NON-FIRST ORDER SPECTRA p. 173 - web.uvic.ca
Transcript of NON-FIRST ORDER SPECTRA p. 173 - web.uvic.ca
NON-FIRST ORDER SPECTRA
>CH---CH2-t d
is only true if (in Hz) > 10J
This is easier to achieve with large fields:
1.5 1.3 = 0.2 ppm J = 7 Hz (constant with field)
At 60MHz, 0.2ppm = 12Hz, so /J = 12/7 = ~2
At 250MHz, 0.2ppm = 50Hz, so /J = 50/7 = ~7
At 600MHz, 0.2ppm = 120Hz, so /J = 120/7 = ~17
p. 173
If <10J, spectra are NOT as predictable
1st order
not 1st order
singlet when =0
p. 174
60MHz
p. 175Brucine hydrate
CH3O are now readily resolved300MHz
p. 175
300MHz
600MHz
Higher field, more 1st orderp. 175
still triplets, but inside lines > outside lines
p. 176
Δ ~ 6
triplets hardly recognizable, outside lines v small
p. 176
no longer recognizable
p. 176
1st order
not 1st order
Spin system labels p. 177
Assumes ‘infinite’ resolution…actual lines will merge
NOMENCLATURE: Spin system notation
If 1st order conditions, i.e. > 10J, use letterswell apart in alphabet (these are arbitrary but accepted)
HF = AX; ClPF2 = AX2 CH3CH2Br = A2X3
If not 1st order, use letters close in alphabet
ClCH2CH2OH = A2B2X
F2P(CH3) = AM2X3
NOTE: the order is arbitrary, e.g. A2M3X aboveand does not imply ANY additional information
p. 178
Different nuclei by DEFINITION satisfythe 1st order requirement!
CHEMICAL EQUIVALENCE: use same letter
CH3CH2Br = A2X3 the two A atoms are the sameand the three X atoms are sameas each other, BUT A = X
C C
HA
FX
FX
HA
In this example, the two F areCHEMICALLY EQUIVALENT
and the two H areCHEMICALLY EQUIVALENT
BUT…
p. 179
C C
HA
FX
FX
HA
The two H’s (or the two F’s) are
NOT MAGNETICALLY EQUIVALENT
C C
HA
FX
FX
HA
3J(HAFX)
2J(HAFX)
Because:the coupling constant of H is not the same TO EACH F
3J = 2J
The F’s are MAGNETICALLY INEQUIVALENT
p. 179
same argument for the H’s
C C
HA'
FX'
FX
HA Fx does not have same Jto HA as to HA’2J
3J
so we call this an AA’XX’ system
p. 179
TO TEST MAGNETIC EQUIVALENCE:
F
F'H'
H
A) hydrogens:
Pick a different COUPLEDSPIN ACTIVE NUCLEUS
compare J of this atom toeach of the atoms you wish to test
JFH is not same as JFH’
so H and H’ are MAGNETICALLY INEQUIVALENT
if you get a “YES” (equivalent), test ALL nuclei !!!
p. 179
XHA HA’
HB’HB
HC
Test: HA and HA’
Pick different spin nucleus, HC
is JCA’ = JCA
YES, so test again!!!
NOW PICK HB: JBA is not same as JBA’
so HA and HA’ are NOT equivalentonce you have one NO, no need to test again
Test HB and HB’ NOT equivalent using HA or HA’
= AA’BB’C system
p. 180
OCH3
COOH
H
H
H
H
Pick one H
are J’s same to both H
NO, so H are not equiv
Likewise pick H and test H again NO
6.8
.8
Chem shifts: ~ 1ppm (300Hz),
J=8, so (/J) >10
so AA’XX’
p. 180
Symmetry and its effects on NMR spectra
What is symmetry?
“A property of a physical system that allows thesystem to remain unchanged by a specific physicalor mathematical transformation, such as translationor rotation”
Or more plainly for a molecule:
“A transformation of a molecule that duplicates thetype and spatial arrangement of ALL atoms inspace”
p. 181
Transformations that meet this criterion are called‘symmetry operations’ and the features of the moleculethat they operate on are known as ‘symmetry elements’
Symmetry Element Symmetry Operation Symbol
1. Identity ‘do nothing’ E
2. Inversion centre invert the molecule ithrough a point
3. Mirror plane reflect through a plane σ
4. Proper axis of rotation rotate about an axis by (360/n)° Cn
5. Improper axis of rotation rotate about an axis by (360/n)° Snand reflect in a plane perpendicular to the axis
All molecules possess the symmetry element E and many we commonly encounter also possess rotation axes (Cn) or mirror planes (σ); fewer contain inversion centres (i) or improper rotation axes (Sn).
p. 181
Example 1BF3 is a trigonal planar molecule with several symmetry elements
including the BF3 plane itself (not shown):
The C3 axis makes all 3 of the 19F nuclei equivalent and only one
signal is observed in the 19F NMR.
F
F
C3
C2
C2
C2 F
p. 182
Example 2Naphthalene is a flat aromatic molecule with rotation axes (shown),
mirror planes that include the C2 axes (not shown for clarity) and an
inversion centre marked with a star (*) :
These symmetry elements result in only 3 unique 13C nuclei and 2
unique types of 1H nuclei.
2
2
2
p. 182
Example 3Water has the familiar bent structure and possesses a C2 axis as shown
and two mirror planes: the plane of the page and one perpendicular to
the page that bisects the H-O-H angle (not shown):
p. 182
Example 4Methane is tetrahedral and has mirror planes (σ), proper (Cn) and
improper (Sn) axes of rotation, but NO inversion centre.
Symmetry Elements:
E, 6 σ, 4 C3, E, 3 σ, C3 E, σ E
3 C2, 3 S4
p. 182
The molecule on the far right is recognizable as a CHIRAL molecule
because it cannot be superimposed on its mirror image.
If the molecule ITSELF had a mirror plane as a symmetry element then
this would not be true. We therefore have a simple test for chirality
based on symmetry elements:
Chiral molecules cannot possess mirror planes, inversion
centres or improper axes of rotation as symmetry
elements!
p. 182
However, chiral molecules CAN possess proper axes of
rotation (Cn) as illustrated below:
2
2
2
2
C2
S,SR,R
H
CO2H
2
R,S
p. 183
How does symmetry affect NMR spectra?
Basically, the presence of a symmetry element in a molecule reduces
the number of resonances because carrying out the symmetry
operations renders some atoms equivalent to others.
Pay close attention to ANY reduction in the number of observed
resonances. Symmetry can be an enormous help in solving structures
if you understand the relationship between the symmetry elements
present and the number of resonances.
p. 183
Make symmetry your friend!
Beware of CHIRAL compounds!!!
H
D
H'
C
A
B
H'
H
D
C
A
B
D
H'
H
C
A
B
A-H C-D B-H’ C-H B-D A-H’
so H and H’ are never the same, they have differentchemical shifts, so are an AB or AX and couple to each other, JAX ~ 15Hz, they are called pro-chiralor diastereotopic hydrogens
p. 183
A X
AB ‘quartet’
J > 10
J < 10
C C Cl
HB
HBHA
HA
HA
3J(HBHA)
3J(HBHA)
3J(HBHA) If molecule frozen3J values dependon dihedral anglesand are differentso would be AA’A’’XX’
But at room temp, thereis free rotation, soall J values averageto 7Hz, so A3X2
2
p. 184
Inorganic systems, need to know geometry!
F
PF
FF
F
axial F’s different
from equatorial F’s
If frozen, AX2Y3 but in fact fluxional, so AX5
31P is triplet of quartets sextetLower T
p. 184-5
L
M
L
L
L
L
eq
ax
ax
eq
eq
1
24
5
3
L
M
L
LL
L
eq
ax
ax
eq
eq
1
2
3
4
5
L
M
L
L
L
L
bas
ax1
24
53 bas
basbas
TBP SqPy TBP
ASIDE: Interconversion of 5-coordinate structures Berry Pseudo-rotation
In general, you need to know geometries to predict the spectra
BF3 CH4 PtCl4 PF5 SF6
(squareplanar)
p. 184
PF3(CH3)2 however has three isomers, which all have different spectra
P = q7
F = d7
CH3
PF
FF
CH3
F
PCH3
CH3
F
F
CH3
PF
CH3
F
F
H = dq
P = dt7
F = dd7F = dt7H = ddt
P = dtqq
F = dtqqF = ddqqH = ddtqH = ddtq
p. 185
SPIN DILUTE SYSTEMSPt has one isotope that has I = 1/2 , 195Pt = 34% abundantso ~2/3 Pt has no spin, 1/3 has spin and so couples
PtCl F
F Cltrans
PtCl Cl
F Fcis
19F
66% of the signal does not split as 66% of the Pt is spin-inactive
34% of the signal is split into a doublet as 195Pt is spin-active (I=1/2)
19F
17% 17%66%ratio 1 : 4 : 1
1J(19F195Pt)
p. 186
PtR3P FA
FA Cl
PtR3P Cl
FA FB
2J(trans)2J(cis)
2J(cis)2J(cis)2J(cis) << 2J(trans)
1 4 1 1 4 1
1 4 1
1 1 4 4 1 1
1 2 1 4 8 4 1 2 1
34% of the signalis split by 195Pt
Split by twoequivalent Fnuclei (bothcis to the P)
2J(31P 19Fcis)
1J(31P 195Pt)
31P
1
4
1
34% of the signalis split by 195Pt
31P
Split by one F(trans to P)
1 1 1 1 4 4 4 4 1 1 1 1
1 Split by one F(cis to P)11 1
4
4
Triplet with satellites Doublet of doublets with satellites
1J(31P 195Pt)
2J(31P 19Fcis)
2J(31P 19Ftrans)
3
p. 187-8
ASSIGNMENT 7