NANYANG TECHNOLOGICAL UNIVERSITY

DIVISION OF CHEMISTRY AND BIOLOGICAL CHEMISTRY

SCHOOL OF PHYSICAL & MATHEMATICAL SCIENCES

CM 3011 - Chemical Spectroscopy and Applications

Final Examination Solution Manual

AY2016/2017

Prepared by

EUGENE ONG

November 2017

1

Question 1 (a) Intense IR absorptions at 1744, 1718 and 1696 cm-1 Carbonyl group

So, Compounds 2 and 3 can be excluded.

Sample A: 1744 cm-1 Compound 4

Due to the five-membered ring structure, the ring strain promotes the s-character of C=O carbon

and stronger coupling with C-C stretch, therefore exhibiting a high wavenumber.

Sample B: 1718 cm-1 Compound 1

Close to the base value of ~1715 cm-1 for alkyl ketones.

Sample C: 1696 cm-1 Compound 5

Due to the resonance effect in the α,β-unsaturated aldehyde, the C=O bond of 5 has more single

bond character than 1 and 4, as shown in Figure 1. Hence 5 exhibits a low wavenumber.

Figure 1. Resonance structure for 5.

(b) Compound 7, as a five-membered lactam, has a low wavenumber of 1653 cm-1 due to the

resonance effect, as shown in Figure 2, that causes the C=O bond to have more single bond

character.

Figure 2. Left: Resonance structure for 7. Right: Resonance is not possible for 6.

On the other hand, Compound 6 has a higher wavenumber of 1733 cm-1. This suggests that 6 do

not have any resonance effects, unlike 7. One should realise that the substituted immonium ion

moiety in resonance structure A assumes a planar geometry. So, due to the cage structure of the

bridgehead amide 6, the two connected alkyl groups with N are locked into a position such that

the resonance structure is very strained, thus making resonance extremely unfavourable. This is

consistent with Bredt’s rule, which states that a double bond cannot be placed at the bridgehead

of a bridged ring system unless the rings are large enough. Hence, without any resonance effects,

6 owes its higher wavenumber to the ring strain effect.

(c) Compound 8

Given: Mass = 114, C: 52.61%, H: 8.83%, N: 24.54%

We find that C3 = 60 g/mol, H10 = 10 g/mol, N2 = 28 g/mol, but this does not add up to 114 g/mol.

However, we add an O atom, where the final mass is 114.

Formula: C5H10N2O

U = 2

2

IR: Strong absorption at 1699 cm-1, which belong to a C=O group, possibly an amide. No

absorption for any primary/secondary amine/amide, this means that the N are tertiary

substituted.

13C NMR: 162.0 ppm, confirms that amide group is present.

1H NMR:

2.78 ppm (s, 6H), suggests two equivalent CH3- groups bonded to an electronegative atom (N in

this case).

3.28 ppm (s, 4H), now we have left two carbons, so two equivalent -CH2- groups bonded to N.

The 1H NMR signals strongly suggest that Compound 8 is symmetrical.

Left with U = 1, other carbonyl or olefinic moiety is impossible here, so the only option is a cyclic

structure.

Figure 3. Partial structures leading to Compound 8.

(d) Note that the two isopropyl methyl groups are not, strictly speaking, equivalent as one of them

is closer to the amine group. You can find this by drawing the most stable conformation in the

Newman projection. So, there are two signals for each methyl group.

Also, we assume that the heteroatom-bonded atoms (-NH2 and -OH) do not couple with

neighbouring proton(s) due to fast chemical exchange. 1H NMR: 8 signals. (7 signals if both -NH2 and -OH groups have rapid proton exchange,

especially at high temperatures.) 13C NMR: 5 signals.

As there is a chiral centre in 9, the diastereotopic protons Ha and Ha’ will couple with Hb, giving

rise to doublet of doublets (dd), as illustrated in Figure 4.

Figure 4. Newman projection of Compound 9.

Since Hb also couple with the isopropyl methine proton, we get doublet of doublet of doublets

(ddd).

3

Question 2

(a) Given formula: C6H3Br3O

U = 4 1H NMR: 2 equivalent aromatic protons at ~8.5 ppm. Good to suspect symmetrical

tetrasubstituted benzene. The aromatic protons could be meta or para to each other.

1H NMR: Broad singlet at ~6 ppm. An -OH group.

IR: Sharp absorbance at ~3500 cm-1. Probably an -OH group with two neighbouring bulky

substituents (Two Br in this case).

Figure 5. Partial structures leading to Compound 10.

13C NMR: 4 aromatic carbon signals. Consistent with proposed structure of Compound 10.

(b) We know that 79Br : 81Br ~ 1 : 1 and there are 3 Br atoms in Compound 10.

Hence, the abundance of each molecular ion can be calculated as follows:

M (328): ½ x ½ x ½ = 1/8

M+2 (330): (½ x ½ x ½) + (½ x ½ x ½) + (½ x ½ x ½) = 3/8

M+4 (332): (½ x ½ x ½) + (½ x ½ x ½) + (½ x ½ x ½) = 3/8

M+6 (334): ½ x ½ x ½ = 1/8

Thus, M : M+2 : M+4 : M+6 = 1 : 3 : 3 : 1

See Supporting Information for the EI-MS spectrum.

(c) Given formula: C6H13NO2

U = 1 13C NMR: Signal at ~170 ppm. Ester or amide group.

IR: Strong absorbance between 1800 and 1700 cm-1. C=O stretching. Most likely ester group.

No absorption for any primary/secondary amine/amide, this means that N is tertiary substituted.

1H NMR: Two different alkyl groups can be deduced:

1) ~1.25 ppm (t, 3H) and ~4.25 ppm (q, 2H): CH3-CH2- bonded to an electronegative atom.

(From CBC data book, highly likely to be CH3-CH2-O-CO-)

2) ~2.4 ppm (s, 6H): Two equivalent CH3- groups bonded to N.

3) ~3.25 ppm (s, 2H): Most likely isolated -CH2- group sandwiched between the carbonyl group

and tertiary amine.

4

Figure 6. Partial structures leading to Compound 11.

(d) DEPT-135 spectrum of Compound 11:

Change the following 13C NMR signals:

1) Signal ~170 ppm should disappear.

2) Signal ~60 ppm and one signal at ~45 ppm should be inverted.

3) Signal ~15 ppm and one signal at ~45 ppm should remain the same.

(e) Given formula: C8H5NO2

U = 7

IR: Absorbance at ~2200 cm-1. Nitrile group. Consistent with weak 13C NMR signal at ~120 ppm.

1H NMR: 3 signals at ~7.3 to ~6.8 ppm suggests 3 aromatic protons with coupling constants that

give information about the substitution pattern, as described below:

1) Ha: ~7.22 ppm (dd, 1H), J = 8.0, 1.6 Hz. Ortho and meta coupling with 2 other aromatic

protons.

2) Hb: ~7.03 ppm (d, 1H), J = 1.6 Hz. Meta coupling with 1 aromatic proton.

3) Hc: ~6.86 ppm (d, 1H), J = 8.0 Hz. Ortho coupling with 1 aromatic proton.

The signal at ~6.0 ppm (s, 2H) do not belong to aromatic protons, otherwise the result will be a

monosubstituted benzene, which does not make any sense here. So, we can deduce the

substitution pattern of the three aromatic protons, which is shown in Figure 7.

Figure 7. The positions of aromatic protons in Compound 12.

Left with U = 1 and CH2O2, no carbonyl group is possible since there are no 13C NMR signal and

IR absorption. The only option is a cyclic structure. However, we notice that there are two

adjacent substituents in the benzene ring. A benzodioxole structure fits the bill, as shown in

Figure 8.

Figure 8. Benzodioxole structure of Compound 12.

5

This fully explains the extremely downfield signal at ~6.0 ppm as there are two neighbouring O

atoms to the methylene protons (double inductive effect).

Obviously, the remaining substituent is the nitrile group. So, the structure of Compound 12 is

complete.

Figure 9. Compound 12.

Assignment of protons for Compound 12:

1) Ha: ~7.22 ppm (dd, 1H), J = 8.0, 1.6 Hz.

2) Hb: ~7.03 ppm (d, 1H), J = 1.6 Hz.

3) Hc: ~6.86 ppm (d, 1H), J = 8.0 Hz.

4) Hd: ~6.0 ppm (s, 2H).

(f) Through the α-cleavage of an ether moiety, the following fragment ion (an oxonium ion) can be

obtained, as shown in Figure 10.

Figure 10. Mechanism for the fragmentation of Compound 12.

Indeed, experimentally, the relative intensity of the fragment ion is found to be 100% and is the

only major fragment ion observed. (See Supporting Information for more details.)

Question 3

Both expected 19F and 19F {1H} NMR signals are illustrated in Figure 11.

Note: The labeling may be too small to be observed. Please zoom in for more details.

6

Figure 11. Expected 19F and 19F {1H} NMR signals of the dimeric platinum(IV) complex.

7

Question 4

(a) The phosphorous-gold (I) complex is shown in Figure 12, together with labelled protons.

Figure 12. Labelled protons of the phosphorous-gold (I) complex.

Assignment of protons for the phosphorous-gold (I) complex:

1) Ha: 2.0193 ppm (s, 6H).

2) Hb: 6.2598 ppm (d, 2H), 2JP-Hb = 38.68 Hz.

3) Hc: 7.1627 ppm (dd, 2H), 3JHc-Hd = 7.92 Hz (aromatic ortho coupling), 3JP-Hc = 10.92 Hz.

4) Hd: 7.2725 ppm (t, 2H), J = 6.96 Hz (aromatic ortho coupling).

5) He: 7.3908 ppm (t, 1H), J = 7.36 Hz (aromatic ortho coupling).

(b) The 31P spectrum shows a triplet of triplets signal. The two different coupling constants are

calculated below:

Larger coupling, 2JP-Hb

= {[(4.4114 – 4.1738) + (4.1738 – 3.9340)] /2} x 161 MHz

= 38.43 Hz

Smaller coupling, 3JP-Hc

= {[(4.4786 - 4.4114) + (4.4114 – 4.3447) + (4.2401 – 4.1738) + (4.1738 – 4.1048) + (4.0033 –

3.9340) + (3.9340 - 3.8680)]/6} x 161 MHz

= 10.85 Hz

xxx End of Solution Manual xxx

8

Supporting Information Disclaimer: The following spectra are identical to the given spectra in the final examination.

The spectra are available through the Spectral Databases for Organic Compounds:

http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi.

Compound 8

9

10

11

Compound 9

12

13

Compound 10

14

15

m/z Intensity (%)

16

Compound 11

17

18

19

Compound 12

20

21

m/z Intensity (%)