Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 13, Problem 13.25P
Interpretation Introduction
Interpretation:
The structure of the compound with the formula
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
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The 13C-NMR spectrum of 3-methyl-2-butanol shows signals at d 17.88 (CH3), 18.16 (CH3), 20.01 (CH3), 35.04 (carbon-3), and 72.75 (carbon-2). Account for the fact that each methyl group in this molecule gives a different signal.
Pls explain too
In the 1H NMR spectra of 2-bromopropane (CH3)2CHBr and 1-bromopropane CH3CH2CH2Br, how many signals do you expect to see?
Chapter 13 Solutions
Organic Chemistry
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Prob. 13.4PCh. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10P
Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.24PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Prob. 13.36APCh. 13 - Prob. 13.37APCh. 13 - Prob. 13.38APCh. 13 - Prob. 13.39APCh. 13 - Prob. 13.40APCh. 13 - Prob. 13.41APCh. 13 - Prob. 13.42APCh. 13 - Prob. 13.43APCh. 13 - Prob. 13.44APCh. 13 - Prob. 13.45APCh. 13 - Prob. 13.46APCh. 13 - Prob. 13.47APCh. 13 - Prob. 13.48APCh. 13 - Prob. 13.49APCh. 13 - Prob. 13.50APCh. 13 - Prob. 13.51APCh. 13 - Prob. 13.52APCh. 13 - Prob. 13.53APCh. 13 - Prob. 13.54APCh. 13 - Prob. 13.55APCh. 13 - Prob. 13.56APCh. 13 - Prob. 13.57APCh. 13 - Prob. 13.58APCh. 13 - Prob. 13.59APCh. 13 - Prob. 13.60APCh. 13 - Prob. 13.61APCh. 13 - Prob. 13.62APCh. 13 - Prob. 13.63APCh. 13 - Prob. 13.64APCh. 13 - Prob. 13.65APCh. 13 - Prob. 13.67AP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- Compound 1 has molecular formula C7H15Cl. It shows two signals in the 1H-NMR spectrum, one at 1.08 ppm and one at 1.59 ppm. The relative integrals of these two signals are 3 and 2, respectively. Propose structures for compound 1, explaining how you reach your conclusion.arrow_forwardThe 'H NMR spectrum of compound A (C3H100) has four signals: a multiplet at 8 = 7.25-7.32 ppm (5 H), a singlet at d = 5.17 ppm (1 H), a quartet at d = 4.98 ppm (1 H), and a doublet at ô = 1.49 ppm (3 H). There are 6 signals in its 13C NMR spectrum. The IR spectrum has a broad absorption in the -3200 cm-1 region. Compound A reacts with KMNO4 in a basic solution followed by acidification to give compound B with the molecular formula C7H6O2. Draw structures for compounds A and B.arrow_forwardDeduce the structure of the compound that gives the following 1H, 13C, and IR spectra. Assign all aspects of the 1H and 13C spectra to the structure you propose. Use letters to correlate protons with the signals in the 1H NMR spectrum, and numbers to correlate carbons with the signals in the 13C spectrum. The mass spectrum of this compound shows the molecular ion at m/z 148. 6H 1H 2H 2H 1H 10 8 6 4 2 84 (ppm) CH CH CH3 CH CH 200 180 160 140 120 100 80 60 40 20 100 90 80 70 60 50 40 30 20 10 4000 3000 2000 1500 1000 500 Wavenumber (cm) Transmittance (%)arrow_forward
- The mass spectrum and 13C NMR spectrum of a hydrocarbon are shown. Propose a structure for this hydrocarbon, and explain the spectral data.arrow_forwardPropose a structural formula for the analgesic phenacetin, molecular formula C10H13NO2, based on its 1H-NMR spectrum.arrow_forwardThe 13C-NMR spectrum of 3-methyl-2-butanol shows signals at 17.88 (CH3), 18.16 (CH3), 20.01 (CH3), 35.04 (carbon-3), and 72.75 (carbon-2). Account for the fact that each methyl group in this molecule gives a different signal.arrow_forward
- The 1H-NMR spectrum of compound R, C6H14O, consists of two signals: d 1.1 (doublet) and d 3.6 (septet) in the ratio 6:1. Propose a structural formula for compound R consistent with this informationarrow_forward1) They were given a mysterious chemical by ASP Rinuprasad, coded as Compound X, which was believed to be a forbidden drug smuggled through KLIA airport. Little they know about the chemical from the convict. As competent officers, together they decided to run several spectroscopic and analytical tests to determine the molecular structure of Compound X. They managed to obtain the spectra and data of IR, 'H NMR (400 MHz in CDCI 3), 13C and DEPT (100.6 MHz in CDC13), COSY, HMQC, HMBC, mass spectrometry and elemental analyses. Determine the molecular structure and name of Compound X. IR SPECTRUM 2.5 100 3.0 4.0 5,0 6.0 7.0 8.0 9.0 10 12 15 2000 avenumber cm 1500 000 1000 1Η ΝMR 7.50 7.45 7.40 7.35 7.30 7.25 7.20 7.15 7.10 7.05 7.00 55 50 1.5 100 9.5 9.0 8.5 8.0 7.5 7.0 65 60 4.5 35 3.0 2.5 20 10 0.5arrow_forwardShown below is th carbon NMR spectrum of a compound of the formula C7H14. The compound decolorizes a solution of bromine in dichloromethane. Assign a structure to this hydrocarbon. Correlation of spectrum with number of hydrogen atoms attached to each carbon=arrow_forward
- what is is the structure of a compound of molecular formula c 10 h 14 o 2 that shows a strong ir absorption at 3150-2850 cm − 1 and give the following 1 h nmr absorptions: 1.4 (triplet, 6 h), 4.0 (quarter, 4h), and 6.8 (singlet, 4h) ppm?arrow_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_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
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
Organic Chemistry
Chemistry
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:Cengage Learning
NMR Spectroscopy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=SBir5wUS3Bo;License: Standard YouTube License, CC-BY