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24 Organic and Biological Chemistry Solutions to Exercises 24.96 Determine the empirical formula of the unknown compound and its oxidation product. Use chemical properties to propose possible structures. 68.1 1 mol = 5.6703; 5.6703/1.1375 = 4.98 ≈ 5 13.7 H 1.008 = 13.5913; 13.5913/1.1375 = 11.95 ≈ 12 18.2 = 1.1375/1.1375 = 1 The empirical formula of the unknown is C₅H₁₂O. 69.7 C 1 mol = 5.8035; = ≈ 5 11.7 n 18.6 = 1.1625/1.1625 = The empirical formula of the oxidation product is C₅H₁₀O. The compound is clearly an alcohol. Its slight solubility in water is consistent with the properties expected of a secondary alcohol with a five-carbon chain. The fact that oxidation results in a ketone, rather than an aldehyde or a carboxylic acid, tells us that it is a secondary alcohol. Some reasonable structures for the unknown secondary alcohol are: OH OH OH 24.97 Determine the empirical formula, molar mass, and thus molecular formula of the compound. Confirm with physical data. 66.7 = 5.554 mol C;5.554/1.381 = 4.021 = 4 11.2 H 1.008 1molH = 11.11 mol H;11.11/1.381 = = 8 22.1 = 1.381 mol 1.381/1.381 = The empirical formula is C₄H₈O. Using Equation 10.11 (MM = molar mass): MM = = 71.9g/mol 745