Name_________Key_________________
215 F12-Exam No. 2
Page 2
I. (9 points) For each of the following sets of molecules, rank the molecules in order of most acidic to least
acidic. Compare the underlined H’s for each set.
(a)
H
N H
H
H
N H
H
N C
H
N H
H
CH3O
B
A
A
B
C
most acidic
least acidic
C
3
(b)
O
O
H
H
OCH3
H
H
O
N
H
O
H
A
H
H
B
H
C
O
A
B
most acidic
C
least acidic
3
(c)
H3C
C
CH3
H3C
CH3OH
OH
H3C
C
B
OH
H3C
B
A
H
C
A
most acidic
C
least acidic
3
II. (11 points) Draw in the box below the structures of the three possible conjugate acids of methylguanidine
and explain as to which of the three conjugate acids is most stable and would be expected to form
preferentially. Use drawings of pertinent resonance forms and several words to explain your answer.
Three possible conjugate acid structures:
N
H
N
H3C
N
1.
H
H
H
H
N H
N
H
N H
H3C
H
3.
2.
N H
N
N
H3C
H
H
H
H
H
H
H3C
N H
N
N H
H
methylguanidine
3
The most stable conjugate acid is:
Explanation:
H
H3C
H
N
1
2
H
N H
H
N H
H
H3C
N
3 (circle one that applies)
2
H
N H
H
N H
H
H3C
N H
N
N H
H
In conjugate acid 3, there are three major resonance contributors shown above.
These are either identical or close in the structure and energy and the positive
charge is delocalized among three nitrogen atoms over a wider range, thus
making conjugate acid 3 quite stable.
6
Name_______Key___________________
215 F12-Exam No. 2
Page 3
III. (18 points) Draw in the box below a step-by-step mechanism for the NaOCH2CH3-catayzed formation of
sodium enolate 3 from diethyl malonate (1) and epoxide 2, using the curved-arrow convention.
O
O
O
O
+
O
O
HOCH2CH3
H H
1
Mechanism:
O
O
O
2
O
3
O
O
O
O
H H
+ HOCH2CH3
+ enantiomer
The other enolate or carbanion form
acceptable.
O
CH3CH2O
OCH2CH3
Na
Na
2 pts for each intermediate; 2 pts for each set of mechanistic arrows
1
O
O
NaOCH2CH3
O
H
O
O
O
O
2
O
O
O
O
O
O
O
3
O
O
O
O
O
O
Na
O
18
IV. (14 points) Treatment of aldehyde 4 with NaBH4 followed by acidic work-up results in the formation of
ester 5. Draw in the box below a step-by-step, curved-arrow mechanism for this transformation.
O
O
O
4
Mechanism:
OH
1. NaBH4
H
2. H3O
H2O
O
H H O
5
O
2 pts for each intermediate
2 pts for each set of mechanistic arrows
O
O
O
4
O
H
Na
H
H
B
H
H
O
more
electrophilic
than the ester
C=O carbon
H H
H
O
O
O
H
H
H
O
Note: PhO-H is far
more acidic than RO-H.
O
H H O
O
H
OH
5
O
H H O
14
Name______Key____________________
215 F12-Exam No. 2
Page 4
V. (21 points) The NaH-mediated intramolecular condensation reaction of 6 yields sodium enolate 7 [J. Org.
Chem. 2012, 77, 9628]. Provide in the box below a step-by-step mechanism using the curved-arrow
convention for this transformation.
O
O
O
N
O
NaH
THF
(solvent)
Ph
O
N
Na O
6
O
Ph
+ H2 + HOCH2CH3
7
Mechanism:
O
Na
O
O
H
H
O
O
H
N
N
Ph
O
O
6
O
O
H
N
O
The other enolate or carbanion form
acceptable.
O
CH3CH2O
O
Ph
Ph
N
O
O
Ph
O
O
N
Na O
3 pts for each intermediate; 3 pts for each set of mechanistic arrows
O
Ph
7
21
VI. (10 points). There are, in principle, three different intramolecular aldol condensation reaction products
that could form from dione 8 by its treatment with 2% NaOH/H2O/CH3CH2OH.
(1) (6 points) Draw in the boxes below the structures of these three potential aldol condensation products.
O
O
Ph
B
A
Ph
Ph
8
2% NaOH/H2O
CH3CH2OH
C
O
+
+
+ H2O
Ph
O
O
2
2
2
(2) (4 points) In the box below indicate as to which of these three compounds is expected to be the major
product from the reaction and provide a brief explanation for your choice.
A, B, or C (circle one) would be the major product from the reaction (2 points).
Explanation for your choice (2 points):
Both Enones B and C have the highly strained cyclobutene ring. Accordingly, as the aldol condensation
reaction is thermodynamically driven, the most stable enone A should be the major product.
Name________Key__________________
215 F12-Exam No. 2
Page 5
VII. (17 points) Complete the following reactions by providing in each of the boxes the structure of the
starting compound, intermediate, or product. Indicate stereochemistry for the product/intermediate and if
more than one stereoisomer is formed, draw one structure and write “+ enantiomer” or “+ diastereomer.”
(1) [J. Org. Chem. 2012, in press]
O
HN
N
O
heat
O
OH
4
(2) [Eur. J. Org. Chem. 2011, 319]
O
O
O
O
O
CH3OK
+
OH
CH3OH
(solvent)
CH3O
O
CH3O
OH
4
(3)
O
NaH
(1 mol
equiv)
O
Na
O
O
O
Na
LDA
(1 mol
equiv)
O
Li
O
O
O
Acceptable even w/o Na+
Acceptable even w/o Na+/Li+
3
di-anion or di-enolate
mono-anion or mono-enolate
The other enolate and/or carbanion form(s) acceptable.
Br
Br
O
O
H
O
+ enantiomer
+ LiBr + NaBr
3
3
Name_________Key_________________
215 F12-Exam No. 2
Page 6
VIII. (20 points) Complete the following reactions by providing in each of the boxes the structure of the
intermediate or product. Indicate stereochemistry for the product/intermediate and if more than one
stereoisomer is formed, draw one structure and write “+ enantiomer” or “+ diastereomer.”
(1) [Org. Lett. 2012, 14, 3340]
O
O
1. LHMDS (1.1 mol equiv)*
-40 °C
O Si
I
2.
I
I
O Si
4
*LiN[Si(CH3)3]2: a strong, bulky base
(2) [J. Org. Chem. 2012, 77, 8056]
O
O
O
O
O
NaOCH2CH3
(catalytic)
+ H2O
O
HOCH2CH3
(solvent)
O
O
O
4
(3) [Org. Lett. 2012, 14, 3222]
O
MgBr
CH3O
+
CH3O
N
CH3
O
OCH3
HCl, H2O
BrMg
O
CH3O
CH3O
C12H16O2
N
Need not to indicate
stereochem here.
CH3
4
+
4
H
CH3O N H
CH3 Cl
chelate intermediate
(4) [Org. Biomol. Chem. 2012, 10, 8660]
O
O
O
O
CH3NO2
N
H
K-OC(CH3)3
(catalytic)
CH3OH (solvent)
N
O
C11H10N2O5
HO
H
NO2
+ enantiomer
4