It is well known that strawberries help to reduce minor headaches. The substance A that is responsib — Organic Chemistry Chemistry Question
Spectroscopy in Bioorganic Chemistry
It is well known that strawberries help to reduce minor headaches. The substance A that is responsible for this effect is also used as an aroma substance in bubble gums. However, it does not taste like strawberries!
5.00 g of substance A yield 2.37 g of water and 6.24 L of carbon dioxide (at 303.7 K and 106.3 kPa). In addition, the infrared (IR), the mass (MS), the 1 H–NMR, and the 13 C–NMR spectra of the substance have been recorded:
[VISUAL]
Determine the molecular weight of the substance from the MS spectrum.
Model Answer
152 g mol–1
The molecular weight corresponds to the peak with the highest m/z in the mass spectrum. The smaller peak at 153 g mol–1 is due to molecules with one 13C isotope (8 carbon atoms × 1% 13C isotopes in nature ≈ 8% of the total signal at 152 g mol–1).
Determine the molecular formula of the substance from the elementary analysis.
Model Answer
C8H8O3
Calculations:
n(H) = 2 × m(H2O) / M(H2O) = 2 × 2.37 g / 18.02 g mol–1 = 0.263 mol
n(C) = p × V(CO2) / (R × T) = (106.3 kPa × 6.24 dm3) / (8.314 J mol–1 K–1 × 303.7 K) = 0.263 mol
n(O) = (m(A) - n(H) × M(H) - n(C) × M(C)) / M(O) = (5.00 g - 0.263 mol × 1.01 g mol–1 - 0.263 mol × 12.01 g mol–1) / 16.00 g mol–1 = 0.098 mol
n(A) = m(A) / M(A) ≈ 5.00 g / (152 g mol–1) = 0.033 mol
N(O) = n(O) / n(A) = 3
N(H) = n(H) / n(A) = 8
N(C) = n(C) / n(A) = 8
Suggest one fragment B (molecular formula and structure(s)) for the signals at m/z = 39 in the MS spectrum. Suggest a probable fragment C (molecular formula and structure(s)) for m/z = 65 that contains B.
Model Answer
B: C3H3+
C: C5H5+
Note that for m/z = 39 only one fragment that has the molecular formula C3H3+ will be chemically meaningful, if the molecule only contains C, H, and O. The same is true for m/z = 65 and C5H5+ and if it has to contain C3H3+. Both fragments are typical of benzenes. Other (non-cyclic) structures of those fragments should also be considered as correct solutions, if they are chemically meaningful.
[VISUAL]
The two groups of signals around 3200 cm–1 and 1700 cm–1 in the IR spectrum are typical of a total of four structural features. Give information about the structures of these four functional groups. What additional information can be given, if the substance contains an –OH group?
Table of IR absorptions:
3800 - 3400 cm–1: O–H (free) (v)
3400 - 3000 cm–1: O–H (hydrogen bond) (v)
3000 - 2600 cm–1: O–H (intramolecular h. bond) (v)
3300 cm–1: C–H in C≡C–H (s)
3100 - 3000 cm–1: C–H in C=C–H (m)
3100 - 3000 cm–1: C–H in C÷C–H (w)
2900 - 2800 cm–1: C–H (alkanes) (s)
2200 cm–1: C≡C (w)
1900 cm–1: C=C=C (m)
1600 cm–1: C=C (w)
1600 - 1500 cm–1: C÷C÷C (s)
1700 cm–1: C=O (s)
Note: The interatomic bond that absorbs the light is bold. The intensities correspond to strong (s), medium (m), weak (w) and varying intensity (v). An aromatic bond is marked by '÷'.
Model Answer
O–H, C–H for the signals around 3200 cm–1,
C=O, benzene for the signals around 1700 cm–1,
the O–H group is involved in a (intra–molecular) hydrogen bond.
(Since it is impossible to distinguish between the signals within these two groups without additional information, the following is not thought to be part of the solution:
Broad peak at 3200 cm–1: C–H
Sharp peak at 2900 cm–1: O–H
Broad peak at 1700 cm–1: C=O
Sharp peaks around 1600 cm–1: benzene)
Assign the total of six signals at 4.0 ppm, 6.5 – 8.0 ppm, and 10.8 ppm in the 1 H–NMR spectrum to moieties that you expect in the unknown substance (consider 16.3 and 16.4).
Model Answer
4.0 ppm: OCH3
6.5 – 8.0 ppm: C6H4
10.8 ppm: OH
This information can directly be obtained from the chemical shift tables.
Assign the signals at 52 ppm, 170 ppm, and 110 – 165 ppm in the 13 C–NMR spectrum to moieties that you expect in the unknown substance (consider 16.3 and 16.4).
Model Answer
52 ppm: CH3
170 ppm: C=O
110 – 165 ppm: C6H4
This information can directly be obtained from the chemical shift tables.
Suggest one molecular structure for the unknown substance. Assign the resonances at 6.8, 6.9, 7.5, and 7.8 ppm in the 1 H–NMR spectrum and the signals at 52 and 161 ppm in the 13 C–NMR spectrum to individual atoms in your solution structure. According to your solution, suggest fragments that explain the signals at m/z=92 and m/z=120 in the MS spectrum. Write down the structural feature that is responsible for the low wave number of the –O–H band.
Model Answer
Methylsalicylate.
The intramolecular hydrogen bond in the figure explains the low wave number of the O–H band. It defines the ortho–position of the substitution as well as the fine splitting of the 1H signals of the aromatic system. The relatively large chemical shifts of the carbon atoms C–8 and C–1 at 52 ppm and 161 ppm are explained by a –I effect of the oxygen they are bonded to.
The assignment of the hydrogen chemical shifts in the aromatic ring is done in the following way: ±M effects define an alternating scheme of positive and negative partial charges at the aromatic ring. H–6 and H–4 have lower chemical shifts than H–5 and H–3. H–4 and H–5 have two neighbouring hydrogen atoms. Their signals are triplets that are shown in the figure. H–3 and H–6 have only one neighbouring hydrogen atom each. Their signals are doublets. All four signals are uniquely assigned by this information.
The signals at m/z = 120 and m/z = 92 are caused by loss of CH3–OH (methanol) or rather CH3–COOH (acetic acid).
[VISUAL]
The substance A is related to a drug widely used against headaches. Write down the chemical structure of this drug.
Model Answer
Acetylsalicylic acid (Aspirin)
[VISUAL]