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Interpretation of more complex spectraPowerPoint Presentation

Interpretation of more complex spectra

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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.

Predict the different chemical shifts1H 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

Chemical equivalence different chemical shifts

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.

J different chemical shifts12 = 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

Chiral Shift Reagents different chemical shifts

Are the hydrogens on the CH different chemical shifts2 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

Non-First Order different chemical shifts1H NMR Spectra

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

AX case AB case A2 case

?

Analysis of AB Case different chemical shifts

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

J different chemical shiftsAB = 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

Top and bottom different chemical shifts

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 different chemical shifts

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

Raccoon different chemical shifts

B= 9 Hz

A = 35.5; 35.5 Hz

JAB= 7.9

JAA = ?

A- B = 35.5-9 = 26.9

Calculated A different chemical shifts2B spectra for different values of JAB

ABX spectrum different chemical shifts

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?

AB different chemical shifts

X

ignore step 6 different chemical shifts

J different chemical shiftsAB 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

J different chemical shiftsAB 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

J different chemical shiftsAB = 15.5

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

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

J different chemical shiftsAB = 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

J different chemical shiftsAB = 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

J different chemical shiftsAB = 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

J different chemical shiftsAB = 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

(7,4,3,1) = different chemical shiftsab+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

- ( different chemical shiftsA - 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;

All values in Hz different chemical shifts

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

Some special cases of ABX like systems different chemical shifts

Deceptively simple spectra:

A = 452 Hz JAB = 8 Hz

B = 452.5 Hz JAX = 0 HZ

X = 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

Deceptively complex spectra: Virtual coupling similar, a deceptively simple spectrum can be obtained

SupposeA = 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

2,6-dimethylquinone and 2,4-dimethylquinone similar, a deceptively simple spectrum can be obtained

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

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