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NMR Assignment

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1 CH-O (deshielded) 4 Olefinic carbons: 2 =CH2. 1 =CH. 1 =C. C10H18O ... Olefinic. 1 CH= 2 CH2= 1H. CH-O. Aliphatic. 6H : 2 Me. Triplet!! 6 H: 1 CH, 2 CH2. OH. HETCOR ... – PowerPoint PPT presentation

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Title: NMR Assignment


1
NMR Assignment
  • Example
  • IPSENOL

2
Unknown Ipsenol
C10H18O
Index of H deficiency C H/2 1 2
3
C13 NMR and DEPT
C-C
CC
CH only
C-O
4
Unknown Ipsenol
C10H18O
Index of H deficiency C H/2 1 2
In 13C All carbons are visible gt no symmetry
  • 5 aliphatic carbons
  • 2 Methyls
  • 1 CH
  • 2 CH2
  • 1 CH-O (deshielded)
  • 4 Olefinic carbons
  • 2 CH2
  • 1 CH
  • 1 C

Counting protons attached to Carbon 17 H
The one Missing might be attached to Heteroatom
OH
5
Proton Spectra Integration
Aliphatic
Olefinic 1 CH 2 CH2
CH-O
6 H 1 CH, 2 CH2 OH
4H
6H 2 Me Triplet!!
1H
1H
6
HETCOR
Me
o
CH
CH-O
o
CH-O
CH
7
Proton Spectra Expansion
dd
Septet?
dd
d
ddd
2 x d
dd
m OH
ddd
CH2
8
COSY
x
o
9
Proton Spectra Analysis
6
6
4
5a
7 cis/trans
6
5
7
H4 not septet
8c
8t
8
3
4
tt
2
1
1
1,1
2
5b
3a
3b
10
HMBC
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