Pure S-lactic acid is reacted with HBr to form 2-bromopropanoic acid CH3 HBr CH3

Question

Pure S-lactic acid is reacted with HBr to form 2-bromopropanoic acid CH3 HBr CH3 HO. HO :QH Br: This reaction produces 10.6 grams of 2-bromopropanoic acid, which was dissolved in 40.0 mL of ethanol to make a solution for polarimetry analysis. The observed optical rotation of this solution was +6.00° in a 1 dm cell. Calculate or determine the following (refer to Section 5.4 in Klein if you need to): Concentration c =| 0 g/mL Specific rotation of product mixture [a]miz Enantiomeric excess ee = | 0 Major enantiomer! R-2-bromopropanoic acid 0 Predominating mechanism ISN1 |X

Explanation / Answer

concentration C = 10.6 / 40.0

concentration C = 0.265 g /mL

specific rotation = observed rotation / l x C

                          = 6 / 1 x 0.265

specific rotation = 22.6o

Enantiomeric excess = (6 / 22.6 ) x 100

Enantiomeric excess = 26.5 %

predominating mechanism = SN2

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