Download Chem 350 Jasperse Ch. 6 Summary of Reaction Types, Ch. 4

Chem 350 Jasperse Ch. 1 Notes
1
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
Chem 350 Jasperse Ch. 4 Notes
2
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
Carbocations
Stability Pattern
Allylic > 3º > 2º > 1º > methyl > alkenyl
(vinyl, aryl)
Electron Electrophilic/
Poor
Acidic
Carbon
Radicals
C
Allylic > 3º > 2º > 1º > methyl > alkenyl
(vinyl, aryl)
Electron Electrophilic/
Poor
Acidic
C
Allylic > alkenyl (vinyl, aryl) > methyl > 1º
> 2º > 3º
Electron Nucleophilic/
Rich
Basic
C
Carbanions
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.
Chem 350 Jasperse Ch. 4 Notes
3
Stability/Reactivity/Selectivity Principles
1. Reactant Stability/Reactivity: The more stable the reactant, the less reactive it will be. In
terms of rates, this means that the more stable the reactant, the slower it will react. (The
concept here is that the more stable the reactant, the more content it is to stay as is, and the less
motivated it is to react and change into something different)
Key note: Often the “reactant” that’s relevant in this context will not be the original
reactant of the reaction, but will be the “reactant” involved in the rate determining step.

Basicity
O
CH2 Na >
NHNa >
A
>
ONa
ONa
C
B
D
Why: As anion stability increases from A to D, the reactivity decreases

Nucleophilicity
O
CH2 Na >
A
NHNa >
>
ONa
C
B
Why: As anion stability increases from A to D, the reactivity decreases

ONa
D
Nucleophilicity
O
SeNa
A
>
>
SNa
>
ONa
C
B
Why: As anion stability increases from A to D, the reactivity decreases

ONa
D
Reactivity toward alkanes via radical halogenation
because F• > Cl• > Br• > I•
F2 > Cl2 > Br2 > I2
Why: Chlorine is more reactive the bromine because chlorine radical is less stable then
bromine radical.

Electrophilicity (Reactivity in SN2, SN1, E2, E1 Reactions)
I
>
Br
>
Why: As carbon-halogen bond stability increases, the reactivity decreases
Cl
Chem 350 Jasperse Ch. 4 Notes
4
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.

Acidity
O
<
CH3
<
NH2
<
OH
OH
Why: Because as the stability of the anion products increases from A to D, the reactivity
of the parent acids increase
O
CH2 Na <
A

<
NHNa
<
ONa
ONa
C
B
D
Reactivity of alkanes toward radical halogenation
H3C
CH3
<
<
<
Why: Because as the stability of the radical produced during the rate-determining-step
increases, the reactivity of the parent alkane increases
•
•
<
1°

•
<
2°
•
<
3°
3° plus resonance
SN1, E1 Reactivity
Br
Br
<
Br
<
Br
<
Why: Because as the stability of the cation produced in the rate-determining step
increases, the reactivity of the parent halide increases as well
+
1°
<
+
2°
<
+
3°
<
+
3° plus resonance
Chem 350 Jasperse Ch. 4 Notes
5
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.)

SN2 Reactivity
Br
3°
Br
<
<
2°
Br
1°
<
Br
1° plus allylic
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.
Chem 350 Jasperse Ch. 6 Notes
Chem 350 Jasperse Ch. 6
6
Summary of Reaction Types, Ch. 4-6, Test 2
1. Radical Halogenation (Ch. 4)
Br
Br2, hv
resonance stabilized>3º>2º>1º>alkenyl
Recognition: X2, hv
Predicting product: Identify which carbon could give the most stable radical, and substitute a Br
for an H on that carbon.
Stereochemistry: Leads to racemic, due to achiral radical intermediate.
Mech: Radical. Be able to draw propagation steps.
• Br
•
Br
H
Br
Br
slow step
• Br
ready
to repeat
first step
+ H-Br
2. SN2 Substitution
Br NaOCH3
+
OCH3
S N2: 1º>2º>3º> alkenyl
Any of a large variety of nuclophiles or electrophiles can work.
Recognition:
A. Anionic Nucleophile, and
B. 1° or 2º alkyl halide
(3º alkyl halides fail, will give E2 upon treatment with Anionic Nucleophile/Base. For 2º alkyl
halides, SN2 is often accompanied by variable amounts of E2.)
Predicting product: Replace the halide with the anion nucleophile
Stereochemistry: Leads to Inversion of Configuration
Mech: Be able to draw completely. Only one concerted step!
Br OCH3
OCH3 + Br
SN2: 1º>2 º>3 º> alkenyl
3. E2 Reactions.
Br
Mech:
Br
NaOCH3
+ H OCH3
H
H
E2: 3º>2º>1º> alkenyl
Recognition:
OCH3
+ H OCH3 + Br
A. Anionic Nucleophile/Base, and
B. 3º or 2º alkyl halide
(1º alkyl halides undergo SN2 instead. For 2º alkyl halides, E2 is often accompanied by variable
amounts of SN2.)
Orientation: The most substituted alkene forms (unless a bulky base is used, ch. 7)
Predicting product: Remove halide and a hydrogen from the neighboring carbon that can give
the most highly substituted alkene. The hydrogen on the neighboring carbon must be trans,
however.
Stereochemistry: Anti elimination. The hydrogen on the neighbor carbon must be trans/anti.
Mech: Concerted. Uses anion. Be able to draw completely. Only one concerted step!
Chem 350 Jasperse Ch. 6 Notes
7
4. SN1 Reactions.
OCH3
Br
HOCH3
S N1: resonance >3º>2º>1º> alkenyl
+ H Br
Recognition:
A. Neutral, weak nucleophile. No anionic nucleophile/base, and
B. 3º or 2º alkyl halide. (Controlled by cation stability).
(1º alkyl halides undergo SN2 instead. For 2º alkyl halides, SN1 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.
H
Br
slow
step
HOCH3
Br
OCH3
OCH3
+ H Br
+ Br
5. E1 Reactions. 3º > 2º > 1º (Controlled by cation stability)
Br
HOCH3
E1: 3º>2º>1º
H
+
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 SN1.)
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.
Br
slow
step
H
H
Br
+ H Br
+ Br
Sorting among SN2, SN1, E2, E1: How do I predict?
Step 1: Check nucleophile/base.
 If neutral, then SN1/E1  mixture of both
 If anionic, then SN2/E2.
Step 2: If anionic, and in the SN2/E2, then Check the substrate.
o 1º  SN2
o 2º  SN2/E2 mixture. Often more SN2, but not reliable…
o 3º  E2
Chem 350 Jasperse Ch. 6 Notes
8
6.16 Comparing SN2 vs SN1
1
2
3
4
5
6
7
8
Nucleophile
Substrate
Allylic effect…
Leaving Group
Solvent
Rate Law
Stereochemistry
(on chiral, normally 2º R-X)
Ions
Rearrangements
SN1
Neutral, weak
3º R-X > 2º R-X
Allylic Helps
I > Br > Cl
Polar needed
K[RX]
Racemization
SN2
Anionic, strong
1º R-X > 2º R-X
Allylic helps
I > Br > Cl
Non-factor
k[RX][Anion]
Inversion
Cationic
Problem at times
Anionic
Never
6.21 Comparing E2 vs E1
1
2
3
4
5
6
7
8
9
Nucleophile/Base
Substrate
Allylic effect…
Leaving Group
Solvent
Rate Law
Stereochemistry
Ions
Rearrangements
Orientation
E1
Neutral, weak, acidic
3º R-X > 2º R-X
Allylic Helps
I > Br > Cl
Polar needed
K[RX]
Non-selective
Cationic
Problem at times
Zaitsev’s Rule: Prefer
more substituted alkene
E2
Anionic, strong, basic
3º RX > 2º RX > 1º RX
Non-factor
I > Br > Cl
Non-factor
k[RX][Anion]
Trans requirement
Anionic
Never
Zaitsev’s Rule: Prefer more
Substituted alkene (assuming
trans requirement permits)
Comparing SN2 vs SN1 vs E2 vs E1: How Do I Predict Which Happens When?
Step 1: Check nucleophile/base.
 If neutral, then SN1/E1  mixture of both
 If anionic, then SN2/E2.
Step 2: If anionic, and in the SN2/E2 pool, then Check the substrate.
o 1º  SN2
o 2º  SN2/E2 mixture. Often more SN2, but not reliable…
o 3º  E2
Notes:
1º R-X
3º R-X
2º R-X
SN2 only
E2 (anionic) or
SN1/E1 (neutral/acidic)
mixtures common
No E2 or SN1/E1 (cation too
lousy for SN1/E1; SN2 too fast
for E2 to compete)
No SN2 (sterics too lousy)
Chem 350 Jasperse Ch. 7 Notes
9
Ch. 7 Structure and Synthesis of Alkenes
C. E-Z Nomenclature (7-5)
 Each carbon of an alkene has two attachments.
1. Identify which of the two attachments on the left alkene carbon has higher priority.
2. Then identify which attachment on the right alkene carbon has higher priority.
 “Z” (“zusammen” = “together”): the priority attachments are cis
 “E” (“entgegan = “opposite”): the priority attachments are trans
A
A
A
B
B
Z ("together")
B
A
B
Z
B
A
E (opposite)
Z
2 common attachments
no Stereo
When does E/Z apply?
1. If either alkene carbon has two common attachments, than stereo doesn’t apply
2. For tri- or tetrasubstituted alkenes (3 or 4 non-hydrogen attachments), E/Z must be
used if there is stereochemistry
3. For di-substituted alkenes (one H on each alkene carbon), either E/Z or cis/trans
designation can be used
7.7 Alkene Stability Pattern
C
C
C C
C
C
tetrasubbed
>
C
C
C
H
> C C >
C C
C
H
H
C
trans
tridisubbed
subbed
C
H
C C >
C
H
1,1disubbed
C
C
C C >
H
H
cis
disubbed
C
H
H
H
>
C C
C C
H
H
H
H
monounsubbed
subbed
di-subbed
A. Increasing Substitution (# of non-hydrogens directly attached to alkene carbons) 
Increased Stability
 Why? Electronic Reasons.
o Alkene carbons are somewhat electron poor due to the inferior overlap of pi
bonds. (One carbon doesn’t really “get” as much of the other carbon’s electron
as is the case in a nice sigma bond).
o Since alkyl groups are electron donors, they stabilize electron-deficient alkene
carbons.
o Analogous to why electron-donating alkyls give the 3º > 2º > 1º stability pattern
for cations and radicals
B. Trans is more stable than cis for 1,2-disubstituted alkenes
 Why?
o Steric Reasons
Chem 350 Jasperse Ch. 7 Notes
10
Reaction Mechanisms (see p. 310)
A. Recognizing/Classifying as Radical, Cationic, or Anionic
1. Radical
 initiation requires both energy (either hv or ) and a weak, breakable heteroatomheteroatom 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.
Chem 350 Jasperse Ch. 8 Notes
11
Summary of Alkene Reactions, Ch. 8.
Memorize Reaction, Orientation where Appropriate, Stereochemistry where
Appropriate, and Mechanism where Appropriate.
-all are drawn using 1-methylcyclohexene as a prototype alkene, because both orientation and
stereochemistry effects are readily apparent.
1
HBr
Br
Orientation
Stereo
Mechanism
Markovnikov
None
Be able to
draw
completely
(no peroxides)
2
CH3
H
HBr
Anti-Markovnikov Nonselective. Be able to
Both cis
draw
and trans
propagation
steps.
peroxides
Br
both cis and trans
3
H2O, H+
4
CH3
OH
CH3
OH
1. Hg(OAc)2, H2O
Markovnikov
None
Be able to
draw
completely
Markovnikov
None
Not
responsible
2. NaBH4
5
CH3
H
1. BH3•THF
2. H2O2, NaOH
6
Anti-Markovnikov Cis
Not
responsible
Markovnikov
None
Not
responsible
None
Cis
Not
responsible
OH
CH3
OR
1. Hg(OAc)2, ROH
2. NaBH4
7
H2, Pt
D
CH3
H
H
D
Chem 350 Jasperse Ch. 8 Notes
8
9
10
12
Br2
CH3
Br
(or Cl2)
H
Br
Br2, H2O
CH3
OH
(or Cl2)
H
Br
Orientation
Stereo
Mechanism
None
Trans
Be able to
draw
completely
Markovnikov
Trans
Be able to
draw
completely
None
Cis
Not
responsible
CH3
PhCO3H
O
H
11
CH3
OH
CH3CO3H
12
Trans
None
Cis
Not
responsible
None
None
Not
responsible
None
None
Not
responsible
H
OH
H2O
CH3
OH
OsO4, H2O2
Be able to
draw
acidcatalyzed
epoxide
hydrolysis
None
OH
H
1. O3
13
O
H 2. Me2S
H
O
Note: H-bearing alkene carbon
ends up as aldehyde.
KMnO4
14
O
H
OH
H-bearing alkene carbon
ends as carboxylic acid
O
Chem 350 Jasperse Ch. 8 Notes
13
Summary of Mechanisms, Ch. 7 + 8.
Alkene Synthesis and Reactions.
1
HBr
Br
(no peroxides)
H Br
H
H
Protonate
H
Note: For unsymmetrical alkenes,
protonation occurs at the less
substituted alkene carbon so that
the more stable cation forms
(3º > 2º > 1º), in keeping with the
product stability-reactivity principle
Br
Br
H
Cation
Capture
H
+ Br
CH3
H
3º
2
CH3
H
vs.
H
H
2º
CH3
H
HBr
peroxides
Br
both cis and trans
Br
H
Brominate
Br
H
Note 1: For unsymmetrical alkenes,
bromination occurs at the less
substituted alkene carbon so that
the more stable radical forms
(3º > 2º > 1º), in keeping with the
product stability-reactivity principle
H Br
H
Hydrogen
Transfer
Br
3º
H
Br
H
CH3
Br
+
Note 2: Hydrogenation of
the radical comes from either
face, thus cis/trans
mixture results
CH3
H
top
Br
CH3
Br
vs.
cis
Br
H
2º
bottom
H
H
CH3
Br
H
trans H
3
H2O, H+
CH3
OH
H
H
Protonate
H
H
H
OH2
O
Cation
Capture
H
H
H -H
Deprotonate
OH
H
H
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)
Chem 350 Jasperse Ch. 8 Notes
8
14
Br2
CH3
Br
(or Cl2)
H
Br
Br Br
Br
H
Br2, H2O
CH3
OH
(or Cl2)
H
Br
Br Br
Br
H
Br
Br
Cation
Capture
H
9
Br
H
H
3 Notes
1. Cation intermediate is cyclic
bromonium (or chloronium) ion
2. The nucleophile captures the
bromonium ion via backside attack
-this leads to the trans stereochemistry
3. The nucleophile attacks the bromonium
ion at the *more* substituted carbon
H
O
H
Br
H
OH2
Cation
Capture
-H
OH
Br
H
4 Notes
1. Cation intermediate is cyclic bromonium (or chloronium) ion
2. The nucleophile captures the bromonium ion via backside attack (ala SN2)
-this leads to the trans stereochemistry
3. The nucleophile attacks the bromonium ion at the *more* substituted carbon
-this explains the orientation (Markovnikov)
a. There is more + charge at the more substituted carbon
b. The Br-C bond to the more substituted carbon is a lot weaker
CH3
Br
H
More
Substituted
End
H
O
H
Br
H
OH
-H
Br
H
Less
Substituted
End
Br
Br
H
O
H H
-H
OH
H
4. Alcohols can function in the same way that water does, resulting in an ether OR rather than
alcohol OH.
Chem 350 Jasperse Ch. 8 Notes
15
Chapter 7 Reactions and Mechanisms, Review
E2
On
R-X,
Normal
Base
Br
Mech:
NaOCH3
(Normal
base)
Br
CH3
+ H OCH3
H
OCH3
H
+ H OCH3 + Br
Notes
1. Trans hydrogen required for E2
2. Zaytsev elimination with normal bases
3. For 3º R-X, E2 only. But with 2º R-X, SN2 competes (and usually prevails)
4. Lots of “normal base” anions.
E2,
On
R-X, Bulky
Base
Br
Mech:
NEt3 or
KOC(CH3)3
Br H2
C
NEt3
+ Et3NH Br
H
(Bulky
bases)
Notes:
1. Hoffman elimination with Bulky Bases
2. E2 dominates over SN2 for not only 3º R-X but also 2º R-X
3. Memorize NEt3 and KOC(CH3)3 as bulky bases.
AcidCatalyzed
E1Elimination
Of
Alcohols
OH
H2SO4
+ H OH
Mech
OH2
OH
H2SO4
Protonation
-H2O
+ HSO4
Elimination
Deprotonation
H
H
+ OH2
Notes:
1. Zaytsev elimination
2. Cationic intermediate means 3º > 2º > 1º
3. 3-Step mechanism
HSO4
H
+ H2SO4
Chem 350 Jasperse Ch. 10 Notes
16
Summary of Alcohol Syntheses, Ch. 10 (and Review of Old Ones).
1
2
Na
R OH
R ONa
1. Na
R OH
R O R'
2. R'-X


Potassium (K) analogous.
Key way to convert alcohol to
alkoxide, reactive as SN2 nucleophile
and E2 base.


Alkoxide formation-SN2 route to ether
The electrophile R'-X must be SN2
reactive, preferably 1º with a good
leaving group
Mech?
3
Mg
R Br
O
4
H
RMgBr
H
O
6
R
R"
O
R
R
2. H3O+
ester
2. H3O+
(or carbonyl
chloride)
H
OH
2. H3O+
R
R'
R'
3º alcohol
R
H
1 carbon Mech
chain
extension
Mech
R'
OH
2º alcohol
R
R"
R'
OH
3º alcohol
2. H3O+
R'MgBr
H
R'
OH
1º alcohol
1. R(R")CO
R"
R'
3º alcohol
OH
H
R'MgBr
R
1. R'MgBr
OR
1. RCHO
R'MgBr
R'
2º alcohol
1. R'MgBr
ketone
7
OH
1. H2CO
2. H3O+
H
R'
2º alcohol
1. R'MgBr
R
H
+
aldehyde 2. H3O
R'MgBr
H
2. H3O+
formaldehyde
5
OH
1. R'MgBr
O
-Li is analogous for making RLi,
which also act analogously.
-MgBr is spectator: R
is key.
1. RCO2R
R'
R
2. H3O+
R'
OH
3º alcohol
All three Mech
R groups
can
be
different.
At least 2 Mech
R groups
must be
the same
Chem 350 Jasperse Ch. 10 Notes
17
H H
1. R'MgBr
O
8
O
9
NaBH4
R
H
aldehyde
O
10
R
R'
R"
or
or
ketone
O
11
R
ester
H
H
1º alcohol
1. LiAlH4
OH
Mech
R
R"
H
2º alcohol
NaBH4 will
Mech
not react with
esters
H
H
1º alcohol
Review Routes to Alcohols
10
R
OH
Markovnikov
R
OH
1. Hg(OAc)2, H2O
11
R
1. BH3-THF
12
R
2. H2O2, NaOH
13
R X
NaOH
R OH
Markovnikov
R
2. NaBH4
R
Mech
R
R
H2O, H+
2-Carbon
chain
extension
Mech
OH
2. H3O+
OH
H H
1º alcohol
OH
2. H3O+
1. LiAlH4
OR
R'
2. H3O+
2. H3O+
CH3OH
H H
O
R'MgBr
1. LiAlH4
CH3OH
NaBH4
1.
H H
1º alcohol
2. H3O+
ethylene
oxide
OH
OH
anti-Markovnikov
SN2 mech, needs 1º or 2º system
and an excellent leaving group
Chem 350 Jasperse Ch. 10 Notes
18
Summary of Mechanisms, Ch. 10
R Br
•
+ Mg
2+
R• + Br + Mg
R• + Br + Mg•
•
•
For Test:
O
1.
OH
1. Z
R
R'
aldehyde
or ketone
or formaldehyde
R
2.
OR'
R
H 3O
Z
OH
R'
R
R
2. H3O
Z
R'
+ HOR'
Z
Z may be R
(RMgBr)
or H
(LiAlH4)
O
O
Z
OH
esters
or
acid chlorides
R
(NaBH4 or LiAlH4)
O
Cl
mech:
(RMgBr)
Z
1. Z
R
R'
or H
R'
O
Z
Z may be R
O
mech:
R
2. H3O
O
Z
R
OR'
Z
OR'
R
Z
Z
OH
R
O
1. R
3.
R
Z
Z
Z
OH
R
2. H3O
Z
O
H 3O
O
mech:
R
R
O
H3O
R
OH
Not for Test
Chem 350 Jasperse Ch. 11 Notes
Summary of Alcohol Reactions, Ch. 11.
???
1
R OH + NaZ
R ONa + HZ


Acid-Base

2
3
Na
R OH
R OH
R ONa
1. Na
R O R'
2. R'-X
OH
4
O
PCC
R
H
H
1º Alcohols Only
R
H
Aldehydes


Potassium (K) analogous.
Key way to convert alcohol to
alkoxide, reactive as SN2 nucleophile
and E2 base.


Alkoxide formation-SN2 route to ether
The electrophile R'-X must be SN2
reactive, preferably 1º with a good
leaving group

Key access to aldehydes, which are
useful for more Grignard chemistry.
Note difference between PCC and
H2CrO4
PCC does not react with 2º alcohols
very rapidly


OH
5
O
H2CrO4
R
R
H
2º Alcohols Only
R
R
Ketones
Deprotonation by a base.
Controlled by relative stability of
RO versus Z .
Consider relative electronegativity and
whether either anion is resonance
stabilized.


Key access to ketones.
PCC does not react very fast with 2º
alcohols


Note difference between
PCC and H2CrO4 when reacting with
1º alcohols.


HI, HCl analogous
Converts alcohol into a bromide that
can be used in Grignards, E2 reactions
Cation mechanism
Usually not method of choice for 1º, 2º
alcohols
H2CrO4 = Na2Cr2O7, H2SO4 or CrO3/H2O
OH
6
8
R
O
H2CrO4
H
H
1º Alcohols Only
HBr
R OH
3º alcohols
R
OH
Acids
R Br
Mech: Be able to draw!


19
Chem 350 Jasperse Ch. 11 Notes
9
PBr3
R OH

R Br
1º or 2º alcohols



1. PBr3 or HBr
10
R OH
RMgBr
2. Mg
Review Reactions
Br
HBr
13
R
14
R
peroxides
R H
Br2, hv
OH
16
Quick 2-step conversion of alcohol
into a nucleophilic Grignard


Markovnikov addition

anti-Markovnikov addition

Radical mechanism, 3º > 2º > 1º

Zaytsev elimination
R
HBr
15
Converts alcohol into a bromide that
can be used in Grignards, E2, SN2
reactions
Inversion of stereochem
Not good for 3º alcohols
R
R
Br
R
R Br
H2SO4, heat
R
R
Mechanisms for ROH RBr Reactions
R-OH
3º
HBr
R-Br
3º mostly, sometimes 1º
H Br
HBr Mech for 3º ROH:
R
R OH
R
OH2
+ Br
R-OH
PBr3
R-Br
R
R OH
OH2 +
Br
1º, 2º
1º, 2º
Br
H
PBr2
Mech: R OH
R
O
Br
PBr2
R-Br
+ H2O
H Br
HBr Mech for 1º ROH:
Br
Br R + HO-PBr2
R-Br + H2O
20