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EBK ORGANIC CHEMISTRY
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- The ¹H NMR spectrum of a compound with the molecular formula C7H₁5Cl exhibits signals with relative integration 9:3:2:1. Propose a structure for this compound. 4-chloroheptane 3-chloro-2,2-dimethylpentane 3-chloro-2,4-dimethylpentane O2-chloro-2,3,3-trimethylbutanearrow_forwardCompounds Y and Z are isomers with the molecular formula C10H12O. The IR spectrum of each compound shows a strong absorption band near 1710 cm−1 . The 1H NMR spectra of Y and Z are given below. Propose structures for Y and Z.arrow_forwardFollowing are 1H-NMR spectra for compounds B (C6H12O2) and C (C6H10O). Upon warming in dilute acid, compound B is converted to compound C. Deduce the structural formulas for compounds B and C.arrow_forward
- Treatment of a hydrocarbon A (molecular formula C9H18) with Br2 in the presence of light forms alkyl halides B and C, both having molecular formula C9H17Br. Reaction of either B or C with KOC(CH3)3 forms compound D (C9H16) as the major product. Ozonolysis of D forms cyclohexanone and acetone. Identify the structures of A–D.arrow_forwardDeduce the structures of compounds A and B, two of the major components of jasmine oil, from the given data. Compound A: C9H10O2; IR absorptions at 3091–2895 and 1743 cm-1; 1H NMR signals at 2.06 (singlet, 3 H), 5.08 (singlet, 2 H), and 7.33 (broad singlet, 5 H) ppm. Compound B: C14H12O2; IR absorptions at 3091–2953 and 1718 cm-1; 1H NMR signals at 5.35 (singlet, 2 H) and 7.26–8.15 (multiplets, 10 H) ppm.arrow_forwardAn unknown compound has a molecular formula of C4H6O2. Its IR spectrum shows absorptions at 3095, 1762, 1254, and 1118 cm -1. It exhibits the following signals in its 1H NMR spectrum (ppm): 2.12 (singlet,3H), 4.55 (doublets of doublets, 1H), 4.85 (doublet of doublets, 1H), 7.25 (doublets of doublets, 1H); and the following signals in its 13C NMR spectrum (ppm): 20.8, 100.4, 141.2, 168.0. Draw the structure of the unknown compoundarrow_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_forwardCompound P has molecular formula C5H9ClO2. Deduce the structure of P from its 1H and 13C NMR spectra.arrow_forwardIn an aqueous solution containing sodium bicarbonate, aniline reacts quickly withbromine to give 2,4,6-tribromoaniline. Nitration of aniline requires very strong conditions,however, and the yields (mostly m-nitroaniline) are poor.(a) What conditions are used for nitration, and what form of aniline is present under theseconditions?arrow_forward
- 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 attemptarrow_forwardReaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment 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 the structure of R? We will learn about this reaction in Chapter 18.arrow_forwardA 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?arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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