Instructions:
(1) (1) THREE OPTIONS:
[i] [preferred] Print the exam, and do it in the normal way;
[ii] Use an iPad;
[iii] If you don’t have a printer or iPad, write out the answers on paper, carefully indicating
which answer goes with which problem.
(2) Sign the Honor Statement in the space below it.
(3) Show your work, and place your name on each page.
(4) The 50-minute limit is working time: start your clock after downloading, printing, etc.
(5) Use of molecular models is encouraged; otherwise this exam is closed-book.
(6) There are four problems.
(7) Exam 2 is an “assignment” on CANVAS; when finished, scan or photograph your completed
exam, and upload it there.
Scoring
Problem Possible Earned
1. 25
2. 30
3. 25
4. 20
Total 100
Honor Statement: I have neither given nor received unauthorized help on this exam.
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(1) (a) Propose a structural reason why 2 is more stable than 1, 3, or 4.
(b) All compounds shown above have the formula C11H14. Use your answer to (a) to
suggest why 2 must be more stable than any other single-bonded compound with this
formula. Show that your argument is rigorously true.
AlBr3
1 2 (93%) 3 (5%) 4 (2%)
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(2) This question concerns the axial/equatorial preferences of two different substituents on two
different six-membered rings.
(a) For each of the four cases 5-8, should the substituent be more stable in the axial position, or
the equatorial positon? Concisely, why?
(b) Reaction of a glycosyl fluoride (9) with a phenol (10) in the presence of a Lewis acid gives
predominantly the axial O-glycoside 11. When the temperature is raised, 11 is converted to
the equatorial C-glycoside 12. Concisely explain the axial/equatorial preferences in 11 and
12.
[c] In the Kochetkov amination, a sugar such as 2’-fucosyllactose (13) reacts with NH3 or NH4+
to give an aminoglycoside (14), entirely in the equatorial form, regardless of protonation
state. Propose an explanation for this equatorial preference. (The answer to this question is
not known: your answer will be graded on logical consistency.)
X X
Z
Z
? 5: X = CH2, Z = C6H5
6: X = CH2, Z = OCH3
7: X = O, Z = C6H5
8: X = O, Z = OCH3
X
Z
O
F
(RO)n +
OH
R
9 10
O
O
(RO)n
R 11
O
(RO)n
HO
R
12
O
+
OH OH
OH
RO
HO
13
“NH3” O
OH
OH
RO
HO NH2
14
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This page left blank intentionally: work space for problem 2.
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(3) Reaction of the tetrahydrocinnolines 15 with dimethyl maleate (16) gives the [3 + 2]
cycloadducts 17 and 18. The major product is 17 rather than 18: concisely explain this
preference.
HN
N
R1
R2
+
CO2Me
CO2Me
15 16
HN N
R1
R2
MeO2C
MeO2C
N
R1
R2
MeO2C
MeO2C
H N
R1
R2
MeO2C
MeO2C
H
17 (major) 18 (minor)
+
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This page deliberately left blank: work space for problem 3.
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(4) A plant used in traditional Chinese medicine yielded a tetracyclic enone whose structure
was first assigned as 19. After investigation, the structure was re-assigned as 20. Suggest
a structural reason why structure 20 might be more plausible than 19.
O
HO
H H H
OH
19
HO
H H
OH
O
H
20
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