Please read directions carefully and answer the following questions completely.
You may write directly on the problem set, but be sure to staple the pages
together. This problem set is due in class at the beginning of lecture by 9:05 AM
on the date indicated above.
Name:
Section TF:
Section Day/Time:
Collaborators:
Question 1
Predict the major product of each of the following Diels-Alder reactions. Be sure to consider both
stereochemistry and regiochemistry carefully.
CO2Me
Ph
MeO2C
(a)
+
Ph
H
(b)
+
(+/-)
NO2
NO2
(+/-)
NMe2
H
O
(c)
+
NMe2
OMe
(+/-)
H
O
(d)
O
O
MeO
H
+
H
(+/-)
NO2
(e)
+
NO2
(+/-)
NC
CN
H
(f)
H
(+/-)
Question 2
Indicate what diene and dienophile could be used to make the following compounds. Be sure to consider
stereochemistry carefully.
MeO
CO2Me
CO2Me
MeO
(a)
+
CO2Me
MeO2C
(+/-)
CN
CN
+
(b)
NO2
NO2
(+/-)
O
CO2Et
(c)
CO2Et
O
+
EtO2C
CO2Et
(+/-)
CO2Et
CO2Et
(d)
+
EtO2C
CO2Et
(+/-)
+
(e)
(+/-)
O
H
(f)
H
(+/-)
O
Question 3
(a) Determine whether each of the following molecules is aromatic, anti-aromatic, or non-aromatic.
Assume planarity for all fully conjugated systems.
(i)
anti-aromatic
(ii)
BH
aromatic
H
B
(iii)
non-aromatic
(iv)
aromatic
N
H
NH2
O
(v)
N
aromatic
aromatic
(vi)
N
H
N
O
N
(vii)
aromatic
(viii)
N
aromatic
N
NH
NH2
(ix)
aromatic
(x)
N
N
H
(xi)
non-aromatic
H
anti-aromatic
(xii)
H
aromatic
Question 3 continued
(b) Explain each of the following observations.
(i) Although most diazo compounds are unstable, the following diazo compound is remarkably stable.
N
N
N
N
This diazo compound has a resonance
structure in which the five-membered
ring is aromatic. This stabilizes the
diazo compound.
aromatic!
(ii) Experiments at very low temperatures have shown that there are two distinct di-deuterocyclobutadienes
where the deuteriums are on adjacent carbons. These compounds are rectangular rather than square.
(As part of your explanation, provide the structures of the two compounds.)
D
D
If the bonds in cyclobutadiene were delocalized, then
cyclobutadiene would be antiaromatic. The lower
energy state has localized single and double bonds,
and thus a rectangular geometry. This means that
there are two possible di-deuterated compounds with
the deuteriums on adjacent carbons.
D
D
(iii) Compared to most hydrocarbons, the following hydrocarbon has a remarkably low pKa.
CH3
CH2
_
CH2
H+
aromatic!
The conjugate base is
especially stable
because it has an
aromatic resonance
structure. This makes
the hydrocarbon
remarkably acidic.
(iv) Hydrocarbons are typically non-polar, but azulene, shown below, is a deep blue hydrocarbon with an
unusually large dipole moment.
A significant dipolar resonance structure
exists in which both rings are aromatic,
giving an overall dipole to the molecule.
azulene
aromatic!
aromatic!
Question 4
(a) When the radiolabeled hydrocarbon shown below is subjected to the following reaction, a mixture of
isomeric products is obtained. Propose a mechanism for the formation of the products.
CH3
1. NaNH2, NH3
2. Cl
CH3
+
CH3
+
= C13
CH3
A
B
C
NH2
H
becomes
aromatic
Cl
CH3
compound B
Cl
CH3
compound A
compound C
Cl
CH3
(b) (i) Circle the reaction that you predict will proceed faster and explain your answer briefly.
EtOH
OEt
I
EtOH
+ HI
I
+ HI
EtO
reaction A
reaction B
In these SN1 reactions, both compounds would form allylic carbocations, but reaction B would need to
become an anti-aromatic intermediate. This is highly unfavorable, so reaction A would be faster.
SN1
I
reaction B
SN1
I
Booo!
Anti-aromatic
reaction A
Nice!
Allylic carbocation
(ii) The major product of reaction A would actually be an isomer of the product shown in part (i). Predict
the major product, and explain briefly why it is the major product.
OEt
major product
This is the thermodynamic product because this conjugated diene is more stable than the
non-conjugated diene shown in part (i)
Question 5
(a) For the following unsymmetrical conjugated diene, there are two potential sets of 1,2 and 1,4-addition
products. Propose curved-arrow mechanisms to account for the four possible products and label each
product as either a kinetic or thermodynamic product. Finally, give a one-sentence explanation for which
pair is more likely to form. (You may disregard stereochemistry.)
D 1,2-addition
DBr
D
pathway A
products
Br
1,4-addition
thermodynamic product
Br
kinetic product
+
This pair came from a 3° and 2° allylic carbocation pair, and this is
more stable than the alternate 2° and 1° allylic carbocation pair.
D
Br
D
D
3° cation
Br
Br
pathway A
2° cation
D
D
Br
Br
1,2-addition
kinetic product
1,4-addition
thermodynamic product
(tri-subbed alkene)
2° cation
D
D
1,2-addition
kinetic product
pathway B
D
Br
Br
1° cation
Br
Br
D
D
Br
1,4-addition
thermodynamic
product
(tri-subbed alkene)
(b) An isomer of 2-hexen-4-yne is partially reduced to give an intermediate compound which reacts with a
diene to give the final product shown below. Draw the structures of the starting material, the intermediate
compound, and the required dienophile. Clearly show all stereochemistry and the reaction conditions for
each step.
option 1:
H2, Lindlar's catalyst
option 1:
option 2:
Na°, NH3 liquid
reagent
option 2:
an isomer of 2-hexen-4-yne
structure of the diene
H
Cl
H
H3C
Cl
CH3
Cl
structure of the dienophile
Cl
H
H
Question 6
Provide multistep syntheses for each of the desired products using the indicated starting materials.
(a)
Any organic and inorganic reagents
with six or fewer carbons
Starting Materials
OH
O
O
O
H
H
OH
HO
+
H2
H
OH
HO
Pd/C
(Diels-Alder)
(+/-)
(+/-)
OH
H+ or
HO- cat.
OH
O
OH
(+/-)
Desired Product
O
(b)
Any other organic
and inorganic
reagents
+
OMe
Starting Materials
O
O
OMe
H
1. O3
H
2. Me2S
O
(Diels-Alder)
OMe
O
O
(+/-)
OMe
(+/-)
NaBH4, MeOH
OH
OTs
TsCl,
HO
(+/-)
O
OMe
N3
NaN3
TsO
pyr
N3
(+/-)
O
OMe
(+/-)
O
OMe
Desired Product
Question 7
(a) (i) Some nitriles are capable of acting as Diels-Alder dienophiles. The compound below undergoes
an intramolecular Diels-Alder reaction when heated. Draw a curved-arrow mechanism that leads to the
most likely product of this intramolecular Diels-Alder reaction.
O
Me
N
O
NMe
C
N
N
(ii) The product of this reaction is actually not stable as drawn and immediately tautomerizes. Draw the
actual final product of the reaction. (Hint: remember tautomerization from your study of alkyne reactions,
and note that nitrogen as well as oxygen compounds can behave in this way.) What is the driving force
for this seemingly disfavored tautomerization (i.e., why is it actually more favorable in the “enol” form
than the “keto” form)?
sp3 carbon
separating
three distinct
pi-systems
O
3-atom
conjugation
By making this an sp2 carbon,
the compound becomes a
huge, almost completely
O
conjugated system.
NMe
NMe
tautomerizes
N
NH
2-atom
pi-system
6-atom
conjugation
This nitrogen version
of an enol is actually
called an enamine.
The driving force for this seemingly "reversed" keto-enol tautomerization is the formation of a
larger conjugated system involving the nitrogen lone pair, the alkene, and the carbonyl.
Remember from lecture: "when there's a choice, choose conjugation"!
(b) Cyclodecapentaene ([10]annulene) is non-aromatic, yet 1,6-methano[10]annulene is aromatic. Explain.
If cyclodecapentaene were truly planar, then the two hydrogens in
the middle would need to occupy the same space. Because this is
impossible, the molecule contorts out of planarity to accomodate
these hydrogens. This is an example of steric inhibition of resonance.
cyclodecapentaene
(also called [10]annulene)
HH
HH
H
H
1,6-methano[10]annulene
(aerial view)
The bridge in 1,6-methano[10]annulene
allows the ten sp2-hybridized carbons to be
planar, thus creating the necessary
conjugation of 10 -electrons that satisfies
the criteria for aromaticity.
1,6-methano[10]annulene
(side view)
Question 8
As you have seen, Diels-Alder reactions can be favored in either the forward or reverse direction, depending
on the substrate structures and conditions. Propose a curved-arrow mechanism for each of the interesting
transformations shown below, and provide a brief explanation of the thermodynamic driving force in each
case.
(a)
reverse Diels-Alder
The thermodynamic driving force is the relief of ring strain.
(b)
O
+
+
O
The thermodynamic
O
driving force for the
reverse cycloaddition (the
Ph
cheleotopic extrusion)
Ph
Ph
that follows the DielsAlder is the expulsion of
gas and the creation of
Ph
an aromatic ring.
Ph
Ph
O
cheleotropic
extrusion
C
O
(carbon monoxide)
+
CO