Compound A (C1,H14) shows prominent peaks in its mass spectrum at m/z 134 and 91. Compound B (also C10H14) shows prominent peaks at m/z 134 and 119. On vigorous oxidation with chromic acid, compound A gave benzoic acid, but compound B was nonreactive. From this information, deduce the structures of both compounds, and then draw the structure of A.

Compound A (C1,H14) shows prominent peaks in its mass spectrum at m/z 134 and 91. Compound B (also C10H14) shows prominent peaks at m/z 134 and 119. On vigorous oxidation with chromic acid, compound A gave benzoic acid, but compound B was nonreactive. From this information, deduce the structures of both compounds, and then draw the structure of A.

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter13: Nuclear Magnetic Resonance Spectroscopy

Section: Chapter Questions

Problem 13.17P: Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each...

See similar textbooks

Similar questions

These are the 1H RMN and IR spectrums of an A compound with formula C10H13NO2. When A reacts with NaOH in water and heat, B compound is formed, with formula C10H11NO. What are the structures of A and B?
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.
Compound A has molecular formula C7H7X. Its 1H-NMR spectrum shows a singlet at 2.25 ppm and two doublets, one at 7.28 ppm and one at 7.39 ppm. The singlet has an integral of three and the doublets each have an integral of two. The mass spectrum of A shows a peak at m/z = 126 and another peak at m/z = 128; the relative height of the two peaks is 3:1 respectively. - Identify what atom X is, explaining your reasoning - Identify Compound A, explaining your reasoning
Compound 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.
Compound C has a molecular ion in its mass spectrum at 146 and a prominent absorption in its IR spectrum at 1762 cm. C shows the following 'H NMR spectral data: 1.47 (doublet, 3 H), 2.07 (singlet, 6 H), and 6.84 (quartet, 1 H) ppm. What is the structure of C?
Following are IR and 1H-NMR spectra of compound D. The mass spectrum of compound D shows a molecular ion peak at m/z 136, a base peak at m/z 107, and other prominent peaks at m/z 118 and 59. Q.) Propose structural formulas for ions in the mass spectrum at m/z 118, 107, and 59.
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) ppm
Compound 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:
A 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 attempt
A 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?
Compound C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below. Suggest a structure for C and explain your reasoning.
Isomer set #1: CH3OCCH,CH3 CH3COCH,CH3 a b Compound D, CHgO2, shows a downfield singlet in its 'H-NMR spectrum. Is it compound a or b? Isomer set #2: CH3OCCH2CH3 CH3COCH2CH3 a b. Compound D, C,H8O2, shows a downficld quartet in its 'H-NMR spectrum. Is it compound a or b?