A. Recognizing/Classifying as Radical, Cationic, or Anionic
1. Radical
initiation requires both energy (either hv or Δ) and a weak, breakable heteroatom-heteroatom
bond
o Cl-Cl, Br-Br, O-O (peroxide), N-Br, etc..
2 Guides for That are Usually Reliable:
hv radical mechanism
peroxides radical mechanism
2. Anionic
a strong anion/base appears in the recipe
no strong acids should appear in the recipe
mechanisms should involve anionic intermediates and reactants, not strongly cationic ones
- (except for do-nothing spectators like metal cations)
The first step in the mechanism will involve the strong anion/base that appears in the recipe
3. Cationic
a strong acid/electrophile appears in the recipe
no strong anion/base should appear in the recipe
mechanisms should involve cationic intermediates and reactants, not strongly anionic ones
- (except for do-nothing spectators like halide or hydrogen sulfate anions)
The first step in the mechanism will involve the acid that appears in the recipe. The last step
will often involve a deprotonation step. Often the main step occurs in between the proton-
on and proton-off steps
B. Miscellaneous Mechanism Tips
1. Keep track of hydrogens on reacting carbons
2. Each step in a mechanism must balance
3. The types of intermediates involved (cation, anion, or radical) should be consistent with
the reaction classification above
a. If the reaction is cationic, don’t show anionic intermediates
b. If the reaction is anionic, don’t show cationic intermediates
4. Usually conditions are ionic.
5. Use a reactive species, whether strong anion or an acid, to start the first step
a. If acidic, first step will involve protonation of the organic
b. If anionic, the first step will involve the anion attacking the organic.
6. While it isn’t always easy to figure out what is a good mechanism, you should often be
able to eliminate an unreasonable mechanism.
Chapter 7 Reactions and Mechanisms, Review
E
On
R-X,
Normal
Base
Notes
1. Trans hydrogen required for E
2. Zaytsev elimination with normal bases
3. For 3º R-X, E2 only. But with 2º R-X, S
N
2 competes (and usually prevails)
4. Lots of “normal base” anions.
E2,
On
R-X, Bulky
Base
Notes:
1. Hoffman elimination with Bulky Bases
2. E2 dominates over S
N
2 for not only 3º R-X but also 2º R-X
3. Memorize NEt
3
and KOC(CH
3
3
as bulky bases.
Acid-
Catalyzed
E1-
Elimination
Of
Alcohols
Notes:
1. Zaytsev elimination
2. Cationic intermediate means 3º > 2º > 1º
3. 3 - Step mechanism
CH
3
Br
OCH
3
H
H H OCH
3
Br
NaOCH
3
H OCH
3
Mech:
(Normal
base)
Br
NEt
3
or
KOC(CH
3
3
(Bulky
bases)
H
2
C
Mech: Br
H
NEt
3
3
NH Br
OH
H
2
SO
4
+H OH
H
2
SO
4
+ HSO
4
+ OH
2
- H
2
O
HSO
4
+ H
2
SO
4
Protonation
Elimination
Deprotonation
OH
OH
2
H
H
H
Mech
Orientation Stereo Mechanism
None Trans Be able to
draw
completely
Markovnikov Trans Be able to
draw
completely
None Cis Not
responsible
None Trans
Be able to
draw
acid-
catalyzed
epoxide
hydrolysis
None Cis Not
responsible
None None Not
responsible
None None Not
responsible
Br
CH
3
H
Br
Br
2
(or Cl
2
OH
CH
3
H
Br
Br
2
, H
2
O
(or Cl
2
O
CH
3
H
PhCO
3
H
OH
CH
3
H
OH
CH
3
CO
3
H
H
2
O
OH
CH
3
OH
H
OsO
4
, H
2
O
2
O
H
H
O
1. O
3
- Me
2
S
Note: H-bearing alkene carbon
ends up as aldehyde.
O
H
OH
O
KMnO 4
H-bearing alkene carbon
ends as carboxylic acid
Summary of Mechanisms, Ch. 7 + 8.
Alkene Synthesis and Reactions.
Note: For unsymmetrical alkenes, protonation again occurs at the less substituted end of the
alkene, in order to produce the more stable radical intermediate (3º > 2º > 1º)
HBr Br
(no peroxides)
Br
H
H
H Br
H
H
Br
Protonate
Cation
Capture
H
Note: For unsymmetrical alkenes,
protonation occurs at the less
substituted alkene carbon so that
the more stable cation forms
( 3 º > 2 º > 1 º), in keeping with the
product stability-reactivity principle
CH
3
H
H
vs.
H
CH
3
H
H
CH
3
Br
both cis and trans
HBr
peroxides
H
H
H
H
Br
Brominate Hydrogen
Transfer
Br
Br
H Br
Br
Note 1 : For unsymmetrical alkenes,
bromination occurs at the less
substituted alkene carbon so that
the more stable radical forms
( 3 º > 2 º > 1 º), in keeping with the
product stability-reactivity principle
Note 2 : Hydrogenation of
the radical comes from either
face, thus cis/trans
mixture results
CH
3
H
Br
vs.
H
CH
3
Br
H
Br
top
bottom
H
Br
H
Br
H
CH
3
CH
3
H
cis
trans
OH
CH
3
H
2
O, H
O
H
H
H
H
Protonate Cation
Capture
H
H
OH
2
H
H
- H
OH
H
H
Deprotonate
Ch. 15 Conjugated Systems
The General Stabilization Effect of Conjugation (Section 15.1, 2, 3, 8, 9)
Conjugated
(more stable)
Isolated
(less
stable)
Notes:
1 Cations
2 Radicals
3 Anions
4 Dienes
5 Ethers An N or O next to a double bond becomes
sp
2
. An isolated N or O is sp
3
6 Amines
7 Esters
8 Amides Very special, chapter 23, all of biochemistry,
proteins, enzymes, etc.
9 Oxyanions
(Carboxylates)
Very special, chapter 21
10 Carbanions
(Enolates)
Very special, chapter 22
11 Aromatics Very special, chapters 16 + 17
Conjugation: Anything that is or can be sp
2
hybridized is stabilized when next to π bonds.
- oxygens, nitrogens, cations, radicals, and anions
Notes:
1. Any atom that can be sp
2
will be sp
2
when next to a double bond
2. “Conjugation” is when sp
2
centers are joined in an uninterrupted series of 3 or more, such that
an uninterrupted series of p-orbitals is possible
3. Any sp
2
center has one p orbital
O
sp
2
, not sp
3
O
sp
3
N
H
sp
2
H
N
sp
3
O
sp
2
O
O
O
sp
3
N
H
sp
2
O
H
N
sp
3
O
sp
2
O
O
O
O
sp
3
sp
2
O
sp
3
O
Impact of Conjugation
4. Stability: Conjugation is stabilizing because of p-orbital overlap (Sections 15.2, 4, 7)
- Note: In the allyl family, resonance = conjugation
One p Two p’s Three p’s Four p’s Six p’s in circuit
Unstabilized
π-bond
Allyl type Butadiene type Aromatic
Isolated C=C
C=O
C=N
5. Reactivity: Conjugation-induced stability impacts reactivity (Sections 15.4-7)
- If the product of a rate-determining step is stabilized, the reaction rate will go faster
(product stability-reactivity principle)
o Common when allylic cations, radicals, or carbanions are involved
- If the reactant in the rate-determining step is stabilized, the reaction rate will go slower
(reactant stability-reactivity principle)
o Why aromatics are so much less reactive
o Why ester, amide, and acid carbonyls are less electrophilic than aldehydes or ketones
6. Molecular shape (Sections 15.3, 8, 9)
- The p-orbitals must be aligned in parallel for max overlap and max stability
- The sp
2
centers must be coplanar
7. Bond Length: Bonds that look like singles but are actually between conjugated sp
2
centers are
shorter than ordinary single bonds
- In amides, esters, and acids, the bond between the carbonyl and the heteroatom is shortened
8. Bond Strength: Bonds that look like singles but are actually between conjugated sp
2
centers
are stronger than ordinary single bonds
O
O
O
O
NH
2
O
OH
O
OR
O
All four sp
2
carbons must be flat for the p's to align
O
NH
2
- 33 A
normal
double
- 54 A
normal
single
- 48 A = Shortened
and Strengthened
conjugated single
Shortened
and Strengthened
O
OH
O
OR
Shortened
and Strengthened
Shortened
and Strengthened
Impact of Allylic Cation Resonance on Reactivity and Product Formation
1. Rates : Resonance/conjugation stability enhances rates when cation formation is rate-
determining
2. Product Distribution: Product mixtures often result if an allylic cation is asymmetric.
- The two different resonance structures can lead to different products.
- When two isomeric products can form from an allylic cation, consider two things:
1. Which product is more stable?
- This will impact “product stability control” = “thermodynamic control” =
“equilibrium control”
- To assess product stability, focus on the alkene substitution
2. Which resonance form of the cation would have made a larger contribution?
- This will often favor “kinetic control”, in which a product which may not ultimately
be the most stable forms preferentially
3. Position of Cation Formation : When a conjugated diene is protonated, consider which site of
protonation would give the best allylic cation.
Sections 15.5,6 1,2 vs. 1,4 Addition to Conjugated Dienes: “Kinetic” vs. “Thermodynamic”
Control
Note: “Thermodynamic Control” = “Product-Stability Control” = “Equilibrium Control”
This is when the most stable of two possible products predominates. Either of two factors can
cause this:
o Transition State: The most stable product is formed fastest via the most stable
transition state (normally true, but not always)
o Equilibrium : Even if the most stable product is not formed fastest, if the two products
can equilibrate, then equilibrium will favor the most stable product
Kinetic Control: If the less stable of two possible products predominates.
This will always require that for some reason the less stable product forms via a better transition
state (transition-state stability/reactivity principle). Common factors:
o Charge distribution in an allylic cation or radical. The position of charge in the major
resonance contributor may lead to one product, even though it may not give the most
stable product.
o Proximity of reactants. In an H-X addition to a diene, often the halide anion is closer to
one end of the allylic cation than the other, resulting in “1,2 addition” over “1,
addition”.
o Steric factors. With a bulky E2 base, for example, the transition state leading to what
would be the more stable Zaytsev alkene has steric problems, so it gives the Hoffman
alkene instead.
15.7 Allylic/Benzylic Radicals
Stability Factors for Radicals:
1. Isolated versus Conjugated/Allylic: Conjugation stabilizes
2. Substitution: More highly substituted are more stable.
- Conjugation/allylic is more important than the substitution pattern of an isolated cation
Impact of Radical Resonance on Reactivity and Product Formation
1. Rates :
2. Product Distribution: Product mixtures often result if an allylic radical is asymmetric.
3. Position of Radical Formation
Section 15.10 S
N
2 on Allylic, Benzylic Systems Are Really Fast
Ex.
Why? Because the backside-attack transition-state is stabilized by conjugation!
(Transition state-stability-reactivity principle).
1. Neither the product nor the reactant has conjugation, so it’s hard to see why conjugation should
apply
2. However, in the 5-coordinate T-state the reactive carbon is sp
2
hybridized
the nucleophile and the electrophile are essentially on opposite ends of a temporary p-orbital.
3. That transient sp
2
hybridization in the transition-state is stabilized by π-overlap with the adjacent
p-bond.
4. This stabilization of the transition-state lowers the activation barrier and greatly accelerates
reaction
Br H
NaOCH
3
H OCH
3
Br H
NaOCH
3
H OCH
3
Slow, and contaminated by competing E 2
Fast and Clean
15 min
10 hours
100 % yield
80 % yield
Br H
H
H
3
CO
Br
H OCH
3
