Concept explainers
Interpretation:
The structure is to be proposed for the compound
Concept introduction:
舧 Nuclear Magnetic Resonance (NMR) is one of the most capable analytical techniques used for determining the
舧 Few elements, such as
舧 In
舧 Induced magnetic field consists of electricity generated from movement in a magnetic field.
舧
舧 In
舧 The splitting of the molecules is determined by the
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- Deduce 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_forwardAlpha-phellandrene, C₁0H16, is a pleasant-smelling chiral compound that may be obtained from the leaves of eucalyptus trees. It has the molecular formula C₁0H16 and a UV absorption maximum at 263 nm. On hydrogenation with a palladium catalyst, an achiral hydrocarbon with the formula C10H20 is obtained. Ozonolysis of alpha-phellandrene, followed by treatment with zinc and acetic acid, produces the following two compounds: CHIC-CH HOCHSCHICH CH(CH3)2 Propose a structure for alpha-phellandrene.arrow_forwardAlpha-phellandrene, C₁0H₁6, is a pleasant-smelling chiral compound that may be obtained from the leaves of eucalyptus trees. It has the molecular formula C₁0H16 and a UV absorption maximum at 263 nm. On hydrogenation with a palladium catalyst, an achiral hydrocarbon with the formula C₁0H20 is obtained. Ozonolysis of alpha-phellandrene, followed by treatment with zinc and acetic acid, produces the following two compounds: 0 || CH3C-CH 0 || ● Propose a structure for alpha-phellandrene. (+ || HCCH₂CHCH I CH(CH3)2 • You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one. n [ ]# ?arrow_forward
- Treatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppmarrow_forwardIdentify products A and B from the given 1H NMR data. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1H NMR spectrum: 2.2 (singlet, 3H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A?arrow_forwardThe 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?arrow_forward
- The treatment of (CH3)2C = CHCH2Br with H2O forms B (molecular formula C5H10O) as one of the products. Determine the structure of B from its H NMR and IR spectra.arrow_forward1,3,5,7-Cyclooctatetraene, C8H8, is an unusual hydrocarbon in that it reacts with exactly 2 equivalents of potassium to give A, C8H8K2, which can be isolated as a white solid. A exhibits a single proton NMR signal. Draw the structure of A.arrow_forwardTreatment of W with CH3Li, followed by CH3I, affords compound Y(C7H14O) as the major product. Y shows a strong absorption in its IRspectrum at 1713 cm−1, and its 1H NMR spectrum is given below. (a)Propose a structure for Y. (b) Draw a stepwise mechanism for theconversion of W to Y.arrow_forward
- -Ocimene is a pleasant-smelling hydrocarbon found in the leaves of certain herbs. It has the molecular formula C10H16 and a UV absorption maximum at 232 nm. On hydrogenation with a palladium catalyst, 2,6-dimethyloctane is obtained. Ozonolysis of -ocimene, followed by treatment with zinc and acetic acid, produces the following four fragments: (a) How many double bonds does -ocimene have? (b) Is -ocimene conjugated or nonconjugated? (c) Propose a structure for -ocimene. (d) Write the reactions, showing starting material and products.arrow_forwardCompound 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_forwardCompound X (molecular formula C10H12O) was treated with NH2NH2, -OH to yield compound Y (molecular formula C10H14). Based on the 1H NMR spectra of X and Y given below, what are the structures of X and Y?arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning