Interpretation of more complex spectra. Diasteriotopic nuclei: chemically different nuclei with...

Interpretation of more complex spectra

Diasteriotopic nuclei: chemically different nuclei with different chemical shiftsTwo 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.

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

Next

Predict the 1H spectrum of

F

F

H

H

H = 2 ppm

F = 99 ppm

JHH = 5 Hz

JHF = 40, 10 Hz

JFF = 50 Hz

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

Magnetic equivalenceTwo 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.

H1

H2

H3

H4

H5

H6

H1 = H4 = 6.22 ppm

H2 = H3 = 6.53

H5 = H6 = 5.85

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

Chemical shift reagents

Europium dipivaloylmethane Eu(dpm)

La(fod)

O HO

O HO

C3F7

Chiral Shift Reagents

O

O

Eu

3

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

Are the hydrogens on the CH2 group equivalent?

CH3CH2OCH(CH3)2

Always equivalent?

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

Non-First Order 1H NMR Spectra

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

AX case AB case A2 case

?

Analysis of AB Case

JAB JAB

4 3 2 1

= [(1- 4)(2 - 3)]1/2 = A - B

½(A + B) = C

C = (1+ 2+ 3+ 4)/4

Cthree unknowns: A, B, JAB

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

A2X A2B Case A3

A2 X

A2B

A3

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

Top and bottom 9 transitions can be observed, 1 weak

Numbering the lines from the B nucleus

8 7 6 5 4 3 2 1

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?

40.7 39.1 32.5 31.9 15.6 9.0 6.7 0

Raccoon

B = 9 Hz

A = 35.5; 35.5 Hz

JAB = 7.9

JAA = ?

A- B = 35.5-9 = 26.9

Calculated A2B spectra for different values of JAB

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?

X

AB

ignore step 6

56.4

12

3

45

6

78

9

10

1112

64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

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

56.4

12

3

45

6

78

9

10

1112

64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

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

56.4

12

3

45

6

78

9

10

1112

64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

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

X = (9+10+11+12)/4 = 183.9

JAB = 15.5

56.4

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3

45

6

78

9

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1112

64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

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

JAB = 15.5;

X = 183.9

(A + B) = 157.5

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

56.4

1

3

5

7

9

10

1112

64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

828

64

5

JAB = 15.5

X = 183.9

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

56.4

1

3

5

7

9

10

1112

64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

828

4 6

(A + B) = 157.5

(JAX+JBX) = 12.7

5

JAB = 15.45

X = 183.9

2M = (4D+2 –JAB

2) ½ ; 2N = (4D-2 –JAB

2) 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

56.4

1

3

5

7

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64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

828

4 6

(A + B) = 157.5

(JAX+JBX) = 12.7

D+ = 11.4

D- = 9.35

(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 frequencythan ab- or a – sign if the reverse is true

56.4

1

3

5

7

9

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64.77279.380.4

83.794.698.9179.6

181.1185.1

189.7

1. (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;

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

All values in Hz

72.2

Some special cases of ABX like systems

Deceptively simple spectra:

A = 452 Hz JAB = 8 HzB = 452.5 Hz JAX = 0 HZX = 473 Hz JBX = 3 HZ

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

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

Ha Hb Hx

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

Deceptively complex spectra: Virtual coupling

This is an example of an ABX system

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

CH2X

CH2X

O

O

H1

H4

5,6

2,3