homework spectroscopy

Questions and problems of Nuclear Magnetic Resonance

University of Barcelona Gabriel Mario

1. How many signals would you expect in the 1H-NMR spectrum of PF3? and in the 19F-NMR or in the 31P-NMR spectra?

1H-NMR

In the 1H-NMR spectrum of PF3, there is no signals because there are not protons

19F-NMR

In the 19F-NMR spectrum of PF3: 3 identical 19Fs = 1 signal At 1 bond: 19F= 1 31P ! = 2NPIP+1=2*1*1/2+1=2 = Doublet

31P-NMR

In the 31P-NMR spectrum of PF3: 131P = 1 signal At 1 bond: 31P = 3 19Fs ! = 2NFIF+1=2*3*1/2+1=4 = Quadruplet

2. Which of the following nuclei ( ! ! ! ! !" ! !" ! !!" !! ) are active

in NMR? Nuclei I Abundance % Active in NMR

!!" 0 99.96 No !! 1 0.015 No !"!"# " 33.8 Si !"!" 0 94.93 No ! !!

!" 0 98.9 No !!

!" 3 19.9 Si

3. What structural information could you get from the following findings?

a) The 1H-NMR spectrum of a compound shows a triplet centered at =

4.00 ppm.

If I=1/2 and ! = 3 If I=1 and ! = 3 = N I ! 1

= ! !! ! !

! ! ! 1

! ! ! ! ! !

! ! !!

= N I+ 1

= !! ! 12 !

= 3 12 1

= !1 The triplet shown in the spectrum of

1H-NMR with I=1/2, describe a coupling with 2 atoms nuclei

The triplet shown in the spectrum of 1H-NMR with I=1, describe a coupling with 1 atoms nuclei

PF F

F

P19F 19F

19F

31PF F

F



b) For a phosphorous compound (H3PO3), the 1H-NMR and the 31P-NMR

spectra show a doublet with identical coupling constant.

Phosphorous compound (H3PO3)

For 1H-NMR and 31P-NMR; I=1/2 and ! = 2

! ! ! ! I+ 1

= 12

= !2 1

2 (1 ! )

! = !

For 1H-NMR and 31P-NMR, there is only coupling with 1 atom (H-P and P-H), because the Oxygen

cannot be detected.

c) The 1H-NMR spectrum of a compound whose formula is PH4F shows only doublet, but in its 19F-NMR only a singlet was detected.

PH4F For

1H-NMR; I=1/2 and ! = 2

For 19F-NMR; I=1/2 and ! = 2

= N I+ 1

= 12

= ! 1

2 !1 2)

= 1

= N I+ 1

= 12 !

! ! !! ! !

2 (1 ! !

! ! !!

The doublet shown in the spectrum of 1H-NMR with I=1/2, describe a coupling with 1 atoms nuclei (H-P)

A singlet shown in the spectrum of 19F-NMR with I=1/2, describe a coupling with 0 atoms nuclei (no coupling)

PO O

O

H

phosphonic acid

H H

PH4+ F-

phosphonium fluoride



4. For the following compounds:

A

B

C

D

E

F

a) Indicate the multiplicity of the signals due to the protons in blue.

EXERCISE RESULTS A

At 3 bond: 1Ha= 6 1Hb ! = 2NH

bI1H+1=2*3*1/2+1= 4 = Quadruplet

B

4JH,N = 0: No coupling

C


bI1H+1=2*2*1/2+1= 3 = Triplet

C C

H

CH3

H

H3C C CH3

O H3C CH2 OCH3

H3C

C C

COOCH3

CH3

H NO2

H

CH2 OC(O)CH3

C C

Ha

C

Ha

C

Hb

Hb

Hb Hb

Hb

Hb

123

C

C

O

C

Ha

Ha

Ha

123

C C OCH3

Ha

Ha

Ha

Hb

Hb

1 23



D


bI1H+1=2*3*1/2+1= 4 = Quadruplet

E


bIHb+1=2*1*1/2+1= 2 =

Doublet

F

4JH,N = 0: No coupling

b) Try to match the protons marked in blue with the appropriate chemical shift that appears in the table.

a b c d e f (in ppm) 1.20 2.40 5.30 8.21 5.01 6.72 Exercise C B A E F D

C

C C

COOCH3

CH3

Ha

Hb

Hb

Hb1

2

3

C

CC

N

Ha

OO

Hb

12

2

3

3

C O

Ha

Ha

C

O

CH3



5. Estimate the multiplicity and the chemical shift of the protons that appear in blue or red in the following compounds:

A

B

C

C

D

F

EXERCISE RESULTS A


bI1H+1=2*3*1/2+1= 4 = Quadruplet At 3 bond: 1Ha= 11Hc ! = 2NH

cI1H+1=2*1*1/2+1= 2 = Doublet ! ! !25 ! !! ! ! ! ! ! ! = ! !!" ! ! !! ! !!" ! !! !! ! !! ! !=

!! !! B


bI1H+1=2*1*1/2+1= 2 = Doublet ! ! ! !!" ! !! ! (! ! ! ! ! ! !!" ! ! !! ! !! ! !! !! ! !!" ! !!

!! !!"

H C C C

H

H H

H H

O

C C

H

H H

O

C C C

H

H

O H

H

H

H C C C

H

H H

H O

O

C

H

H

H

C C C C

H

H H

H

O C

C

H

C

H

H H

H

H

O

HO

O

H

C C

H

H

H

H

OH

Hb C C C

Hb

Hb Ha

Ha Hc

O123

2 3

C C

Ha

Ha Hb

O1

2 3



C

4JH,N = 0: No coupling ! ! ! !!" ! !! ! ! ! ! ! ! ! ! !!" ! ! !!1!! ! !! !! ! !!" ! !!

!! !!"

D


bI1H+1=2*3*1/2+1= 4 = Quadruplet ! ! ! !25 ! !! ! ! ! ! ! ! ! ! !!" ! ! !! ! !!" ! !! !! ! !! ! !!

!! !!

E


bI1H+1=2*2*1/2+1= 3 = Triplet At 3 bond: 1Hb= 2 1Ha ! = 2NH

aI1H+1=2*2*1/2+1= 3 = Triplet ! ! ! !!" ! !! ! ! ! !) = 0,23 + (1,7) + (0,47) !

! ! !! ! !! !! !! ! ! ! !!" ! ! !! ! !! ! !! !! ! !!" ! !!

!! ! !! ! !! !! ,1 F


bI1H+1=2*2*1/2+1= 3 = Triplet At 3 bond: 1Hb= 2 1Ha ! = 2NH

aI1H+1=2*2*1/2+1= 3 = Triplet At 3 bond: 1Hb= 1 1Hc ! = 2NH

cI1H+1=2*1*1/2+1= 2 = Doublet ! ! ! .!" ! !! ! !!) ! ! ! !!" ! ! !!1,85) + (! !!" ! +

(2,56) !! !! !!! ! ! ! !!" ! ! !! ! !!" ) ! !!0!!" ! !!

!! ! !!" ! !! !! !!!

C C C

Ha

Ha

O H

H

H

Hb C C C

Hb

Hb Ha

Ha O

O

C

H

H

H

123

C C C C

Ha

Ha Hb

Hb

O C

C

H

C

H

H H

H

H

O

HO

O

H

12 3

C C

Ha

Ha

Hb

Hb

O

12

3

Hc2 1



6. For each one of the following compounds: A

B

C

D

a) Indicate the number of signals expected in the 1H-NMR spectra together with their relative intensities.

b) Estimate their multiplicity and c) Calculate the chemical shifts and

EXERCISE RESULTS

A

In the 1H-NMR spectrum: 2 different 1Hs = 2 signals

Relative intensity = 2:6 = 1:3 At 3 bond: 1Ha= 3 1Hb ! = 2NH

bI1H+1=2*3*1/2+1= 4 = Quadruplet = ! !!" ! !! !"# ! !! !"#$% ! !! !"# ! ! ! ! ,!" ! ! !! ! ! !! !! ! ! !!" ! !! !! ! !!! ! !! !! !! !

B


Relative intensity = 1:3:3 At 3 bond: 1Ha= 3 1Hb ! = 2NH

bI1H+1=2*3*1/2+1= 4 = Quadruplet At 3 bond: 1Ha= 1 19F ! = 2NFI

1H+1=2*1*1/2+1= 2 = Doublet

! ! ! !!" ! !! !"# ! !! !"#$% ! !! !"# ! ! ! 5,25 + (-0,26) + (-1,02) + (0,44) = 4,41

H3C

C C

H

CH3

H

H3C

C C

CH3

F

H

H3C

C C

H

CHO

H

H3C

C C

CHO

H

H

C

C C

Ha

CH3

Ha

Hb

Hb

Hb1

2

3

C

C C

C

F

Ha

Hb

Hb

Hb1

22

3

Hc

Hc

Hc



C


Relative intensity = 1:1:1:3 At 3 bond: 1Ha= 3 1Hc ! = 2NH

cI1H+1=2*3*1/2+1= 4 = Quadruplet At 3 bond: 1Ha= 1 1Hd ! = 2NH

bI1H+1=2*0*1/2+1= 2 = Singlet (Aldehyde) At 3 bond: 1Hb= 1 1Hd ! = 2NH

bI1H+1=2*0*1/2+1= 2 = Singlet (Aldehyde) ! ! ! !!" ! !! !"# ! !! !"#$% ! !! !"# ! ! ! ! ! ! !!" ! ! !! ! ! !! !! ! !!" ! !! !! ! !!! ! !! !! !!" ! ! ! ! ! !!" ! ! !! ! ! !! !! ! ! !!" ! !! !! ! !!" ! !!

!! !!" D


Relative intensity = 1:1:1:3 At 3 bond: 1Ha= 3 1Hc ! = 2NH

cI1H+1=2*3*1/2+1= 4 = Quadruplet At 3 bond: 1Ha= 1 1Hb ! = 2NH

bI1H+1=2*1*1/2+1= 2 = Doublet At 3 bond: 1Hb= 1 1Hd ! = 2NH

dI1H+1=2*0*1/2+1= 2 = Singlet (Aldehyde) ! ! ! !!" ! !! !"# ! !! !"#$% ! !! !"# ! ! ! ! ! ! !!" ! ! !! ! !!" ! + (0! !! !! ! !!! ! !! 6,!" ! ! ! ! ! !!" ! !! ! ! !!" ! !! !(0) + (! !!" ! !!

!! !!"

C

C C

Hb

C

Ha

Hc

Hc

Hc

1

3

2

Hd

O

1

2

3

C

C C

C

Hb

Ha

Hc

Hc

Hc

1

2

3

2 3

Hd

O

2

1



d) Is 3JH,H of A greater or smaller than 3JH,H of D? Explain.

The 3JH,H of exercise D is greater th an A, because the presence of a double bond in the path from Ha to Hb.

7. Among the compounds shown below select those, which may exhibit one or more broad bands in the 1H-NMR spectrum and explain why.

A

B

C

D

2 Broad Bands 3 Broad Bands 1 Broad Bands 1 Broad Bands All of the compounds exhibit one or more broad bands in the 1H-NMR spectrum, because they can undergo to proton exchange, generated by the concentration, kinetics of NMR or the presence of the solvent.

H2N CH

CH3

C

O

OH

N

NH

CH2

HC C

OH

O

H2N

SH

N

H

homework spectroscopy

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