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- Identify the lettered compounds in the following reaction scheme.Compounds F, G, and K are isomers of molecular formula C13H18O. Howcould 1H NMR spectroscopy distinguish these three compounds fromeach other?Resveratrol is an antioxidant found in the skin of red grapes. Itsanticancer, anti-inflammatory, and various cardiovascular effects areunder active investigation. (a) Draw all resonance structures for theradical that results from homolysis of the OH bond shown in red. (b)Explain why homolysis of this OH bond is preferred to homolysis ofeither OH bond in the other benzene ring.Tropone is an unusually basic carbonyl (C=O) compound. When it is treated with one equivalent of the strong acid HBF4, it forms A, C7H7OBF4. Draw the strucutre of A as its most stable resonance structure.
- Electrophilic aromatic substitution usually occurs at the 1-position of naphthalene, also called the a position. Predict themajor products of the reactions of naphthalene with the following reagents. f) fuming sulfuric acidSome of the following compounds show aromatic properties and orhers do not. Predict whic ones are likely to be aromativ and explain why they are aromatic or not (A, B and C)Identify the structures of D and E, isomers of molecular formulaC6H12O2, from their IR and 1H NMR data. Signals at 1.35 and 1.60 ppm inthe 1H NMR spectrum of D and 1.90 ppm in the 1H NMR spectrum of Eare multiplets.
- An unknown compound A (molecular formula C7H14O) was treated withNaBH4 in CH3OH to form compound B (molecular formula C7H16O).Compound A has a strong absorption in its IR spectrum at 1716 cm−1.Compound B has a strong absorption in its IR spectrum at 3600−3200cm−1. The 1H NMR spectra of A and B are given. What are the structuresof A and B?Compound 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?How is compound A related to compounds B–E? Choose fromenantiomers, diastereomers, constitutional isomers, or identicalmolecules.
- Hexahelicene seems a poor candidate for optical activity because all its carbon atomsare sp2hybrids and presumably flat. Nevertheless, hexahelicene has been synthesized and separated into enantiomers. Its optical rotation is enormous: [a]D = 3700°.Explain why hexahelicene is optically active, and speculate as to why the rotation isso largeIdentify each one Which one is aromatic antiacromatic or nonaromatic ( need explaination)Under certain reaction conditions, 2,3-dibromobutane reacts with twoequivalents of base to give three products, each of which contains twonew π bonds. Product A has two sp hybridized carbon atoms, product Bhas one sp hybridized carbon atom, and product C has none. What arethe structures of A, B, and C?