# Solutions - Chemistry at Caltech

```Chemistry 143 Problem Set 3 Solutions
April 29, 2015
6
Problem 1
7
(a) The "frozen-out" spectrum of cis-decalin at -20 °C shows five signals. This structure has a C2 axis
5
8 ax
C2 axis of symmetry as shown. Thus, (C1,C5), (C2,C6), (C3,C7), (C4,C8) and (C9,C10) are
1 eq
9
equivalent pairs of carbons. In the top structure, it can be seen that C8 is attached to the other
10
ring with an equatorial bond (in bold), while C5 is attached with an axial bond.
2
3
4
Upon inversion of both chairs, the bottom structure will be formed with C8' axial and C5'
equatorial. Therefore, as the sample is warmed up to facilitate chair inversion, the signals for
7'
(C4,C8) and (C1,C5) will exchange. Similarly, the (C2,C6) signal will exchange with the
8'
(C3,C7) signal. Also, C9 and C10 exchange with each other, but they were already chemically
ax
6'
equivalent. So the sharp peak at 37 ppm can be assigned to (C9,C10) which displays no
2'
1'
9'
10'
broadening during the variable temperature experiment.
3'
4'
eq 5'
(b) In the 13C NMR at -20 °C, the ppm values are roughly 37, 33, 28, 26 and 22 ppm (call them
a, b, c, d, and e respectively). The two broad peaks at +20 are at ~ 29.5 ppm and 25 ppm). So b
coalesces with d and c with e each with a coalescence temperature (Tc) of about +5 °C = 278K.
during chair inversion:
kinv =
π
•(7ppm•20Hz/ppm) = 2.0836x1010•278K•e-[ΔG†/(.001987•278K)]
√2
Solving for ΔG† for the b:d coalescence: ΔG† = 13.1 kcal/mole
(C1,C5)
(C2,C6)
(C9,C10)
(C4,C8)
(C3,C7)
Solving for ΔG† for the c:e coalescence: ΔG† ~ 13.2 kcal/mole
JAX
Problem 2
13
Dioxane C satellites show AA'XX' pattern due to the magnetic inequivalence
HX
fast
O
of the O-13CH212CH2-O subunit in the molecules containing exactly one 13C atom.
Although the chair flip of dioxane is expected to be fast at room temperature, JAX HA 13C
O
≠ JAX' which can be clearly seen because the trans-relationship of JAX' contributes
HX'
HA'
some diaxial J value (~10Hz) to the weighted average of J values over all possible
1
J
conformations. One could easily obtain the JCH for dioxane
AX'
(~ 142.8 Hz = 1923.40 - 1780.57) which is in accord with the 1JCH value of an ether
(140 Hz for CH3OCH3).
JAX
HA
O
13C
O
HA'
HX
JAX' HX'
Problem 3
(a) The "frozen-out" spectrum of chloromethyltriptycene at -25 °C shows four signals for the
quaternary aromatic carbons. This structure has a mirror plane of symmetry. Thus, (C17,C19) and
(C18,C20) are equivalent pairs of carbons and are seen with twice the intensity of C15 and C16.
16
As the sample is warmed up to facilitate rotation of the CH2Cl group, the signal for C15 will
15
exchange with the (C17,C19) signal. Similarly, the C16 signal will exchange with the (C18,C20)
20
18
signal. This represents a 2:1 coalescence which although is not rigorously adherent to the ideal 1:1
19
17
coalescence formula kC = (π/√2)•Δν, we said in class that it is still a reasonable approximation.
H
H
(b) In the 68 °C spectrum, two lines are seen under fast exchange. The line positions for these two
lines fall near the midpoint between the lines A/B and C/D in the low temperature spectrum.
Cl
(c) In the 13C NMR at -25 °C, the ppm values are roughly 149.0, 147.4, 146.8 and 144.3 ppm (for
during CH2Cl rotation:
A, B, C and D respectively). Signal C coalesces with D at about +17 °C = 290K. At 290K, the
coalescence of A and B is a little past the flat-topped Tc. Although the equation is not rigorously
(C15)
(C17,C19)
(C16)
(C18,C20)
correct for a 2:1 coalescence, it serves as a very good approximation for ΔG†.
krot ~ (π/√2)•(2.5ppm•25Hz/ppm) = 2.0836x1010•290K•e-[ΔG†/(.001987•290K)]
Solving for ΔG†: ΔG† ~ 14.1 kcal/mole
Problem 4
(a) If the aryl ring rotates, the tert-butyl groups will exchange.
(b) The Cp rings are diastereotopic because the phosphorus is a chiral center. In order for
the Cp rings to exchange, the phosphorus chiral center must invert. There is no other way to
make the Cp rings become equivalent.
However, for part (a), it is possible that the tert-butyl groups might also exchange upon
inversion of the phosphorus. If the variable temperature NMR gives us two different ΔG†
energies, then it would provide strong evidence for two separate dynamic processes. But,
if we get the same ΔG† from both calculations, then it could be just P inversion that is
responsible for the NMR data.
(c) Can we say that because we see two different Tc temperatures, then it must be two different
processes (two different ΔG†'s)? No. because Δν's are different.
(d) Just calculate for each set of peaks and see what we get. Since the ΔG† values are close,
it is very possible that P inversion achieves both coalescences.
Cp rings: k = (π/√2)•(125Hz) =
NH
Zr
H P
(b)
tBu
(a)
tBu
(b)
Solving for ΔG†: ΔG† = 12.9 kcal/mole
Zr
H
N
H P
tBu groups: k = (π/√2)•(25Hz) = 2.0836x1010•263K•e-[ΔG†/(.001987•263K)]
tBu
(a)
Solving for ΔG†: ΔG† = 13.2 kcal/mole
CH3
1H
Me's: k = (π/√2)•(25Hz) = 2.0836x1010•221K•e-[ΔG†/(.001987•221K)]
Solving for ΔG†: ΔG† = 11.0 kcal/mole
13C
Me's: k = (π/√2)•(250Hz) = 2.0836x1010•243K•e-[ΔG†/(.001987•243K)]
Solving for ΔG†: ΔG† = 11.1 kcal/mole
tBu
tBu
2.0836x1010•273K•e-[ΔG†/(.001987•273K)]
Problem 5
(a) In order for the methyl groups to be inequivalent, they need to be in different
environments. Since the gold is coordinates by bonding to the pi-orbital of the
double bond, the substituents on the double bond that is coordinated by the gold
would be displayed front and back (the two H's) which are equivalent by symmetry
in the structure at right. However, the other pi-bond of the allene displays its
substituents left and right (the methyls) which are inequivalent in the structure
at right.
(b) In some way, the dimethylallene unit rotates and the gold atom finds itself
coordinated to the other side of the C=CH2 subunit of the allene. This achieves
exchange of the methyl group positions. We see coalescence at about -52 °C
for the 1H signals of the methyl groups (Δν = 0.05 ppm = 25 Hz) and at about
-30 °C for the 13C signals (Δν = 2.0 ppm = 250 Hz).
tBu
CH3
C
C
P
Au+
C
HH
```