Alkene-Alkyne Homework Answer Key, CH343, Spring 2010
1. Show how the following compounds can be synthesized in high yield from the indicated starting materials,
and any other reagents. All of these require a minimum of a carbon-carbon bond forming reaction, followed by
a functional group transformation. Attempted one-step reactions will receive NO credit.
Product
Starting Material
OH
a.
1. BH3-THF
CH3
CH3
Ph3P
CH3
2. NaOH, H2O2
H2O
O
Wittig
To make this alcohol, you could add water to an alkene in an Anti-Markovnikov sense, using
hydroboration oxidation. To make the double bond in that specific place, you need to use a Wittig
reaction.
b.
OH
OH
CH3
OsO4
CH3
H2O2
Ph3P
CH3
O
Wittig
A diol is most easily made by oxidizing an alkene with OsO4/H2O2, or dilute KMnO4. Again, to make
the alkene in that specific location, you need to use a Wittig reaction.
c.
O
Dilute KMnO4
H3C
PTC
O
H3C
C
C
1. NaNH2
2.
H3C
C
CH
Cl
A diketone on adjacent carbons can be made by oxidizing an alkyne. The alkyne can be made by
alkylating a smaller alkyne, which is an old carbon-carbon bond-forming reaction.
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d.
O
C
HO
HO
C
O
O
O
KMnO4, H2O
(hot. conc.)
OH
C
C
OH
OH
heat
+
heat
O
The Diels-Alder reaction gives us the carbons we need. Oxidation of the alkene makes two more
carboxylic acids.
e.
H3C
Br
HBr
ROOR, heat
H3C
CH2
H2C=PPh3
O
H3C
H
You need to add a carbon, and put the halogen on the end. Addition of H-Br to an alkene in an AntiMarkovnikov fashion requires peroxides. You can make the alkene from the aldehyde stating material
directly with a Wittig reaction.
f.
O
N
O
O
CH3
CH2I2
Zn-Cu
N
O
O
CH3
N
heat
CH3
O
We can only start with things with one ring, We need to make a three-membered ring and six-membered
ring. Our best six-membered ring-making reaction so far is the Diels-Alder reaction. If we chose
appropriate starting materials, we can make the bicyclic ring system directly. Using the Simmons-Smith
reaction gives us the three-membered ring.
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3. Fill in the needed reagents in the following reaction scheme. It is part of a synthesis of Longifolene, which is
found in turpentine.
H3C O
O
H3C O
O
1. CH3MgBr
H2SO4
2. H3O+
O
heat
H3C
HO
CH3
O
CH3
O
CH3
+
H2O, H
O
H3C O
CH3COOOH
Strong base
heat
(Chapter 22)
H3C
H3C
O
Br
H3C
CH3
OH
O
CH3
H2SO4
CHBr3, KOH
O
O
CH3
CH3
Br
heat
CH3
3
3. Propose a reasonable structure for the following sets of spectra. Explain what information you obtained from
each spectrum. You will obtain information from each spectrum! Label the peaks in the NMR spectra with the
corresponding hydrogens and carbons in the structure.
Mass Spectrum
MW = 82, so probably no N. No indication of Cl or Br.
IR Spectrum
sp2 C-H at 3100, sp3 C-H at 2900, C=C at 1600
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Carbon NMR Spectrum
H2C
CH3
H2C
CH3
sp2 C’s at 145 and 115 ppm. sp3 C at 20 ppm. Only three types of C implies symmetry.
Proton NMR Spectrum
H2C
CH3
H2C
CH3
Alkene H’s are around 5. Two H’s on the CH2 are not equivalent (why?), so they show up as separate peaks.
The methyls are equivalent, so they show up as a peak.
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