EBK ORGANIC CHEMISTRY STUDY GUIDE AND S
6th Edition
ISBN: 9781319385415
Author: PARISE
Publisher: VST
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 17, Problem 17.36AP
Interpretation Introduction
Interpretation:
The structural basis for the given observations is to be stated.
Concept introduction:
Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.
Shown 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=
Treatment of 2-methylpropanenitrile [(CH3)2CHCN] withCH3CH2CH2MgBr, followed by aqueous acid, affords compound V, whichhas molecular formula C7H14O. V has a strong absorption in its IRspectrum 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), and2.60 (septet, 1 H) ppm. What is the structure of V?
Chapter 17 Solutions
EBK ORGANIC CHEMISTRY STUDY GUIDE AND S
Ch. 17 - Prob. 17.1PCh. 17 - Prob. 17.2PCh. 17 - Prob. 17.3PCh. 17 - Prob. 17.4PCh. 17 - Prob. 17.5PCh. 17 - Prob. 17.6PCh. 17 - Prob. 17.7PCh. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Prob. 17.10P
Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22APCh. 17 - Prob. 17.23APCh. 17 - Prob. 17.24APCh. 17 - Prob. 17.25APCh. 17 - Prob. 17.26APCh. 17 - Prob. 17.27APCh. 17 - Prob. 17.28APCh. 17 - Prob. 17.29APCh. 17 - Prob. 17.30APCh. 17 - Prob. 17.31APCh. 17 - Prob. 17.32APCh. 17 - Prob. 17.33APCh. 17 - Prob. 17.35APCh. 17 - Prob. 17.36APCh. 17 - Prob. 17.37APCh. 17 - Prob. 17.38APCh. 17 - Prob. 17.39APCh. 17 - Prob. 17.40APCh. 17 - Prob. 17.41APCh. 17 - Prob. 17.42APCh. 17 - Prob. 17.43APCh. 17 - Prob. 17.44APCh. 17 - Prob. 17.45APCh. 17 - Prob. 17.46APCh. 17 - Prob. 17.47APCh. 17 - Prob. 17.48APCh. 17 - Prob. 17.49APCh. 17 - Prob. 17.50APCh. 17 - Prob. 17.51APCh. 17 - Prob. 17.52APCh. 17 - Prob. 17.53APCh. 17 - Prob. 17.54APCh. 17 - Prob. 17.55APCh. 17 - Prob. 17.56APCh. 17 - Prob. 17.57APCh. 17 - Prob. 17.58APCh. 17 - Prob. 17.59AP
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
- Treatment 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_forwardFollowing is the 1H-NMR spectrum of compound O, molecular formula C7H12. Compound O reacts with bromine in carbon tetrachloride to give a compound with the molecular formula C7H12Br2. The 13C-NMR spectrum of compound O shows signals at d 150.12, 106.43, 35.44, 28.36, and 26.36. Deduce the structural formula of compound O.arrow_forwardDeduce the structure of compound C.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_forwardThe 1H NMR spectra of two carboxylic acids with molecular formula C3H5O2Cl are shown below. Identify the carboxylic acids. (The “offset” notation means that the farthest-left signal has been moved to the right by the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of 9.8 + 2.4 = 12.2 ppm.)arrow_forwardThe 13C NMR spectrum of 1-bromo-3-chloropropane contains peaks at δ 30, δ 35, and δ 43. Assign these signals to the appropriate carbons.arrow_forward
- Predict the ¹H NMR spectrum of diethoxymethane.arrow_forwardThe 'H NMR spectrum of 1,2-dimethoxyethane (CH;OCH,CH2OCH3) recorded on a 300 MHz NMR spectrometer consists of signals at 1017 Hz and 1065 Hz downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) At what frequency would each absorption occur if the spectrum were recorded on a 500 MHz NMR spectrometer?arrow_forwardThe 1H NMR spectra of two carboxylic acids with molecular formula C3H5O2Cl are shown below. Identify the carboxylic acids. (The “offset” notation means that the farthest-left signal has been moved to the rightby the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of 9.8 + 2.4 = 12.2 ppm.)arrow_forward
- Propose a structure given the 1H and 13C NMR spectra of the unknown compound. Assign chemical shifts to corresponding hydrogen and carbon atoms Molecular Formula: C5H10O3arrow_forwardA compound (L) with the molecular formula C9H10 reacts with bromine and gives an IR absorption spectrum that includes the following absorption peaks: 3035 cm ¹(m), 3020 cm ¹(m), 2925 cm ¹(m), 2853 cm ¹(w), 1640 cm ¹1(m), 990 cm ¹(s), 915 cm ¹(s), 740 cm ¹(s), 695 cm ¹(s). The ¹H NMR spectrum of L consists of: Doublet 3.1 ppm (2H) Multiplet 5.1 ppm Multiplet 7.1 (5H) ppm Multiplet 4.8 ppm Multiplet 5.8 ppm The UV spectrum shows a maximum at 255 nm. Propose a structure for compound L. Edit Drawing harrow_forwardThe functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound containing no nitrogen exhibits strong, broad absorption across the 2500-3300 cm-1 region, accompanied by 2200 (w) and 1715 (s) cm-1bands.Relative absorption intensity: (s)=strong, (m)=medium, (w)=weak. What functional class(es) does the compound belong to? List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1. The functional class(es) of this compound is(are)fill in the blank 1.(Enter letters from the table below, in any order, with no spaces or commas.) a. alkane (List only if no other functional class applies.) b. alkene h. amine c. terminal alkyne i. aldehyde or ketone d. internal alkyne j. carboxylic acid e. arene k. ester f. alcohol l. nitrile g. etherarrow_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