Question

1.Identify the compound whose spectral data are shown.

2. Describe the coupling observed and show the splitting diagram for each hydrogen.

3. Determine the value of each coupling constants.

4. Label and assign each 1 H and 13 C signal to your answer.

100 3 % of Base Peak 90 80 100 120 m/z 5 SA % Transmittance --1689 -3047 --14197 --1273 -957 E9EE- -702 3000 1000 4000 'H NMR 300 MHz 2000 Wavenumber (cm-1) 2720 2700 Hz 2600 2580 Hz 2440 2420 Hz 2200 2180 Hz 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 ppm 13C/DEPT NMR 75.5 MHz CH↑ CH/KHz ? CH2P solvent 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 ppm

Answer 1

ANSWER:

Table resume of spectral data

	Signal	Observation
Mass	121 m/z	odd molecular ion => compound with N
IR	3047 cm-1	C-H stretch (aromatic)
1689 cm-1	C=O stretch (conjugated)
1600 cm-1	C=C stretch (aromatic)
1273 cm-1	C-N stretch (aromatic)
1H-NMR	2.5 ppm (3H, s)	CH3-C=O
	7.3 ppm (1H, ddd)	CH3-C=O
	8.1 ppm (3H, dt)	CH3-C=O
	8.6 ppm (3H, dd)	CH3-C=O
	9.0 ppm (3H, dd)	CH3-C=O
13-C-NMR	196 ppm (>C<)	C=O
	154 ppm (CH)	=C-H (aromatic)
	149 ppm (CH)	=C-H (aromatic)
	136 ppm (CH)	=C-H (aromatic)
	132 ppm (>C>)	=C< (aromatic)
	154 ppm (CH3)	CH3-C=O (aromatic)

According to spectral data the compound has the following properties:

• An aromatic ring with only four aromatic protons
• A ketone group
• A methyl group attached to carbonylic carbon

The compound appears to have 7 C (from 13C-NMR), 7 H (from 1H-NMR), 1 O (from IR) and 1 N (from mass). This is in corcondance with the molecular ion at 121 m/z

MW = 7xC + 7xH + 1xN + 1xO

MW = 7x12 + 7x1 + 1x14 + 1x16 = 121

Then, the molecular formula of compound is

C7H7NO

And the unsaturation degree is

2xC +2 - H+N DOU = 2x7+2 - 7+1 = 5

• The compound has 5 unsaturations: one from ketone group, and the other fours of aromatic ring (six-membered ring)

The compound is has only 7 C and two of them are out of the ring (C=O and -CH3). And there is only four aromatic protons and the ring must be a six-membered ring (according DOU). Then, the N atom is part of aromatic ring and there are three possible structures:

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We can discard structure C, because in this case we will observe only 2 type of aromatic H and 3 of aromatic C. Then our compound is A or B.

Aromatic carbons adjacent to N has a higher chemical shits than other aromatic carbons. Accroding 13C data, those carbons (149 and 154 ppm) are tertiary carbons (each one has an H). Then, the compound is the structure B:

136.8.1 190 123.7.3 Y320 149, 8.8 154.0.0

Splitting parrtern of protons and coupling constants

• Coupling constant are calculated measured the distance (in Hz) between splitted peaks

CH3 Couplings pen 'SH (c) (A) 11 L C-B A-C ded dd 9,0 NZHD 7,3 A-D c-D D-B . (A) Splitting » dd 8.6 dd Bodd cddd 2 نے الله نے bc= 2,25 Hz . lo Si JB, D = 4,81 Hz JBC 2 1,75 Hz A, D = 0,90 Hz 4 36D 37,94 H2 - 1 - 50624,81 HZ hili ll ll SA, p=0, 90 HP soi JC,D 37,94 Hz GAJA, C = 2,25 Hz i tipi se na 1,75 HZ