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You must label all the NMR peaks and the main IR peaks. If you are turning in your work on a separate sheet of paper you must answer the questions in the order I asked them and staple your papers. It is due Wednesday at 10 am. No late or unstapled quizzes will be accepted.
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Question & Answer: You must label all the NMR peaks and the main IR peaks. If you are turning in your work on…..
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Expert Answer
Find double bond equivalence: DBE = C+1-H/2-X/2+N/2
DBE = 9+1-4:
DBE = 5. This means total 5 double bonds in the compound in the compound. One should be aromatic group.
If we see 1H nmr and IR there is an aldehyde group and ethelene present on the aromatic ring. A singlet at 7.18 shows both substituted in meta position to each other.
| Signal | Chemical shift | Multiplicity | No. of Hydrogens |
| a | 5.25 | Dd, J = 16, 13 | 1H (alkene) |
| b | 5.76 | Dd J = 16, 0.9 | 1H (alkene) |
| c | 6.72 | Dd J = 13, 0.9 | 1H (alkene) |
| d | 7.18 | singlet | aromatic |
| e | 7.48 | triplet | aromatic |
| f | 7.76 | doublet | aromatic |
| g | 7.88 | doublet | aromatic |
| h | 9.88 | singlet | Aldehyde proton |
IR values:
1700- aromatic aldehyde group
1600, 1200- alkene group
750, 2900- aromatic SP2 hydrogens
