Question

Use the spectra below to determine the molecular formula and chemical structure of the unknown compound.

MASS 142 % of Base Peak 1147 40 50 70 80 5 100 % Transmittance Jari- 3000 1000 4000 'H NMR 300 MERE 2000 Wavenumber (cm) 3.0 25 20 15 Ppm 4.0 3.5 C/DEPT NMR 75.5 MHz ppm 60 50 40 80 70 90 100 110 120 170 160 150 140 130

Answer 1

ANSWER:

First we made a table with the relevant information of spectra

Spectrum	Signal / correlation	Observation
Mass	m/z 114	Molecular ion
IR	1728 cm-1	C=O group
1H	4.02 ppm (t, 2H)	-CH2-O-
2.28 ppm (t, 2H)	-CH2-C(O)-
1.60 ppm (m, 4H)	-CH2-
1.35 ppm (m, 2H)	-CH2-
13C / Dept / HETCOR	178 ppm (q)	-COOR
64 ppm (CH2, 4.02 ppm)	-CH2-O-
34 ppm (CH2, 2.28 ppm)	-CH2-C(O)-
28 ppm (CH2, 1.60 ppm)	-CH2-
25 ppm (CH2, 1.35 ppm)	-CH2-
24 ppm (CH2, 1.60 ppm)	-CH2-
COSY	1.35 ppm / 1.60 ppm	-CH2-CH2-
1.60 ppm / 2.28 ppm	CH2-CH2-C(O)-
1.60 ppm / 4.02 ppm	CH2-CH2-O-

From the spectra data we get taht the compound is a lactone (cyclic ester), this information is obtained from:

• The IR spectrum indicates the presence of a carboyl group (signal at 1728 cm-1)
• All protons in the 1H-NMR and carbons in 13C-Dept are -CH2- groups (this indicates a ring).
• The ester group is confirmed by the chemical shift at 173 ppm (C=O), 64 ppm (-CH2-O-) and 4.02 ppm (-CH2-O-)

According the mass spectrum and 1H,13C-NMR, the molecular formula of compound is C6H10O2:

• There are six type of carbons in the 13C-NMR and ten protons (four types of protons) in the 1H-NMR
• The rule of 13, allows to calculate the formula:
  • Molecular ion = 114
  • Compound has two oxygen atoms (one carbonyl and one -OR group), then

$C_{m}H_{m+n}O_{2}$

CmHm+n = 114 - 32 = 82

82 = m + + = 6+ + 13

C6H1002

• This formual constains two unsaturations (DOU). This accords with the carbonyl group and a ring

2C + 2 - H 2 x 6+2 - 10 DOU =

Then, the strunture of the compound is

34/2,28 H2C173 24/1.60 HzC CH264/4.02 HC-CH228/1,60 25/1.35