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NMR Assignment

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

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  1. NMR Assignment Example : IPSENOL

  2. Unknown: Ipsenol C10H18O Index of H deficiency: C –H/2 + 1 =2

  3. C-C CH and CH3 UP CH2 Down C=C C13 NMR and DEPT CH only C-O

  4. 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

  5. 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

  6. o x =CH2 Me o =CH CH-O HETCOR o CH-O =CH

  7. 2 s? d dd Proton Spectra: Expansion Septet? dd d ddd 2 x d dd m + OH ddd CH2

  8. =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

  9. 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

  10. 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

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