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NMR Assignment . Example : IPSENOL. Unknown: Ipsenol. C 10 H 18 O. Index of H deficiency: C –H/2 + 1 = 2. C-C. CH and CH 3 UP. CH 2 Down. C=C. C13 NMR and DEPT. CH only. C-O. C 10 H 18 O. Unknown: Ipsenol. Index of H deficiency: C –H/2 + 1 = 2.

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nmr assignment

NMR Assignment

Example :

IPSENOL

unknown ipsenol
Unknown: Ipsenol

C10H18O

Index of H deficiency: C –H/2 + 1 =2

c13 nmr and dept

C-C

CH and CH3 UP

CH2 Down

C=C

C13 NMR and DEPT

CH only

C-O

unknown ipsenol4

C10H18O

Unknown: Ipsenol

Index of H deficiency: C –H/2 + 1 =2

In 13C: All carbons are visible => 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

proton spectra integration

Aliphatic

Proton Spectra: Integration

Olefinic

1 CH=

2 CH2=

CH-O

6 H:

1 CH, 2 CH2

OH

4H

6H : 2 Me

Triplet!!

1H

1H

hetcor

o

x

=CH2

Me

o

=CH

CH-O

HETCOR

o

CH-O

=CH

proton spectra expansion

2 s?

d

dd

Proton Spectra: Expansion

Septet?

dd

d

ddd

2 x d

dd

m + OH

ddd

CH2

slide8

=CH2

Me

=CH

CH-O

o

CH2 x

Me

C

o

CH2

CH-OH

CH2

CH-O

CH o

=CH2

Me

Me

CH=CH2 o

=CH

x

o

COSY
proton spectra analysis

6’

6’

Proton Spectra: Analysis

7 cis/trans

4

5a

6

5

7

H4 : not septet

8c

8t

8

3

4

tt

2

1’

1

2

1,1’

5b

3a

3b

slide10

x

Me

o

CH=

CH-O

o

CH2 x

o

C

CH

CH2

CH2 o

CH-OH

CH2

CH o

Me

Me

o

x

x

=CH

=CH

=C

=C

HMBC