Chem 230-001 Lecture 23 Page- 1 Hydroboration/ oxidation
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Racemic trans-2-methylcyclohexanol may be neatly synthesized from 1-methyl1-cyclohexene by using a reagent called borane, BH3.
0.33 equivs.
BH3, O(CH2CH3)2
H2O2
H2O
OH-
B
B(OH)3
+
OEt2
OH
-
Let’s discuss the details of this reaction.
-
Take heart! There is nothing new to remember. You can predict the products
of the borane reagent with any alkene based on what you already know!
-
BH3, Borane exists as B2H6, diborane, which is a gas that readily dissolves in
ether
B2H6 + O(CH2CH3)2 → BH3-O(CH2CH3)2
-
If you draw the Lewis structure you have to put a negative formal charge on B
and a positive formal charge on O.
-
in ether the equilibrium below exists
O
O
H B H
H
-
H B H
H
-
thus releasing a little bit of borane at a time in solution
-
borane is a powerful Lewis acid, borane is trigonal
-
***why?
-
It reacts with the alkene in the following manner. Why?
‡
H
CH3
H B H
H
H
H
H B H
H
CH3
H2B
H
transition state
-
The first usage of the first B-H bond in borane generates another Lewis acid
which can react with two more alkenes.
Chem 230-001 Lecture 23 Page- 2 -
The addition is anti-Markovnikov. Why?
-
It is anti-Markovnikov for the same reason that the addition of HX to the double
bond is Markovnikov!
-
The addition reaction is syn.
o Syn is the opposite of anti.
o This means that the B and the H go on the same side of the double bond.
o Syn addition has stereochemical consequences.
Details of the oxidation reaction
H
H
Et2O
R
B
H
R
RR
R
HR
O
O-OH
-
HOO H R
B
H
R
RR
R
HR
O-OH
H
O
R
R
R
H
B OOH + OH
H
RR
R
HR
H
B
RR
R
HR
H
R
O
O
B
+ OH
R
R
H
O
R
R
R
R
The last intermediate hydrolyzes. Instructor will show you how in class.
Asymmetric synthesis of trans-2-methylcyclohexanol
-
H. C. Brown discovered the following reagent
BH
α-pinene
2
α-pinene is chiral and optically pure.
-
It reacts with 1-methyl-1-hexene, followed by oxidation to make chiral and
optically pure 2-methylcyclohexanol (one enantiomer). ***Why?
Cationic polymerization
-
Poly means many.
-
Mer means body.
-
Polymers are long chains.
o We discussed these in when we talked about long-chain alkanes
Chem 230-001 Lecture 23 Page- 3 -
Think about the feasibility of forming long chains by the following cationic
reaction, ***what conditions would you need? Would you need a high or a
low concentration of styrene? What would the product be?
OH
F3C S O
O (cat)
?
CH2Cl2
Br
Br2/ CH3CH2OH
OCH3
KOtBu
HOtBu
OCH3
The better nucleophile attacks the most substituted carbon of the epibromonium ion.
- do you know why the Br(-) was eliminated instead of the (-)OCH2CH3 group?
After all both are nucleophiles and both could be leaving groups.
Br
H
-
O
Can you come up with a reason for the selectivity of the last substitution
reaction?
Mercuration
-
This reaction is never run on a large scale.
-
This reaction is not environmentally friendly.
-
I am presenting this reaction primarily for historical reasons. You need to
know the organic cannon to take the MCAT and other pre-professional
entrance examinations.
1) Hg(OAc)2, CH3CO2H
H
2) NaBH4, EtOH
OH
Chem 230-001 Lecture 23 Page- 4 -
You may equate NaBH4 to the borane reagents we discussed previously. This
would be a mistake. They are very different.
-
The borane reagents BH3(-)OR2(+) are Lewis acids.
-
NaBH4 is a base.
-
It is usually used in alcoholic solvent to perform reduction
-
It reacts with the alcohol in following manner
o BH4(-) + HOCH2CH3 → H2 + BH3-OCH2CH3
o This goes to a point but does not consume all the BH4(-) reagent.
There are still reactive B-H sites to perform reduction.
O
R
Hg(OAc)2
H2O
R
O
O
O
Hg
O
+
O
+
O
O
NaBH4
EtOH
O
R
Hg(0) +
Hg
OH2
H2O
R
O
R
Hg
OH
OH
O
+
HO
Net hydration across the double bond. Regiochemistry is Markovikov.
Addition of other reagents across the double bond.
O
O
H-O-H
A
H-O-O-H
B
HO
R
C
O O
CF3
E
O
O
H
HO
H
R
D
O O
CF3
F
A: Hydrogen oxide, AKA water, B: hydrogen peroxide
C: a carboxylic acid, D: a peroxyacid
E: Trifluoroacetic acid, F: peroxytrifluoroacetic acid, trifluoroperacetic acid
-
Like the Br-Br bond the O-O bond is weak also.
o Br (-) is a good leaving group.
Chem 230-001 Lecture 23 Page- 5 -
The reagents above have the tendency to react in the following manner
-
Which one of the reagents shown above would you suspect would be more
likely to undergo heterolytic bond cleavage?
O
O
O
O
H
O
O
F
F
O
O
O
+
O
H
F
H
O
+
O
H
F F
F
Br
Br Br
+ Br
-
Because (+)OH is so high in energy these dissociations do not occur.
-
But you can think about the these reagents in terms of their reactivity with the
C-C double in these terms for example:
Br
Br Br
+
+
O
O
H
+
O O
Br
O
+
CF3
HO
CF3
epoxide
-
The double bonds in the phenyl ring are exceptionally stable. More on this
later.
-
The product epoxide is strained and hence high in energy, but the weak ROOR bond in the starting material provide enough of an energetic advantage to
make the three membered ring.
-
There is not an open cation on the reaction path
-
We don’t detect products from the rearrangement of the putative naked cation.
Chem 230-001 Lecture 23 Page- 6 -
O
O
CO2H
7
Oleic acid
O
O
H
7
CO2H
7
7
+
20 °C
O
O
O
H
O
CO2H
7
7
H
7
no free carbocation
CO2H
7
trans-product
is not observed
Asymmetric addition of :O::
- Sharpless epoxidation
O H
TiOiPr4
diethyltartrate
tBuOOH
CH2Cl2
O H
O
one enantiomer
-
need an allylic alcohol
-
CH2=CH–CH2-OH
-
The reagent that accomplishes the epoxidation is optically pure and chiral
-
Barry Sharpless recently was awarded the Nobel Prize for the development of
this chemistry and chemistry similar to it.
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http://www.nobel.se/chemistry/laureates/2001/public.html