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Chapter 17 Solutions
ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
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- Ketones undergo a reduction when treated with sodium borohydride, NaBH4. What is the structure of the compound produced by reaction of 2-butanone with NaBH4 if it has an IR absorption at 3400 cm-1 and M+=74 in the mass spectrum?arrow_forwardAddition of m-xylene to the strongly acidic solvent HF/SbF5 at 45C gives a new species, which shows 1H-NMR resonances at 2.88 (3H), 3.00 (3H), 4.67 (2H), 7.93 (1H), 7.83 (1H), and 8.68 (1H). Assign a structure to the species giving this spectrum.arrow_forwardWhen compound X (C15H17N) is treated with bezenesulfonyl chloride and aqueous potassium hydroxide, no apparent change occurs. Acidification of the mixture gives a clear solution. Propose a structure for X based on the given 1H NMR spectra. Explain your answer by interpreting the spectraarrow_forward
- Treatment of benzoic acid (C6H5CO2H) with NaOH followed by 1-iodo-3methylbutane forms H. H has a molecular ion at 192 and IR absorptions at 3064, 3035, 2960−2872, and 1721 cm−1. Propose a structure for H.arrow_forwardTreatment of ketone A with ethynyllithium (HC≡CLi) followed by D3O+ afforded a compound B of molecular formula C12H13DO3, which gave an IR absorption at approximately 1715 cm−1. What is the structure of B and how is it formed?arrow_forwardAn unknown compound C3H2NCl shows moderately strong IR absorptions around 1650 cm-1 and 2200 cm-1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ 5.9 and δ 7.1. Propose a structure consistent with this data?arrow_forward
- Compounds W and X are isomers. They have the molecular formula C9H8O. The IR spectrum ofeach compound shows a strong absorption band near 1715 cm−1. Oxidation of either compound withhot, basic potassium permanganate followed by acidification yields phthalic acid. The proton NMRspectrum of W shows a multiplet at δ 7.3 and a singlet at δ 3.4. The proton NMR spectrum of Xshows a multiplet at δ 7.5, a triplet at δ 3.1, and a triplet at δ 2.5. Propose structures for W and X.arrow_forwardReaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed bytreatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is thestructure of R?arrow_forwardCompounds 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_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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