Summary of Alkene Reactions, Ch. 8. Memorize Reaction

33 Pages·2006·2.03 MB·English

by  Craig Jasperse

Chem 350 Jasperse Ch. 8 Handouts 1
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
1
HBr Br
Markovnikov None Be able to
(no peroxides) draw
completely

2 CH
3
HBr H
Anti-Markovnikov Nonselective. Be able to
Both cis draw
peroxides
Br
and trans propagation
both cis and trans
steps.

3 CH
H O, H+ OH3
2 Markovnikov None Be able to
draw

completely

4 CH
1. Hg(OAc) , H O 3
2 2 OH
Markovnikov None Not
2. NaBH4 responsible

5 CH
1. BH •THF 3
3 H
Anti-Markovnikov Cis Not
2. H2O2, NaOH OH responsible

6 CH
1. Hg(OAc) , ROH 3
2 OR
Markovnikov None Not
2. NaBH4 responsible

7 CH
3
H2, Pt H None Cis Not
responsible
D H
D
Chem 350 Jasperse Ch. 8 Handouts 2

Orientation Stereo Mechanism
8 CH
Br2 Br3 None Trans Be able to
draw
(or Cl ) H
2
Br completely

9 CH
Br2, H2O OH3 Markovnikov Trans Be able to
draw
(or Cl ) H
2
Br completely

10 CH
3
PhCO H
3 None Cis Not
O
responsible
H

11 CH Be able to
CH CO H 3
3 3 OH
None Trans draw
H O H acid-
2
OH catalyzed
epoxide
hydrolysis

12 CH
3
OsO4, H2O2 OH None Cis Not
OH responsible
H

13 1. O
3
None None Not
O
H2. Me2S responsible
H O
Note: H-bearing alkene carbon
ends up as aldehyde.

14 KMnO4
None None Not
O
H responsible
OH O
H-bearing alkene carbon
ends as carboxylic acid
Chem 350 Jasperse Ch. 8 Handouts 3
Summary of Mechanisms, Ch. 7 + 8.
Alkene Synthesis and Reactions.

1
HBr Br
(no peroxides)

Note: For unsymmetrical alkenes,
H Br Br Br protonation occurs at the less
H substituted alkene carbon so that
H Protonate Cation H
the more stable cation forms
H Capture H
(3º > 2º > 1º), in keeping with the
+ Br product stability-reactivity principle
CH CH
3 3
vs. H
H
H H
3º 2º

2 CH
3
HBr H
peroxides
Br
both cis and trans

Br H Br H Note 2: Hydrogenation of
H Brominate Br Hydrogen Br + Br tfhaece r,a tdhiucas l cciso/mtraens sfrom either
H Transfer H mixture results CH
3
H
Note 1: For unsymmetrical alkenes, top Br
bromination occurs at the less CH CH
3 3 cis H
substituted alkene carbon so that
vs. Br bottom
the more stable radical forms Br H
Br
(3º > 2º > 1º), in keeping with the
product stability-reactivity principle 3º H 2º H H CH3
Br
trans H

3 CH
H O, H+ OH3
2

H
H OH2 O OH
H -H
H
H Protonate Cation H H
H Capture H Deprotonate 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 Handouts 4

4 CH
1. Hg(OAc) , H O 3
2 2 OH
2. NaBH4

H
HgOAc OH2 O OH
HgOAc H -H
H Cation HgOAc HgOAc
- OAc H Capture H Deprotonate H
NaBH
4
Hg(OAc)
2
OH
H
H

5 CH
1. BH •THF 3
3 H
2. H2O2, NaOH OH

Notes
H CHH3 H2O2, NaOH CHH3 a. concerted addition of B-H across C=C
-explains the cis stereochemistry
BH2 BH2 OH b. the B-H addition is Markovnikov; the
B is !+, the H is !-
c. The H O , NaOH process is complex,
2 2
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.

6 CH
1. Hg(OAc) , ROH 3
2 OR
2. NaBH4

H
HgOAc HgOAc HOCH3 O CH -H OCH3
3
H Cation HgOAc HgOAc
- OAc H Capture H Deprotonate H
NaBH
4
Hg(OAc)
2
OCH
3
H
H
Chem 350 Jasperse Ch. 8 Handouts 5

8 CH
Br2 Br3
(or Cl ) H
2
Br

3 Notes
Br Br Br Br
1. Cation intermediate is cyclic
Br
H Br bromonium (or chloronium) ion
H CCaatpiotunre H 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

9 CH
Br2, H2O OH3
(or Cl ) H
2
Br

H
Br Br Br OH2 O H -H OH
H Br
Cation Br
H Capture H 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
O
More H -H OH
Substituted Br
End Br
H
CH3 H
Br
Less
Substituted
H
End
Br
Br
H -H OH
O
H
H 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 Handouts 6

10 CH
3
PhCO H
3
O
H

ONE CH3 " #+ H
O
! " STEP! ! + ! ! O O
!
! Nioons " Ph #$ Ph
Ph Carbonyl-hydrogen
Hydrogen-bonded reactant
Notes
1. Complex arrow pushing
2. No ions required
3. The carbonyl oxygen picks up the hydrogen, leading directly to a neutral carboxylic acid
-The peracid is already pre-organized for this' via internal H-bonding between carbonyl and H

11 CH
CH CO H 3
3 3 OH
H O H
2
OH

H
OH OSTNEEP! COH3 H COHH3 OH2 O H -H OH
O O No Cation OH OH
ions H H Capture H H
CH3
Notes:
a. The nucleophile (water) attacks from the more substituted end of the protonated epoxide
 More δ+ charge there
 The C-O bond to the more substituted end is much weaker
b. The nucleophile adds via S 2-like backside attack. Inversion at the top stereocenter, but not the
N
bottom, explains the trans stereochemistry.

12 CH
3
OsO4, H2O2 OH
OH
H

O O Ccios nacdedritteiodn CHO3 OsO H2O CHO3H + HO OsO Os (VI)
Os O O OH HO O
O O Osmate
H H
Ester
Osmium
Os (VIII) Os (VI) Hydrolysis H2O2 Reoxidation
O O
+ H O
Os 2
O O
Os (VIII)
Chem 350 Jasperse Ch. 8 Handouts 7
Chapter 7 Reactions and Mechanisms, Review

E2
Br Br
Mech:
On CH
NaOCH 3
3
R-X,
Normal (Normal +H OCH3 H OCH3
base) H + H OCH + Br
Base 3

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, S 2 competes (and usually prevails)
N
4. Lots of “normal base” anions.

EO2n, Br NKOEtC3 (oCrH ) Mech: BrHC2 NEt3 + Et NH Br
3 3 H 3
R-X, Bulky
(Bulky
Base
bases)

Notes:
1. Hoffman elimination with Bulky Bases
2. E2 dominates over S 2 for not only 3º R-X but also 2º R-X
N
3. Memorize NEt and KOC(CH ) as bulky bases.
3 3 3

Acid- OH
Catalyzed H2SO4
+H OH
E1-

Elimination

Of
Mech
Alcohols
OH
2
OH H2SO4 -H2O Deprotonation
H
Protonation Elimination H
+ HSO4 H HSO
4
+ H SO
+ OH2 2 4

Notes:
1. Zaytsev elimination
2. Cationic intermediate means 3º > 2º > 1º
3. 3-Step mechanism
Chem 350 Jasperse Ch. 8 Handouts 8
Ch. 8 Reactions of Alkenes

8-1,2 Introduction

CH
3
CH
3 B
+ A B Addition Reaction
A
H
H
1. Thermodynamics: Usually exothermic
 1 π + 1 σ  2 σ bonds
2. Kinetics: π bond is exposed and accessible

Generic Electrophilic Addition Mechanism
CH CH CH
CH3 Cation 3 Cation 3 3
A B B vs A
+ B
!+ !" Formation A Capture A B
H
A B H E H F H
or
CH3 CH CH3
A Doesn't Happen 3 B
Because
+ B A
Inferior Cation A
H H
C Product E
H
Forms D

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 9

8.3 H-X Hydrogen Halide Addition: Ionic/Cationic Addition in the Absence of Peroxides
(Reaction 1)
H X
H X
General: C C C C

Orientation Stereo Mechanism
1
HBr Br
Markovnikov None Be able to
(no peroxides) draw
completely

Markovnikov’s Rule (For Predicting Products): When H-X (or any unsymmetrical species
Aδ+Bδ-) adds to an unsymmetrical alkene:
o the H+ (or Aδ+) adds to the less substituted carbon (the one with more H’s)
o the X- (or Bδ-) adds to the more substituted carbon (the one with more non-H’s).
o Note: Markovnikov’s rule does not apply if either the alkene or the atoms that are adding
are symmetrical

Examples, Predict the Products.
Does Markovnikov’s
Rule matter?
1 Br Yes
HBr

2 HCl Cl Yes

3 HI No
I

4 Br No
HBr

5 Yes
HBr
Br

6 !+ !- Cl Yes
I Cl
I
Chem 350 Jasperse Ch. 8 Handouts 10

Mechanism
H Br Br Br
H
H Protonate Cation H
H Capture H
+ Br
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

Br
H Br
H Br H Markovnikov Product
Slow Step
2º 2º
or
For unsymmetrical alkenes,
protonation occurs at the
H
less substituted alkene carbon H
so that the more stable cation forms
Br Br
(3º > 2º > 1º), in keeping with the anti-Markovnikov Product
product stability-reactivity principle 1º 1º

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.

Br
HBr
Draw the mechanis for the following reaction:
Br
H2C H Br H3C + Br H3C