CHEM 234, Spring 2010
Second Midterm
ANSWER
PRINTED
FIRST NAME
KEY
PRINTED
LAST NAME
Person on your LEFT (or Aisle)
Ian R. Gould
ASU ID or
Posting ID
Person on your RIGHT (or Aisle)
peri3
nomen
1__________/10.........................9__________/18.......................
• PRINT YOUR NAME ON EACH PAGE!
mxn 1
2__________/20........................
• READ THE DIRECTIONS CAREFULLY!
D/A
3__________/16..........................
• USE BLANK PAGES AS SCRATCH PAPER
reactions
4__________/16..........................
work on blank pages will not be graded...
k vs Th
5__________/34
.........................
•WRITE CLEARLY!
retro
6__________/25.........................
• MOLECULAR MODELS ARE ALLOWED
peri1
7__________/18.........................
• DO NOT USE RED INK
peri2
8__________/18.........................
• DON'T CHEAT, USE COMMON SENSE!
Total (incl Extra)________/175+5
Extra Credit_____/5
H
He
Li Be
B
N
O
F
Ne
Na Mg
Al Si P
S
Cl
Ar
Ga Ge As Se Br
In Sn Sb Te I
K
Ca
Rb Sr
Cs Ba
Sc Ti V
Y
Cr Mn Fe Co Ni Cu Zn
Zr Nb Mo Tc Ru Rh Pd Ag Cd
Lu Hf Ta W
small range
range of values
broad peak
Re Os Ir Pt Au Hg
O H
C N
N H
C O
Tl Pb Bi Po At
H
C
C
Rn
Me/Et
1600–1660
R C OH
Approximate Coupling
Constants, J (Hz), for
1
H NMR Spectra
H H
H
~7
H
C C
OR
1735
CH
H
~10
~8
H
1600
O
NR2
H
O
C CH3
–H2C NR2
7
140
6
120
5
100
Aromatic
C CH
4
80
3
60
2
40
–OCH2–
R C N
R2C CR2
RC
H
~15
C C
1500
C CH2
8
160
~2
H
C
–OCH2–
NMR Correlation Charts
H
H
1650
2000
~2
C C
O
Aromatic Ar H
mainly 8 - 6.5
O
C
~2.7
1680
C
1710
2500
9
180
t-Bu/Me
N
–H2C X
10
200
~1.1
~2.9
C C
C
3000
11
220 O
~0.95
i-Pr/Me
C
C
2200
amine R NH2 variable and condition
alcohol R OH dependent, ca. 2 - 6 δ
(δ, ppm)
Et/Me
~2.6
O
broad ~3000
O
C H
~1.4
H
O
C O H
O
R C OH
~0.9
C
N
C
2850–2960
3500
Me/Me
H
broad with spikes ~3300
broad ~3300
Me/Me
O
2200
O H
H/Me
Xe
C
2720–2820
2 peaks
3000–
3100
N H
Kr
H
C
3300
~1.0
Infrared Correlation Chart
H
Gauche
Eclipsing
H/H
usually
strong
O
C H
(cm-1)
C
Interaction Energies, kcal/mol
CR
Alkyl
3Y > 2Y > 1Y
1
20
0
0
Alkyl 3Y > 2Y > 1Y
C X
C NR2
-2-
CHEMISTRY 234, Spring 2010 MIDTERM #2
NAME
Question 1 (10 pts.) Give the IUPAC name for the following compound. Be sure to use
cis/trans, E/Z or R/S where appropriate.
OEt
2
1
3
4
OH
(2S)-ethoxyhept-(4E)-en-(3R)-ol
6
5
grading -2pts each error
Question 2 (20 pts) GIve a curved-arrow pushing mechanism for the following reaction. Be sure
to indicate the Lewis acids/bases (LA/LB) and Bronsted acids/bases (BA/BB) as appropriate
conc. HBr
LB/BB
O
heat
Ph
H–Br
Br
O
LA
Br
Br
+
Ph
+ H2O
LA/BA
LB
LB
Ph
H
LA
Br
must be SN2 first
H–Br
O
+
LA/BA
must be SN1 second
H
Br
Ph
LB/BB H
grading: mainly half right = half points etc
1 pts if SN1 first then SN2 second
18 pts if SN2 in second step
18 pts if SN1 in first step
no LA/BA and LB/BB then -5 pts
only LA/BA and LB/BB 4 pts
O
H
-3-
CHEMISTRY 234, Spring 2010 MIDTERM #2
NAME
Question 3 (16 pts) Give the missing products or reactants in the following Diels-Alder reactions.
In the missing product, indicate both the relative and the absolute stereochemistry.
CHO
heat
a)
+
(±)
MeO
CHO
grading = 8pts each
Ph
b)
CF3
MeO
Ph
heat
+
CF3
Ph
Ph
CF3
CF3
Question 4 (16 pts.) For the following reactions, provide the missing MAJOR REACTION
PRODUCT. Indicate stereochemistry where appropriate.
grading = 8pts each
Br
Ph
Excess PhMgBr
a)
some partial credit examples
Br
Ph
Br
Ph
4pts
0pts
Ph
4pts
Ph
Ph
Br
6pts
Br
Ph
Ph
HO
1. MCPBA
Ph
b)
Ph
Ph
O Ph
2. EtOH / HCl (cat.)
Ph
Ph
OH
6pts
EtO
4pts
Ph
OEt
some partial credit examples
HO
6pts
OEt
-4-
CHEMISTRY 234, Spring 2010 MIDTERM #2
NAME
Question 5 (34 pts.) For the following reaction, give a full curved-arrow pushing mechanism
for formation of BOTH products and indicate the Lewis acid and base at each step (LA or
LB) and whether they are also Bronsted acids and bases (BA or BB).
Include all reasonable resonance contributors for any intermediates
AND INDICATE THE MAJOR RESONANCE CONTRIBUTOR IF APPROPRIATE!!
H
Br
LA/BA
Br
HBr
Br
+
A
Br
LB
H
MAJOR
B
H
grading 14 pts, half right = half points etc
LA
Br LB
b) Indicate which product, A or B, would be formed under thermodynamically controlled
conditions and which would be formed under kinetically controlled conditions and give a BRIEF
explanation of the role of temperature in determining kinetic and thermodynamic control
kinetic = A, thermodynamic = B, B is more likely to be formed at HIGH temperature since higher
temperature enables REVERSIBLE reactions and allows the reaction to explore the entire
reaction energy surface and find the lowest energy product, at low temperatures the reactions are
irreversible and the fastest formed product is the major product
grading 10 pts, key words highlighted in blue
c) Draw an energy diagram for formation of both A and B ON THE SAME DIAGRAM, clearly
indicate which curve refers to formation of A and which to formation of B
Energy
Br
A
grading 10 pts, half right = half points etc
Br
B
reaction coordinate
Extra Credit Question (5 pts.) Which kind of structures were discussed as being carconogenic
poly-aromatics
from weekly work # 7
THF
Ethers
Dienes
-5-
CHEMISTRY 234, Spring 2010 MIDTERM #2
NAME
Question 6 (25 pts.) Show how you would make the target compond on the right from the
starting compound on the left. Show reagents and conditions where appropriate, and the
structures of important intermediate compounds. Do not show any (arrow pushing) mechanisms.
O
approximate grading/pts scheme in blue
NBS, hν 2pts
Br
2pts
Br
O–
3pts
+Na
1pt
2pts
2pts
Na+ –H
Na+ –OMe
4pts
1. BH3.THF
3pts
3pts
OH
2. –OH/H2O2
3pts
Question 7 (18 pts)
The product of ONE of the following two reactions A or B is allowed and the other is forbidden
a) draw the curved arrow-pushing that describes the bond making/breaking for both reactions
A
Ph
B
heat
2pts
Ph
Ph
Ph
heat
2pts
Ph
Ph
Ph
Ph
b) draw a picture of the F.M.O. relevant to these reactions on top of each structure below
3pts
Ph
Ph
3pts
Ph
Ph
c) Using F.M.O. theory, briefly explain which reaction has the allowed product, A or B, include
the terms disrotatory and/or conrotatory as appropriate
both reactions are disrotatory ring closure reactions, disrotatory ring closure in A leads to a
bonding situation for the new sigma-bond, disrotatory ring closing in B results in an antibonding situation for the new sigma-bond, reaction A is allowed
8pts
-6-
CHEMISTRY 234, Spring 2010 MIDTERM #2
NAME
Question 8 (18 pts) Shown below is the product of a cycloaddition reaction between two
reactants A and B. In this question you are eventually going to determine whether THIS product
is allowed or forbidden.
FIRST, some questions about the product that is provided.... Me
Me
Et
Et
CF3
CF3
4pts
Me
heat
+
CF3
CF3
A
B
Me
a) Do the curved arrow-pushing that describes bond making and breaking
6
b) Give the number of electrons that are involved in this reaction
2pts
c) For the product SHOWN (which may or may not be allowed, remember), was the reaction
2pts
suprafacial
suprafacial or antarafacial as far as the reactant A was concerned?
d) For the product SHOWN (which may or may not be allowed, remember), was the reaction
suprafacial
suprafacial or antarafacial as far as the reactant B was concerned?
2pts
NOW, some questions about the ALLOWED reaction for these reactants
e) Would an ALLOWED reaction between reactants A and B proceed via a Huckel or a Mobius
transition state. Give a BRIEF explanation
this is a 6-electron reaction, the lowest energy transition state is a cyclic Huckel loop since this is
aromatic and relatively stable
6pts, key words highlighted in blue
f) Is the product shown above allowed or forbidden?
allowed
2pts
Question 9 (18 pts)
a) Give the curved arrow-pushing and the allowed product for the following cycloaddition
reaction. Pay attention to stereochemistry
H
H
H
2pts
6pts
CF3
C H
C
CF3
heat
(±)
Me N
Me N
+
F3 C
O
O
CF
3
b) ON TOP of the structures as indicated, draw the requested F.M.O.s and give the total number
of π-molecular orbitals and electrons associated with the π-system for each structure.
Me
3pts
F3 C
H
3pts
N
C
CF3
O
H
draw the LUMO
draw the HOMO
total # of π-M.O.s
total # of π-M.O.s
2
1pt
3
for this structure =
for this structure =
1pt
total # of electrons in the πsystem for this structure =
4
1pt
total # of electrons in the πsystem for this structure =
2
1pt