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