Organic Chemistry
8th Edition
ISBN: 9781305580350
Author: William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher: Cengage Learning
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Chapter 22, Problem 22.10P
Addition of m-xylene to the strongly acidic solvent HF/SbF5 at –45°C gives a new species, which shows 1H-NMR resonances at δ 2.88 (3H), 3.00 (3H), 4.67 (2H), 7.93 (1H), 7.83 (1H), and 8.68 (1H). Assign a structure to the species giving this spectrum.
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Addition of m-xylene to the strongly acidic solvent HF/SbF5 at -45°C gives a new species, which shows 1H-NMR resonances at d 2.88 (3H), 3.00 (3H), 4.67 (2H), 7.93 (1H), 7.83 (1H), and 8.68 (1H). Assign a structure to the species giving this spectrum.
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) ppm
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.
Chapter 22 Solutions
Organic Chemistry
Ch. 22.1 - Prob. 22.1PCh. 22.1 - Write a structural formula for the product from...Ch. 22.1 - Prob. 22.3PCh. 22.2 - Prob. 22.4PCh. 22.2 - Predict the major produce(s) of each electrophilic...Ch. 22.3 - In SN2 reactions of haloalkanes, the order of...Ch. 22 - Prob. 22.8PCh. 22 - Prob. 22.9PCh. 22 - Addition of m-xylene to the strongly acidic...Ch. 22 - Addition of tert-butylbenzene to the strongly...
Ch. 22 - What product do you predict from the reaction of...Ch. 22 - Other groups besides H+ can act as leaving groups...Ch. 22 - Prob. 22.14PCh. 22 - Prob. 22.15PCh. 22 - Prob. 22.16PCh. 22 - Prob. 22.17PCh. 22 - Suggest a reason why the nitroso group, N=O, is...Ch. 22 - Prob. 22.19PCh. 22 - Prob. 22.20PCh. 22 - The following molecules each contain two aromatic...Ch. 22 - Prob. 22.22PCh. 22 - Prob. 22.23PCh. 22 - The insecticide DDT is prepared by the following...Ch. 22 - Prob. 22.25PCh. 22 - Prob. 22.26PCh. 22 - Prob. 22.27PCh. 22 - Prob. 22.28PCh. 22 - Prob. 22.29PCh. 22 - Prob. 22.32PCh. 22 - Show how to prepare each compound from...Ch. 22 - Prob. 22.34PCh. 22 - Show reagents and conditions to bring about the...Ch. 22 - Prob. 22.36PCh. 22 - Propose a synthesis for each compound from...Ch. 22 - The first widely used herbicide for the control of...Ch. 22 - The first widely used herbicide for the control of...Ch. 22 - Prob. 22.40PCh. 22 - Prob. 22.41PCh. 22 - Prob. 22.42PCh. 22 - Prob. 22.43PCh. 22 - Cancer of the prostate is the second leading cause...Ch. 22 - Prob. 22.45PCh. 22 - Prob. 22.46PCh. 22 - Prob. 22.47PCh. 22 - When certain aromatic compounds are treated with...Ch. 22 - Prob. 22.49PCh. 22 - Following is the structure of miconazole, the...Ch. 22 - Prob. 22.51PCh. 22 - Prob. 22.52PCh. 22 - Prob. 22.53PCh. 22 - Show how the antidepressant venlafaxine (Effexor)...Ch. 22 - Prob. 22.57PCh. 22 - Given this retrosynthetic analysis, propose a...Ch. 22 - Prob. 22.59PCh. 22 - Prob. 22.60PCh. 22 - Prob. 22.61PCh. 22 - A newer generation of antipsychotics, among them...Ch. 22 - Prob. 22.63P
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- 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 (t, 9H), 1.22 (q, 2H) ppmarrow_forwardAs reaction of (CH3)2CO with LIC≡CH followed by H2O affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600−3200, 3303, 2938, and 2120 cm−1. D shows the following 1H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?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
- Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.arrow_forwardAddition of mesitylene to the strongly acidic solvent HF / SbF5 at -45 °C gives a new species which shows a 1H-NMR resonance at δ 4.67 (2H). Draw the structure of this species.arrow_forwardCompounds 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_forward
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