Transcript NMR Spectroscopy

Chem 2600 Spring 2006
Prof. Peter W. Dibble
Lab: Greg Patenaude
Lab E-810 Ext. 2308 Office: E-858 Ext. 2305; NMR
D771B Ext. 2314; Email: [email protected]
Class we b site (for downloading class materials): classes.uleth.ca/200601/chem2600a/ or follow the
links at classes.uleth.ca
Prerequisite: Chem 2500 or permission of instructor. If yo u do no t have the prerequisite yo u will be
deregistered.
Text: "Organic Chemistry" 2nd. edition, by Dr. Thomas N. Sorrell. In addition to the text yo u will find
it very useful to obtain a molecular model kit. You will be allowed to use these kits during all exams in
this and (my) future courses. These kits are also useful for Chem 2810. Darling model kits (so named
after their creator, not because they are cute) are available from the PCB Club (the Physics, Chemistry
and Biology un dergrad uat e society), E-859. The boo kstore sells a different kind of kit for ca. $50.00.
Course Outline: (in approx. order, time permitting)
NMR
Subs titution Reac tions of Alcohols
Elimination Reac tions
Addition reactions of Alkenes and Alkyne s
Addition reactions of conjugated dienes
The Chemistry of Benzene
Addition Reac tions of Aldehyd es and Ketones
Reac tions of Carboxylic acid deriva tives
Organometallic chemistry
Free Radicals
Elementary Mass spectroscopy
Special Topics (Drug development)
Chapter in Te xt
13
7
8
9
10
12
14
Ev aluation:
%
Assignments, [email protected]
10
2 Midterms @ 10
20
(7-9) pm Tuesdays, Feb. 8th and March 8 th L1060)
Lab
30
(50% constitutes a pass)
Final Exam
40
(Tentatively, Thur sday, April 19th, 9-12am)
You will be permitted the use of model kits on your exams.
The exercizes and assignments are meant to be partial preparation for the midterms. Ignore
them at your great peril. I will be releasing my exam bank wi th answ ers as pdf files. In addition, I will
try to post or emai thos e text questions that I feel are representative of wh at yo u sho uld kno w.
NOTE: Failure to sit for an exam will result in a mark of zero unless a valid reason for absence
is presented. Notify me as soon as possible if yo u are going to miss an exam. If you miss a midterm for
a valid reason, that portion of yo ur grade will be added to the final. Students must pass bot h the lab
and lecture portions of the course (i.e. a good lab mark will not pull a failing lecture mark up to a
pass).
IF YOU ARE CAUGHT CHEATING ON ANY EXAM YOU WILL BE ASSIGNED A
GRADE OF F FOR THE COURSE INSTANTLY AND A LETTER DESCRIBING
YOUR OFFENSE WILL BE PLACED IN YOUR STUDENT FILE. TWO SUCH
LETTERS IS GROUNDS FOR EXPULSION FROM THE UNIVERSITY. Check out
the calendar, page 63.
Nuclear Magnetic Resonance
Spectroscopy
O
CH 2
7
6
5
4
PPM
3
2
CH 2 CH 3
1
A proton NMR spectrum. Information from peaks:
magnitude (integration), position and multiplicity.
0
Signal…
Lines…
Multiplicity…
How does NMR it work?
Nuclei behave as if they spin…
Nuclear spin is quantized and
described by the quantum no. I,
where I = 0, 1/2, 1, 3/2, 2, ….
Spin 1/2 nuclei: 1H (proton), 13C, 19F, 31P
Spin 0 nuclei: 12C, 16O
Spin 1 nuclei: 14N, 2H
Spin 5/2: 17O
Spin 3: 10B
In a given sample of a compound in solution, spins are random
and their fields…
In the presence of an externally applied magnetic field (Ho), a
nucleus will adopt 2I + 1 orientations with differing energies.
For the proton, there are two “spin states”.
Ho
+1 / 2
-1 / 2
Zeeman splitting
-1 / 2
² E = h
E
Ho
+1 / 2
Important:
The relative number of nuclei in the different spin states
(proton, 60 MHz) is:
E
N
upper
N lower
e
kT
1,000,000

1,000,009
h
E  h 
Ho
2


Ho
2
The magnetogyric ratio, , is a physical
constant for each nucleus.
1H
2H
13C
19F
31P
So ∆E depends on…
Resonance occurs when…
267.53 radians/Tesla*
41.1
67.28
251.7
108.3
* 1 Tesla = 10,000 Gauss.
For the proton nucleus:
•


Ho
14.10 kG
58.74 kG
117,500 kG

60 MHz
250 MHz
500 MHz
But…
the actual resonance frequency for a given nucleus
depends on its…
LOCAL CHEMICAL ENVIRONMENT
i.e. magnetic interactions within the molecule.
These local magnetic effects are due to:
•  electrons
•  electrons
• other nuclei - especially protons
 electrons
H
In the locale of the proton,
the field lines are opposed
to the applied magnetic
field. This has the effect
of “shielding” the proton
from the full Ho.
C
He
e
Ho
Chemical shielding has the effect of moving signals to the
right on an NMR spectrum. We refer to this as…
Deshielding - downfield
7
6
5
4
PPM
Shielding - upfield
3
Chemical Shift
2
1
0
The magnitude of He is proportional to the electron
density in the  bond.
He
Consider…
Ho
CH 3
CH 3
H3 C
O
CH 3
H3 C
C
CH 3
CH 3
H3 C
Si
CH 3
CH 3
Tetramethylsilane
aka TMS
Because almost all proton resonance frequencies are found
downfield of the TMS signal, TMS is used as an internal
reference. All peak positions are measured as frequencies in Hz
downfield of TMS, with TMS at zero.
There is a problem however..
 is proportional to Ho
Ho = 60 MHz,  = 162 Hz downfield of TMS
Ho = 100 MHz,  = 270 Hz downfield of TMS
Solution to this problem…
We define the d scale:
d
 signal downfi eld of TMS
spectrometer frequency in MHz
 ppm
162Hz
270Hz
d

 2.70ppm
60MHz 100MHz
Chemical shift correlates well with electronegativity…
H3 C
X
F
EN
4.0
dppm 
O
3.5

CHCl 3
CH 2 Cl 2
CH 3 Cl
7 .2 7
5 .3 0
3 .0 5
CH 2 Br
3 .3 0
Cl
3.1

CH2 CH 2 Br
1 .6 9
Br
2.8

I
2.5

H
2.1

CH2 CH 2 CH 2 Br
1 .2 5
TMS
1.8

CH2 CH 2 CH 3
1 .2 0
Electron Effects - Diamagnetic Anisotropy
Alkenes…
5.70
e
H
H
H3 C
H
H
C
H
C
H
1.65
5.59
1.65
5.59
5.03
1.71
H
1.96
4.97
1.96
1.90
1.90
4.00
4.60
6.40
O
O
6.05
Ho
Vinyl protons are downfield,
4.5 - 6 ppm.
H
OCH 3
5.80
H
H
6.43
3.76
Aldehydes
7.45
O
H3 C
e
C
2.20
O
7.54
H
9.72
7.45
R
C
O
H
Alkynes
H
e
1.80
H3 C
C
C
7.81
C
H 1.82
C
7.81
H
9.87
7.26
e
7.26
7.26
7.26
7.26
7.26
H3 C
H
2.35
7.06
7.06
7.14
7.14
7.07
X-ray Structure of
the 3C
Naphthodifuran
Cyclophane
Chemical shift equivalence
… of atoms or groups.
Groups or nuclei are shift equivalent if they can be
exchanged by a bond rotation without changing the structure
of the molecule.
H
C
H
H
H3 C
C
CH 3
CH 3
These atoms/groups are said to be “homotopic”.
Homotopic atoms/groups are always shift equivalent.
Shift equivalence can also be determined by symmetry properties
of the molecule.
A proper axis of rotation…
H3 C
C
CH 3
CH 3
H
H
C
H
CH 3
CH 3
H
H
Atoms/groups that can be reflected in an internal mirror plane
of symmetry, but not exchanged by bond rotation or a proper
axis of symmetry are “enantiotopic”.
O
O
C
C
H
H
H
H
H
H
H
H3 C
H2
C
Cl
H
H
H
H
C
C
H
H
Cl
For our current purposes, enantiotopic protons are always shift
equivalent.
It is easy to recognize atoms/groups that are constitutionally
different.
O
C
CH 3
H
H
H H
H
H
C
C
H2
CH 2
C
HO
C
H
H
H
Protons that are not constitutionally different, that cannot
be exchanged by bond rotation or by molecular
symmetry are “diastereotopic”.
CH 3
Cl
H
H
H
H
H3 C
H
H
H
H
C
H H
CH 3
H
Cl
C
Cl
Cl
H
Cl
Diastereotopic atoms/groups are NOT shift equivalent,
except by coincidence (i.e. they happen to have the same
chemical shift by accident).
O
CH 2 CH 3
CH 2
7
6
5
4
PPM
Raw integrals: 8.5:3.5:3.4:5.0
3
2
1
0
Other nuclei - spin-spin coupling
Consider the vicinal protons in the molecule:
HA
C
Cl
Cl
HM
C
Br
In the absence of any influence by HM,
HA will resonate at A.
Br
But proton HM generates its own magnetic field which will affect
A. Is this field aligned with Ho or against Ho i.e. will it shield or
deshield HA?
Remember that the population of the two spin states are
nearly equal. In the total of all molecules, half of the HM
protons will be aligned with Ho and half against.
So in half the molecules, HM shields HA and in half HM deshields
HA. We therefore see two peaks for HA, of equal intensity; one
vA
upfield of A and one downfield of A.
We call this type of signal a … “doublet”.
We say that HA is “split” into a doublet by HM.
This effect is referred to as … “spin-spin”
coupling.
The distance between the two peaks in
Hz is J, the coupling constant.
J AM
Important aspects of coupling:
• coupling always goes both ways. If HA is split by HM,
then HM must also be split by HA and J must be equal in
both cases.
• coupling is a through-bond and not a through-space
effect.
• coupling between shift-equivalent nuclei is not observed.
A few words about J.
The magnitude of J can be extremely useful in
determining structure. It depends on:
• number of bonds between nuclei
• type of bonds between nuclei
• type of nuclei
• conformation
Because coupling is a through bond effect, the magnitude of
J depends on the number of bonds between coupling nuclei.
One bond couplings are larger than two bond couplings, two
larger than three. In proton-proton couplings, four bond
couplings are not usually observed.
Geminal
H
Vicinal
J3 = 7 Hz*
H H
Cl
H
H
Cl
H
Cl
Long range
J4 = 0 Hz
J2 = 12-14 Hz
H H
Cl
H
Cl
H
* In conformationally averaged systems.
Cl
Type of bonds…
-bonds transmit coupling effects more effectively than  bonds.
H1
H1
H1
H2
H2
H2
H3
H3
H3
H4
H4
H4
H
H
H
CH 3
The magnitude of vicinal couplings depends strongly on the
overlap between adjacent C-H bonds.
C
H
HH
H
H
C
H
Cl
H Cl
H
H
H
H
Cl
H
H
Cl
C
C
H
H
H
H Cl
H
H
C
H
H
C
H
H
H
H
H
Cl
H
H
H
H
Consider the following molecule:
HA HX
C
Cl
Cl
C
HY
Cl
The CH2 has one neighbouring proton in equal proportions of
the up and down spin states. This resonance will therefore
appear as a …
And HA…
Possible spin combinations: HX HY
So HA will appear as a three line pattern: a triplet, with peak
intensities in a 1:2:1 ratio where the lines are separated by J Hz.
The central line will occur at the resonance frequency of HA.
NOTE: The SIGNAL INTEGRATION tells you about the
number of protons that give rise to a particular signal.
MULTIPLICITY tells you about the number of …
NEIGHBOURING NUCLEI.
The n + 1 rule: for simple aliphatic systems, the number
of lines in a given signal is n+1 where n is the no. of
neighbouring protons.
O
C
H 3 CH 2 C
Cl
In this molecule, the CH3 protons will appear as a …
The CH2 protons have three neighbours.
The spin combinations are…
Giving rise to a quartet with
peak intensities of 1:3:3:1.
The combination of a 2 proton quartet and a three proton
triplet is characteristic of the presence of an ethyl group.
2
1
PPM
0
The isopropyl group.
H
H3 C
3
Cl
C
CH 3
2
PPM
1
0
The n-propyl group.
3
CH 3 CH 2 CH 2 Cl
2
PPM
1
0
The t-butyl group.
H2
C
Cl
C
H3 C
3
2
PPM
CH 3
CH 3
1
0
Chemical exchange processes - protons attached to O and N.
(Alcohols, phenols, carboxylic acids, amines - but not amides)
Unlike most spectroscopic methods, the acquisition of signal
in NMR spectroscopy takes about three seconds (proton).
In that time, protons attached to O or N can be transferred
from one molecule to another via the autoionization
process…
H
O
H3 C
H
O
H3 C
H
O
O
H3 C
H3 C
H
H
O
O
H3 C
H
H
H3 C
O
H3 C
H
O
H3 C
O
H3 C
H
H
O
H3 C
O
H3 C
H
H
O
H3 C
H
H3 C
O
H
H3 C
O
H
H3 C
O
H3 C
O
If the rate of exchange is slow compared to the time scale of
the NMR experiment (I.e. three seconds) then the spectrum
is that expected of CH3OH. Under these conditions, vicinal
OH:CH coupling is observed. This is rarely the case.
If the rate of exchange is comparable to the NMR time scale,
then one observes the exchanging proton in a range of
environments and at a range of chemical shift postions.
HHH
H
HHH
H
H
O CH
O
3
H3 C
H
Under these conditions, the OH peak is broad and
coupling is not observed…
H
The rate of exchange is catalyzed by acids and bases,
depends on solvent, temperature, concentration, purity, and
lunar phase.
Exchangeable protons…
• do not couple
• have variable shift positions
• are observed as broad peaks
• exchange with D2O
OH
Alcohols 1-5 ppm
Phenols 3.5-6 ppm
Carboxylic acids 10-12 ppm
My personal record 13 ppm.
NH (amines) 0.5 - 5 ppm
3
2
PPM
1
0
2
1
PPM
0
Occasionally, carboxylic acid protons resonances are so
broad …
..that the only way to tell that they are there is to integrate
the baseline.
D2O exchange…
Exchangeable protons on a molecule will exchange
with exchangeable protons on other molecules…
D2 O
O
H3 C
H
This can be very useful.
HOD
O
H3 C
D
Coupling in non n+1 systems.
Substituted benzenes.
1,4
X
These always appear as
perfectly symmetrical
patterns that look like
two doublets or a
quartet.
CN
Br
Y
Symmetrical 1,2 disubstituted benzenes
These always appear as a perfectly symmetrical
pattern with a lot of fine structure …
X
Cl
Cl
X
1,2,4- trisubstituted systems and the dreaded tree diagrams.
CN
H6
H2
H5
Cl
OH
J5,6 = 6 Hz
J2,6 = 2 Hz
J2,5 = 0 Hz
6
2
5
Coupling in non n+1 systems.
Monosubstituted benzenes.
X
If the substituent is an alkyl group or halogen,
then all five protons tend to show up in the same
place as either a singlet or a somewhat broad
singlet.
If X is a strong electron-withdrawing group (carbonyl, nitro,
sulfonic acid) then the ortho and para protons will be pulled
downfield.
X
If X is a strong electron-donating group (OH, OR, NH2)
then the ortho and para protons are pushed upfield.
OH