Download CHAPTER 8 - REACTION EXAMPLES (Based on the 6th edition of

CHAPTER 8 - REACTION EXAMPLES
(Based on the 6th edition of the textbook)
SOLVED PROBLEM 8-1, p. 329
?
a)
Br
Analysis: Markovnikov addition of HBr to the C=C bond. Use HBr:
HBr
Br
Complement:
?
Br
Analysis: Anti-Markovnikov addition of HBr to the C=C bond. Use HBr in the presence of peroxides
HBr
ROOR
CH 3
OH
b)
?
Br
CH 3
Br
Retrosynthetic analysis, or retrosynthesis:
CH 3
Br
HBr
H 2SO 4
peroxides
H 3O
Recommended problem: 8-4, p. 330
CH 3
OH
Concentrated acid favors E1,
dilute acid favors hydration
OXYMERCURATION - DEMERCURATION, ALSO CALLED OXYMERCURATION - REDUCTION
Hg(AcO)2
NaBH 4
a)
HO
Markovnikov alcohol
H 2O
OH
Hg(AcO)2
b)
NaBH 4
Markovnikov alcohol
without rearrangement
H 2O
Contrast with addition of water in the presence of strong acid:
H 3O
OH
H 2O
3o cation
2o cation
Rearranged alcohol
OH
Hg(AcO)2
c)
NaBH 4
CH 3
H 2O
HYDROBORATION - OXIDATION SEQUENCE. An effective way to make Anti-Markovnikov alcohols.
Water adds to the double bond with syn-stereochemistry.
B2H 6
a)
H 2O 2
Anti-Markovnikov alcohol
or:
BH 3 / THF
OH
SOLVED PROBLEM 8-3, p. 339
BH 3 / THF
H 2O 2
OH
CH 3
H
H
OH
CH 3
H
OH
H
CH 3
CH 3
+
or:
OH
OH
trans isomer
CHIRAL!
Enantiomer
Problem 8-15 (b)
BH 3, THF
NaBH 4
+ enantiomer
H
H
OH
OH
Can I make B from A by hydroboration - oxidation?
CH 3
?
Notice that the methyl and the alcohol
group are cis to each other.
OH
B
A
Analysis:
BH 3, THF
CH 3
H
NaBH 4
H
A
OH
OH
Hydroborationoxidation yields the
trans alcohol.
H
or:
BH 3, THF
NaBH 4
+
OH
Answer: NO
Recommended: 8-15 (c), p. 342
CH 3
CH 3
OH
ADDITION OF CHLORINE OR BROMINE TO THE C=C BOND.
CL2
Cl
Cl
a)
+
solvent
Cl
Cl
Anti addition yields the trans product, which is chiral.
Therefore the enantiomer also forms.
Br
H
b)
H
C
C
CH 3
H 3C
H
H 3C
Br2
solvent
S
C
S
Br
+
C
Br
H
enantiomer
CH 3
Br
Br
c)
H 3C
H
C
C
CH 3
H
H 3C
H
Br2
solvent
R
C
Br
S
Br
meso compound (no enantiomer)
C
H
CH 3
Br
Br
H 3C
Br2
d)
solvent
Br
Recommended: 8-17, p. 345
Br
Br
+
enantiomer
Variation with water:
OH
H 3C
OH
Br
Br2
+
enantiomer
H 2O
Br
Recommended: 8-5 and 8-6, p. 347
CATALYTIC HYDROGENATION - Transformation of alkenes into alkanes (syn addition of hydrogen).
H2
Pt, Pd, or Ni
a)
CH 3
H2
b)
Pd
CH 3
H2
c)
Pd
Recommended: 8-23, p. 350
SYN HYDROXYLATION - Syn addition of OH / OH to the C=C bond.
KMnO 4
cold, dilute
CH 3
OH
OH
CH 3
MILD OXIDATION
OsO 4
H 2O 2
CH 3
OH
OH
CH 3
meso form, no enantiomer
ANTI HYDROXYLATION SEQUENCE- Anti addition of OH / OH to the C=C bond.
MCPCBA
a)
OH
H 3O
O
+
metachloro
perbenzoic acid
enantiomer
OH
epoxide
(3-membered cyclic ether)
b)
H
H
C
MCPBA
C
H 3C
CH 3
O
H
H
H 3C
CH 3
H 3O
HO
H
C
H 3C
C
CH 3
MCPBA
H
O
H
CH 3
H 3C
d)
H 3O
H
HO
H
CH 3
R S
H
H 3C
OH
meso
CH 3
CH 3
O
OH
chiral
trans epoxide
MCPBA
CH 3
H
H
H 3C
cis epoxide
c)
R R
H 3O
OH
OH
Recommended for syn and anti hydroxilation: 8-34 (all), p. 359.
+
enantiomer
OXIDATIVE CLEAVAGE: Strong oxidation with potassium permanganate.
In this reaction each of the sp2 carbons involved in the pi bond gets oxidized to its maximum possible
oxidation state. Refer to notes set # 20 (Oxidation and Reduction in Organic Chemistry) to find out what
these states are.
O
KMnO 4
a)
hot,
concentrated
CO 2
O
KMnO 4
b)
+
OH
D
HO
OH
+
O
O
O
c)
KMnO 4
OH
D
OH
OH
or:
HO
O
O
O
O
d)
KMnO 4
CH 3
D
OH
O
e)
O
KMnO 4
D
2
OH
or:
H 3C
O
OXIDATIVE CLEAVAGE: Ozonolysis
In this reaction each of the sp2 carbons involved in the pi bond gets oxidized either to aldehyde or ketone,
depending on whether it ends up at the end of a carbon chain or in the middle after the pi bond cleaves.
If the oxidized carbon ends up at the end of a carbon chain it becomes an aldehyde, otherwise it becomes
a ketone.
O
O
1) O 3
a)
2) Me2S
Me = methyl
+
H
H
H
formaldehyde
acetaldehyde
O
H
1) O 3
b)
+
H
2) Me2S
O
propionaldehyde
acetaldehyde
O
c)
1) O 3
CHO
CHO
2) Me2S
or:
H
H
O
O
O
d)
e)
1) O 3
CH 3
2) Me2S
CHO
1) O 3
O
H
or:
H 3C
O
2
2) Me2S
Recommended strong oxidation (oxidative cleavage) with KMnO4 and with ozone: 8-7, 8-36, and
8-37, p. 362-363.
Recommended problems from the end of the chapter: 47 (all), 49 ( a-f ), 58 (all), 63.