Organic Chemistry - Standalone book
10th Edition
ISBN: 9780073511214
Author: Francis A Carey Dr., Robert M. Giuliano
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 12.18, Problem 20P
Give an explanation for each of the following observations:
Compound A has six
electrons but is not
Compound B has six
Compound C has
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
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)
Based on the structures given below, which of the following statements is entirely true?
N
Quinoline
NH
Isoquinoline Indole
Isoindole
All four compounds are non-aromatic, with 12 pi-electrons each and are equal in basicity.
All four compounds are aromatic, with 10 pi-electrons each and are equal in basicity.
Only quinoline and isoquinoline having 10 pi-electrons each are aromatic while indole and isoindole are antiaromatic having 8 pi-electrons each.
All four compounds are aromatic, with 10 pi-electrons each, and with quinoline and isoquinoline being stronger bases than indole and isoindole.
The nitrogen atoms of quinoline and isoquinoline are sp2 hybridized while the nitrogen atoms of indole and isoindole are sp³ hybridized
Previous
Select the most accurate statement about the following compound.
The compound is aromatic because there are 4n+2 pi electrons.
O The compound is aromatic because it is completely conjugated
throughout the entire ring.
O The compound is non-aromatic (not aromatic) because it is not planar.
O The compound is anti-aromatic because there are 4n pi electrons.
Chapter 12 Solutions
Organic Chemistry - Standalone book
Ch. 12.2 - Write structural formulas for toluene (C6H5CH3)...Ch. 12.3 - Prob. 2PCh. 12.5 - Prob. 3PCh. 12.5 - Prob. 4PCh. 12.6 - Prob. 5PCh. 12.6 - Chrysene is an aromatic hydrocarbon found in coal...Ch. 12.8 - Prob. 7PCh. 12.9 - As measured by their first-order rate constants,...Ch. 12.9 - Give the structure of the principal organic...Ch. 12.9 - Prob. 10P
Ch. 12.10 - Prob. 11PCh. 12.11 - Prob. 12PCh. 12.12 - Prob. 13PCh. 12.13 - Prob. 14PCh. 12.13 - Prob. 15PCh. 12.15 - The regioselectivity of Birch reduction of...Ch. 12.16 - Prob. 17PCh. 12.17 - Both cyclooctatetraene and styrene have the...Ch. 12.17 - Prob. 19PCh. 12.18 - Give an explanation for each of the following...Ch. 12.19 - Prob. 21PCh. 12.19 - What does a comparison of the heats of combustion...Ch. 12.20 - Prob. 23PCh. 12.20 - Prob. 24PCh. 12.20 - Prob. 25PCh. 12.20 - Prob. 26PCh. 12.20 - Prob. 27PCh. 12.20 - Prob. 28PCh. 12.21 - Prob. 29PCh. 12.21 - Prob. 30PCh. 12.22 - Prob. 31PCh. 12.22 - Prob. 32PCh. 12 - Write structural formulas and give the IUPAC names...Ch. 12 - Prob. 34PCh. 12 - Prob. 35PCh. 12 - Prob. 36PCh. 12 - Prob. 37PCh. 12 - Acridine is a heterocyclic aromatic compound...Ch. 12 - Prob. 39PCh. 12 - Prob. 40PCh. 12 - Prob. 41PCh. 12 - Evaluate each of the following processes applied...Ch. 12 - Prob. 43PCh. 12 - Prob. 44PCh. 12 - Prob. 45PCh. 12 - Prob. 46PCh. 12 - Anthracene undergoes a DielsAlder reaction with...Ch. 12 - Prob. 48PCh. 12 - Prob. 49PCh. 12 - The relative rates of reaction of ethane, toluene,...Ch. 12 - Both 1,2-dihydronaphthalene and...Ch. 12 - Prob. 52PCh. 12 - Prob. 53PCh. 12 - Prob. 54PCh. 12 - Prob. 55PCh. 12 - Prob. 56PCh. 12 - Each of the following reactions has been described...Ch. 12 - Prob. 58PCh. 12 - A compound was obtained from a natural product and...Ch. 12 - Prob. 60PCh. 12 - Suggest reagents suitable for carrying out each of...Ch. 12 - Prob. 62PCh. 12 - Prob. 63DSPCh. 12 - Prob. 64DSPCh. 12 - Prob. 65DSPCh. 12 - Prob. 66DSP
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
- When 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_forwardCompound A of molecular formula C3H6O shows a noteworthy infrared absorption at 1716 cm-1. Its 1H-NMR spectrum shows one singlet – δ 2.2 (6H) ppm. Its 13C-NMR spectrum has two signals – δ 30, 207 ppm. Suggest a structure for this compound.arrow_forwardCompound A is an aromatic compound with the molecular formula C8H8. When treated with excess BR2, compound A is converted into compound B, with the molecular formula C8H8BR2. Draw the structure of compound a. arrow_forward
- 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.arrow_forwardCompound A has molecular formula C7H7X. Its 1H-NMR spectrum shows a singlet at 2.26 ppm and two doublets, one at 6.95 ppm and one at 7.28 ppm. The singlet has an integral of three and the doublets each have an integral of two. Its 13C-NMR shows five signals. The mass spectrum of A shows a peak at m/z = 170 and another peak at m/z = 172; the relative height of the two peaks is 1:1 respectively. - Identify what atom X is, explaining your reasoning - 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 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 Compound B is treated with sodium ethoxide to generate compound C. The 1H-NMR spectrum of C shows…arrow_forwardAll-cis cyclodecapentaene is a stable molecule that shows a single absorption in its1HNMR spectrum at 5.67 δ. Tell whether it is aromatic, and explain its NMR spectrumarrow_forward
- Compound B has molecular formula C9H12O. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.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_forwardCompound A has molecular formula C7H7X. Its ¹H-NMR spectrum shows a singlet at 2.26 ppm and two doublets, one at 6.95 ppm and one at 7.28 ppm. The singlet has an integral of three and the doublets each have an integral of two. Its 13C- NMR shows five signals. The mass spectrum of A shows a peak at m/z 170 and another peak at m/z = 172; the relative height of the two peaks is 1:1 respectively. - Identify what atom X is, explaining your reasoning - 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 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 Compound B is treated with sodium ethoxide to generate compound C. The ¹H-NMR spectrum of C shows…arrow_forward
- When compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1 H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The 1 H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.arrow_forwardSome of the following compounds show aromatic properties, and others do not predict which ones are likely to be aromatic and explain why they are aromatic or not.arrow_forwardA and B are isomeric dicarbonyl compounds of the molecular formula C5H&O2. The 'H NMR spectrum of A contains a singlet at 2.05 ppm and another singlet at 5.40 ppm. The 'H NMR spectrum of B contains three signals: a singlet at 2.3 ppm, a triplet at 1.10 ppm and a quartet at 2.70 ppm. Suggest structures for A and B and draw them in their respective boxes below. 1st attemptarrow_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