Structure & Reactivity

Nuclear Magnetic Resonance Spectroscopy

NMR18.  Solutions to Selected Problems.

 

Problem NMR2.1.

a) 10 ppm     b) 64 ppm    c) 158 ppm

 

Problem NMR2.2.

a) 63 ppm    b) 201 ppm    c) 71 ppm

 

Problem NMR3.1.

 

Problem NMR3.2.

a)  pentane, CH3CH2CH2CH2CH3, or hexane, CH3CH2CH2CH2CH2CH3

b)  heptane, CH3CH2CH2CH2CH2CH2CH3, or octane, CH3CH2CH2CH2CH2CH2CH2CH3

c)  nonane, CH3CH2CH2CH2CH2CH2CH2CH2CH3, or decane, CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

 

Problem NMR4.1.

 

Problem NMR4.2.

a) sp3        b) sp2        c)  sp2       d)  sp3       e) sp3         f) sp2   

 

Problem NMR5.1.

 

Problem NMR5.2.

 

Problem NMR6.1.

 

Problem NMR6.2.

a)

b)

c)

 

Problem NMR8.1.

a) H-Csp3        b) H-Csp3           c) H-Csp2           d) H-Csp3          e) H-Csp2          f)  H-Csp2         g)  H-Csp3  

 

Problem NMR8.2.

a) carbon        b) carbon        c) nitrogen        d) oxygen          e) carbon        f)  oxygen         g)  nitrogen

 

Problem NMR8.3.

a) aromatic        b) aromatic        c) alkene        d) alkene        e) alkene        f) aromatic

 

Problem NMR8.4.

a)  H-CAr-C        b) H-CAr-C         c) H-CAr-N         d)  C=O         e)  C=O         f)  H-CAr-N         g) H-CAr-N

 

Problem NMR8.5.

a)  3.4 ppm: Ha        1.5 ppm: Hb, Hc        0.9 ppm: Hd

b)  7.4 ppm: Hd        6.9 ppm: Hb, Hc        3.7 ppm: Ha

c) 10.1 ppm: Ha        7.9 ppm: Hb         7.6 ppm: Hd        7.5 ppm: Hc

d)        7.9 ppm: Hb         7.6 ppm: Hd        7.4 ppm: Hc        2.5 ppm: Ha  

 

Problem NMR8.6.

a)

b)

c)

d)

 

Problem NMR9.1.

a)  0.9 ppm, 6H; 1.0 ppm, 6H (1:1 ratio)

b) 0.9 ppm, 6H; 1.0 ppm, 12H (1:2 ratio)

c)  0.9 ppm, 6H; 1.0 ppm, 16H (3:8 ratio)

d) 0.9 ppm, 6H; 1.0 ppm, 10H (3:5 ratio)

e) 0.9 ppm, 6H; 1.0 ppm, 8H (3:4 ratio)

 

Problem NMR9.2.

a)

b)

c)

d)

 

Problem NMR9.3.

a)  5:2:3

 

Problem NMR10.1.

(There will be some variation in the shift depending on the rest of the structure; this is just an estimate.)

a)  3.5 ppm, quartet, 2H        b)  1.5 ppm, sextet, 2H        c) 2.6 ppm, septet, 1H

d) 2.3 ppm, quintet, 1H        e)  2.2 ppm, quartet, 2H        f)  1.7 ppm, nonet, 1H

 

Problem NMR10.2.

(There will be some variation in the shift depending on the rest of the structure; this is just an estimate.)

a)  7.8 ppm, doublet, 1H        b) 8.4 ppm, singlet, 1H        c) 6.7 ppm, singlet, 1H

d)  7.2 ppm, troplet, 1H        e)  6.9 ppm, doublet, 1H

 

Problem NMR10.3.

 

Problem NMR10.4.

3.4 ppm, doublet: CH-CH2-O

2.1 ppm, singlet: OH

1.7 ppm, nonet:  (CH3)2CHCH2

0.9 ppm, doublet:  CH(CH3)2

 

Problem NMR10.5.

a)  This is a simulated spectrum.  The peaks at f, e, and c are singlets.  The peak at d is a doublet.  The peaks at a & b are unfortunately coincident, so their multiplicities are obscured, but they would be doublets & triplets, respectively.

b)

c)

d)  The peaks at b and c unfortunately coincide, but they would be a septet and a triplet, respectively.  Otherwise, a is a singlet, d is a triplet, and e is a doublet.

 

NMR 13.1.

a) TBME        b) acetone        c) THF        d) methanol        e) ethyl acetate

 

NMR 14.1.

a) appearance of CH-O near 3.5 ppm (multiplet, 1H)

b) appearance of CH2-O near 3.5 ppm (triplet, 2H); disappearance of =CH near 5-6 ppm (mutiplets, total 3H)

c) appearance of HC=O near 10 ppm (triplet, 1H); disappearance of =CH near 5-6 ppm (mutiplets, total 3H)

d) appearance of HC=O near 10 ppm (triplet, 1H);  disappearance of CH2-O near 3.5 ppm (doublet, 2H)

e) appearance of CH2-O near 3.5 ppm (triplet, 2H)

f) appearance of CH2-O near 4 ppm (triplet, 2H)

g) appearance of =CH near 5-6 ppm (mutiplets, total 2H); disappearance of HC=O near 10 ppm (triplet, 1H);

h) disappearance of =CH near 5-6 ppm (mutiplets, total 2H); appearance of triplet:sextet:triplet pattern between 1-2 ppm

i) appearance of HC=O near 10 ppm (triplet, 1H);  disappearance of CH2-O near 3.5 ppm (doublet, 2H)

j) appearance of CH-O near 3 ppm (singlet, 3H) and 3.5 ppm (sextet, 1H); disappearance of =CH near 5-6 ppm (mutiplets, total 2H)

 

Problem NMR15.1.

a) student 1: Let's use the H2C-O peak of ethyl propanoate at 4 ppm and the acetonitrile methyl at 2 ppm.  The ratio appears to be 2:1, but they represent 2 protons and 3 protons, respectively.  That means the ratio of molecules is 2/2:1/3 = 3:1 ethyl propanoate : acetonitrile.

student 2: We'll use the H2C-O peak of ethyl propanoate at 4 ppm and the H2C-O peak of THF at 3.5 ppm.  The ratio appears to be 3:2, but they represent 2 protons and 4 protons, respectively.  That means the ratio of molecules is 3/2:2/4 = 12:4 = 3:1 ethyl propanoate : THF.

student 3: Look at the H2CCl2 peak of dichloromethane at 5 ppm and the H2C-O peak of THF at 4 ppm.  The ratio appears to be 1:2, and they both represent 2 protons, so the ratio of molecules is 1:2 dichloromethane : ethyl propanoate.

b) student 1: The sample is [1/(1+3)]x100% = 25% acetonitrile.

student 2: The sample is [1/(1+3)]x100% = 25% THF.

student 3: The sample is [1/(1+2)]x100% = 33% acetonitrile.

 

Problem NMR15.2.

The obvious NMR handles are the H-C=O aldehyde proton at 10 ppm for benzaldehyde and the alcohol-adjacent H-C-O proton at 4.5 ppm for 1-phenylpropanol. 

Each of those peaks represents one proton, so the integral ratio of 1:2 suggests a ratio of bezaldehyde to 1-phenylpropanol of of 1:2.  That translates into 33% benzaldehyde, 67% 1-phenylpropanol.

 

Problem NMR15.3.

The NMR handles here are the H-C=O aldehyde proton at 10 ppm for benzaldehyde and the alcohol-adjacent H2C-O protons near 5 ppm for benzyl alcohol. 

In this case, we need to correct for the differing numbers of protons represented by each peak: 1H for the aldehyde peak but 2H for the alcohol one.  The integral ratio of 1:6 therefore suggests a ratio of bezaldehyde to 1-phenylpropanol of of 1:3.  That translates into 25% benzaldehyde, 75% 1-phenylpropanol.

 

Problem NMR15.4.

We could use the peak corresponding to the O-CH-C=O proton above 5 ppm for the repeat unit and the peak for the CH2-O proton in the initiator/end group near 3.5 ppm.  The integral ratio is 24:1, but they represent different numbers of hydrogens, so the repeat unit to end group ratio is really 24/1:1/2, or 48:1.  The degree of polymerization is 48.

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use./p>

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

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