h nmr lecture
DESCRIPTION
spectroscopyTRANSCRIPT
Rule of Thirteen
•When the formula of the unknown is not given, you are in trouble…..unless
you know the Rule of Thirteen
•Molecular weight of the compound is M
M/13 = n + r/13
where n is the integer derived from dividing M/13, and r is the remander
Note: do not use the calculator for this rule
•Base formula (not necesseraly the formula of your given compound): CnHn+r
•Parent formula: CnH2n+2
IHD: [2n+2-(n+r)]/2 =(n-r+2)/2
•Before you go on with the IHD calculations, check on the presence of
heteroatoms:
For example, if there is an oxygen present, then O=16
Substitute it in your base formula by subtracting oxygen and adding a
molecular fragment with the same weight (C=12 and 4H=4)
Examples
O
The molecular weight of the unknown aldehyde is 134g/molPropose several structures
Subtract oxygen134-16=118
Rule of 13: 118/13= 9+1/13Base formula is C9H10OParent formula is C9H20IHD = 20-10/2=5benzene ring and a double bond=aldehyde
O
Nuclear Magnetic Resonance
•Atoms with nuclei containing an odd number of protons or neutrons (1H,13C, 14N, 19F, and 31P) have a magnetic moment µ induced by the nuclear
spin.
•Hydrogen atoms are more than 99% 1H. Other isotopes have special
names: 2H = deuterium, 3H = Tritium (radioactive).
•Spin quantum number I=1/2, meaning their magnetic moments can only
have two equal but opposite values +µ and -µ.
•In the applied magnetic field, they can be oriented parallel ! (m=+1/2)
or anti-parallel " (m=-1/2). The energy is slightly different.
S
A spinning nucleus with it's magnetic field alinged with the magnetic field of a magnet
N
S
N
!- spin state,
favorable,
lower energyN
S !- spin state,unfavorable,lower energy
A spinning nucleus with it's magnetic field alinged against the magnetic field of a magnet
S
N
Nuclear Magnetic Resonance and Relaxation
Applied magnetic field (field strength H0 and flow density B0)
A proton experiences an ‘impact’ generated by a high frequency transmitter
A resulting movement proceeds in two directions (directional quantization):
o Direction of the external field (low energy)
o Direction opposite to the applied field (high energy)
o When the frequency of the transmitter equals to the precession
frequency, an inversion of the magnetic moment occurs (Nuclear Magnetic
Resonance).
o The process of returning to the lower energy position is called
relaxation.
S
N
E2
E1
#E = E2 – E1
high frequency
transmitter ($0)
• Number of peaks: number of non-equivalent protons in the sample
• Chemical shift: kinds of protons in the molecule
• J-coupling: protons that are near other protons.
• Integrals: the ratio of each kind of proton in our sample.
Interpreting SpectraThe Chemical Shift (%) Scale
Simple presentation:
•Choose a standard sample, an NMR of that standard
•Measure its absorbance frequency.
• We then measure the frequency of our sample and subtract its frequency from that
of the standard.
•We then divide by the frequency of the standard. This gives a number called the
“chemical shift,” %
•% does not depend on the magnetic field strength.
• Why not?
•Lets examine the same sample at 300 and 600 MHz field
•On 300 MHz instrument
•standard absorbs at 300,000,000 Hz (300 MHz)
•sample absorbs at 300,000,300 Hz.
•The difference is 300 Hz 300/300,000,000 = 1/1,000,000
•1 part per million (or 1 ppm).
•On 600 MHz instrument
•standard absorbs at 600,000,000 Hz (600 MHz)
•sample absorbs at 600,000,600 Hz. (the frequency of sample increases proportionally)
•The difference is now 600 Hz, 600,000,000 (600/600,000,000 = 1/1,000,000, = 1 ppm).
CH3
Si
H3C
H3C CH3
Chemical Shifts
•Most protons appear between 0 and 12 ppm.
! ppm
TMS
CH3CH3
RONR2
CH3OCH3
RO
HR
R R
HH
RO
Ph CH3
HR
Cl
CH3
Ph
OH
OH
R
NHR
Upfield regionof the spectrumDownfield region
of the spectrum
012345678910
CH3HO(R)
1112
OH
RO
TMS = Me Si
Me
Me
Me
Inductive electron withdrawal by neighboring electronegative groups.
0.8ppm 3.0ppm 5.3ppm 7.3ppm
Chemical Shifts
H C
H
H
H
H C
H
H
Cl
H C
H
Cl
Cl
H C
Cl
Cl
Cl
H C
H
H
H
I C
H
H
H
Cl C
H
H
H
HO C
H
H
H
F C
H
H
H
Br C
H
H
H
electronegativity
of X
2.1 2.5 2.8 3.1 3.5 4.0
chemical shifts, ppm0.23 2.16 2.68 3.05 3.40 4.26
0123
PPM
O
Cl
012345
PPM
O Cl
O
01234
PPM
O
Hybridization effect
•More S-character to the adjacent carbon, more electronegative, stronger is the deshielding
effect on the H’s of the methyl group
C CH3 CH3CH3
0.2ppm!! 1.6-1.9ppm 2.1-2.5ppm
H
C CH3
H
CH3
CH3
H
Wrong!!!
•In this case, your expectations for chemicals shifts of protons attached to the carbons
of sp3, sp2, and sp hybridization should increase in the same order. Right?
Magnetic Anisotropy Effect
C
C
H
H
H
applied field Ho
enhancement of applied magnetic field shielding of applied magnetic field
+
+
(-) (-) C C
+
+
(-) (-) C CH H
(-)
(-)
+ +
6.6 -8.5 ppm 4.9-6.2 ppm 2.4-3.1 ppm
Spin-spin splitting or J-coupling
•When the field created by HB reinforces the magnetic field of the NMR machine (B0 ), HA
feels a slightly stronger field.
•When the field created by HB opposes B0, HA feels a slightly weaker field.
• So, we see two signals for HA depending on the alignment of HB. The same is true for HB, it
can feel either a slightly stronger or weaker field due to HA’s presence. So, rather than see a
single line for each of these protons, we see two lines for each.
C C
HBHA
HA HB
HA is split into two lines because it feels the magnetic field of HB.
HB is split into two lines because it feels the magnetic field of HA.
For this line, HB is lined up with the magnetic field
(adds to the overall magnetic field, so the line
comes at higher frequency)
For this line, HB is lined up against the magnetic field(subtracts from the overall magnetic field, so the line
comes at lower frequency)
C C
HBHA
HA HB
HA is split into two lines because it feels the magnetic field of HB.
HB is split into two lines because it feels the magnetic field of HA.
For this line, HB is lined up with the magnetic field
(adds to the overall magnetic field, so the line
comes at higher frequency)
For this line, HB is lined up against the magnetic field(subtracts from the overall magnetic field, so the line
comes at lower frequency)
HB
HA
HB
HA
•When there is more than one proton splitting a neighboring proton, we get more lines.
More on J-coupling
C C
HBHA
HA'HA + HA' HB
HA and HA' appear at the same chemical shift because they are
in identical environments They are also split into two lines (called a doublet) because they
feel the magnetic field of HB.
HB is split into three lines because it feels the magnetic
field of HA and HA'
Note that the signal produced by HA + HA' is twice the size
of that produced by HB
C C
HBHA
HA'HA + HA' HB
HA and HA' appear at the same chemical shift because they are
in identical environments They are also split into two lines (called a doublet) because they
feel the magnetic field of HB.
HB is split into three lines because it feels the magnetic
field of HA and HA'
Note that the signal produced by HA + HA' is twice the size
of that produced by HB
HB
HA’
HB
HA
HB
HA’
HB
HA
HA
HB
HA’
HA
HB
HA’
HA
HB
HA’
HA
HB
HA’
Splitting Pattern (Spin – Spin Coupling)
•Splitting can occur via space or binding electrons.
•Protons on the same carbon are called geminal: rarely observed
•Protons on adjacent carbons are called vicinal
CC C
H
H
H
H
1
2
1
2
3
geminal splitting vicinal splitting
Coupling Constants: Alkenes
•Coupling constants in alkenes differs depending on whether the protons are cis or trans to
each other.
•In a terminal alkene the cis and trans protons are NOT Equivalent!!!
• One is on the same side as the substituent, the other is on the opposite side.
•The coupling of trans protons to each other is typically ~ 16 Hz.
•The coupling of cis protons is a little smaller~ 12 Hz.
HA
Now, let's "turn on" HA - HX coupling. This splits the single line into two lines that are 16 Hz appart
If uncoupled, HA would appear as a singlet where the dashed line indicates
Now, let's "turn on" HA - HM coupling. This splits each of the two new lines into two lines that are 12 Hz appart for a total of four lines
12 Hz
16 Hz
12 Hz
HA
HM
HX
12Hz coupling
16 Hz coupling
HO
H
H
H
HO
H
H
H
HA
Now, let's "turn on" HA - HX coupling. This splits the single line into two lines that are 16 Hz appart
If uncoupled, HA would appear as a singlet where the dashed line indicates
Now, let's "turn on" HA - HM coupling. This splits each of the two new lines into two lines that are 12 Hz appart for a total of four lines
12 Hz
16 Hz
12 Hz
HA
HM
HX
12Hz coupling
16 Hz coupling
Common Splitting Patterns for Aromatic Compounds
Chemical Shifts 7.00 –8.3ppm
• Monosubstituted aromatic rings
012345678
PPM
Cl
Ha
Hb
Ha
Hb
Hc
dd
dd
t
a b
c
Disubstituted aromatic rings (X=Y)
012345678
PPM
Cl
Cl
Ha
Ha
Hb
Hb
d
d
a b
012345678
PPM
Cl
Ha
Cl
Hb
Hc
Hb
s
d
t
a
b c
012345678
PPM
Cl
Ha
Ha
Ha
Ha
Cl
a
s
Disubstituted aromatic rings (X,Y)
OH
Ha
Hb
NO2
Hb
Had
d
a b
0246810
PPM
0246810
PPM
OH
NO2
Ha
Hd
Hc
Hb
dd
pt
pt
dda bd c
0246810
PPM
OH
Ha
NO2
Hb
Hc
Hd
ddmight look like singlet or pt
ddd
pt
dddab dc
RO
012345
PPM
01234
PPM
X (no hydrogens)
Spinning bands
•Occur in very concentrated samples due to the coupling to 13C.
•Appear as very short and very symmetrical peaks on both sides of a peak.
•Will disappear under dilution.
Integrals
•Represent the ratio of each kind of proton.
•The heights of lines are proportional to the intensity of the signal.
3H'S
3H'S
2 H'S
O
O H H
O CH3
O
H3C O
O
Good websites for NMR tutorials:
http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/problems_fs_start.html
http://www.nd.edu/~smithgrp/structure/workbook.html
Summary:
•Information that can be obtained from H-NMR:
•Number of peaks: number of magnetically non-equivalent H’s
•Chemical shifts: chemical and magnetic environment of the given hydrogen
•Window: 0-15 ppm, signals to the left are downfield or deshielded hydrogens
•Signals to the right are upfield or shielded hydrogens
•The more electron density is on the H, the more shielded it is.
•Splitting patterns: the number of H’s on adjacent carbons
•Determined by N+1 rule for freely rotating systems, where N is the number of
hydrogens on adjacent carbons
•Most common patterns:
•Isolated ethyl group: a quartet and a triplet
•Isolated isopropyl group: a septet and a tall doublet
•Para-substituted aromatic ring: two doublets at 6.5-8.5 ppm
•Meta-substituted aromatic ring: a distinct singlet in aromatic region
•Acidic protons are usually unresolved wide signals and will change their position
with concentration and in the presence of water
•Integrations: area under the peak, represents the relative number of hydrogens
under the peak.
•If the area under the peak is not a whole number, multiply all the integrations by
an integer to obtain a whole number for all peaks.
Solving an NMR problem: C10H12O
Solving an NMR problem: C8H10BrN Solving an NMR problem: C
7H12O3
Solving an NMR problem: C10H12O3
Solving an NMR problem: C8H8O2
Solving an NMR problem: C8H11N Solving an NMR problem: ExampleC
6H5N
Solving an NMR problem: C8H8O2
Solving an NMR problem: C9H11O2