Concept explainers
Interpretation:
The product obtained from the Wittig reaction has to be identified.
Concept introduction:
Alkyl bromide reacts with triphenylphosphine to form triphenylphosphonium salt. The salt is further treated with base to from phosphonium ylide. Phosphonium ylide reacts with
The
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Chapter 16 Solutions
EBK ORGANIC CHEMISTRY
- Aldehydes and ketones react with thiols to yield thioacetals just as they react with alcohols to yield acetals. Predict the product of the following reaction, and propose a mechanism:arrow_forwardKetones 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_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_forward
- Enamines normally react with methyl iodide to give two products: one arising from alkylation at nitrogen and the second arising from alkylation at carbon. For example, Heating the mixture of C-alkylation and N-alkylation products gives only the product from C-alkylation. Propose a mechanism for this isomerization.arrow_forwardCompound H (C8H6O3) gives a precipitate when treated with hydroxylamine in aqueous ethanol and a silver mirror when treated with Tollens solution. Following is its 1H-NMR spectrum. Deduce the structure of compound H.arrow_forwardThymol (molecular formula C10H14O) is the major component of the oil of thyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and 1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose a possible structure for thymol.arrow_forward
- Identify products A and B from the given 1H NMR data. Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1H) ppm. What is the structure of B?arrow_forwardWhen compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.arrow_forwardThymol (molecular formula C10H14O) is the major component of the oil of thyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and 1585 cm–1. The 1H NMR spectrum of thymol is given below. Propose a possible structure for thymol.arrow_forward
- When acetonitrile is treated with concentrated sulfuric acid and tert-butanol, followed by water, a product is formed whose 'H NMR spectrum exhibits the following three signals: singlet, 1.3 ppm, 9 H; singlet, 2.0 ppm, 3 H; and broad singlet, 8.2 ppm, 1 H. Its IR spectrum exhibits one broad absorption of medium intensity between 3300 and 3500 cm-1, and a narrow, intense absorption near 1650 cm¯1. A key intermediate is shown. Draw the structure of the product, and draw the complete, detailed mechanism for the reaction. CH3 CH3 H2SO4 (conc) H2O ? + H3C H3C H3C-C=Ñ-c `CH3 CH3 HOarrow_forwardThymol (molecular formula C10H14O) is the major component of the oil ofthyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose apossible structure for thymol.arrow_forwardWhen compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1 H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The 1 H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning