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Concept explainers
(a)
Interpretation:
The reason as to why sodium salt of cyclopentadiene shows a single peak in NMR spectrum is to be stated.
Concept introduction:
Many nuclei and electrons have spin; due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an
(b)
Interpretation:
The reason as to why there is a reduction in the chemical shift value of methyl group in the given compound is to be stated.
Concept introduction:
Many nuclei and electrons have spin; due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an angular momentum related to the spin. The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus. The chemical shift value is dependent on its surrounding protons.
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Chapter 16 Solutions
EBK ORGANIC CHEMISTRY
- Compounds A and B are isomers having the molecular formula C4H8O3. Identify A and B on the basis of their 1H NMR spectra.Compound A: δ 1.3 (3H, triplet); 3.6 (2H, quartet); 4.1 (2H, singlet); 11.1 (1H, broad singlet)Compound B: δ 2.6 (2H, triplet); 3.4 (3H, singlet); 3.7 (2H triplet); 11.3 (1H, broad singlet)arrow_forwardNMR of p-iodonitrobenzene Provide the annotated NMR spectra with a drawn structure of your product and clear identification of which protons and carbons in the structure gives rise to which peak in the spectra.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
- Treatment of 2-methylpropanenitrile [(CH3)2CHCN] with CH3CH2CH2MgBr, followed by aqueous acid, affords compound V, which has molecular formula C7H14O. V has a strong absorption in its IR spectrum at 1713 cm−1, and gives the following 1H NMR data: 0.91 (triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and 2.60 (septet, 1 H) ppm. What is the structure of V? We will learn about this reaction in Chapter 20.arrow_forwardThere are several isomeric alcohols and ethers of molecular formula C5H12O. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. Isomer A: δ = 0.92 (t, 7.8 Hz, 3 H), 1.20 (s, 6H), 1.49 (q, 7.8 Hz, 2H), 1.85 (s, 1H) ppm Isomer B: δ = 1.19 (s, 9 H), 3.21 (s, 3H) ppmarrow_forwardHow could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forward
- When the 1î-NMR spectrum of acetone, CH3COCH3, is recorded on an instrument operating at 200 MHz, a single sharp resonance at 2.1î is seen. (a) How many hertz downfield from TMS does the acetone resonance correspond to? (b) If the 1î-NMR spectrum of acetone were recorded at 500 MHz, what would the position of the absorption be in î units? (c) How many hertz downfield from TMS does this 500 MHz resonance correspond to?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_forwardCompound C has the molecular formula C5H8O. The IR, 1H, 13C, and DEPT NMR spectra of this compound are shown below. (a) Calculate the double bond equivalent of compound C and briefly explain what the values obtains represents. Interpret the IR spectrum. (b) Based on the information provided, determine the structure of compound D.arrow_forward
- Describe how utlraviolet (UV) spectroscopy could distinguish propiophenone from 1-phenyl-2-propanone and p-ethylbenzaldehyde. Consider their respective wavelengths at their maximum UV absorbances, (λ max).arrow_forwardConsider the following three compounds: CH,=CH, CH,=CH-CH=CH, CH,=CH-CH=CH-CH=CH, (1) (2) (3) Compound 1 contains a simple isolated carbon-carbon double bond, but the other two have conjugated double bonds. (i) Which compound will have the largest absorption (max) Value? (ii) Give a reason for your answer in part (b)(i).arrow_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
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