Problem Set #4 - Chemistry at Caltech

Chemistry 143
Problem Set #4
Due May 15, 2015
Problem 1. From the T1 spectra below of ethyl diazoacetate in CHCl3 using d2 times from
0.1875 to 96 seconds, determine the T1 of each proton in the molecule.
Problem 2. For this selective population transfer experiment on 40% ethyl formate in CDCl3,
the 13C satellites of the –CHO peak were identified in the 300 MHz 1H NMR spectrum. A normal
one-scan 13C NMR spectrum acquired without BB 1H decoupling is shown (Spectrum 1).
Immediately after irradiation of the downfield satellite of the –CHO for 5 seconds to achieve
presaturation of this peak, a one-scan 13C NMR spectrum was acquired affording Spectrum 2.
The same experiment performed by instead presaturating the upfield satellite gave Spectrum 3.
a) Draw the energy level diagram presented in class for the 13C-1H spin system and label the
population excesses and deficiencies for each of the four levels. Show how presaturation of one
of the 13C satellites in the 1H NMR spectrum changes the populations in this diagram. (Hint:
Presaturation achieves equalization of the populations across the transition corresponding to the
irradiated peak.) Under the optimum choice of molecule and conditions, what intensities of the
two signals in the 13C spectrum represent the maximum degree of polarization transfer possible
for this experiment. (Assume the original intensities are +2, +2.)
b) Describe briefly the characteristics of the sample that would enable the above experiment.
Problem 3 For the unknown C6H10O are shown the 1H NMR (200 MHz) spectrum (a) and at 50
MHz, the 13C NMR (50 MHz) spectrum and INEPT spectrum (shown in the upfield region (b)
and downfield region (c)).
a) For each of the 5 carbon signals, use the INEPT spectrum to determine if it is a Cq, CH, CH2
or CH3 group. (Recall that you are able to distinguish CH and CH2 using the line separations
and anticipated JCH values).
(b) Given that the integrations of the 1H spectrum are 1:2:2:2:3, what is the structure of this
unknown?
!
"
!
"
-3
+2
CH3:
!!
!"
""
+9
+8
0
-7
-8
+15
!!!
!!"
!""
"""
121
-4
!!!
!!"
!""
"""
CH2:
+4
!!
!"
""
11
+5
!!
!"
""
CH:
INEPT with
phase cycling
!
"
SPI
expt.
Normal
13 C
!!!
!!"
!""
"""
Problem 4. Shown at right are the expected
intensities for CH, CH2, and CH3 groups that
would be produced by the SPI and INEPT
experiments under ideal conditions. As the
INEPT experiment is generally acquired with
phase cycling, it can be seen that the
contribution to the multiplets from the natural
13
C magnetization is removed in this manner.
For the CH group, we derived in class that the
SPI experiment would produce one line with
intensity of -2ΔH+2ΔC and the other line with
intensity of +2ΔH+2ΔC. If we set ΔH=2 and
ΔC=0.5 we get the intensities shown.
The SPI intensities for the CH2 and CH3
groups can be generated by considering the net
polarization transfer from the protons
participating for each line in the multiplet:
+13
+12
+12
133 1
-11
-9
-12
-12
intensity = (probability of state)*(net polarization transfer from H’s + natural 13C polarization)
An excerpt from “NMR and Chemistry,” 2nd
edition by J. W. Akitt describes the SPI
experiment performed on acetone. In the
proton spectrum of acetone (not shown), one
methyl satellite signal corresponding to
CH313C(O)CH3 is inverted with a selective π
pulse and the carbonyl carbon is observed in
the 13C NMR without broadband decoupling.
Show how the calculated intensities can be
derived.
Selective population inversion
Pulse techniques allow us to change spin populations drastically and
controllably simply by subjecting the nuclei to an inverting 180° pulse.
The resulting return to equilibrium is accompanied by significant, and in
some cases, very large, changes in intensity of the lines due to coupled
transitions, magnified if the magnetogyric ratio of the inverted spin is
greater than that for the observed spin. The method is particularly useful
for assisting in the observation of quaternary carbon atoms with long
relaxation times, since the effective relaxation time in such an experiment
is that of the inverted proton and this is usually much shorter than that for
carbon. An example is given at right for the carbonyl carbon of acetone.
The proton spectrum of acetone contains two sets of 13C satellites, one
pair widely spaced and due to coupling to the directly bonded 13C in the
methyl groups, and one with a much smaller coupling constant due to the
two-bond coupling between the methyl protons and the carbonyl 13C. The
carbonyl 13C resonance is therefore split into a septet with intensities
1:6:15:20:15:6:1 by this coupling. If we invert the proton spins
corresponding to one of the inner 13C satellites by irradiating this with a
long, selective 180° pulse and follow this immediately with a pulse at the
13
C frequency to generate a 13C FID, we find that the carbonyl carbon
intensities have been greatly perturbed and now have the intensities
+25:+102:+135:+20:-105:-90:-23, which is a useful gain in intensity.
Problem 5. The normal 119Sn NMR spectrum of EtSn(OAc)3 displays nine lines due to the
coupling of the 119Sn nucleus (I=1/2) to all five protons of the ethyl group. The use of the
INEPT pulse sequence allows a significant enhancement of the signal to be obtained.
Unfortunately, the J coupling constants of 119Sn to the methyl (3JSn-H = 242 Hz) and methylene
(2JSn-H = 118 Hz) are not the same so a single chosen value of τ cannot suffice for all hydrogens
of the ethyl group.
Give the expected multiplet intensities
for the normal 119Sn NMR spectrum of
EtSn(OAc)3.
Indicate which peaks
would be too close to resolve (closer than
10 Hz).
Now, considering the spectra at right,
two 119Sn INEPT experiments were
performed with the τ values of 0.0021
sec and 0.0050 seconds. Of the two
spectra shown at right (a and b) which
spectrum corresponds to which τ value?
Show how the 1H net magnetization
vectors of a proton on the methyl group
and a proton on the methylene group
would evolve during the τ - 180°Sn/180°H
- τ part of each INEPT experiment. For
each, what degree of polarization transfer
is expected ?
MHSn!
CH2:
#=
0.0021 s
H x'
y'
y'
x'
x'
#=
0.0021 s
x'
#=
0.0050 s
x'
#=
0.0050 s
180°
H x'
y'
y'
y'
180°
MHSn"
Sn
x'
x'
x'
#=
0.0050 s
x'
#=
0.0050 s
180°
H x'
y'
y'
y'
y'
180°
MHSn"
x'
y'
Sn
x'
y'
CH3:
y'
180°
x'
MHSn!
#=
0.0021 s
H x'
y'
MHSn"
CH2:
x'
180°
y'
MHSn!
y'
Sn
x'
CH3:
y'
180°
MHSn"
MHSn!
#=
0.0021 s
180°
Sn
x'
x'
x'
Problem 6. To aid in the identification of an unknown compound containing only five carbons,
an INEPT spectrum (spectrum b) was taken with the standard setting of τ = 0.002 sec.
Comparison of spectrum b against the 13C NMR acquired with 1H BB decoupling (spectrum a)
showed strong enhancement of four of the five carbons.
a) Assign the # of attached hydrogens and the
JCH value to each of the four enhanced carbons.
b) When the INEPT experiment on this
molecule is performed with t = 0.001 sec,
spectrum c is obtained. Explain (1) why the
peak near 70 ppm shows a strong enhancement
for this setting of τ, and (2) why it didn’t show
up in spectrum b. Assign the # of attached
hydrogens and the JCH value for this signal in the
table.
# of Attached
Hydrogens
JCH (Hz)
80.9 ppm
70.6 ppm
see below
66.1 ppm
28.8 ppm
22.1 ppm
c) Using the 1H NMR (CDCl3, 300 MHz) below along with the INEPT results, determine the
structure of this unknown (molecular formula C5H8O) and assign the 13C carbon chemical shifts.
Problem 6.
C 5 H8 O
(a) 13C NMR
(b) INEPT τ = 2ms
(b) INEPT τ = 1ms
CDCl3 (75 MHz)