A and B, isomers of molecular formula C3H5Cl3, are formed by the radical chlorination of a dihalide C of molecular formula C3H6Cl2. a.Identify the structures of A and B from the following 1H NMR data: Compound A: singlet at 2.23 and singlet at 4.04 ppm Compound B: doublet at 1.69, multiplet at 4.34, and doublet at 5.85 ppm b.What is the structure of C?

A and B, isomers of molecular formula C3H5Cl3, are formed by the radical chlorination of a dihalide C of molecular formula C3H6Cl2. a.Identify the structures of A and B from the following 1H NMR data: Compound A: singlet at 2.23 and singlet at 4.04 ppm Compound B: doublet at 1.69, multiplet at 4.34, and doublet at 5.85 ppm b.What is the structure of C?

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...

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When 2-bromo-3,3-dimethylbutane is treated with K+ -OC(CH3)3, a single product T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4, the major product U has the same molecular formula. Given the following 1H NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T.1H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz),and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm1H NMR of U: 1.60 (singlet) ppm
Reaction of tert-butyl pentyl ether [CH3CH2CH2CH2CH2OC(CH3)3] with HBr forms 1-bromopentane (CH3CH2CH2CH2CH2Br) and compound B. B has a molecular ion in its mass spectrum at 56 and gives peaks in its IR spectrum at 3150−3000, 3000−2850, and 1650 cm−1. Propose a structure for B, and draw a stepwise mechanism that accounts for its formation.
Four isomers A-D with the formula C5H12O exhibit different reactivity patterns as indicated below. Isomer A reacts with PCC and CrO3 to provide identical products. Isomer B reacts with PCC and CrO3 to provide different products. Isomers C and D do not react with either PCC or CrO3. Isomers A, B, and C readily react with NaH but D does not show any reactivity with NaH. The 1H NMR splitting patterns for these isomers are as follows. Isomer A: δ 0.91 (d, 3H), 0.90 (d, 3H), 1.18 (d, 3H), 1.92 (dqq, 1H), 3.38 (dq, 1H), 3.58 (bs, 1H). Isomer B: δ 1.25 (s, 9H), 3.45 (s, 2H), 3.65 (bs, 1H). Isomer C: δ 0.90 (t, 3H), 1.44 (q, 2H), 1.24 (s, 6H), 3.65 (bs, 1H). Isomer D: δ 1.10 (t, 3H), 1.13 (d, 6H), 3.19 (septet, 1H), 3.50 (q, 2H). Based on all of this information, provide the structures of A-D. (Note: If there is a chiral center on any of these molecules, assume the stereochemistry as “R”. dqq=doublet of a quartet of a quartet, bs=broad singlet)
You have a sample of a compound of molecular formula C11H15NO2, which has a benzene ring substituted by two groups, (CH3)2N− and −CO2CH2CH3, and exhibits the given 13C NMR. What disubstituted benzene isomer corresponds to these 13C data?
You have a sample of a compound of molecular formula C11H15NO2, which has a benzene ring substituted by two groups, (CH3)2N – and – CO2CH2CH3, and exhibits the given 13C NMR. What disubstituted benzene isomer corresponds to these 13C data?
When 2-bromo-3,3-dimethylbutane is treated with K+ −OC(CH3)3, a singleproduct T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4, the major product U has thesame molecular formula. Given the following 1H NMR data, what are thestructures of T and U? Explain in detail the splitting patterns observedfor the three split signals in T.1H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet ofdoublets, 1 H, J = 18, 10 Hz) ppm1H NMR of U: 1.60 (singlet) ppm Additional problems on the spectroscopy of alkenes are given in Chapters A–C:Mass spectrometry: A.16b, A.20, A.23Infrared spectroscopy: B.5, B.7(A), B.12c, B17a, B.18cNuclear magnetic resonance spectroscopy: C.12a; C.15d, e; C.29d; C.32d;C.37; C.38d, f; C.43i, j; C.44; C.45; C.49d, f; C.50b; C.51c; C.55
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?
Although benzene itself absorbs at 128 ppm in its 13C NMR spectrum, the carbons of substituted benzenes absorb either upeld or downeld from this value depending on the substituent. Explain the observed values for the carbon ortho to the given substituent in the monosubstituted benzene derivatives X and Y.
Although benzene itself absorbs at 128 ppm in its 13C NMR spectrum, the carbons of substituted benzenes absorb either upfield or downfield from this value depending on the substituent. Explain the observed values for the carbon ortho to the given substituent in the monosubstituted benzene derivatives X and Y.
(a) The 'H-NMR spectrum of cyclobutanone shows two signals - signal A at 3.00 ppm and signal B at 1.95 ppm. Give the multiplicity of each signal. cyclobutanone (b) When cyclobutanone is treated with D20 and NaOD, the only signal observable in the 1H-NMR is a singlet at 2.00 ppm. Explain why this is the case. [Note: Deuterium atoms do not display signals in the TH-NMR spectrum]
Assume that you have a compound with the formula C4H8O. a) How many double bonds and/or rings does your compound contain? b) If your compound shows an infrared absorption peak at 1715 cm-1, what functional group does it have? c) If your compound shows a single 1H NMR absorption peak at 2.1 δ, what is its structure?
The IR spectrum of compound A with a molecular formula of C5H12O is shown below. Compound A is oxidized to give compound B, a ketone with a molecular formula of C5H10O. When compound A is heated with H2SO4, compounds C and D are obtained. Considerably more D is obtained than C.Reaction of compound C with O3, followed by treatment with dimethyl sulfide, gives two products: formaldehyde and compound E, with a molecular formula of C4H8O. Reaction of compound D with O3, followed by treatment with dimethyl sulfide, gives two products: compound F, with a molecular formula of C3H6O, and compound G, with a molecular formula of C2H4O. What are the structures of compounds A through G?