Massachusetts Institute of Technology
Organic Chemistry 5.13
Wednesday, October 26, 2005
Prof. Timothy F. Jamison
Hour Exam #2
Name
SOLUTIONS
_________________________________________________
(please both print and sign your name)
Official Recitation Instructor
Directions:
____________________________________
Closed book exam, no books, notebooks, notes, etc. allowed.
Calculators are not permitted for this exam. However, rulers and molecular model
sets are permitted.
Please read through the entire exam before beginning, in order to make sure that
you have all the pages and in order to gauge the relative difficulty of each
question. Budget your time accordingly.
Show all of your work if you wish to receive partial credit. You should have 7
pages total: 5 exam pages including this page and 2 blank pages for scratchwork.
Question:
Grader:
1. ________/
42 points (page 2)
_______
1. ________/
30 points (page 3)
_______
2. ________/
28 points
_______
100 points
_______
Total: _________/
1
1. (72 points total, 3points per box) in each box below, draw the structure
of the reagent or major product of the reaction, where appropriate. If no reaction occurs,
put a large X in the box. Clearly indicate the double bond geometry and relative
stereochemistry of the major product, where appropriate.
(a.)
1. NaH
2. PhCH2Br
CH3OH
OH
OH
1. NaHCO3
2. CH3l
(c.)
1. NaOH
2. CH2=CHCH2CI
t-BuSH
Me
O
Me
D
+
H
O
O
O
Me
OMe
O H
O
H
Me
H
O
O
H
m-CPBA
(draw the structure)
H
O O
S
NaBH4
Cl
Me
tBu
DMDO (excess)
(draw the structure)
Me
OMe
Hg(OAc)
MeOH
O
O
Me
Me
t-BuSH
Hg(OAc)2
(g-h.)
(i-n.)
Ph
OCH3
1. NaH
2. CH3l
(b.)
(d-f.)
CH3O
H
O3
(o.)
OH
OH
Me
H
O
NaBH4
H
O
Me
H
(CH3)2S
O
O
O
H
H
Me
H
O
O
O
H
H
Figure by MIT OCW.
O
(o).
+
OMe
D
O
Me
H
O
(p).
+
O
C
(q).
Me
O
O
Cy = Cydohexyl =
H O
O
D
Me
Me
H O
O
H O
Me
hv
Dis
Et
Ph
H O
O
Me
(r).
OMe
Cy
D
+
Me
Me
Ph
Et
Me
(s)
Me
hv
Dis
Ph
Ph
Et
Me
Me
(t-u)
Me
(v-w)
CO2Me
+
C N O
hv
D
Me
con
Dis
Me
Me
CO2Me
D
CO2Me
CO2Me
C10H12O4
CO2Me
(x)
Et
+ Me
CO2Me
D
Ph
MeO2C
3
CO2Me
C14H18O4
D
N
O
Me
N
Ph
or
Me
O
CO2Me
Figure by MIT OCW.
2. (28 points total) In a Nazarov Cyclization (below), treatment of a dienone with a strong
Lewis acid effects a thermal 4π electrocyclic ring closure, giving intermediate A, and an aqueous
workup affords the final product (B), the thermodynamically most stable cyclopentenone.
TiCl4
O
O
TiCl4
O
C
TiCl4
H2O
O
B
A
Figure by MIT OCW.
a. In the diagram below, draw the π atomic orbitals (by shading the lobes appropriately) that
represent the π system of C (the precursor to A) in the reaction above (2 points each).
b. Write the number of nodes in the box to the left of each orbital array (1 point each).
c. For the ground state of C, draw the electron population for each orbital on the line to the
right of each orbital array. Clearly indicate whether each electron is “spin up” or “spin
down”. If there are no electrons in a given orbital, leave the line blank (1 point each).
# of Nodes
Electron
population
Orbitals
E
1 point per box
2 points per orbital array
1 point each
Figure by MIT OCW.
4
4. (continued)
d. (4 points each) For the example of the Nazarov cyclization below, in the indicated
boxes draw the direct product of the electrocyclic ring closure and the
cyclopentenone final product after the aqueous workup. In both cases, clearly
indicate stereochemistry and double bond geometry, as appropriate.
O TiCl4
O
Me
Me
TiCl4
Me
H
H2O
Me
Me
Me Me
direct cyclization product
Me
O
Me
cyclopentenone
LESS STABLE CYCLOPENTENONES (2 Pts PARTIAL CREDIT):
H
O
O
Me
Me
Me
Me
Me
Me
Figure by MIT OCW.
5