Chapter 3: Coupling Constants

Transcript Chapter 3: Coupling Constants

Coupling Constants (J)

• • Coupling constants are a very important and useful feature of an NMR spectrum Importantly, coupling constants identifies pairs of nuclei that are chemically bonded to each other

1 H 1 H

• Multiplicity identifies the number of protons (or other nuclei) that are chemical bonded to the other nuclei • The magnitude of the coupling constants identifies the coupling partner, and provides information on dihedral angles, hydrogen bonds, the number of intervening bonds, and the type of coupled nuclei ( 1 H, 13 C, 15 N, 19 F, etc.)

Coupling Constants (J)

-

spin-spin coupling, scalar coupling or J-coupling

Random tumbling of molecules averages through-space effect of nuclear magnets to zero

B o

b a b 1

B o

b a a

random tumbling leads to no interaction between the spin-states despite the small magnetic fields

a b 2

Coupling Constants (J)

-

spin-spin coupling, scalar coupling or J-coupling

Instead, nuclear spin state is communicated through bonding electrons

Energy of electron spin states are degenerate in absence of nuclear spin With a nuclear spin, the electron spin opposite to nuclear spin is lower energy Number of possible energy states of nuclear electron spin pairs increases with the number of nuclear spins

Spin state is “sensed” through bonds resulting in higher or lower energy

aligned or anti-aligned with magnetic field

Coupling Constants

Energy level of a nuclei are affected by covalently-bonded neighbors spin-states

Mixing of Spin Systems One and Two

b b ab

Spin System One

b a a b a a b a b a

Spin System Two

Coupling Constants

Mixing of energy levels results in additional transitions – peaks are split

+J/4

b b

S J (Hz) J (Hz)

-J/4 a b b a

+J/4

a a

I S

Spin-States of covalently-bonded nuclei want to be aligned

The magnitude of the separation is called

coupling constant

(

) and has units of Hz

Coupling Constants

• • • •

Through-bond interaction that results in the splitting of a single peak into multiple peaks of various intensities

  Spacing in

hertz

(hz) between the peaks is a

constant

Independent of magnetic field strength

Multiple coupling interactions may exist

 Increase complexity of splitting pattern Coupling can range from one-bond to five-bond   One, two and three bond coupling are most common Longer range coupling usually occur through aromatic systems Coupling can be between heteronuclear and homonuclear spin pairs  Both nuclei need to be NMR active i.e.

12 C does not cause splitting

1 H 1 H 1 H 1 H 13 C

one-bond three-bond four-bond

1 H 1 H

five-bond

Coupling Constants

•  

Splitting pattern depends on the number of equivalent atoms bonded to the nuclei

Determines the number of possible spin-pair combinations and energy levels Each peak intensity in the splitting pattern is determined by the number of spin pairs of equivalent energy

Coupling Constants

•   

Splitting pattern follows Pascal’s triangle

Number of peaks and relative peak intensity determined by the number of attached nuclei Peak separation determined by coupling constant (J) Negative coupling  reverse relative energy levels

1 1 1 1 2 1 1 3 3 1 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

3 attached nuclei

Pascal’s triangle

3 3 1 J J J 1

Quartet

Relative Intensity

Coupling Constants

Common NMR Splitting Patterns

singlet doublet triplet quartet pentet 1:1 1:2:1 1:3:3:1 1:4:6:4:1

Coupling Rules:

1. equivalent nuclei do not interact 2. coupling constants decreases with separation ( typically # 3 bonds) 3. multiplicity given by number of attached equivalent protons (n+1) 4. multiple spin systems 5.

 multiplicity  (n a +1)(n b +1) Relative peak heights/area follows Pascal’s triangle 6. Coupling constant are independent of applied field strength 7. Coupling constants can be negative IMPORTANT: Coupling constant pattern allow for the identification of bonded nuclei.

Coupling Constants

Common NMR Splitting Patterns

Coupling Constants

•   

Coupling only occurs between non-equivalent nuclei

Chemical shift equivalence Magnetic equivalence For no coupling to occur, nuclei has to be BOTH chemical shift and magnetic equivalent The CH 3 protons (H 1 , H 2 , H 3 ) are in identical environments, are equivalent, and are not coupled to one another The H a and H b protons are in different environments (proximity to Cl), are not equivalent, and are coupled

Coupling Constants

•

Rules for Chemical Shift Equivalence:

Nuclei are interchangeable by symmetry operation i.

Rotation about symmetric axis (C n ) ii.

Inversion at a center of symmetry (i) iii. reflection at a plane of symmetry ( s ) iv. Higher orders of rotation about an axis followed by reflection in a plane normal to this axis (S n ) v.

Symmetry element (axis, center or plane) must be symmetry element for entire molecule

Examples of Chemical Shift Equivalent Nuclei

Ha Ha Ha Ha Ha C Hb C C Ha Ha Ha Hb Ha Ha Ha 180 o Ha Symmetry planes

Coupling Constants

•

Rules for Chemical Shift Equivalence:

Nuclei are interchangeable by a rapid process i.

> once in about 10 -3 seconds ii.

Rotation about a bond, interconversion of ring pucker, etc.

Examples of Chemical Shift Equivalent Nuclei

Ha Ha Rapid exchange Ha Ha Rapid exchange

Coupling Constants

• •

Magnetic Equivalence: Nuclei must first be chemical shift equivalent

Must couple equally to each nucleus in every other set of chemically equivalent nuclei i.

need to examine geometrical relationships ii.

the bond distance and angles from each nucleus to another chemical set must be identical iii. Nuclei can be interchanged through a reflection plane passing through the nuclei from the other chemical set and a perpendicular to a line joining the chemical shift equivalent nuclei Chemical shift equivalent, but not magnetic equivalent

Examples of Non-magnetically equivalent nuclei

Hb Hb' 3 3 J ab Ha Ha' ≠ 3 J a’b J ab’ ≠ 3 J a’b’ Cl Cl Fa Ha C C Fa' Fa Ha' Fa' C C Ha Fa Ha' Ha' C C Ha Fa' 3 3 J HaFa ≠ 3 J Ha’Fa J HaFa’ ≠ 3 J Ha’Fa’ 3 3 J HaHc ≠ 3 J HaHc’ J HbHc ≠ 3 J HbHc’ 3 3 J HaHc ≠ 3 J HbHc J HaHc’ ≠ 3 J HbHc’

Coupling Constants

• •

Magnetic Equivalence:

Non-magnetically equivalent nuclei may lead to second order effects and very complex splitting patterns

Second order effects will be discussed later i.

Due to small chemical shift differences between coupled nuclei ( Dn ~ J) http://www.chem.wisc.edu/areas/reich/chem605/index.htm

Coupling Constants

Multiple Spin Systems

multiplicity  (n a +1)(n b +1) What is the splitting pattern for CH 2 ?

Hb Cl H Ha C C C Cl H Ha 3 J Hb = 6 Hz Ha

1 1 1 1 2 1 1 3 3 1 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 Down-field resonance split into quartet

3 J Hb = 6 Hz 3 J Ha = 7 Hz Coupling to H b splits the CH 2 resonance into a doublet separated by 6 Hz

up-field resonance split into quartet

Coupling to H a splits each doublet into a quartet separated by 7 Hz

Coupling Constants

3 3 J Hb J Ha = 7 Hz = 7 Hz 1.5

1.0

0.5

0.0

3.0

2.5

2.0

5.0

4.5

4.0

3.5

Looks like a pentet!

Intensities don’t follow Pascal’s triangle (1 4 6 4 1)

2.00

3 J Hb = 6 Hz 3 J Ha = 3 Hz 0.5

0.0

1.5

1.0

2.5

2.0

3.5

3.0

Looks like a sextet!

Intensities don’t follow Pascal’s triangle (1 5 10 10 5 1)

2.00

Occurs because of overlap of peaks within the splitting pattern

Coupling Constants

Coupling Constants Provide Connectivity Information

– chemical shifts identify what functional groups are present Hb Cl C H C Cl H Ha C Ha Ha

NMR Peaks for coupled nuclei share the same coupling constants

Integral:

CH 6 Hz 6 Hz

7 Hz CH 2 6 Hz 6 Hz 7 Hz

CH 3 7 Hz 7 Hz

Coupling Constants

Deconvoluting a spin system

– determining the J-values – determining the multiplicities present

J coupling analysis:

ii.

Is the pattern symmetric about the center?

Assign integral intensity to each line, outer lines assigned to 1 iii. Are the intensities symmetric about the center?

iv. Add up the assigned intensities – Sum must be 2 n , n = number of nuclei – Ex: sum = 16, n = 4 v.

Separation of outer most lines is a coupling constant – Relative intensity determines the number of coupled nuclei – – Ex: intensity ratio: 1:2, 2 coupled nuclei 1 st splitting pattern is a triplet (1:2:1) vi.

vii.

Draw the first coupling pattern Account for

all

the peaks in the spin pattern by repeatedly matching the 1 st splitting pattern viii.

Smallest coupling constant has been assigned

Coupling Constants

Deconvoluting a spin system

– determining the J-values – determining the multiplicities present

J coupling analysis:

ix. Coupling pattern is reduced to the center lines of the 1 st splitting pattern.

Repeat process – Ex: sum = 8, n = 3 – – Ex: intensity ratio: 1:1, 1 coupled nuclei 2 nd splitting pattern is a doublet (1:1) xi. Repeat until singlet is generated

Coupling Constants

Demo ACD C+H NMR Viewer software

–

first order coupling constants

Coupling Constants

Description of Spin System

– each unique set of spins is assigned a letter from the alphabet  the total number of nuclei in the set are indicated as a subscript – the relative chemical shift difference is represented by separation in the alphabet sequence   Large chemical shift differences are represented by AX or AMX ( n AX Small chemical shift differences are represented by AB ( n AB < 5J AB ) >> J AX )  Can also have mixed systems: ABX  magnetically in equivalent nuclei are differentiated by a single quote: AA’XX’ or brackets [AX] 2 Hx Ha Cl CH 2 ClCHCl 2 A 2 X system CH 3 CH 2 R A 3 X 2 system CH 3 CH 2 F A 2 M 2 X system Hx' Cl Ha' [AX] 2 or AA’XX’ system AB system

A M X A M X TMS

A M X J(AM) J(AX) J(AX) J(AM) J(MX) J(AM) J(AX) J(MX) J(AX)

J(AM) = 4 Hz J(AX) = 2.5 Hz J(MX) = 6 Hz

Coupling Constants (J) Observed splitting is a result of this electron-nucleus hyperfine interaction

• Coupling is measured in hertz (Hz)   Range from 0.05 Hz to thousands of Hz o o o o Can be positive or negative 1 J C-H 1 J A-X and many other one-bond coupling are positive is negative if g are opposite sign 2 J H-H 3 J H-H in sp 3 CH 2 groups are commonly negative is always positive

reversed

For an AX system, J AX is negative if the energy of the A state is lower when X has the same spin as A (

)

The spin states and transitions are swapped

reversed reversed

Coupling Constants (J) Measure the Relative Sign of Coupling Constants

• Multiple experimental approaches (different NMR pulse sequences) or simulations

E. COSY – two-dimensional NMR experiment

cross peaks identify which chemical shifts are coupled

Coupling Constants (J) Measure the Relative Sign of Coupling Constants

• The cross-peak patterns identifies the coupling constant sign and magnitude

Based on the slopes of the diagonal line drawn through coupling pattern

3 J AX and 3 J BX have the same sign 3 J AB opposite sign of 3 J AX and 3 J BX

Yellow-highlighted regions are expanded

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: Number of bonds  Bond order (single, double triple) 3 J AB 4 J AC 5 J AB 9.4 Hz 1.1 Hz 0.9 Hz  3 J HH Hz 8 Angles between bonds 3 J HH 11.6 & 19.1 Hz 3 J HH 9.1 Hz

trans

3 J HH ~ 17 Hz

cis

3 J HH ~10 Hz

geminal

2 J HH ~2.5 Hz

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: − dihedral angle

Fixed or average conformation

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: − Cyclohexanes dihedral angles

Fixed or average conformation

3 J 3 J aa ee 9-12 Hz or 3 J ea 3-4 Hz 3 J aa >> 3 J ee , 3 J ea Dual Karplus curves for the axial and equatorial protons

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: − Cyclohexanes dihedral angles

examples

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: − Cyclopentanes dihedral angles

Fixed or average conformation

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: Comparison between Cyclohexanes and Cyclopentanes

In chair cyclohexane, only one vicinal coupling can be large (>7 Hz) In cyclopentane, two or three vicinal coupling can be large (>7 Hz)

 Because of range of cyclopentane conformations, vicinal couplings are variable: J cis > J trans and J cis > J trans  Only in rigid cyclopentanes can a stereochemistry be defined: J cis > J trans

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: − Cyclobutanes are flatter than cyclopentanes, so: J cis > J trans unless structure features induce strong puckering of the ring or electronegative substituents are present  Cyclopropanes are rigidly fixed, so J cis > J trans is always true

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: Orientation −

unless structure features induce strong electronegative substituents are present puckering of the ring or

Since methyl groups can freely rotate, the observed coupling is the average of the three individual coupling constants −

Internal hydrogen bonds may lead to constrained conformations and distinct different coupling constants

Coupling Constants (J)

Magnitude of the splitting is dependent on:  Electronegativity of Substituents

3 J H-H coupling constant decreases as electronegativity increases 3 J H-H decreases even more with two electronegative substituents

Coupling Constants (J)

Magnitude of the splitting is dependent on:  Electronegativity of Substituents

3 J H-H coupling constant decreases as electronegativity of substituents increases for cycloalkenes 3 J H-H coupling constant decreases as electronegativity of substituents increases for alkenes

Coupling Constants (J)

Magnitude of the splitting is dependent on:  Ring Size −

Coupling constants decrease as ring size gets smaller

−

Coupling constants also decrease as ring is formed and gets smaller

Coupling Constants (J)

Magnitude of the splitting is dependent on:  Bond order −

Coupling constant decreases as bond order decreases

3 J H-H = 8.65 x (n bond order) + 1.66

 Heterocycles –

Heterocycles have smaller coupling constants compared to hydrocarbons systems

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: Proportional to g a g b 1 J C-H 125 Hz 1 J N-H 95 Hz 

character of bonding orbital –

Increases with increasing s-character in C-H bond

2 J F-H 48.2 Hz

Coupling Constants (J)

•  Magnitude of the splitting is dependent on: Attenuated as the number of bonds increase –

Usually requires conjugated systems (aromatic, allylic, propargylic, allenic) or favorable geometric alignment (W-coupling)

–

Not usually seen over more than 4 to 5 bonds (acetylenes and allenes)

Coupling Constants (J)

Magnitude of the splitting is dependent on:  Geminal protons (H-C-H) fall into two major groups –

Unstrained sp 3 CH 2 protons: 2 J H-H -12 Hz

–

Vinyl sp 2 CH protons: 2 J H-H 2 Hz

Coupling Constants (J)

Magnitude of the splitting is dependent on:  Geminal protons coupling constants are effected by the electronic effects of substituents –

Based on the interaction between the filled and empty orbitals of the CH 2 fragment

Note: opposite trend

Coupling Constants (J)

Magnitude of the splitting is dependent on electronic effects:   In acyclic and unstrained ring systems: 2 J H-H ~ -10 to -13 Hz When CH 2 is substituted with a becomes more negative: 2 J H-H p -acceptor, like carbonyl or cyano coupling ~ -16 to -25 Hz −

Reliable and can help with structure assignments

 Conjugated aryl, alkene and alkyne substituents also makes coupling becomes more negative

Coupling Constants (J)

Magnitude of the splitting is dependent on electronic effects:    In unsaturated carbons: 2 J H-H ~ 2.5 Hz Electronegative substituents (F,O) behave as effect with 2 J H-H close to zero p -acceptors with a negative Electropositive substituents (Si, Li) behave as effect with 2 J H-H p -donors with a negative  Oxygen substituents can behave as a strong s -acceptor and strong (lone pair), both positive effects leading to a large 2 J H-H p -donor or as a strong p acceptor leading to large negative coupling

Coupling Constants (J)

Magnitude of the splitting is dependent on electronic effects:  Summary of effects, do s and p donors s and p acceptors have opposite effects on coupling, as

Coupling Constants (J)

Coupling Constants

• • • •

Weak coupling or first-order approximation

Up to now, we have assumed the frequency difference (chemical shift) between the coupled nuclei is large

Dn >> J Second order effects come into play when this assumption is no longer valid i.

Dn < 5J Second order effects lead to very complex splitting patterns that are difficult, if not impossible to interpret manually and leads to incorrect chemical shifts and coupling constants Interpreting NMR spectra with second-order effects usually requires software

Coupling Constants (J)

Second-Order Effects (Strong Coupling)

– occurs when chemical shift differences is similar in magnitude to coupling constants ( Dn /J < 5)  chemical shifts and coupling constants have similar energy and intermingle  results from mixing of the equivalent ab and ba spin states  

none of the transitions are purely one nuclei described by quantum mechanical wave functions

AB spin system

Coupling Constants (J)

Second-Order Effects (Strong Coupling)

 perturbs peak intensity and position

AB spin system



as chemical shift differences decrease, intensity of outer lines become weaker and internal lines become stronger



the multiplet leans towards each other (“roof” effect) which increases as chemical shift difference decreases

Coupling Constants (J)

Second-Order Effects (Strong Coupling)

 becomes easier to interpret at higher magnetic field strengths Higher field increases Dn /

Coupling Constants (J)

Second-Order Effects (Strong Coupling)

– hierarchy of coupling constants with increasing second-order effects 1. AX and all other first order systems (AX 2 , AMX, A 3 X 2 , etc.) 2. AB i.

ii.

Line intensities start to lean

can be measured, d can be calculated 3. AB 2 i.

ii.

9. Etc.

Extra lines Both

and d have to be calculated 4. ABX, ABX 2 , ABX 3

J AB

can be measured, everything else requires calculation 5. ABC 6.

Both

and d have to be determined from computer simulation AA’XX’ 7.

ii.

Do not become first order even at high magnetic fields Both

and d have to be determined from computer simulation AA’BB’ AA’BB’X

Coupling Constants (J)

Second-Order Effects (Strong Coupling)

– general effect of strong couplings on NMR spectra 1.

Line intensities are no longer integral ratios, no longer follow Pascal’s triangle 2. Line positions are no longer symmetrically related to chemical shift position i.

Multiplet center may no longer be chemical shift (AB and higher) 3. Some or all coupling constants can no longer be obtained from the line separations (ABX and higher) 4. The signs of coupling constants affect the line positions and intensities (ABX and higher) 5. Additional lines over the number predicted by simple coupling rules appear i.

Peaks with intensities of 2 or more are split into individual components More lines then the expected triplet for the boxed CH 2 pair

Coupling Constants (J)

Second-Order Effects (Strong Coupling)

– general effect of strong couplings on NMR spectra 6. Coupling between equivalent nuclei (

J AA

’ position or

J XX’

) affects line count and i.

Second order effects appear even if Dn /

is large for groups of magnetically non-equivalent protons with identical chemical shifts which are coupled i.

Do not get simpler at higher fields 7. Computer analysis becomes mandatory to extract accurate

(ABC and higher) and d values 8. Ultimately spectra become so complex that the only useful information is integration, chemical shift and general appearance.

Coupling Constants (J)

Second-Order Effects

– as the chemical shifts coalesce  intensity of outer lines decrease  inner peaks eventually collapse to singlet  nuclei become chemically and magnetically equivalent

Weaker outer lines may be overlooked and interpreted as a doublet May be misinterpreted as a quartet

AB spin system

Coupling Constants (J)

Second-Order Effects (AB)

– analysis of second-order splitting patterns 

remember:

resonance positions are also perturbed  separation between outer lines and inner lines (a-b, c-d) yields coupling constant 

J AB

= ( n a n b ) = ( n c n d )  true chemical shift is not the doublet centers     n center Dn AB = ½( n b + n c ) = √ ( n a n d ) ( n b n c ) n A = n center + ½ Dn AB n B = n center ½ Dn AB d

Coupling Constants (J)

Second-Order Effects (AB 2 )

– as the chemical shifts coalesce  line intensities no longer follow simple rules  arithmetic average of the line positions no longer give true chemical shifts 

J AB

can still be measured directly from spectrum  none of the line separation correspond to

J AB

 additional lines appear

AB

spin system

Note: splitting of intense lines

Coupling Constants (J)

Second-Order Effects (AB 2 )

– four A lines n 1 – n 4 and four B lines n 5 – calculation of n A , n B , and

J AB

is simple: – n 8 and the very weak combination line n 9 – how to report an AB 2 spin system in a journal manuscript:  report the two chemical shifts as an AB 2 multiplete (m): 2.63, 2.69 (AB 2 m, 3H,

J AB

= 12.2 Hz)

Coupling Constants (J)

Second-Order Effects (AB 2 )

– unique features of second-order splitting pattern for

AB 2

system  Spectrum depends

only

on the ratio Dn /

 lines 1 to 4 correspond to the one proton part (

)  lines 5 to 8 correspond to the two-proton part (

B 2

)  line 5 ( n 5 ) is the most intense line  lines 5 and 6 often do not split up  when Dn /

J << 1

, the spectrum appears nearly symmetrical  lines 1,2, 8 (n 1 , n 2 , n 8 ) become very weak  looks like a distorted triplet with 1:10:1 area ratio 

J AB

and

J BB

do not affect the spectrum

Coupling Constants (J)

Second-Order Effects (ABX)

– most complex spin-system that can still be manually analyzed –

ABX

has a common appearance   

– unsymmetrical 8-line pattern that integrates to 2 protons

AB X

– 4 doublets with the same separation

J AB

with strong leaning – symmetric 6-line pattern that integrates to 1 proton 

– 5 th and 6 th lines are small and not often seen,

apparent doublet of doublet



J AB

and n

are directly measurable from spectrum n

-

center of peaks

AB



J AX, J BX ,

and n

need to be calculated

X

Coupling Constants (J)

Second-Order Effects (ABX)

– Many ABX patterns are sufficiently close to AMX ( n AB >>

J AB

) 

first-order solution has an excellent chance of being correct A

doublet of doublet separation is

J AX

J BX

Center doublets and get

pattern – First, identify the distorted doublet of doublets for both

and

– Remove the splitting (identify the center of each doublet)

which leaves an

pattern – Solve 

pattern as before to get

J AB

, n A , and n B

large errors when J AX and J BX are very different or

AB small compared to J AB

Coupling Constants (J)

Second-Order Effects (ABX)

– Correct analysis of ABX patterns 

Reverse the order of extracting coupling constants to approximate solution

Identify the two

quartets

J ab+

J ab-

– First, identify the two

quartets  

separation between the four pairs of lines are identical tall inner line associated with shorter outer line (leaning)

Coupling Constants (J)

Second-Order Effects (ABX)

– Correct choice of ab quartet – Incorrect choice of ab quartet

Coupling Constants (J)

Second-Order Effects (ABX)

– Solve the two ab quartets  

Treat as normal AB patterns and obtain four chemical shifts (

a+ ,

b+ ,

a ,

b ) Don’t know which half is a and which is b -

two possible solutions

Coupling Constants (J)

Second-Order Effects (ABX)

– Solution 1 and Solution 2 –

depends on the relative sign of J AX and J BX

 

Solution 1: J AX and J BX

same sign

Solution 2: J AX and J BX

different sign Swap the a & b labels

Coupling Constants (J)

Second-Order Effects (ABX)

– Which solution is the correct one?

–

Several criteria can be used:

Magnitude of the couplings – one solution may give dubious (very large or very small) couplings Signs of coupling constants – the signs can sometimes be predicted and rule out a solution  all vicinal

3 J

couplings are positive, 3.

 geminal

2 J

couplings at sp 3 carbons are usually negative   CH X CH A H B –

J AX

and

J BX

have the same sign CH A CH B H X –

J AX

and

J BX

have different signs Analysis of the X-part – the intensities of the lines in the X-part are always different –

most reliable way to identify the correct solution

Two different X patterns depending on relative sign of

J AX

and

J BX

Coupling Constants (J)

Second-Order Effects (ABX)

– Effective of relative sign of J AX and J BX on

pattern

J AX

Solution 1 and

J BX

same sign

J AX

Solution 2 and

J BX

different sign

Coupling Constants (J)

Second-Order Effects (ABX)

–

pattern from

ABX

spin system as a function of changing n AB

Coupling Constants (J)

Second-Order Effects (ABX)

–

pattern from

ABX

spin system as a function of the relative sign and Magnitude of

J AX

and

J BX

J AX

and J

same sign

J AX

and J

different sign

Coupling Constants (J)

Demo ACD C+H NMR Viewer software

–