5. Cyclopropanation Reactions – SYN additions always (cannot form a three-membered
ring on both faces at the same time!)
+
Carbene Reactions:
:CR2 Neutral
Only 6 e- (e- poor – no octet)
Sp2 hybridized, empty p orbital
Both a Lewis Acid and a Lewis Base
Reagents:
a. CH2N2
Diazomethane
H
C
N N
H
Reactive species:
CH2N2
Draw the products:
Ex.
CH2N2
Ex.
CH2N2
b. CHCl3, KOH
Cl
Cl
C
H
OH
Cl
CHCl3, KOH
Cl
Cl
Draw the product:
1
Ex.
CHCl3, KOH
Ex.
CHCl3, KOH
c. Carbenoid Reaction: Simmons-Smith Reaction
(“:CR2”)
CH2I2, Zn(Cu)
H
H
Draw the product:
Ex.
CH3
CH(CH3)I2, Zn(Cu)
H
Ex.
CH2CH3
CH(CH2CH3)I2, Zn(Cu)
H
Ex.
CH3
C(CH3)2I2, Zn(Cu)
CH3
6. Epoxide Formation
mCPBA
O
+
O
mCPBA = meta-chloroperbenzoic acid
O
Cl
O
OH
2
Ex.
mCPBA
mCPBA is an electrophilic oxidizing agent (e- poor) and reacts with the most electronrich alkene present.
Ex.
1 equiv. mCPBA
More alkyl groups on an alkene make the alkene more electron rich!
7. Hydrogenation – Addition of H, H – SYN Addition
H
H2
H
+
Pd/C or PtO2
H
H
Same thing (just a more subtle product drawing):
H2
Pd/C or PtO2
+
Draw the product:
Ex.
H2
Pd/C
Ex.
H2
Pd/C
This is a “Heterogeneous Reaction” – the Catalyst is suspended, not dissolved in the
solution. The reaction occurs on the solid catalyst:
a. Saturate the Catalyst with H2
3
add H2
H
H H
H H H
b. Add the alkene – it attaches to the catalyst, exposing one face of alkene to H atoms
H
H H
H H H
H
H H
H H H
c. First H atom transfers…
H
H H
H H H
H
H H
H
H H
d. Second H atom is transferred, releasing the new saturated system to the solution.
H
H H
H
H H
H H
H H
H H
This reaction, like the other alkene reactions, does not affect
aromatic rings.
Remember – aromatic rings are generally (although not exclusively – remember
ferrocene) six-membered rings with alternating double and single bonds. Cyclohexane,
cyclohexene and cyclohexadiene are NOT aromatic, although they all seem to look alike!
8. Dihydroxylation – Addition of OH, OH (1,2 diol)
SYN Addition
4
OH
OH
OH
+
OH
There are two sets of reagents you can choose to use:
KMnO4, NaOH, H2O – cheap but pH dependent (low yields)
O
O
Mn
O
OH
O
OH
cyclic manganate intermediate
OR
1. OsO4
2. NaHSO3, H2O
Osmium tetraoxide is expensive and toxic but gives high yields for product formation.
O
O
Os
O
O
OH
OH
cyclic osmate intermediate
Both form cyclic 5-membered ring intermediates via the transition metal – since 5membered rings cannot form simultaneously on both the top AND bottom of a double
bond, the dihydroxylation must occur SYN.
Draw the product:
Ex.
1. OsO4
2. NaHSO3, H2O
Ex.
KMnO4, NaOH, H2O
5
Ex.
1. OsO 4
2. NaHSO3, H2O
Now try:
Ex.
1. OsO 4
2. NaHSO3, H2O
(END SYN REACTIONS)
9. Addition of X2 (X = Cl, Br) – ANTI addition
X2
X
X
Used in the “old” days as a qualitative test for alkenes (or alkynes, Ch. 8), when bromine
is used. Br2 is a red liquid.
If red bromine is added to a solution with an unknown compound and the solution stays
red, then the bromine had nothing to react with – no alkene in that compound.
Br2 + No Alkene à Br2 still – RED COLOR
If bromine is added to a solution with an unknown compound and the solution turns
colorless, then the bromine reacted which means the unknown had a functional group
that Br2 reacts with (ALKENE!). Once bromine reacts, then there will be no more Br2,
hence no more red color.
Br2 + Alkene à New Product (No Br2) – NO COLOR
Draw the products:
Ex.
Ex.
Br2
Cl2
6
Ex.
Cl2
Mechanism: This reaction involves the addition of a nucleophilic, electron-rich alkene to
X2, a POLARIZABLE compound.
So: Let’s look at bond connections first:
Step 1: the e- rich alkene approaches the X-X bond, causing the X-X to polarize,
creating an electrophile for the nucleophilic alkene to react with, forming a halonium ion:
δ+
δ-
X-X
X
In the next step, the halide attacks:
X
X
X
X
The intermediate that forms is not a carbocation this time. In this reaction, a
three-membered HALONIUM ION forms (that’s generic for halide but is
specifically a bromonium or chloronium ion and yes, it looks like the
mercurinium ion!). How do we know this is the intermediate and not a carbocation?
X
The physical results tell us this. ALL products from this particular reaction have ANTI
stereochemistry, where X and X always add to opposite faces of alkene. If a
carbocation had been formed, it would be planar and easily accessible from both top and
bottom and the second halide could then attack from either top or bottom, resulting in
both SYN and ANTI products.
This. Never. Occurs. Ever.
Why couldn’t a carbocation form?
Halides are too electron rich to be bonded next to a positively charged carbon atom,
without reacting with it!
X
X
X
7
And, AGAIN, why ANTI?
Once the intermediate halonium ion has formed, the three-membered ring has
effectively blocked one face of the alkene’s reaction site, forcing the second halide to
attack from the opposite face. This always results in a stereospecific ANTI addition
product.
Now – go back and add in stereochemistry… add in the wedges/dashes…
The Side View:
top face blocked!
H
Br
H
Br
Br
H
Br
Br
H
H
Br
H
10. Halohydrin Formation – Addn of OH, X to alkene
Markovnikov, ANTI
Ex.
X
X2
H2O
OH
Alternatively, if you substitute an alcohol for the water and make an ether instead:
Ex.
X
X2
CH3OH
OCH3
Draw the Products:
Ex.
Br2
H2O
Ex.
Cl2
H2O
Ex.
Br2
CH3OH
8
Mechanism:
OH
Br2
H2O
Br
Top View:
H
Br
Br
O
H
Br
O
H
H
Br
Br
O
H
Br
Side View: Look at that ANTI addition occurring…
H
CH3
Br
Br
Br
H
CH3
H-O-H
Br
H
H
O
CH3
H
Br
Br
H
O
CH3
H
So – Why “ANTI”?
One face is blocked by halonium ion, just like in the addition of just X2:
Br
H
CH3
.
And Now - Why “Markovnikov”?
9
First, recall than any positive charge, full or partial, is more stable on a more
substituted carbon. Markovnikov means the reaction proceeds through the most stable
intermediate, full or partial charge. Which end is more substituted?
Br
CH3
H
The more substituted end of the halonium ion breaks the C-X bond more easily because
that carbon atom forms a partial positive charge that would be more stable (lower in
energy, lower activation barrier), and thus the addition of water occurs there, faster,
just like in oxymercuration, with a mercurinium ion.
Br
H
CH3
Nuc
The more substituted side would form a more stable partial charge.
More stable intermediate = Markovnikov
And now the last reaction type:
11. Oxidative Cleavage of Alkenes – requires the breaking of both the pi AND sigma
bonds of alkenes, thus fracturing the carbon skeleton into pieces.
R1
R3
R1
R2
R4
R2
O
+
O
R3
R4
There are two sets of reagents you must know how to work with:
a. Ozonolysis – (with reductive work-up)
-Conversion of alkenes to aldehydes and/or ketones
-Uses Ozone (O3) – formed from oxygen in an ozonator
eO2
O3
O
O
O
Reagent:
O
1. O3
O
O
2. Zn, H3O+
10
Aldehydes and Ketones form with ozonolysis:
R1
H
R1
R2
R4
R2
R1
H
R1
R2
H
R2
O
+
O
O
+
O
H
R4
H
H
Draw the products (BOTH):
Ex.
1. O3
2. Zn, H3O+
Ex.
1. O3
2. Zn, H3O+
Ex.
1. O3
2. Zn, H3O+
b. KMnO4, H2O or H3O+
• Conversion of alkenes to ketones/carboxylic acids/CO2 (H2CO3)
• pH dependent reaction – if the reaction pH becomes too HIGH, dihydroxylation
occurs!
General Equations:
R1
R2
R3
R1
R4
R2
O
+
O
R3
R4
11
R1
R3
R1
R2
H
R2
+
O
R3
O
H
R3
O
R1
H
R1
R2
H
R2
O
+
OH
O
O
H
H
OH
OH
Draw the products:
Ex.
KMnO4, H2O
Ex.
KMnO4, H3O+
Ex.
KMnO4, H3O+
Now: Both of them…
Ex.
1. O3
2. Zn, H3O+
Ex.
12
KMnO4, H3O+
Ex.
KMnO4,NaOH, H2O
Finally…What about:
1. O3
2. Zn, H3O+
13