*20. The two isomeric compounds with the formula C10H14 have NMR spectra shown below. Make no attempt to interpret the aromatic proton area between 7.1 and 7.3 ppm except to determine the number of protons attached to the aromatic ring. Draw the structures of the isomers. C10H14 5.10 7.5 7.0 1.10 2.15 3.04 2.94 4.0 2.0 1.0 (a) C10H14

*20. The two isomeric compounds with the formula C10H14 have NMR spectra shown below. Make no attempt to interpret the aromatic proton area between 7.1 and 7.3 ppm except to determine the number of protons attached to the aromatic ring. Draw the structures of the isomers. C10H14 5.10 7.5 7.0 1.10 2.15 3.04 2.94 4.0 2.0 1.0 (a) C10H14

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter15: Benzene And Aromaticity

Section15.SE: Something Extra

Problem 15VC: 1, 6-Methanonaphthalene has an interesting 1H NMR spectrum in which the eight hydrogens around the...

See similar textbooks

Similar questions

1, 6-Methanonaphthalene has an interesting 1H NMR spectrum in which the eight hydrogens around the perimeter absorb at 6.9 to 7.3 Î´, while the two CH2 protons absorb at -0.5 Î´. Tell whether it is aromatic, and explain its NMR spectrum.
Using your knowledge of general chemistry and organic chemistry, answer the following questions: Predict the relative acidities within each of the following groups, then identify the conjugate bases, and finally predict their relative leaving group abilities HClO3 and HClO2 PH3 and H2S [NH4]+ and [H3O]+ Explain why the trifluoromethane sulfonate (TfO-) ion is a better leaving group than the methanesulfonate (MsO-) ion.
3 The 1H- and 13C-NMR data of an ester of molecular formula C6H10O2 are given below. Also shown are the COSY and HETCOR NMR spectra of the ester. Identify the ester, explaining how you reach your conclusion. 1H-NMR: 7.20-6.90 (1H), 5.85 (1H), 4.16 (2H), 1.88 (3H), 1.31 (3H) ppm 13C-NMR: 166.7, 144.5, 123.0 , 60.2, 18.0, 14.3 ppm
Propose a structure, carefully outlining how you arrived at that answer
24. NMR spectra of two isomeric compounds with formula C|0H,2O. Their infrared spectra show strong bands close to 1715 cm*1. Do not attempt to interpret the aromatic proton region between 7.1 and 7.4 ppm, unless to determine the number of protons attached to the aromatic ring. draw the structures of the compounds. note: point out the structures on the chart for better understanding
There is a Compound called A with the formula C13H20N2O2. In the proton NMR two triplets at 2.8 and 4.3 are coupled to each other. Similarly, a triplet at 1.1 and a quartet at 2.6 are coupled to each other. the carbonyl group appearing at 1669 cm -1 in the IR spectrum has an unusually low value. Please provide the structure of A.
As we will learn in Chapter 20, 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. 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?
As we will learn in Chapter 17, 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?
Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively. - Identify compound B, explaining your reasoning
25. NMR spectra (Figures a, b, c and d) of four isomers are shown. with formula Cl0HuOr Their infrared spectra show strong bands close to 1735 cm'1, do not interpret the aromatic proton region between 7.0 and 7.5 ppm, except to determine the number of protons attached to the aromatic ring. Draw as structures of compounds.
When an NMR spectrum is taken of the molecule below, there is the expected 5H in the aromatic region, triplet (3H) and quartet (2H) upfield of the aromatic region. After adding a drop of acidic D2O, the spectrum changes to one that just has 5H aromatic protons and an upfield 3H singlet. What happens? Explain with mechanisms step by step and include charges and lone pairs of all compounds involved.
The 1H NMR spectra for two esters with molecular formula C8H8O2 are shown next. Which of the esters is hydrolyzed more rapidly in an aqueous solution with a pH of 10?