Transcript 13.9 Spin-Spin Splitting

13.9 Spin-Spin Splitting
SPIN-SPIN SPLITTING
Often a group of hydrogens will appear as a multiplet
rather than as a single peak.
Multiplets are named as follows:
Singlet
Doublet
Triplet
Quartet
Quintet
Septet
Octet
Nonet
This happens because of interaction with neighboring
hydrogens and is called SPIN-SPIN SPLITTING.
1,1,2-Trichloroethane
integral = 2
Cl H
H C C Cl
integral = 1
triplet
Cl H
doublet
n + 1 RULE
this hydrogen’s peak
is split by its two neighbors
these hydrogens are
split by their single
neighbor
H
H
H
H
C
C
C
C
H
two neighbors
n+1 = 3
triplet
H
one neighbor
n+1 = 2
doublet
MULTIPLETS
singlet
doublet
triplet
quartet
quintet
sextet
septet
Some Common Patterns
SOME COMMON SPLITTING PATTERNS
X CH CH Y
(x=y)
CH-CH2
X CH2 CH2 Y
(x=y)
CH-CH3
-CH2-CH3
CH3
CH
CH
3
tert-butyl group
CH 3
H3C
C
CH 3
Cl
9 equivalent protons = singlet
EXCEPTIONS TO THE N+1 RULE
IMPORTANT !
1)
2)
Protons that are equivalent by symmetry
usually do not split one another
X CH CH Y
X CH2 CH2 Y
no splitting if x=y
no splitting if x=y
Protons in the same group
usually do not split one another
H
C H
H
H
or
C
H
SOME EXAMPLE SPECTRA
WITH SPLITTING
NMR Spectrum of Bromoethane
Br CH2CH3
NMR Spectrum of 2-Nitropropane
H
CH3
C
CH3
+
N
O
O-
1:6:15:20:16:6:1
in higher multiplets the outer peaks
are often nearly lost in the baseline
NMR Spectrum of Acetaldehyde
O
CH3 C
offset = 2.0 ppm
H
The propyl group
CH3-CH2-CH2-X
Can you predict the splitting patterns for this compound?
INTENSITIES OF
MULTIPLET PEAKS
PASCAL’S TRIANGLE
PASCAL’S TRIANGLE
Intensities of
multiplet peaks
1
The interior
1 1
entries are
the sums of
1 2 1
the two
numbers
immediately
1
3
3
1
above.
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
singlet
doublet
triplet
quartet
quintet
sextet
septet
octet
THE ORIGIN OF
SPIN-SPIN SPLITTING
HOW IT HAPPENS
THE CHEMICAL SHIFT OF PROTON HA IS
AFFECTED BY THE SPIN OF ITS NEIGHBORS
aligned with Bo
+1/2
50 % of
molecules
opposed to Bo
-1/2
H
HA
H
HA
C
C
C
C
50 % of
molecules
Bo
downfield
neighbor aligned
upfield
neighbor opposed
At any given time about half of the molecules in solution will
have spin +1/2 and the other half will have spin -1/2.
SPIN ARRANGEMENTS
one neighbor
n+1 = 2
doublet
one neighbor
n+1 = 2
doublet
H
H
H
H
C
C
C
C
yellow spins
blue spins
The resonance positions (splitting) of a given hydrogen is
affected by the possible spins of its neighbor.
SPIN ARRANGEMENTS
two neighbors
n+1 = 3
triplet
one neighbor
n+1 = 2
doublet
H
H
H
H
C
C
C
C
H
H
methine spins
methylene spins
SPIN ARRANGEMENTS
three neighbors
n+1 = 4
quartet
H
H
C
C
H
H
H
two neighbors
n+1 = 3
triplet
H
H
C
C
H
H
methylene spins
methyl spins
H
13.10
The Coupling Constant
THE COUPLING CONSTANT
H H
J
J
C C H
J
H H
J
J
J
The coupling constant is the distance J (measured in Hz)
between the peaks in a multiplet.
J is a measure of the amount of interaction between the
two sets of hydrogens creating the multiplet.
FIELD COMPARISON
100 MHz
100 Hz
Coupling constants are
constant - they do not
change at different
field strengths
6
5
4
3
2
1
400 Hz
Separation
is larger
3
7.5 Hz
J = 7.5 Hz
200 MHz
The shift is
dependant
on the field
200 Hz
200 Hz
7.5 Hz
J = 7.5 Hz
2
1
ppm
Why buy a higher
field instrument?
Spectra are
simplified!
Overlapping
multiplets are
separated.
Second-order
effects are
minimized.
50 MHz
J = 7.5 Hz
3
2
1
2
1
2
1
100 MHz
J = 7.5 Hz
3
200 MHz
J = 7.5 Hz
3
NOTATION FOR COUPLING CONSTANTS
The most commonly encountered type of coupling is
between hydrogens on adjacent carbon atoms.
3J
H H
C C
This is sometimes called vicinal coupling.
It is designated 3J since three bonds
intervene between the two hydrogens.
Another type of coupling that can also occur in special
cases is
2J or geminal coupling
H
( most often 2J = 0 )
C H
Geminal coupling does not occur when
2J
the two hydrogens are equivalent due to
rotations around the other two bonds.
LONG RANGE COUPLINGS
Couplings larger than 2J or 3J also exist, but operate
only in special situations, especially in unsaturated
systems.
Couplings larger than 3J (e.g., 4J, 5J, etc) are usually
called “long-range coupling.”
SOME REPRESENTATIVE COUPLING CONSTANTS
H H
vicinal
C C
6 to 8 Hz
three bond
3J
11 to 18 Hz
three bond
3J
6 to 15 Hz
three bond
3J
0 to 5 Hz
two bond
H
C C
trans
cis
geminal
H
H
H
C C
H
C
H
2J
H
H
cis
6 to 12 Hz
trans
4 to 8 Hz
three bond
3J
4 to 10 Hz
three bond
3J
0 to 3 Hz
four bond
4J
0 to 3 Hz
four bond
4J
H
C
C H
H
C C
C H
H C C C
H
Couplings that occur at distances greater than three bonds are
called long-range couplings and they are usually small (<3 Hz)
13.11 NMR Spectra of
Carbonyl Compounds
• Anisotropy in carbonyl compounds
• Anisotropy deshields C-H on aldehydes:
9-10 ppm
• Anisotropy also deshields methylene and
methyl groups next to C=O: 2.0 - 2.5 ppm
• Methylene groups directly attached to
oxygen appear near 4.0 ppm
1
2-Butanone (Methyl Ethyl Ketone)
60 MHz Spectrum
O
CH3 C CH2CH3
2-butanone, 300 MHz
spectrum
WWU Chemistry
2
Ethyl Acetate
Compare the methylene shift to that of Methyl Ethyl Ketone (previous slide).
O
CH3 C O CH2CH3
t-Butyl Methyl Ketone
3
(3,3-dimethyl-2-butanone)
O CH3
CH3
C C CH3
CH3
4
Phenylethyl Acetate
O
CH2CH2 O C CH3
5
Ethyl Succinate
O
O
CH3CH2 O C CH2CH2 C O CH2CH3
6
a-Chloropropionic Acid
O
CH3
CH C OH
Cl
13.12 and 13.13
Alkenes, Alkynes and
Aromatic Compounds
CHEMICAL SHIFTS
Alkenes and alkynes
• vinyl protons appear between
5 to 6.5 ppm (anisotropy)
• methylene and methyl groups next to
a double bond appear at about 1.5 to
2.0 ppm
• for terminal alkynes, proton appears
near 2 ppm
BENZENE RING HYDROGENS
Ring current causes protons attached to the
ring to appear in the range of 7 to 8 ppm.
Protons in a methyl or methylene group
attached to the ring appear in the range
of 2 to 2.5 ppm.
NMR Spectrum of Toluene
5
CH3
3
THE EFFECT OF CARBONYL SUBSTITUENTS
When a carbonyl group is attached to the ring the
o- and p- protons are deshielded by the anisotropic
field of C=O
O
H
C
R
R
H
H
C
O
H
Only the o- protons are in range for this effect.
Acetophenone (90 MHz)
O
C
CH3
H
H
2
3
deshielded
3
NMR Spectrum of
1-iodo-4-methoxybenzene
3
OCH3
I
CHCl3 impurity
2
2
NMR Spectrum of
1-bromo-4-ethoxybenzene
Br
OCH2CH3
4
2
3
THE p-DISUBSTITUTED PATTERN CHANGES AS THE
TWO GROUPS BECOME MORE AND MORE SIMILAR
All peaks move closer.
Outer peaks get smaller …………………..… and finally disappear.
Inner peaks get taller…………………………. and finally merge.
X
X
Y
X'
X=Y
X ~ X’
X
X
X=X
all H
equivalent
same groups
NMR Spectrum of
1-amino-4-ethoxybenzene
4
H2N
OCH2CH3
2
2
3
NMR Spectrum of p-Xylene
(1,4-dimethylbenzene)
6
CH3
4
CH3
13.14 Hydroxyl
and Amino Protons
Hydroxyl and Amino Protons
Hydroxyl and amino protons can appear
almost anywhere in the spectrum (H-bonding).
These absorptions are usually broader than
other proton peaks and can often be identified
because of this fact.
Carboxylic acid protons generally appear far
downfield near 11 to 12 ppm.
SPIN-SPIN DECOUPLING BY EXCHANGE
In alcohols coupling between the O-H hydrogen and
those on adjacent carbon atoms is usually not seen.
C
O
This is due to rapid exchange of
OH protons between the various
alcohol molecules in the solution.
H
H
The OH peak is usually broad.
In ultrapure alcohols, however,
coupling will sometimes be seen.
NMR Spectrum of Ethanol
CH3CH2 OH
3
2
1
1-propanol
CH3CH2CH2 OH
13.16 Unequal Couplings
Tree Diagrams
WHERE DOES THE N+1 RULE WORK ?
The n+1 rule works only for protons in aliphatic chains
and rings, and then under special conditions.
There are two requirements for the n+1 rule to work:
1) All 3J values must be the same all along the chain.
2) There must be free rotation or inversion (rings) to
make all of the hydrogens on a single carbon be
nearly equivalent.
The typical situation
where the n+1 rule
applies.
H
H
H
C C C
H
3J
H
a
H
= 3J b
Hydrogens can
interchange their
positions by
rotations about
the C-C bonds.
WHAT HAPPENS WHEN THE J VALUES ARE NOT EQUAL ?
H
3J
a
=
3J
b
H
H
C C C
H
3J
H
a
H
3J
b
In this situation each coupling must be considered
independently of the other.
A “splitting tree” is constructed as shown on the
next slide.
CONSTRUCTING A TREE DIAGRAM
( SUPPOSE 3Ja = 7 Hz and 3Jb = 3 Hz )
H
H
H
The largest J value
is usually used first.
-CH2-CH2-CH2-
C C C
H
H
C C C
3J
H
a=
H
H
7
H
H
H
H
3J
triplet of triplets
b
=3
WHEN BOTH 3J VALUES ARE THE SAME
The n+1 rule is followed
-CH2-CH2-CH2-
n+1 = (4 + 1) = 5
….. because of overlapping legs
you get the quintet predicted by
the n+1 rule.
2-PHENYLPROPANAL
A case where there are unequal J values.
Spectrum of 2-Phenylpropanal
a
b
d
CH3 CH CHO
J = 7 Hz
TMS
c
J = 2 Hz
a
c
d
b
7 Hz
2 Hz
CH3 CH CHO
Rather than the expected
quintet …..
the methine hydrogen
is split by two different
3J values.
3J
ANALYSIS
OF METHINE
HYDROGEN’S
SPLITTING
1
3J
quartet of doublets
= 7 Hz
2
quartet
by -CH3
= 2 Hz doublet
by -CHO
2-PHENYLPROPANAL
• Adjacent protons are three bonds away from
each other: 3J, often = 7 Hz
• The aldehyde proton d has a 3J = 2 Hz
coupling to the single proton b
• the methyl protons a have a 3J = 7 Hz
coupling to proton b
• proton b is a quartet of doublets
VINYL ACETATE
COUPLING CONSTANTS
PROTONS ON C=C DOUBLE BONDS
PROTONS ON C=C DOUBLE BONDS
• In alkenes,
3J-cis
= 8 Hz
• In alkenes, 3J-trans = 16 Hz
• In alkenes, when protons
are on the same carbon,
2J-geminal = 0-2 Hz
H
H
H
H
H
H
NMR Spectrum of Vinyl
Acetate
O
CH3 C
O CH CH2
Analysis of Vinyl Acetate
O
HC
CH3 C
3J-trans
HB
O
C C
HC
3J
> 3J-cis > 2J-gem
HB
HA
3J
BC
trans
BC
trans
3J
AC
cis
HA
3J
AC
cis
2J
AB
2J
AB
gem
gem
OVERVIEW
TYPES OF INFORMATION
FROM THE NMR SPECTRUM
1. Each different type of hydrogen gives a peak
or group of peaks (multiplet).
2. The chemical shift (d, in ppm) gives a clue as
to the type of hydrogen generating the peak
(alkane, alkene, benzene, aldehyde, etc.)
3. The integral gives the relative numbers of each
type of hydrogen.
4. Spin-spin splitting gives the number of hydrogens
on adjacent carbons.
5. The coupling constant J also gives information
about the arrangement of the atoms involved.
SPECTROSCOPY IS A POWERFUL TOOL
Generally, with only three pieces of data
1) empirical formula (or % composition)
2) infrared spectrum
3) NMR spectrum
a chemist can often figure out the complete
structure of an unknown molecule.
EACH TECHNIQUE YIELDS VALUABLE DATA
FORMULA
Gives the relative numbers of C and H
and other atoms
INFRARED SPECTRUM
Reveals the types of bonds that are present.
NMR SPECTRUM
Reveals the environment of each hydrogen
and the relative numbers of each type.