Interpretation:
The structure for an alcohol with molecular formula
Concept introduction:
Few elements such as
Chemical shift is the measurement on
NMR resonance generally splits into
Two hydrogen on adjacent atoms splits into three peaks, a triplet and so on.
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Organic Chemistry
- (a) Compound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning. (b) Compound B has molecular formula C8H6O2. The IR, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for B and explain your reasoning. (c) Compound C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for C and explain your reasoning.arrow_forward(a) Compound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning. (b) Compound B has molecular formula C8H6O2. The IR, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for B and explain your reasoning. (c) Compound C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for C and explain your reasoning.arrow_forward(a) Compound A has molecular formula C9H18O but shows only one singlet in the 1H-NMR spectrum. Suggest a structure for A and explain your reasoning. (b) Compound B has molecular formula C10H14. The IR, mass, 1H-NMR, and 13C-NMR speca are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for B and explain your reasoning.arrow_forward
- Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.arrow_forwardCompound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning.arrow_forwardFollowing is the 1H-NMR spectrum of compound O, molecular formula C7H12. Compound O reacts with bromine in carbon tetrachloride to give a compound with the molecular formula C7H12Br2. The 13C-NMR spectrum of compound O shows signals at d 150.12, 106.43, 35.44, 28.36, and 26.36. Deduce the structural formula of compound O.arrow_forward
- There are several isomeric alkanes of molecular formula C6H14.Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. Isomer A: δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm Isomer B: δ = 0.84 (t, 3 H), 0.86 (s, 9H), 1.22 (q, 2H) ppmarrow_forwardThe two isomeric compounds with the formula C3H5ClO2 have NMR spectra shown in 4a and 4b below. The downfield protons appearing in the NMR spectra at about 12.1 and 11.5 ppm, respectively, are shown as insets. Determine the structure of the two isomersarrow_forwardTreatment of alcohol A (molecular formula C5H12O) with CrO3, H2SO4, and H2O affords B with molecular formula C5H10O, which gives an IR absorption at 1718 cm−1. The 1H NMR spectrum of B contains the following signals: 1.10 (doublet, 6 H), 2.14 (singlet, 3 H), and 2.58 (septet, 1 H) ppm. What are the structures of A and B?arrow_forward
- Compound X of the molecular formula C7H10 has the 13C NMR spectrum (5 signals) shown below. On treatment with excess H2/Pt (catalytic hydrogenation), X is converted to methylcyclohexane. Propose a structure for X and justify your reasoning by clearly labeling each carbon signal and write out the reaction. 200 180 160 140 120 100 80 60 40 20arrow_forwardThere are several isomeric alkanes of molecular formula C6H14. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. Isomer A: δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm Isomer B: δ = 0.84 (t, 3 H), 0.86 (t, 9H), 1.22 (q, 2H) ppmarrow_forwardA hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?arrow_forward