Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 14, Problem 40P
Consider these reactions:
The intermediate A is a covalently bonded compound that has typical
What are the structures of A and B?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Identify products A and B from the given 1H NMR data.
Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1H) ppm. What is the structure of B?
Identify products A and B from the given 1H NMR data.a. Treatment of CH2 = CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1H NMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A?b. Treatment of acetone [(CH3)2C = O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?
Identify products A and B from the given 1H NMR data.
a.Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1H NMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A?
b.Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?
Chapter 14 Solutions
Organic Chemistry
Ch. 14 - PRACTICE PROBLEM 14.1 Provide a name for each of...Ch. 14 - Prob. 2PPCh. 14 - Prob. 3PPCh. 14 - Practice Problem 14.4 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.5 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.6 1,3,5-Cycloheptatriene is...Ch. 14 - Prob. 7PPCh. 14 - Prob. 8PPCh. 14 - Practice Problem 14.9 In 1967 R. Breslow (of...Ch. 14 - Prob. 10PP
Ch. 14 - Practice Problem 14.11 In addition to a signal...Ch. 14 - PRACTICE PROBLEM 14.12
Azulene has an appreciable...Ch. 14 - Practice Problem 14.13 (a) The -Sh group is...Ch. 14 - Practice Problem 14.14
Explain how NMR...Ch. 14 - PRACTICE PROBLEM 14.15 Four benzenoid compounds,...Ch. 14 - Prob. 16PCh. 14 - Write structural formulas and give acceptable...Ch. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Which of the hydrogen atoms shown below is more...Ch. 14 - 14.22 The rings below are joined by a double bond...Ch. 14 - Prob. 23PCh. 14 - 14.24 (a) In 1960 T. Katz (Columbia University)...Ch. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - 14.27 5-Chloro-1,3-cyclopentadiene (below)...Ch. 14 - Prob. 28PCh. 14 - Furan possesses less aromatic character than...Ch. 14 - 14.30 For each of the pairs below, predict...Ch. 14 - Assign structures to each of the compounds A, B,...Ch. 14 - Prob. 32PCh. 14 - Give a structure for compound F that is consistent...Ch. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - The IR and 1H NMR spectra for compound X(C8H10)...Ch. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - 14.39 Given the following information, predict the...Ch. 14 - Consider these reactions: The intermediate A is a...Ch. 14 - Prob. 41PCh. 14 - Compound E has the spectral features given below....Ch. 14 - Draw all of the molecular orbitals for...Ch. 14 - Prob. 1LGPCh. 14 - Prob. 2LGPCh. 14 - 3. The NMR signals for the aromatic hydrogens of...Ch. 14 - Prob. 4LGPCh. 14 - Prob. 5LGP
Additional Science Textbook Solutions
Find more solutions based on key concepts
3. What is free-fall, and why does it make you weightless? Briefly describe why astronauts are weightless in th...
The Cosmic Perspective (8th Edition)
Explain all answers clearly, with complete sentences and proper essay structure if needed. An asterisk(*) desig...
Cosmic Perspective Fundamentals
a. Draw the mechanism for the following reaction if it a involves specific-base catalysis. b. Draw the mechanis...
Organic Chemistry (8th Edition)
Water in nature exists in three different phases: solid, liquid, and vapor (gas). Indicate the relative magnitu...
Fundamentals Of Thermodynamics
What type of cut would separate the brain into anterior and posterior parts?
Anatomy & Physiology (6th Edition)
Ten changes observed in the world needs to be describes. The changes which involve chemistry needs to be identi...
Living By Chemistry: First Edition Textbook
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 compound C (molecular formula C4H8O) with C6H5MgBr, followed by H2O, affords compound D (molecular formula C10H14O). Compound D has a strong peak in its IR spectrum at 3600–3200 cm−1. The 1H NMR spectral data of C and D are given. What are the structures of C and D? Compound C signals at 1.3 (singlet, 6 H) and 2.4 (singlet, 2 H) ppm Compound D signals at 1.2 (singlet, 6 H), 1.6 (singlet, 1 H), 2.7 (singlet, 2 H), and 7.2 (multiplet, 5 H) ppmarrow_forwardIdentify products A and B from the given 1H NMR data. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1H NMR spectrum: 2.2 (singlet, 3H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A?arrow_forwardCompound A with molecular formula C6H10 has two peaks in its 1H NMR spectrum, both of which are singlets (with ratio 9 : 1). Compound A reacts with an acidic aqueous solution containing mercuric sulfate to form compound B, which gives a positive iodoform test and has an 1H NMR spectrum that shows two singlets (with ratio 3 : 1). Identify A and B.arrow_forward
- Compounds 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_forwardWhen compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3Hdoublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identifycompounds A and B.arrow_forwardTreatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppmarrow_forward
- Compound A has molecular formula C8H8O. Its 1H-NMR spectrum shows a singlet of integral 3 at 2.58 ppm and a multiplet of integral 5 at 7.84 ppm. Its 13C-NMR shows six signals. The IR spectrum of A shows a noteworthy stretch at 1698 cm-1. - Identify Compound A, explaining your reasoning Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows three singlets, one at 2.52 pm of integral 3, one at 6.67 ppm of integral 1, and one at 7.94 ppm of integral 2. The 13C-NMR spectrum of B shows six signals. - Identify compound B, explaining your reasoning Compound B is treated with methylmagnesium bromide and then aqueous acid to generate compound C. The 1H-NMR spectrum of C shows four singlets, one at 1.42 ppm of integral 6, one at 2.3 ppm of integral 1, one at 8.52 ppm of integral 2, and one at 8.97 ppm of integral 1. The 13C-NMR spectrum of C shows six signals. The IR spectrum of C shows a noteworthy broad stretch at 3326 cm-1. -…arrow_forwardCompound X (molecular formula C10H120) was treated with NH2NH2, ¯OH to yield compound Y (molecular formula C10H14). Match the 1H NMR spectra of X and Y to the corresponding structures of X and Y. Compound NH2NH2 Compound 'H NMR of X 6 H OH Y 1 H 5H 8. 6. 4 ppm or H NMR of Y 6 H 2H 5H 1 H multiplet multiplet 8. 6. 4. 3. 1 nnm 2. 2. 3, O:arrow_forwardstep by step show details how to do this question？I am confusedarrow_forward
- Use the observed 1H NMR data to decide whether C and its dianion are aromatic, antiaromatic, or not aromatic. C shows NMR signals at –4.25 (6 H) and 8.14–8.67 (10 H) ppm. The dianion of C shows NMR signals at –3 (10 H) and 21 (6 H) ppm. Why are the signals shifted upfield (or downfield) to such a large extent?arrow_forwardCompound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily in 10% Na2CO3 and 10% NaOH. When these alkaline solutions are acidified with 10% HCl, compound I is recovered unchanged. Given this information and its 1H-NMR spectrum, deduce the structure of compound I.arrow_forwardWhen compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.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