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Interpretation of more complex spectra. Diasteriotopic nuclei: chemically different nuclei with different chemical shifts Two nuclei or groups attached to the same atom that are present in environments that are not related to each other by an axis of symmetry.

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slide5

Diasteriotopic nuclei: chemically different nuclei with different chemical shifts

Two nuclei or groups attached to the same atom that are present in environments that are not related to each other by an axis of symmetry.

Operational definition: If sequentially replacing the nuclei in question with an isotopic nucleus leads to diasteriomers, the nuclei in question are diasteriotopic.

slide6

Predict the 1H spectrum of CH2F2 spin of F is = ½

Next

Predict the 1H spectrum of

  • H = 2 ppm
  • F = 99 ppm

JHH = 5 Hz

JHF = 40, 10 Hz

JFF = 50 Hz

slide8

Chemical equivalence

Two nuclei related by an element of symmetry including a plane of symmetry in an achiral environment are chemically identical.

Magnetic equivalence

Two chemically identical nuclei may not necessarily be magnetically equivalent. Two chemically equivalent nuclei will have the same chemical shift but may become magnetically non-equivalent if they are coupled to the same nucleus with different coupling constants.

slide9

J12 = J34 = 5.1 Hz

J13 = J24 = 1.3

J14 = 1.95

J15 = J46 = -0.42

J16 = J45 = -0.18

J23 = 1.95

J25 = J36 = 1.4

J26 = J35 =1.4

J5,6 = 0.1

H1 = H4 = 6.22 ppm

H2 = H3 = 6.53

H5 = H6 = 5.85

slide10

Chemical shift reagents

Europium dipivaloylmethane Eu(dpm)

La(fod)

slide17

Are the hydrogens on the CH2 group equivalent?

CH3CH2OCH(CH3)2

Always equivalent?

Enantiotopic nuclei:

Enantiotopic nuclei: two nuclei that are related to each other by a plane of symmetry but not an axis of symmetry.

These nuclei have the same chemical shift in an achiral environment but have different chemical shifts in a chiral environment

Operational definition: If replacing one hydrogen by deuterium leads to a new asymmetric carbon atom, the two hydrogens are enantiotopic

slide18

Non-First Order 1H NMR Spectra

Cases where  (chemical shift) < 10 J (coupling constant)

AX case  AB case  A2 case

?

slide21

Analysis of AB Case

C

three unknowns: A, B, JAB

JAB

JAB

4 3 2 1

 = [(1- 4)(2 - 3)]1/2 = A - B

½(A + B) = C

C = (1+ 2+ 3+ 4)/4

slide22

JAB = 247.5-231.5 = 16; JAB = 211.5-195.7= 15.8; JAB (avg)= 15.9 Hz

2C = 2(1+ 2+ 3+ 4)/4; 2C = (247.5+231.5+211.5+195.7)/2 =443.1 = A+ B

A - B = [(1- 4)(2 - 3)]1/2 =[(247.5-195.7)(231.5-195.7)]1/2 = 32.2

A+ B = 443.1; A - B = 32.2; A = 237.6; B = 205.5; ∆ = 32.2

slide23

A2X A2B Case A3

A3

X

A2

A2B

slide24

Top and bottom

9 transitions can be observed, 1 weak

Numbering the lines from the B nucleus

A2B

8 7 6 5 4 3 2 1

slide25

Top and bottom

9 transitions can be observed, 1 weak

Numbering the lines from the B nucleus

B = 3 = 9 Hz

A = 1/2( 5 + 8) ½(31.9+39.1) = 35.5

JAB= 1/3( 1 - 4 + 6 - 8) = 1/3(0-15.6+32.5-40.7) = 7.9

How do you know that you have analyzed a spectrum correctly?

8 7 6 5 4 3 2 1

40.7 39.1 32.5 31.9 15.6 9.0 6.7 0

slide26

Raccoon

B= 9 Hz

A = 35.5; 35.5 Hz

JAB= 7.9

JAA = ?

A- B = 35.5-9 = 26.9

slide28

ABX spectrum

What do you need to know to analyze an ABX spectrum?

You need to know:

3 chemical shifts: A, B, X

3 coupling constants: JAB, JAX, JBX

What does an ABX spectrum look like?

slide29

AB

X

slide31

JAB appears four times in the AB portion of the spectrum

8-6 = 98.9-83.7 = 15.2

5-2 = 80.4-64.7 = 15.7

3-1 =72-56.4 = 15.6 JAB = 15.45

7-4 = 94.6-79.3 = 15.3

slide32

JAB appears four times in the AB portion of the spectrum

8-6 = 98.9-83.7 = 15.2

5-2 = 80.4-64.7 = 15.7

3-1 =72-56.4 = 15.6 JAB = 15.45

7-4 = 94.6-79.3 = 15.3

slide33

JAB = 15.5

The chemical shift of X is the average of all X lines

X =(9+10+11+12)/4 = 183.9

slide34

JAB = 15.5

X = 183.9

The average of all AB lines is equal to ½(A + B)

(A + B) = 2*(1+2+3+4+5+6+7+8)/8 = 157.5

slide35

JAB = 15.5;

X = 183.9

(A + B) = 157.5

6

5

4

8

2

8

½ (JAX+JBX) = the separation of the centers of the two AB quartets

(JAX+JBX) = 2[(8+6+5+2)/4-(7+4+3+1)/4]= 12.7

slide36

JAB = 15.5

X = 183.9

5

6

4

(A + B) = 157.5

(JAX+JBX) = 12.7

8

2

8

2D+ = separation of the 1st and 3rd lines of first AB quartet;

2D- = separation of the 1st and 3rd lines of second AB quartet;

Chose 2D+ as the larger value

D+ = ½(4-1) =11.45; ½(7-3)= 11.3; D+av = 11.4

D- = ½(6-2) =9.2; ½(5-8) = 9.5 D-av = 9.35

slide37

JAB = 15.45

X = 183.9

6

4

(A + B) = 157.5

(JAX+JBX) = 12.7

D+ = 11.4

D- = 9.35

8

2

8

2M = (4D+2 –JAB2) ½ ; 2N = (4D-2 –JAB2) ½

M = ½(4*11.42 – 15.52) ½ = 8.36

N = ½(4*9.352 – 15.52) ½ = 5.23

1. A - B = M+N = 13.6; ½(JAX –JBX) = M-N = 3.13

2. A - B = M-N = 3.13; ½(JAX –JBX) = M+N= 13.6

slide38

(7,4,3,1) = ab+quartet because D+ = 11.3; ab+ centered at 75.6

(8,6,5,2) = ab-quartet because D- = 9.5; ab- centered at 81.9

Assign (JAX+JBX) a + sign if ab+ is centered at a higher frequency

than ab- or a – sign if the reverse is true

slide39

(A - B) = 13.6; ½(JAX-JBX) = 3.13; (JAX –JBX) = 6.26

  • (A + B) = 157.5 (JAX+JBX) = -12.7
  • A = 85.65 JAX = -3.2;
  • B = 71.85 JBX = -9.5;
  • 2. (A - B) = 3.13; ½(JAX-JBX) = 13.6; (JAX –JBX) = 27.2
  • (A + B) = 157.5 (JAX+JBX) = -12.7
  • A = 80.3 JAX = 7.25;
  • B = 77.2 JBX = -19.95;
slide40

All values in Hz

72.2

JAB =J12= 15.45

X = 3 =183.9

Solution 1

A = 1 = 85.65; JAX = J13 = -3.2

B = 2 = 71.85; JBX = J23 = -9.5

Solution 2

A = 1 = 80.3; JAX = J13 = 7.25

B = 2 = 77.2 ; JBX = J23 = -19.95

slide41

Some special cases of ABX like systems

Deceptively simple spectra:

A = 452 Hz JAB = 8 Hz

B = 452.5 Hz JAX = 0 HZ

X = 473 Hz JBX = 3 HZ

slide42

When two chemical shifts are fortuitously very similar, an ABX case can give a deceptively simple spectrum.

Only by simulating the spectrum can you convince yourself the the structure is more complex than the nmr suggests

slide43

Whenever the chemical shifts of the A and B nuclei are very similar, a deceptively simple spectrum can be obtained

100 MHz; 100, 101, 700, 8, 6 10 Hz

slide44

Deceptively complex spectra: Virtual coupling

SupposeA = 100 Hz; B = 104 Hz; X = 200 Hz

and JAX = 0 Hz; JBX = 8; JAB = 6 Hz;

Predict the first order spectrum for X

This is an example of an ABX system

slide45

2,6-dimethylquinone and 2,4-dimethylquinone

1 = 4 = 6

2 = 3 = 5 = 6 = 2

J12 = J13= J45=J46=6

J14 = 1;

J15 = J16 = J24=J34 =2

J23 = J56 = -10

J25=J26= J35=J36 =0