Concept explainers
Interpretation:
The 13CNMR spectrum of homotropilidene taken at room temperature shows three peaks.
Concept introduction:
Carbon-13 (C13) nuclear magnetic resonance (most commonly known as carbon-13 NMR or 13C NMR or sometimes simply referred to as carbon NMR) is the application of nuclear magnetic resonance (NMR) spectroscopy to carbon. It is analogous to proton NMR (1HNMR) and allows the identification of carbon atoms in an organic molecule just as proton NMR identifies hydrogen atoms. As such 13CNMR is an important tool in chemical structure elucidation in organic chemistry. 13CNMR detects only the 13C isotope of carbon, whose
13C chemical shifts follow the same principles as those of 1H, although the typical range of chemical shifts is much larger than for 1H.
Want to see the full answer?
Check out a sample textbook solutionChapter 30 Solutions
Organic Chemistry
- Analyze the 13C-NMR spectrum of C8H9NO given below and draw the structure of the compound.arrow_forwardChoose the data that best matches this partial structure; the hydrogens represented by the data are in bold: H H H Od, 2H, 4.0 ppm Os, 2H, 7.1 ppm Od, 2H, 6.8 ppm Ot, 2H, 6.9 ppm Harrow_forwardDescribe the 1H NMR spectrum of each compound. State how many NMR signals are present, the splitting pattern for each signal, and the approximate chemical shift.arrow_forward
- Sketch the 1H NMR spectra of the following compounds.arrow_forwardDraw the H1 NMR spectra of ethylcyclopropane. Draw the chemical structure and predict the proton splitting and chemical shifts.arrow_forwardidentify the compound with molecular formula C2H6O that gives this 1H NMR spectrum.arrow_forward
- A compound has the molecular formula: C4H8O2 and gives the following C-13 NMR spectrum. Provide the most likely functional groups for each signal. nmrsim presentation 1 1 C:Bruken Topspin3.5pl7 examdata -170.7658 150 100 d 170.8 (O CH3) ; 60.4 (OCH2); 20.8, 14.1 (2 x C=0) 8170.8 (0 CH₂); 60.4 (OCH₂); 20.8, 14.1 (2 x CH3). 8 170.8 (CH3); 60.4 (C-0); 20.8, 14.1 (2 x OCH₂) 8170.8 (C=0); 60.4 (OCH₂); 20.8, 14.1 (2 x CH3). 8170.8 (C) ; 60.4 (O CH2); 20.8, 14.1 (2 x O CH3) -60.4293 50 -20.7902 -14.1385 [ppm] [+]arrow_forwardI have an NMR spectrum. There are 5 peaks. There is a doublet at 9.6 ppm (integrates to 2H) where aldehydes usually arise, don't know why there is a doublet. There is another doublet at 6.8 ppm, again integrating to 2H. There is a singlet at 4.5 ppm, integrating to 1H. There is a quartet at 3.4 ppm integrating to 2H, and lastly there is triplet at 1.2 ppm integrating to 3H. i recofnize the ethyl group but don't understand the peak at 9.6 unless it is not an aldehyde.arrow_forwardWhich of the following compounds is consistent with the following proton decoupled 13C NMR spectrum 2 OH HO B 11 PPM OH III 30 20 IV OHarrow_forward
- The 1H-NMR spectrum of 1,3-propanediol (HO-CH2-CH2-CH2-OH) shows a quintet at 1.81 ppm, a singlet at 2.75 pm, and a triplet at 3.83 ppm. Assign each signal to the protons it corresponds to in the molecule. Explain the splitting pattern observed for each signal.arrow_forwardHow could 1H NMR spectroscopy be used to distinguish betweencompounds X and Y?arrow_forwardWhich structure is consistent with the 13C NMR spectrum shown?arrow_forward
- EBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENTOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning