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Supplemental Information. 1H spectrum of 2-iodopropane (from Handbook of Proton NMR Spectra and Data ). Supplemental Information. 1H spectrum of p-xylene (from Sigma-Aldrich website) Compare the aromatic signal with that of bisphenol-A. Homework (Due on 10/26).

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supplemental information
Supplemental Information
  • 1H spectrum of 2-iodopropane (from Handbook of Proton NMR Spectra and Data)
supplemental information1
Supplemental Information
  • 1H spectrum of p-xylene (from Sigma-Aldrich website)
    • Compare the aromatic signal with that of bisphenol-A
homework due on 10 26
Homework (Due on 10/26)
  • Using the spectra provided to you (taken on 400MHz spectrometer), measure 1JCH for proton “f”. Please remember to use proper units for your result.
  • Please draw the structure of the molecule and assign all the carbons and protons by the numbers and letters that I labeled on 1H and 13C spectra.
    • Hint: (1) this is a dicyclic molecule with formula C10H18O (2) one piece of the puzzle is a carbon bond sequence 7-4-5 (see labeling on the 13C spectrum). (3) there is an alcohol group, and the alcohol protons typically do not give cross peaks on COSY, HMQC, and HMBC spectra.
  • A useful approach to the problem may be: (1) use 1H, 13C, DEPT and HMQC to construct “bricks” or individual groups (CH, CH2, CH3, quaternary carbons); (2) use COSY to build some obvious local connections; (3) use HMBC to build final connections
  • Please explain:
    • Why is proton peak “f” a singlet
    • Why is proton signal “a” split into 4 peaks
    • The major feature of proton signal “d” is a triplet. Why?
    • Which protons on your molecule are contributing to proton signal group “c”
    • Why do proton f and carbon 4 not have a correlation peak
    • Why do proton a and carbon 7 not have a correlation peak
    • On page 10, the peaks at (1.02, 20.4) and (0.82, 20) are narrowly-splitting doublets even though their corresponding proton peaks are singlets. Why? What J-coupling constants might you extract here?
  • These questions will help you verify your solution. If your molecule can’t seem to explain any of the questions, your solution is likely wrong
  • Polymer797dd students and visitors please send answers to me. Chem552 students please send answers to Jim Chambers.
  • Note that some proton groups are overlapping, so I integrated them together.

a b c d e f g

13 c and dept 135
13C and DEPT-135

1 2 3 4 5 6 7 8 9 10

HMQC (right figure is a close-up view of the left)
  • HMQC tells us that both the proton signal group “c” and “e” are actually from 3 different groups.
  • Two CH2 carbons are bonded to non-equivalent protons.
  • The cross peaks are only arising from 2JHH and 3JHH. (There are no 4JHH cross peaks)


  • All C-H pairs related by 2J have peaks.
  • No all C-H pairs related by 3J have peaks (some do; some don’t).
hmbc close up view 1
HMBC – Close-Up View 1
  • In this signal region, some 2J and 3J signals overlap with 1J signals. So I circled 1J peaks with red circles and 2J and 3J peaks with green circles to make the job easier for you. Peaks not circled are all from 2J and 3J couplings.

HMBC – Close-Up View 2

  • Note the subtle 13C chemical shift difference between carbon 8 and 9 cross peaks.