Question

NMR Exercise Lab Determine the structure of the following unknown organic compounds from the 'H and C NM below. When supplying spectroscopic information for parts b-c below, be sure to give units appropriately when identifying specific peaks in a spectrum that are associated with specific portions of the structural formula that you give in part a. (a) Provide a structural formula which is consistent with all of the given information; give the degrees of unsaturation (DoU). (b) For the 15'C NMR spectrum, make a table in which you assign, chemical shift of each C in the structural formula that you gave in part a. Redraw the structure by this table and label each carbon with a number in the structural formula that as best as possible, the you gave in part a. How many different Cs are there? (c) For the H NMR spectrum, make a table in which you assign, as best as possible, the mula that you gave in part a. Also give in chemical shift of each H in the structural fo your table the number of H's (integration value), and the multiplicity (singlet, doublet, triplet, etc.) for each set of H's labeled in your structural formula in part a

Answer 1

Answer question A:

MASS SPECTRUM

• The mass spectrum of the unknowm compound indicates that the molecular formula is of the unknown compound is C₇H₈O. It is possible dertermine the degrees of unsaturation (DoU) of the molecule whit this information:

$DoU=\frac{2C+2-H}{2}=\frac{2 \left (7 \right ) + 2-8}{2}=4$

• The 4 unsaturation degrees could be due to C=C bonds, C=O bonds, aromatic rings, etc.

IR SPECTRUM

• The IR spectrum of the unknowm compound show the following signals:

Signal	Wavenumber (cm-1)	Intensity	Assignation
A	3500-3250	strong broad band	O-H stretch absorption for alcohol
B	3100-3000	strong band	C-H stretch absorption for aromatic rings
C	3000-2800	strong band	C-H stretch absorption for C sp3
D	2000-1800	small bands	aromatic overtones
E	approx. 1600	small band	C=C stretch absorption for aromatic rings
F	approx. 1200	strong band	C-O stretch absorption

E C B IR Spectrum (lkquid fim) F A 3326 800 1200 1600 2000 3000 4000 v (cm)

• This data indicates that the unknown compound has an aromatic ring and an alcohol group. then, the 4 unsaturation degrees find previously are due to an aromatic ring.

¹³C-NMR AND ¹H-NMR ESPECTRA

• The ¹³C-NMR spectra showed the following information:

δ (ppm)	DEPT	Assignation
65	-CH2-	-CH2- group adjacent to an oxygen atom
127	CH/CH3	=CH- group of an aromatic ring
127.6	CH/CH3	=CH- group of an aromatic ring
128.5	CH/CH3	=CH- group of an aromatic ring
141	>C<	Quaternary carbon of an aromatic ring

• The ¹H-NMR spectra gives the following information:

δ (ppm)	Multiplicity	Integration	Assignation
2.60	singlet	1H	Proton of -OH group
4.75	singlet	2H	-CH2- group adjacent to an oxygen atom
7.40	singlet	5H	aromatic protons

• The NMR data indicates the presence of a benzene ring monosubstituted:
  • There is 5 aromatic protons
  • There is 4 types of aromatic carbons: 1 quaternary and 3 CH (this means that two carbons of the ring are equivalents to other two C)

R D A A C

• As the molecular formula is C₇H₈O and there must be and -CH2-group adjacent to and oxygen and one -OH group. Then the compound is:

Он

Answer question B:

• The compound have five types of carbons:

δ (ppm)	Carbon	Assignation
65	1	7 1 OH 2 3 5 4
127	3 / 7
127.6	5
128.5	4 / 6
141	2

Answer question C:

• According to ¹H-NMR spectra the compound have three types of protons:

Proton	δ (ppm)	Multiplicity	Integration	Assignation
1	2.60	singlet	1H	4 3 ннн 5 Н. ОН он, н вн Н 7
2 / 3	4.75	singlet	2H
4 - 8	7.40	singlet	5H

• However, the five aromatic protons are not exactly equivalents. They are very similiar and they appears as a singlet in ¹H-NMR spectra:
  • Protons 4 and 8 are equivalents
  • Protons 5 and 7 are equivalents