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Chemistry 143 Problem Set 1 Solutions
April 10, 2015
Problem 1 The structures 1b and 2b are optically active (chiral) porphyrin derivatives for use in
enantioselective epoxidation. The threitol-strapped groups are chiral - one can verify that they all have the
same (S,S)- stereochemistry. Therefore, they each have a C 2 rotational axis of symmetry. (If either had a
mirror plane of symmetry, then it would not be chiral). In 1b, the C2 axis of symmetry makes Ha equivalent
with its neighbor on the same double
bond linkage. In 2b, the C 2 axis of
C2 axis
symmetry makes one double bond
linkage equivalent to the other on the
Hb
other side of the porphyrin ring.
Hb
Therefore, in 1b, we expect the
chemically equivalent protons Ha to be
Ha
Ha
Hb
Ha
observed as a singlet (Spectrum B).
C2 axis
But in 2b, we expect to see H a and H b
Hb
inequivalent and coupling to each
Ha
other (Spectrum A). JACS 1995, 117,
692.
Problem 2 (CH3)4Sn (150 MHz, CDCl3)
Outer Lines: -1237.23-(-1573.36) = 336.1 Hz
Inner Lines: -1244.60-(-1565.99) = 321.4 Hz
(µ119Sn/µ117Sn) = (-1.0409/-0.9949) = 1.046 = (336.1/321.4)
115Sn
peak is just a tiny blip (maybe next year I'll get a better spectrum).
(µ119Sn/µ115Sn)* 336.1 = (-1.0409/-0.9132)*336.1 = 294.9 Hz
ALL I=0 Sn isotopes:
8.58% 119Sn(CH3)4
7.61% 117Sn(CH3)4
0.35% 115Sn(CH3)4 ?
Problem 3
b 2.02 dd, 2H, J = 6.7, 8.5 Hz
c 1.62 dt, 1H, J = 5.1, 6.7 Hz
d 1.20 dt, 1H, J = 5.1, 8.5 Hz
1J(119Sn-C)
~ 336.1 Hz
1J(117Sn-C)
CO2Me
CO2Me
Hd
Hc
~ 321.4 Hz
Hb
Hb
CO2Me
CO2Me
A2MX
Pople Notations: A4
The cis-diester is the only structure that we would expect to give three resonances.
Hx
Hx'
Ha CO2Me
Ha'
CO2Me
AA'XX'
Problem 4 Unknown Cyano Compound C10H8NO2Cl IR shows 2253.7 cm-1 (CN group) and 1732 cm-1
(C=O group). 1H NMR (500 MHz) and 13C (60.9 MHz).
Step 1: Unsaturation # = (2*C-H+N-Cl+2)/2 = (20-8+1-1+2)/2 = 7
Step 2: In addition to C=O (165ppm) and CN (117.9 ppm), six sp2 carbons indicate aryl ring.1+2+4=7
Step 3: No Symmetry in 13C NMR (10 lines = 10 C)
a
7.90
ddd, 1H, J = 0.5, 2.0, 8.0 Hz
Step 4: 1H Spin system:
8.0 Hz
H
6.0 Hz
b
7.59
ddd, 1H, J = 2.0, 7.0, 8.0 Hz
C
O
O
2
Ha
C
H2
CN
Cl
Hd
7.0 Hz
Hb
8.0 Hz
Hc
c
7.56
ddd, 1H, J = 0.5, 2.0, 8.0 Hz
d
7.47
ddd, 1H, J = 2.0, 7.0, 8.0 Hz
e
4.57
t, 2H, J = 6.0 Hz
f
3.04
t, 2H, J = 6.0 Hz
Step 5: Connect the -Cl, -CN and -COO- groups to the spin system with consideration of chemical shifts:
The Aryl group is not directly attached to oxygen or the ipso-carbon would be 150-160 ppm.
4.57 ppm - CH2 attached to oxygen (60 ppm 13C shift).
3.04 ppm - attached to Cyano, (17 ppm 13C shift), not chloro (more downfield in 1H and 13C NMR).
Problem 5 (a) The vicinal coupling constants (3JHH)
of a cyclopentadienyl ring are all nearly zero.
Reaction of 1 with NaBH 4 produces the alcohol 2
which bears a chiral center at the new -CH(OH)Me
carbon. This renders the four cyclopentadienyl
protons diastereotopic (although only one pair is
resolved in the 1H NMR. But in the 13C NMR, all
five carbons of the top Cp ring are inequivalent.
diastereotopic (inequivalent)
H
OH
diastereotopic
(inequivalent)
C
Fe
CH3
2
Problem
6
The
main
peak
(singlet)
is
from
the
3J
194Pt-194Pt isotopic species. The other peaks are from the 195Pt-194Pt species.
1J
2J
S
When one Pt nucleus is 195Pt, each 31P nucleus couples with different J value.
Ph3P 195Pt 194Pt PPh3
The 195Pt nucleus will maintain the same spin quantum state during the
CO CO
acquisition of the 31P scan. So we have a superposition of the spectra (two
(each 31P nucleus couples to 195Pt)
overlapping AB patterns) for:
(I = +1/2)
(I = -1/2)
S (I = 0)
S (I = 0)
It appears from the leaning of the
AB patterns that the peaks
picked in red go together and
likewise the peaks picked in green
also go together. That would
indicate that the J values 1JPt-P and
2J
Pt-P are the same sign.
From the reference:
1J
Pt-P = 3161 Hz
2J
Pt-P = 122 Hz
3J
P-P = 149 Hz
Ph3P
195Pt
CO
194Pt
PPh3
CO
Ph3P
195Pt
CO
1J
Pt-P
2J
Pt-P
3J
P-P
194Pt
CO
PPh3