CHEM 203
HOMEWORK 6
Alkynes, halides, rad. halogenation
1. Many insects communicate by releasing volatile compounds ("pheromones") that convey specific messages
when they bind to appropriate receptors located on the antennae of other members of the same species. The
judicious use of pheromones provides an environmentally benign method of insect control. An effort to
determine the structure of the sex pheromone ("muscalure") of the female housefly involved the operations
outlined below. Provide the structures of A – D and write accurate mechanisms for their formation.
CH2=CH–(CH2)5–CH3
CH3–(CH2)12–C
HBr, radical
C–H
A
initiator
H2, Lindlar
1. NaNH2
catalyst
B
2. A
R—R
2 R•
C: identical to
muscalure
Na, liq. NH3
H—Br
D: not the same
as muscalure
R–H + Br •
radical
initiator
Br
Br •
CH2 CH–(CH2)5–CH3
Br
CH2 CH–(CH2)5–CH3
•
CH2 CH2–(CH2)5–CH3 + Br •
A
H—Br
H2, Lindlar
CH3–(CH2)12
(CH2)7–CH3
C C
H
H
C: identical to muscalure
catalyst
CH3–(CH2)12—C
C–(CH2)7–CH3
B
Na, liq. NH3
CH3–(CH2)12
H
C C
H
(CH2)7–CH3
D: not the same as muscalure
mechanisms for the formation of C and D:
B = R1—C
H—H
Lindlar cat. = Pd(0)
H2
oxidative
addition
C—R2
H Pd(II)
H
hydropalladation
(similar to
hydroboration;
syn addition)
R1
R2
reductive
C C
C
H
Pd(II) elimination
H
CHEM 203
HOMEWORK 6
Na
liq. NH3
p. 2
Na+ + e— (blue solution)
B = R1—C
H NH2
••
•
1
R —C
SET
C—R2
+ NaNH2
C
Na+
R2
R1
••
C
R2
H
H2N H
SET
C
R2
H
C
Na+
H
e—
R1
H
C
•
R1—C
+ NaNH2
C
R2
D
2. Propose a method to convert 1-butyne (structure below) into compounds a. - k. If a compound appears to be
unavailable as the major product of any reaction known to you, answer "inaccessible"
b.
a. OH
OH
H
OH
CH3–CH2–C
H
OH
j.
Cl
H
H
i.
Cl
H2, Lindlar
H2O
(or Na/NH3)
H2SO4
2. H2O2
aq. NaOH
OH
1. OsO4
OH
c.
2. aq. NaHSO3
e.
OH
Br2
H2O
(part a.)
Br
OH
INACCESSIBLE
1. O3
f.
(part a.)
Cl
OH
h.
2. H2O2 / H+
OH
O
O
+
H
OH
OH
OH
f.
Br
OH
H
H
1. BH3
d.
OH
C–H
C–H
b.
(part a.)
OH
e.
O
OH
H
(part a.)
Br
d.
O
k.
a. CH3–CH2–C
OH
c.
H
H
Br
g.
CHEM 203
g.
HOMEWORK 6
NaNH2
CH3–CH2–C
C–H
Na+
CH3–I
Na
(e.g.)
NH3
Br
Br2
H
H
Br
OH
H
1. OsO4
H2, Lindlar
h.
p. 3
(part g.)
2. aq. NaHSO3
H
i.
H
Cl
OH
H
1. OsO4
j.
H
2. aq. NaHSO3
(part g.)
OH
1. O3
k.
H
+ O=CH2
O
2. Zn / H+
(part a.)
OH
Cl
Cl2
(part h.)
H
3. Propose a method for the preparation of compounds a. – h. below starting from appropriate alkenes. Draw a
clear structure of your proposed starting olefin and list all reagents / catalysts, in the correct order, that are
required to induce the desired transformation. Your method must be a good one, i.e., the desired compound
must be the major product of your reaction(s). I a product does not appear to be available by any method
known to you, answer "INACCESSIBLE.". Note: it is understood that chiral compounds will be obtained
as racemic mixtures.
O
Br
Cl
a.
OH
HO
b.
Cl
OCH3
f.
a.
H
Br
O
c.
g. Cl
Br
d.
Cl
Cl
OH
i.
NBS
Cl
b.
e.
I
h.
HCl
Br
hν
j.
Br
CHEM 203
c.
HOMEWORK 6
O
1. O3
2. H2O2, H+
OH
HO
HBr, rad.
d.
initiator
O
NBS
e.
p. 4
Br
OCH3
Br2
f.
hν
CH3OH
g. INACCESSIBLE
Cl
Cl2
h.
hν
I
HI
Br
H
Br
Cl
Cl
H2SO4
j.
i.
OH
H2O
4. Homework 5 indicated that the C–H bonds adjacent to the O atom in diethyl ether are significantly weaker
than those removed from it. Yet, the C–H bonds in n-pentane (the linear isomer) are all approximately of
equal strength. Propose a rationale for the weakness of the O–C–H bonds in diethyl ether.
approx. ΔHdiss (kcal/mole)
CH2–CH–O–CH2–CH3
H
H
CH2–CH–CH–CH2–CH3
H
H
H
lobes of
the σ*H–H CH3
H
••
O
H
C–H
C–H
95
85
C–H
C–H
C–H
98
95
95
it seems likely that hyperconjugative delocalization of a lone
pair of electrons associated with the O atom into the σ* orbital
of neighboring C–H bonds (diagram on the left) results in
weakening of such C–H bonds (reminder: introducing electron
density into an antibonding orbital diminishes the bond order;
i.e., it weakens the bond)
• • hyperconjugation
5. The molecules shown below react with NBS / hν to furnish a single product of mono-bromination. Predict
the structure of such a product, write a detailed mechanism for its formation, and provide a brief rationale
for your prediction.
N
O
a.
N
b.
c.
CHEM 203
a.
HOMEWORK 6
p. 5
NBS
the weakest C–H bonds in this molecule are those adjacent
to the O atom (cf. red C–H bond; see problem 4 above), so:
H
O
hν
O
Br
mechanism:
O
± H+
frequent
H—Br contaminant
in NBS
O
OH
OH
N Br
N Br
O
O
N
H
+ Br—Br
N
Br –
O
O
H—Br
hν
Br
O
O
+ Br •
b.
2 Br •
•
O
Br—Br
Br—Br
H
N
the weakest C–H bonds in this molecule are those adjacent
to the benzene-like ring (cf. benzylic-type red C–H bond), so:
N
Br
H
OH
OH
N Br
N Br
O
O
N
H
± H+
frequent
H—Br contaminant
in NBS
O
N
Br –
+ Br—Br
O
O
H—Br
•
N
N
Br
c.
+ Br •
hν
2 Br •
N
Br—Br
H
N
Br—Br
(stabilized by
resonance)
N
N
NBS
the weakest C–H bond in this molecule is
the tertiary C–H one (cf. red C–H bond), so:
H
hν
N
hν
mechanism:
O
N
NBS
Br
CHEM 203
HOMEWORK 6
p. 6
mechanism:
O
± H+
frequent
H—Br contaminant
in NBS
O
OH
OH
N Br
O
+ Br—Br
N
Br –
N Br
O
N
H
O
O
H—Br
hν
2 Br •
•
Br—Br
H
Br—Br
Br
+ Br •
6. Provide an explanation for the fact that the reaction of 1-butene with NBS / hν furnishes mostly trans-1bromo-2-butene (it is obviously essential to write a detailed mechanism for the reaction).
NBS
Br
hν
O
frequent
H—Br contaminant
in NBS
O
N Br
± H+
OH
O
H—Br
2 Br •
•
hν
Br—Br
H
among the two possible electron distributions,
the one with the more highly substituted double
bond is energetically more favorable (notes of Oct. 8).
Evidently, the radical tends to react with Br2 at the
terminal C atom, so as to maintain the π bond at a
more favorable site:
•
•
+ Br—Br
O
O
this radical is
stabilized by
resonance:
OH
N
Br –
N Br
O
N
H
symbol for
resonance
•
•
symbol for
resonance
Br—Br
Br
+ Br •
CHEM 203
HOMEWORK 6
p. 7
7. Propose a method for the preparation of compounds a. – h. below starting with acetylene and methane as
the sole source of carbon atoms. Provide your answer as a flowchart illustrating all the individual steps (with
appropriate reagents, catalysts, etc.), in the correct order, that are required to obtain the final products. Note:
it is understood that chiral compounds will be obtained as racemic mixtures.
b.
a. OH
H–C C–H
acetylene
H
OH
f.
c.
OH
OH
Br
OH
H
CH4
methane
OH
H
g.
Br
a.
Br
Br
H
Br
H
l.
H
H
OH
m.
i.
Cl
H
Cl
n.
OH
OH
H
H
• this alcohol can be made by hydration of propene, CH2=CH—CH3
• somehow we must connect a molecule of acetylene (2 carbons) to
one of methane to make propene
• propene could be made by semihydrogenation of propyne
• propyne can be made by SN2 methylation of acetylide with, e.g., CH3Cl
OH
j.
Cl
OH
H
H
Br
e.
Cl
h.
OH
Br
k.
Cl
OH
OH
H
H
H
d.
H
OH
o.
SO:
Cl2
CH4
CH3–Cl
hν
H–C C–H
NaNH2
H–C C
Na+
CH3–Cl
H–C C–CH3
H2, Lindlar cat.
H2SO4
(Na / NH3liq OK)
H2O
• this alcohol can be made by hydroboration / oxidation of propene
OH • propene can be made as seen in part a. above
b.
1. BH3
(part a.)
2. H2O2
aq. NaOH
OH
• this diol can be made by osmylation of propene
OH • propene can be made as seen in part a. above
c.
OH
1. OsO4
(part a.)
d.
Cl
2. aq. NaHSO3
OH
OH
• this chlorohydrin can be made from propene
OH
Cl2, H2O
(part a.)
SO:
Cl
OH
SO:
SO:
OH
CHEM 203
HOMEWORK 6
e.
Br
p. 8
• this bromide can be made by radical addition of HBr to propene
HBr
OH
H
f.
H
Br
rad. initiator
(part a.)
• this diol can be made by osmylation of trans-2-butene
• the only method we know to make trans-alkenes is to reduce an alkyne with Na / liq. NH3
• to make trans-2-butene we then need 2-butyne
• 2-butyne can be made by SN2 methylation of propyne with, e.g., CH3Cl
• both propyne and CH3Cl have been made in part a. above
SO:
OH
H–C C–CH3
NaNH2
Na+
C C–CH3
(part a.)
CH3–Cl
(part a.)
Na / liq. NH3
CH3–C C–CH3
OH
H
1. OsO4
2. aq. NaHSO3
Br
H
g.
H
H
OH
• this dibromide can be made by bromination of cis-2-butene
• the only method we know to make cis-alkenes is to reduce an alkyne with H2/Lindlar cat.
• to make cis-2-butene we then need 2-butyne
• 2-butyne can be made as per part f. above
SO:
Br
H2 / Lindlar cat.
CH3–C C–CH3
(part f.)
Br
Br2
H
H
(note: Na / liq. NH3
NOT OK)
Br
• this diol can be made by osmylation of cis-2-butene
OH
H • cis-2-butene can be made as per part g. above
h.
H
OH
Cl
H
i.
Cl
SO:
OH
H
1. OsO4
(part g.)
H
(note: H2 / Lindlar
NOT OK)
H
2. aq. NaHSO3
OH
• this dichloride can be made by chlorination of trans-2-butene
• trans-2-butene can be made as per part f. above
Cl
H
Cl2
H
Cl
• this alcohol can be made by hydration or hydroboration-oxidation of
either trans-2-butene (part f. above) or cis-2-butene (part g. above)
j.
OH
H2SO4
(part f.)
H2O
OH
CHEM 203
HOMEWORK 6
p. 9
• this dibromide can be made only by bromination of 1-butene
Br • 1-butene could be made by semihydrogenation of 1-butyne
• 1-butyne can be made by SN2 ethylation of acetylene with, e.g., CH3–CH2–I
• CH3–CH2–I can be made by addition of HI to ethylene
• ethylene can be made by semihydrogenation of acetylene
k.
Br
SO:
H2, Lindlar cat.
CH2=CH2
H–C C–H
H-I
CH3–CH2-I
(Na / NH3liq OK)
NaNH2
H–C C–H
Na+
H2, Lindlar cat.
CH3–CH2-I
C C–H
CH2–CH2–C C–H
Br2
(Na / NH3liq OK)
Br
Br
Br
H
l.
H
Br
• this dibromide can be made only by bromination of trans-2-pentene
• trans-2-pentene could be made by semihydrogenation of 2-pentyne with Na/liq. NH3
• 2-pentyne can be made by SN2 ethylation of propyne (part a.) with, e.g., CH3–CH2–I
(part k.) or SN2 methylation of 1-buyne (part k.) with, e.g., CH3–Cl (part a.)
SO:
CH2–CH2–C C–H
(part k.)
NaNH2
CH2–CH2–C C
CH3–Cl
Na+
Na / liq. NH3
m.
H
H
NaNH2
CH2–CH2–C C
Na+
CH3–CH2–I
Cl
H
CH2–CH2–C C–CH2–CH3
(part k.)
(part k.)
Cl
Br
SO:
CH2–CH2–C C–H
H
H
• this diol can be made only by osmylation of trans-3-hexene
• trans-3-hexene could be made by semihydrogenation of 3-hexyne with Na/liq. NH3
• 3-hexyne can be made by SN2 ethylation of 1-butyne (part k.) with CH3–CH2–I (part k.)
OH
n.
Br
H-I
(H2, Lindlar
cat. NOT OK)
OH
H
CH2–CH2–C C–CH3
Na / liq. NH3
1. OsO4
(H2, Lindlar
cat. NOT OK)
2. aq. NaHSO3
OH
H
H
OH
• this dichloride can be made only by chlorination of cis-3-hexene
• cis-3-hexene could be made by semihydrogenation of 3-hexyne with H2/Lindlar cat.
• 3-hexyne can be made as per part m.
CHEM 203
10
HOMEWORK 6
p.
H2, Lindlar cat.
CH2–CH2–C C–CH2–CH3
(part m.)
OH
H
o.
H
OH
(Na / liq. NH3
NOT OK)
H
H
• this diol can be made only by osmylation of cis-3-hexene
• cis-3-hexene can be made as seen in part n.
OH
H
1. OsO4
2. aq. NaHSO3
Cl
Cl2
H
OH
Cl