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Identification of Unknown Compounds

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Identification of Unknown Compounds. Compound from Soybean Oil ... 3.60 with a doublet of 2 protons = 3.95 with 1 proton = 4.2 with 2 protons = 5.3 with 4 protons ... – PowerPoint PPT presentation

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Title: Identification of Unknown Compounds


1
Identification of Unknown Compounds
2
Compound from Soybean Oil
1. Hydrogenation Hydrogen gas was consumed.
2. Mass Peaks m/e 74 m/e 294
(molecular ion), and many other peaks 3. IR
Spectrum Absorption peaks at 3.3 ?m 5.8
?m
6.0 ?m and many other peaks 4. NMR
(proton) Spectrum Absorption at ? 5.3
? 3.6
? around 1.02.0 (very
crowded) ? 0.7 The
ratio of peak areas
3
Compound from Soybean Oil
  • Hydrogenation Hydrogen gas was consumed.
  • Mass Peaks m/e 74
  • m/e 292 (molecular ion), and many other
    peaks
  • 3. IR Spectrum
  • at 3.3 ?m
  • 5.8 ?m
  • 6.0 ?m
  • and many other peaks
  • 4. NMR Spectrum
  • Absorption peaks at ? 5.3
  • ? 3.6
  • ? around
    1.02.0 (very crowded)
  • ? 0.7
  • The ratio of peak areas

6.0
? 5.3

3.0
? 3.6
4
Unknown Compound from Pine Tree Lipid
5
Infrared, mass and NMR spectra of an unknown
compound are shown.
Identify the unknown compound
6
Trans-3.5 Dimethoxystilbene
H
H
OCH
3
H
H
H
H
H
H
H
H
OCH
3
7
Trans-3.5 Dimethoxystilbene
d 3.7, 6.2, 6.5, 6.9, and 7.3
Peak ratios 6, 1, 2, 2, and 5
H
H
OCH
3
H
H
H
H
H
H
H
H
OCH
3
8
Prooxidant Compound from Soybean Oil
9
IR and Mass Information
IR Spectrum 2.85 ?m, 3.3 ?m, 3.4-3.5 ?m, 5.8
?m, and 14.5 ?m Mass Spectrum M/e 339,
336, 263, and 262 Elemental analysis No
nitrogen in the molecule
10
Infrared and Mass Spectra
11
Nuclear Magnetic Resonance Spectrum
NMR has 11 absorbency peaks in the ratio of
314224222124 from the upper field to
the lower field ? 0.9 with a triplet of 3
protons ? 1.25 with 14 protons ? 1.50 with 2
protons ? 1.70 with 2 protons ? 2.05 with 4
protons ? 2.35 with 2 protons ? 2.80 with 2
protons ? 3.60 with a doublet of 2 protons ?
3.95 with 1 proton ? 4.2 with 2 protons ? 5.3
with 4 protons
12
Nuclear Magnetic Resonance Spectrum with Proton
Numbers
4
2
4
3
14
2
1
2
2
2
2
13
Sn-?-Linolenin
O
C
H
C
H
C
H
C
C
H
C
H
C
H
C
H
C
H
C
H
C
H
H
O
C
C
H
C
H
C
H
C
H
C
H
C
H
C
H
3
2
2
2
2
2
2
2
2
2
2
2
2
2
C
H
O
H
C
H
O
H
2
14
Sn-?-Linolenin
?2.05
?2.35
?5.3
?4.2
?1.7
O
C
H
C
H
C
H
C
H
C
H
C
H
C
H
C
H
C
H
C
H
C
H
O
C
C
H
C
H
C
H
C
H
C
H
C
H
C
H
3
2
2
2
2
2
2
2
2
2
2
2
2
2
C
H
O
H
?1.5
?2.8
C
H
O
H
2
?3.6
?0.9
?1.25
?3.95
See the NMRs of fatty acid and methyl ester
15
Sn-b-Linolenin
H
C
H
O
2
O
C
H
C
H
C
H
C
H
C
H
C
H
O
C
H
C
H
C
H
C
H
C
C
H
C
H
C
H
C
H
C
H
C
H
C
H
C
H
3
2
2
2
2
2
2
2
2
2
2
2
2
C
H
O
H
2
16
Describe the characteristic features of IR, UV,
NMR, and Mass spectra.
  • H H O
  • CH3 CH2 CH2 -CC-CH2-CC-CH2 CH2-CH2-C-OCH3
    (methyl ester)
  • H H
  • 1 2 3 4 5
    6 3 7 8
  • UV
  • IR
  • NMR The above compound have 8 distinctive groups
    of protons.
  • Write the chemical shifts (d) of the 8
    distinctive proton groups.
  • 1 2 3 4
  • 5 6 7 8
  • Mass spectrum

17
Problem Exercises
  • What are the approximate d (chemical shifts) and
    multiplets (spin spin splits) of CH3-CH2-CHO NMR
    spectrum using a 100 MHz NMR.
  • d Multiplets
  • CH3
  • CH2
  • CHO

18
Problem Exercises
  • The d of a 1H of toluene is 6.0 ppm at 2.35T.
    The reference TMS absorbs 100 MHz (mega hertz) at
    2.35 T. What is the absorption frequency of the
    1H of toluene?
  • What are the d and absorbance frequency of the
    1H of toluene at 200 MHz under the magnetic field
    of 4.70T?
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