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Treatment of compound D with LiAlH4 followed by H2O forms compound E. D shows a molecular ion in its mass spectrum at
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Chapter 25 Solutions
ORGANIC CHEMISTRY
- Compound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily in 10% Na2CO3 and 10% NaOH. When these alkaline solutions are acidified with 10% HCl, compound I is recovered unchanged. Given this information and its 1H-NMR spectrum, deduce the structure of compound I.arrow_forwardA molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a 3J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm 1. Provide the structures of the starting material and product. NaOH H,0 C;H„Br 6H triplet 4H quintet 1H quintet 10 8 Ppmarrow_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
- A molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm1. Provide the structures of the starting material (1 pt) and product (2 pts) NaOH H20 C5H1,Br 6H triplet 4H quintet 1H quintet 10 8 7 6. 4 3 1 HSP-06-347 ppmarrow_forwardHow could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forwardPropose a structure for a compound that displays the given spectroscopic data. The molecular ion in the mass spectrum appears at m/z = 116. IR: 1710 (s) and 3000 (s, broad) cm¹. ¹H NMR: 8 = 0.94 (t, J = 7.0 Hz, 6H), 1.59 (m, 4H), 2.36 (quintet, J = 7.0 Hz, 1H) and 12.04 (broad s, 1H) ppm. 13C NMR: 8 = 11.7, 24.7, 48.7, and 183 ppm. Deduce the structure. Select 11 ||| ||| Q Search Draw C H O • Templates More Question Source: Vollhardt 7e - Organic Chemistry: Structure And Function P W Erase Publisher: W.H. Freeman ENG 4: 4/18arrow_forward
- Reaction of butanenitrile (CH3CH2CH2CN) with methylmagnesium bromide (CH3MgBr), followed by treatment with aqueous acid, forms compound G. G has a molecular ion in its mass spectrum at m/z = 86 and a base peak at m/z = 43. G exhibits a strong absorption in its IR spectrum at 1721 cm−1 and has the 1H NMR spectrum given below. What is the structure of G?arrow_forwardAs reaction of (CH3)2CO with LIC≡CH followed by H2O affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600−3200, 3303, 2938, and 2120 cm−1. D shows the following 1H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?arrow_forwardTreatment of 2-methylpropanenitrile [(CH3)2CHCN] with CH3CH2CH2MgBr, followed by aqueous acid, affords compound V, which has molecular formula C7H14O. V has a strong absorption in its IR spectrum at 1713 cm−1, and gives the following 1H NMR data: 0.91 (triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and 2.60 (septet, 1 H) ppm. What is the structure of V? We will learn about this reaction in Chapter 20.arrow_forward
- Treatment of 2-methylpropanenitrile [(CH3)2CHCN] withCH3CH2CH2MgBr, followed by aqueous acid, affords compound V, whichhas molecular formula C7H14O. V has a strong absorption in its IRspectrum at 1713 cm−1, and gives the following 1H NMR data: 0.91(triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and2.60 (septet, 1 H) ppm. What is the structure of V?arrow_forwardThe NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 . The 1H and 13C peaks are also listedbelow. Provide a full analysis of the NMR spectra for compound 1. correct assignment of NMR spectra of both 13C spectra. correct rationalisation of 13C spectrum1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment. you should be able to assign these peaks to one of two carbon atoms in 1.arrow_forwardC9H10O2: IR absorption at 1718 cm–1. Propose a structure with data given.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning