EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 15, Problem 15.10P
Interpretation Introduction
(a)
Interpretation:
The
Concept introduction:
Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and
Interpretation Introduction
(b)
Interpretation:
The
Concept introduction:
Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. The absorption is a phenomenon in which the matter goes into the bulk phase. The absorption of light in the UV region happens due to the transition of electrons from one level to another. The wavelength of maximum absorption that is
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Chapter 15 Solutions
EBK ORGANIC CHEMISTRY
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42APCh. 15 - Prob. 15.43APCh. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50APCh. 15 - Prob. 15.51APCh. 15 - Prob. 15.52APCh. 15 - Prob. 15.53APCh. 15 - Prob. 15.54APCh. 15 - Prob. 15.55APCh. 15 - Prob. 15.56APCh. 15 - Prob. 15.57APCh. 15 - Prob. 15.58APCh. 15 - Prob. 15.59APCh. 15 - Prob. 15.60APCh. 15 - Prob. 15.61APCh. 15 - Prob. 15.62APCh. 15 - Prob. 15.63APCh. 15 - Prob. 15.64APCh. 15 - Prob. 15.65APCh. 15 - Prob. 15.66APCh. 15 - Prob. 15.67APCh. 15 - Prob. 15.68APCh. 15 - Prob. 15.69APCh. 15 - Prob. 15.70APCh. 15 - Prob. 15.71APCh. 15 - Prob. 15.72APCh. 15 - Prob. 15.73APCh. 15 - Prob. 15.74APCh. 15 - Prob. 15.75APCh. 15 - Prob. 15.76APCh. 15 - Prob. 15.77APCh. 15 - Prob. 15.78APCh. 15 - Prob. 15.79APCh. 15 - Prob. 15.80APCh. 15 - Prob. 15.81APCh. 15 - Prob. 15.82APCh. 15 - Prob. 15.83APCh. 15 - Prob. 15.84APCh. 15 - Prob. 15.85APCh. 15 - Prob. 15.86AP
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- 2. DEDUCE THE STRUCTURE FOR EACH COMPOUND WITH THE INDICATED MOLECULAR FORMULA AND THE 1H-NMR AND IR SPECTRA PROVIDED. (a) Formula: C4H8O2arrow_forwardWhich of the following compounds justifications for your answer. V) are aromatic and anti-aromatic? Give i) ii) iii) iv)arrow_forwardDescribe the characteristic infrared absorption frequencies that would allow you to distinguish between the following pairs of compounds. (a) 2,3-dimethylbut-2-ene and 2,3-dimethylbut-1-ene (b) cyclohexa-1,3-diene and cyclohexa-1,4-dienearrow_forward
- (b) Distinguish each as aromatic or antiaromatic in terms of electronic basis. Explain which one is more stable. (i) Compound A Compound B (ii) Compound C Compound D (iii) Compound E Compound Farrow_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(c) Each of the following compounds below are characterized by a 'H NMR spectrum that consists of only a single peak having the chemical shift indicated. Identify and draw a bond-line structure for each compound, respectively.. (i) C2H3Cl3-at ~ 2.7 ppm (ii) C5H10-at 1.5 ppmarrow_forward
- 8. (a) Benzene derivatives exhibit medium to strong absorption in UV-region. Explain why aniline and phenoxide ion have strong UV-absorptions.arrow_forwardCompound A has prominent infrared absorptions at 1050, 1786, and 1852 cm and shows a single absorption in the proton NMR spectrum at 8 3.00. When heated gently with methanol, compound B, C,H,O,, is obtained. Compound B has IR absorptions at 2500–3000 (broad), 1730, and 1701 cm-, and its proton NMR spectrum in D,0 consists of resonances at & 2.7 (complex splitting) and & 3.7 (a singlet) in the intensity ratio 4:3. Give the structures A and B, omitting stereochemistry.arrow_forwardCompound E with a molecular formula C8H7CIO shows a prominent band in its IR spectrum at 1690 cm-1. 'H NMR spectrum revealed only two major types of protons in the ratio of 5: 2. 2 (i) Calculate the degree of unsaturation for compound E. (ii) By using the IR and 'H NMR data given, deduce the structure of compound E.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) The 'H-NMR spectrum of cyclobutanone shows two signals - signal A at 3.00 ppm and signal B at 1.95 ppm. Give the multiplicity of each signal. cyclobutanone (b) When cyclobutanone is treated with D20 and NaOD, the only signal observable in the 1H-NMR is a singlet at 2.00 ppm. Explain why this is the case. [Note: Deuterium atoms do not display signals in the TH-NMR spectrum]arrow_forwardThe 1H and 13C NMR spectra of compound A, C8H9Br are shown below. Answer the following questions. 1(a) Degree of the unsaturation of this compound is = , 1(b) The derived unsaturation number indicates that compound has ............= 1(c) Two peaks in between 6.5 - 8.0 δ indicate that compound is= 1(d) According to the splitting pattern of the peak at 1.20 δ and 2.58 δ indicates that compound has a .................. group= 1(e) According to the 1H NMR spectrum the number of nonequivalent aromatic proton sets in the compound = 1(f) According to the 13C NMR, the number of nonequivalent carbons in the compound is = 1(g) According to your answer in Q 1(f) the compound has a plane of symmetry Yes or NO = 1(h) The IUAC name for this unknown compound isNOT TOO SURE ABOUT MY ANSWERS, PLEASE CORRECT ME IF I'M WRONGarrow_forward
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