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Chem 350 Jasperse Ch. 8 Handouts
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.
Orientation Stereo Mechanism
Br HBr
(no peroxides)
Markovnikov None Be able to
draw
completely
H
CH
3
Br
both cis and trans
HBr
peroxides
Anti-Markovnikov Nonselective.
Both cis
and trans
Be able to
draw
propagation
steps.
OH
CH
3
H
2
O, H
Markovnikov None Be able to
draw
completely
OH
CH
3
- Hg(OAc) 2
, H
2
O
- NaBH 4
Markovnikov None Not
responsible
H
CH
3
OH
1. BH
3
• THF
2. H
2
O
2
, NaOH
Anti-Markovnikov Cis Not
responsible
OR
CH
3
- Hg(OAc) 2
, ROH
- NaBH 4
Markovnikov None Not
responsible
H
CH
3
H
D
D
H
2
, Pt
None Cis Not
responsible
Chem 350 Jasperse Ch. 8 Handouts
Orientation Stereo Mechanism
Br
CH
3
H
Br
Br 2
(or Cl 2
None Trans Be able to
draw
completely
OH
CH
3
H
Br
Br 2
, H
2
O
(or Cl 2
Markovnikov Trans Be able to
draw
completely
O
CH
3
H
PhCO 3
H
None Cis Not
responsible
OH
CH
3
H
OH
CH
3
CO
3
H
H
2
O
None Trans
Be able to
draw
acid-
catalyzed
epoxide
hydrolysis
OH
CH
3
OH
H
OsO 4
, H
2
O
2
None Cis Not
responsible
O
H
H
O
1. O
3
- Me 2
S
Note: H-bearing alkene carbon
ends up as aldehyde.
None None Not
responsible
O
H
OH
O
KMnO 4
H-bearing alkene carbon
ends as carboxylic acid
None None Not
responsible
Chem 350 Jasperse Ch. 8 Handouts
OH
CH
3
- Hg(OAc) 2
, H
2
O
- NaBH 4
O
H
H
H
Cation
Capture
HgOAc
HgOAc
H
H
- H
OH
H
HgOAc
Deprotonate
HgOAc
NaBH 4
OH
H
H
Hg(OAc) 2
OH
2
H
CH
3
OH
1. BH
3
• THF
2. H
2
O
2
, NaOH
H
CH 3
OH
H
BH 2
H
CH 3
BH 2
H 2
O 2
, NaOH
Notes
a. concerted addition of B-H across C=C
- explains the cis stereochemistry
b. the B-H addition is Markovnikov; the
B is !+, the H is !-
c. The H 2
O 2
, NaOH process is complex,
but replaces the B with OH with complete
retention of stereochem
- the explains why the cis stereochemistry
established in step one is preserved in step 2.
OR
CH
3
- Hg(OAc) 2
, ROH
- NaBH 4
O
H
H
H
Cation
Capture
HgOAc
HgOAc
H
CH
3
- H
OCH
3
H
HgOAc
Deprotonate
HgOAc
NaBH 4
OCH
3
H
H
Hg(OAc) 2
HOCH
3
Chem 350 Jasperse Ch. 8 Handouts
Br
CH
3
H
Br
Br 2
(or Cl 2
H
H
H
Cation
Capture
Br
Br
Br Br Br Br
3 Notes
- Cation intermediate is cyclic
bromonium (or chloronium) ion
- The nucleophile captures the
bromonium ion via backside attack
- this leads to the trans stereochemistry
- The nucleophile attacks the bromonium
ion at the more substituted carbon
OH
CH
3
H
Br
Br 2
, H
2
O
(or Cl 2
H
H
H
Cation
Capture
Br
Br
Br Br O
OH
2
H
H
H
Br
OH
- 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
H
CH
3
H
Br
Br
O
H
H
H
Br
OH
- H
More
Substituted
End
H
Br
O
H
H
H
Br
OH
- H
Less
Substituted
End
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 Handouts
Chapter 7 Reactions and Mechanisms, Review
E
On
R-X,
Normal
Base
CH
3
Br
OCH
3
H
H H OCH
3
Br
NaOCH 3
H OCH
3
Mech:
(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
Br NEt 3
or
KOC(CH
3
3
(Bulky
bases)
H
2
C
Mech: Br
H
NEt 3
NH Br
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
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
Notes:
1. Zaytsev elimination
2. Cationic intermediate means 3º > 2º > 1º
3. 3 - Step mechanism
Chem 350 Jasperse Ch. 8 Handouts
Ch. 8 Reactions of Alkenes
8 - 1,2 Introduction
CH
3
A B
CH
3
B
A
H
H
Addition Reaction
1. Thermodynamics: Usually exothermic
1 π + 1 σ 2 σ bonds
2. Kinetics: π bond is exposed and accessible
Generic Electrophilic Addition Mechanism
CH
3
A B
CH
3
B
A
H
H
CH
3
A
H
+ B
CH
3
A
H
+ B
or
CH
3
B
A
H
vs
CH
3
A
B
H
Cation
Formation
Cation
Capture
CH
3
H
A
A
B
C
D
E
E
F
Doesn't Happen
Because
Inferior Cation
Product
Forms
2 Steps: Cation formation and cation capture
- Cation formation is the slow step
o Cation stability will routinely determine the orientation in the first step
Which is preferred, A B or A C?
- Often the cation is a normal cation B. Sometimes 3-membered ring cations D will be involved.
- In some cases, the cation will be captured by a neutral species (like water), in which case an
extra deprotonation step will be involved
4 Aspects to Watch For
1. Orientation
- Matters only if both of two things are true:
a. The alkene is unsymmetrical, and
b. The electrophile is unsymmetrical
2. Relative Stereochemistry
o Matters only if both the first and the second alkene carbons are transformed into chiral
centers
3. Mechanism
4. Relative Reactivity of Different Alkenes
o Stability of cation formed is key
Chem 350 Jasperse Ch. 8 Handouts
Mechanism
Br
H
H
H Br
H
H
Br
Protonate Cation
Capture
H
o Protonate first
o Capture cation second
o Cation formaton (step 1) is the slow step
Rank the Reactivity of the following toward HBr addition.
3 (2º) 2 (3º) 1 (3º allylic)
Issue: Cation stability
Why Does Markovnikov’s Rule Apply? Product/Stability Reactivity Rule.
o Formation of the most stable carbocation results in Markovnikov orientation
H Br
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
or
H
H
Br
Br
H
Br
H
Br
Markovnikov Product
anti-Markovnikov Product
Slow Step
o This same logic applies anytime something adds to an alkene.
o You want to make the best possible intermediate in the rate-determining step.
Draw the mechanis for the following reaction:
HBr
Br
H
2
C
H Br
H
3
C
H
3
C
Br
Chem 350 Jasperse Ch. 8 Handouts
8.3B Free Radical Addition of HBr with Peroxide Initiator: Anti-Markovnikov Addition (Rxn 2)
H
CH
3
Br
both cis and trans
HBr
peroxides
Anti-Markovnikov Nonselective.
Both cis
and trans
Be able to
draw
propagation
steps.
- Peroxides are radical initiators, and cause the mechanism to shift to a radical mechanism
- With peroxides, the orientation is reversed to anti-Markovnikov: now the Br adds to the less
substituted end and the H adds to the more substituted end of an unsymmetrical alkene
o No peroxides: Br goes to more substituted end
o With peroxides: Br goes to less substituted end
- The anti-Markovnikov radical process works only with HBr, not HCl or HI
- The radical process is faster, and wins when peroxides make it possible. In the absence of
peroxides, the slower cationic process happens.
Mechanism, and Reason for AntiMarkovnikov Orientation
Br
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
or
Br
Br
2 º radical
Br
H
Br
H
1 º radical
Markovnikov Product
anti-Markovnikov Product
Slow Step
H Br
H Br
Examples, Predict the Products.
Does Markovnikov’s
Rule matter?
HBr, peroxides
HBr, no peroxides
Br
Br
Yes
2 HBr, peroxides
HBr, no peroxides
Br
Br
Yes
HBr, peroxides
HBr, no peroxides
Br
Br
No
Chem 350 Jasperse Ch. 8 Handouts
Examples, Predict the Products.
Does Markovnikov’s
Rule matter?
H
2
O, H
OH
Yes
H
2
O, H
HO
Yes
H
2
O, H
OH
No
H
2
O, H
OH
No
H
2
O, H
OH
Yes
Problems with Acid-Catalyzed Addition of Water to Alkenes
1. Alkenes with poor water solubility often don’t add very well.
- Can’t drive the equilibrium strongly to the alcohol side in that case
- Solvent mixtures can often help, but not always good enough
2. Alcohol/Alkene equilibrium sometimes poor
3. Carbocation rearrangements can be a problem
4. The degree of Markovnikov selectivity isn’t always satisfactory
- 99:1 isomer selectivity is a lot nicer than 90:10…
o Especially if you have to purify!
5. Obviously you can’t get the reverse, anti-Markovnikov alcohol products.
Each of these limitations, when they are a problem, can be solved by alternative recipes that
indirectly add H-OH.
Draw the mechanism for the following reaction:
H
2
O, H
HO
H
2
C
O
H
H
3
C
OH
2
H
H
OH
Chem 350 Jasperse Ch. 8 Handouts
8.5 Indirect Markovnikov Addition of H-OH via Oxymercuration/Demercuration. Reaction 4.
General: C C C C
H OH
- Hg(OAc) 2
, H
2
O
- NaBH 4
OH
CH
3
- Hg(OAc) 2
, H
2
O
- NaBH 4
Markovnikov
Stereo:
None
Mech:
Not
responsible
Notes:
1. Often higher yields, cleaner, faster, and easier
2. No restrictions
3. No cation rearrangements
4. Very strong , often superior Markovnikov selectivity
o OH adds to the more substituted end, H to the less substituted end
Does Markovnikov’s
Rule matter?
- Hg(OAc) 2
, H
2
O
- NaBH 4
H
2
O, H
+ OH
OH
Yes
- Hg(OAc) 2
, H
2
O
- NaBH 4
H
2
O, H
HO
HO
Yes
H
2
O/H
vs Oxymercuration/Demercuration: Which should I use?
- Both normally give same product
- For predict-the-product problems, be able to handle either recipe
- For provide-the-right-recipe problems, I will accept either answer.
o H
2
O/H
is easier to write!
- In the real lab, the choice is decided on a case-by-case basis.
o Default to H
2
O/H
o Go to oxymercuration/demercuration when direct acid-catalyzed hydration doesn’t
work as well as you’d like
Chem 350 Jasperse Ch. 8 Handouts
8.7 Indirect anti-Markovnikov Addition of H-OH via Hydroboration/Oxidation. Reaction 5.
C C
C C
H BH
2
Overall pathway:
1. BH
3
• THF 2. H
2
O
2
, NaOH
C C
H OH
"Hydroboration" "Oxidation"
H
CH
3
OH
1. BH
3
• THF
2. H
2
O
2
, NaOH
plus enantiomer
Anti-Markovnikov Cis Not
responsible
Notes:
1. Anti-Markovnikov orientation : the OH ends up on the less substituted end of an
unsymmetrical alkene; the H adds to the more substituted end
2. Cis addition. Both the H and the OH add from the same side.
3. When does cis/trans addition stereochemistry matter?
o Only when both alkene carbons turn into chiral centers in the product.
o If one does but not both, then the relative stereochemistry doesn’t matter
o For Markovnikov additions involving H-Br or H-OH, the H usually adds to a carbon that
already has an H, so that in the product it is not a stereocenter.
o In anti-Markovnikov additions, much more common for both carbons to become chiral
carbons
4. Chiral products are Racemic (two enantiomers form) but not optically active
o When only one chiral center forms (often in the Markovnikov additions), any chiral
product will always be racemic
o When two chiral centers form, as in the example above, of the four possible
stereoisomers, you get only two of them, in racemic mixture.
H
CH
3
OH
1. BH
3
• THF
2. H
2
O
2
, NaOH
H
CH
3
OH
H
CH
3
OH
H
CH
3
OH
A
B
C
D
Cis Addition Enantiomers
Do Form
Trans Addition Enantiomers
Do NOT Form
Examples, Predict the Products.
Does
Markov.
Matter?
Does
Stereo
Matter?
1. BH
3
• THF
2. H
2
O
2
, NaOH
OH
Yes No
Chem 350 Jasperse Ch. 8 Handouts
Does
Markov.
Matter?
Does
Stereo
Matter?
1. BH
3
• THF
2. H
2
O
2
, NaOH
HO
Yes No
1. BH
3
• THF
2. H
2
O
2
, NaOH OH
No No
1. BH
3
• THF
2. H
2
O
2
, NaOH
OH
No No
1. BH
3
- THF
- NaOH, H 2
O
2
- Hg(OAc) 2
, H
2
O
- NaBH 4
H
2
O, H
OH
H
H
3
C
OH
OH
H
Yes Yes
No
No
1. Which starting alkenes would produce the following products following hydroboration-
oxidation? Factor in the stereochemistry of the products in considering what starting materials
would work.
1. BH
3
- THF
- NaOH, H 2
O
2
1. BH
3
- THF
- NaOH, H 2
O
2
Ph
H
H
HO
H
3
C
Ph
OH
H
H
H
3
C
Ph
H
CH
3
H
Ph
CH
3
CH
3
Ph
CH
3
H
HO
H
H
3
C
2. Fill in recipes for converting 1-butene into the three derivatives shown.
1. BH
3
- THF
- NaOH, H 2
O
2
OH
- Hg(OAc) OH 2
, H
2
O
- NaBH 4
H
2
O, H
or
Chem 350 Jasperse Ch. 8 Handouts
8.6 Alkoxymercuration-Demercuration: Markovnikov Addition of H-OR (Reaction 6)
General: C C C C
H OR
- Hg(OAc) 2
, ROH
- NaBH 4
OR
CH
3
- Hg(OAc) 2
, ROH
- NaBH 4
Markovnikov Stereo:
None
Mech:
Not
responsible
Mechanism
OR
CH
3
- Hg(OAc) 2
, ROH
- NaBH 4
O
H
H
H
Cation
Capture
HgOAc
HgOAc
H
CH
3
- H
OCH
3
H
HgOAc
Deprotonate
HgOAc
NaBH 4
OCH
3
H
H
Hg(OAc) 2
HOCH
3
Notes:
1. Everything is the same as with oxymercuration-demercuration to form an alcohol, except
you use an alcohol instead of water
2. This results in an oxygen with it’s spectator carbon chain adding rather than an OH
3. Strong Markovnikov orientation
o The OR adds to the more substituted end of the alkene
o The Hydrogen ends up on the less substituted end of the alkene
4. The mechanisms are analogous.
Examples, Predict the Products.
Does Mark’s
Rule matter?
Does
Stereo?
- Hg(OAc) 2
, CH
3
CH
2
OH
- NaBH 4
O
Yes No
- Hg(OAc) 2
- NaBH 4
OH
O
Yes No
Chem 350 Jasperse Ch. 8 Handouts
Ether Synthesis: Two Routes
1. From Alkene and Alcohol: By Oxymercuration/Demercuration
2. From R-Br and Alkoxide Anion: By S
N
3. Multistep Syntheses : Design Reactants for the Following Conversions
- Note: It is often most strategic to think backward from product to precursor.
- Then think back how you could access the precursor from the starting material.
- There may sometimes be more than one suitable route.
a.
OCH
3
HBr, peroxides
NaOCH 3
b.
- Hg(OAc) 2
, CH
3
OH
- NaBH 4
OCH
3
c.
OH
CH
3
OH
CH
3
1. H
2
SO
4
, heat
2. BH
3
- THF
- NaOH, H 2
O
2
d.
- Br 2
, hv
- NaOCH 3
(or other small, normal base)
e.
- Br 2
, hv
- NEt 3
(or KOCMe 3
f.
OH
CH
3
OH
CH
3
Br
CH
3
- H 2
SO
4
, heat
- HBr
g.
1. H
2
SO
4
, heat
- HBr, peroxides
OH
CH
3
CH
3
Br
