NANYANG TECHNOLOGICAL UNIVERSITY

DIVISION OF CHEMISTRY AND BIOLOGICAL CHEMISTRY

SCHOOL OF PHYSICAL & MATHEMATICAL SCIENCES

CM 3011 - Chemical Spectroscopy and Applications

Final Examination Solution Manual

AY2014/2015

Prepared by

EUGENE ONG

November 2017

1

Question 1 (a) We use the following formula:

�̃� =1

2𝜋𝑐√

𝑘

𝜇 where

1

𝜇=

1

𝑚1+

1

𝑚2

Let �̃�1 be the absorption wavenumber for 12C14N, �̃�2 be the absorption wavenumber for 13C15N,

𝑚1 be the atomic mass for 12C or 13C and 𝑚2 be the atomic mass for 14N or 15N.

First, we find the reduced mass for both scenarios:

1

𝜇1=

1

𝑚1+

1

𝑚2=

1

12+

1

14=

13

84 and

1

𝜇2=

1

𝑚1+

1

𝑚2=

1

13+

1

15=

28

195

Then, we can find �̃�2 by using the ratio method.

�̃�1

�̃�2=

12𝜋𝑐

√𝑘𝜇1

12𝜋𝑐

√𝑘𝜇2

= √

1𝜇1

1𝜇2

= √

138428

195

= √845

784

�̃�2 = �̃�1 ÷ √845

784= 2225 cm−1 ÷ √

845

784= 2143.185 cm−1 ≈ 2143 cm−1

The C≡N stretching wavenumber for 1* is predicted to be 2143 cm-1.

(b) NOT EXAMINABLE

(c) Given: M (133) = 100%, M+2 (135) = 32.5%

From the molecular ion peaks, Compound 5 should contain a Cl atom, since 35Cl : 37Cl ~ 3:1.

Rule of 13: 133/13 = 10 + 3/13

Base formula: C10H13

Modified formula: C10H13 + Cl – C3 + H = C7H14Cl

Nitrogen rule. At least one N atom.

Modified formula: C7H14Cl + N – CH2 = C6H12ClN. Does not match with 1H NMR integration.

We add one O atom.

Modified formula: C6H12ClN + O – CH4 = C5H8ClNO. Still does not match with 1H NMR

integration.

We add another O atom.

Modified formula: C5H8ClNO + O – CH4 = C4H4ClNO2. Match with 1H NMR integration.

U = 3

2

IR: Absorption at 1711 cm-1 indicate carbonyl group. No absorption for any primary/secondary

amine/amide, this means that the N are tertiary substituted.

13C NMR: 179.3 ppm, ester or amide group.

1H NMR: 2.57 (s, 4H), two equivalent -CH2- groups. Good to suspect a symmetrical structure.

The other two carbons must belong to two carbonyl groups. Left with U = 1, other carbonyl or

olefinic moiety is impossible here, so the only option is a cyclic structure.

Figure 1. Partial structures leading to Compound 5.

(d) As shown in Figure 2, due to the magnetic anisotropy of the C≡C triple bond in Compound 7, the

H situated next to the triple bond will be more deshielded, hence resulting in a downfield shift

and thus larger 1H NMR chemical shift than Compound 6.

Figure 2. Magnetic anisotropy of C≡C triple bond in Compound 7.

Question 2

(a) Given formula: C4H6O5

U = 2

IR: Broad absorption at 3500 to 3000 cm-1. O-H stretching.

Strong absorption at ~1700 cm-1. C=O stretching.

1H NMR: Broad signal at ~13.5 ppm suggests two carboxylic groups.

Broad signal at ~5.5 ppm suggests an alcohol group.

13C NMR: Two signals between 180 to 170 ppm belong to two carboxylic groups that are

inequivalent.

1H NMR: ~4.3 ppm (dd, 1H), coupling with two diastereotopic protons.

~2.6 ppm (dd, 1H) and ~2.45 ppm (dd, 1H), belong to two diastereotopic protons.

These signals are characteristic of a 3-spin system.

(+) - Shielded. Upfield shift. Lower δ.

(–) - Deshielded. Downfield shift. Higher δ.

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Figure 3. Partial structures leading to Compound 8.

One possible fragmentation reaction in Compound 8 is the α-cleavage of a carboxylic acid group

that gives an acylium ion of m/z = 45, as shown in Figure 4.

Figure 4. Mechanism for the fragmentation of Compound 8.

Indeed, the fragment ion is observed experimentally and relative intensity is found to be 37.4%.

(See Supporting Information for more details.)

(b) Given formula: C9H8O2

U = 6 1H NMR: 4 signals within 7.3 to 6.9 ppm indicate 4 aromatic protons. This suggests that

Compound 9 is a disubstituted benzene. The signal splitting of the 4 aromatic protons give

information about the substitution pattern, as described below:

1) Two dd signals: Coupling with 1 ortho proton and 1 meta proton.

2) Two td signals: Coupling with 2 ortho protons and 1 meta proton.

The substitution pattern of the aromatic protons and the two unknown substituents is shown in

Figure 5.

Figure 5. The positions of the two unknown substituents of Compound 9.

So, Compound 9 is a ortho-disubstituted benzene.

IR: Strong absorption near 1800 cm-1. Carbonyl group. 13C NMR: ~170 ppm indicates an ester group.

1H NMR: Two triplet signals at ~3.00 ppm and ~2.75 ppm indicate a -CH2-CH2- group.

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

structure. A cyclic structure fits well with the ortho-disubstituted benzene.

4

You are almost complete if you get the structure 9 or 9'. The key to distinguish these two is by

looking at the chemical shift of the methylene group (~2.75 ppm). 9' will give a much downfield

shift since the methylene group is directly bonded to O (From CBC Databook, chemical shift of -

CH2-O-R is ~3.4 ppm, but more in this case, since the methylene group is also affected by the

magnetic anisotropy of the benzene ring).

Figure 6. Partial structures leading to Compound 9.

DEPT-135 spectrum of Compound 9:

Change the following 13C NMR signals:

1) Signals ~170 ppm, ~150 ppm and ~123 ppm should disappear.

2) Signals ~30 ppm and ~25 ppm should be inverted.

Other signals should remain the same.

(c) Given formula: C9H9NO3

U = 6

Hint: Compound 10 is a tetrasubstituted benzene derivative.

M-15: Methyl fragment.

M-43: Most likely -CO-Me fragment.

IR: Two bands between 3500 to 3000 cm-1. Primary amine. Consistent with broad singlet (2H)

in 1H NMR.

13C NMR: ~200 ppm, a ketone. 1H NMR: ~2.5 ppm (s, 3H), methyl group attached to a carbonyl group. Confirms the presence

of -CO-Me group. (From CBC Databook, chemical shift of CH3-CO-Ar is 2.6 ppm)

1H NMR: Two singlets at ~7.0 ppm and ~6.25 ppm represent two aromatic protons that are not

coupled to each other. Therefore, the substitution pattern of the tetrasubstituted benzene

derivative is known, as shown in Figure 7.

Figure 7. Substitution pattern of Compound 10.

Left with C9H9NO3 – H2N – C2H3O – C6H2 = CH2O2 and U = 1, no other carbonyl or olefinic

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

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

Figure 8.

Figure 8. Benzodioxole structure of Compound 10.

This fully explains the extremely downfield signal of the methylene group at ~5.9 ppm as there are two neighbouring O atoms to the methylene protons (double inductive effect).

Obviously, the two remaining substituents belong to a primary amine and a methyl ketone. So,

the structure for Compound 10 is complete.

Figure 9. Compound 10.

(d) The absorption of C=O stretching in IR spectrum is within the region of 1600 to 1700 cm-1, which

is lower than base value of 1715 cm-1. There are two factors that contribute to this phenomenon,

which are resonance effect and hydrogen bonding effect. Resonance with the amine group and

benzene ring decreases the order of C=O bond, as shown in Figure 10. In addition, the carbonyl

forms a hydrogen bond with the neighboring amine, thus weakening the C=O bond. As a result,

these two factors cause the absorption to shift to a lower wavenumber.

Figure 10. Resonance mechanism for Compound 10.

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Question 3 & 4

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Question 5

xxx End of Solution Manual xxx

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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 5

9

10

m/z Intensity (%)

11

Compound 8

12

13

m/z Intensity (%)

14

Compound 9

15

16

17

Compound 10

18

19

m/z Intensity (%)