Question

6) The IR-, 1H- and 13C-spectra of an unknown sample having the formula C8H1602 are shown on the next page. a) What can you conclude about the structure form the 13C NMR data ? b) What functional group is present ? Explain your rational c) Propose a structure that is consistent with the spectroscopic data ?

Answer 1

ANSWER:

• Identification of unsaturation degrees (DOU):

$DOU= \frac{2C+2-H}{2}=\frac{2\pm 8+2-16}{2}=1$

There is only one unsaturation in the compound. It could be a C=C, C=O or a ring.

13C-NMR Data:

Chemical Shift	Assigment	Chemical Shift	Assigment
176 ppm	C=O	22 ppm	CH3, CH2 or CH
68 ppm	C-O	17 ppm	-CH3
40 ppm	C-C=O	10 ppm	-CH3
28 ppm	CH3, CH2 or CH

• The 13C-NMR data indicates that we have only seven types of carbons. As the molecular formula has eight carbons, this means than one signal between 10 and 28 ppm integrates for two carbons.
• The signal at 176 ppm shows the presence of a carbonyl group. The chemical shift indicates that is probably and ester (-COOR) or a carboxylic acid (-COOH).
• The signal at 68 ppm assigned as C-O could be for the ester group (-COOR) or for an hydroxyl group (R-OH).

IR Data:

Signal	Assigment
3000-2900 cm-1	C-H stretch (sp3 carbons)
1720 cm-1	C=O stretch
1200 cm-1	C-O stretch

• The IR signal at 1720 cm-1 confirms the carbonyl group observed in the 13C-NMR spectra.
• The absence of a broad and strong band centered at 3300 cm-1 discard the presence of an hydroxyl group (R-OH)
• In addition, the absence of a broad band centered at 3000 cm-1 discard the presence of an carboxylic group (--COOH)
• Then, according to IR and 13C-NMR data, the unknown compound is ans ester

1H-NMR Data:

Chemical Shift	Integration	Multiplicity	Assigment
5.00 ppm	1H	m	>CH-OR
2.30 ppm	1H	m	>CH-C=O
1.70 ppm	1H	m	-CH2-CH3
1.40 ppm	1H	m	-CH2-CH3
1.25 ppm	6H	d	CH3-CH
1.10 ppm	3H	d	CH3-CH
0.80 ppm	3H	t	CH3-CH2

• Signal at 5.00 ppm is a CH group (integrates for 1H) that is linked to an oxygen atom
• Signal at 2.30 ppm is a CH group (integrates for 1H) that is adjacent to carbonyl group
• Signal at 0.80 ppm is a methyl group (integrates for 3H) attached to a CH2 group (the multiplicity of signal is a triplet). This means that protons at 1.70 ppm and 1.40 pm are the protons of the CH2 group.
• Signal at 1.25 ppm are two methyl groups (integrates for 6H) attached to a CH group (the multiplicity of signal is a doublet). This means that we have an isopropyl group in the molecule.
  • The high value for the chemical shift of methyl groups indicates that they are near to an oxygen atom, then, they must be attached to CH group at 5.00 ppm => The -OR chain of ester is an isopropyl group
• Signal at 1.10 ppm is a methyl group (integrates for 3H) attached to a CH group (the multiplicity of signal is a doublet).

Using all data, we get the structure of compound C8H16O2 as

CH HjC Y O CH CH3