Synthesis Of 1-Bromobutane Lab Report

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Synthesis of 1-Bromobutane I. Conclusion In this experiment we prepare 1-bromobutane from 1-butanol by using Sn2 reaction. By heating the primary alcohol with two reagents: NaBr and H2SO4, an aqueous solution containing the alko-halide and water will be produced. The overall equation for the reaction: H2SO4 + NaBr + CH3CH2CH2CH2OH -------> CH3CH2CH2CH2Br + H2O + NaHSO4 Questions: 3- By changing the source of the halide from NaBr to NaCl: H2SO4 + NaCl + CH3CH2CH2CH2OH -------> CH3CH2CH2CH2Cl + H2O + NaHSO4 2- In refluxing you gently heating the mixture without losing product to evaporation. If the mixture were boiled some of the solvent that contains some products will be gone with evaporation. II. Introduction - Synthesis is an endothermic reaction to generate products from reactants with the help of specific reagents. Primary alcohol goes Sn2 reaction. The OH- is a poor leaving group but adding water will help…show more content…
Methods and Materials - Place 1.33g NaBr, 1.5 ml of water and 20 drops of 1-butanol to a 5-ml round bottom long-necked flask and set a distillation apparatus. Add 22 drops of concentrated H2SO4 and reflux for 55 minutes. For more details about the procedure and how to set the simple distillation apparatus check the book pages 312 to 314. IV. Results - After transferring the distillate to a reaction group the halide form the top layer. The final weight of the halide was 0.17 g. % yield = (experiment yield / Theoretical yield) * 100 Calculating theoretical yield: 1 ml butanol * (0.81 g butanol / 1 ml butanol) * (1 Mole butanol / 74.12 g butanol) * (137.03 bromobutane /1 mole bromobutane) = 1.497 g of 1-bromobutane. 1.33 g of NaBr * (1 mole NaBr / 102.91 g NaBr) * (1 mole bromobutane / 1 mole NaBr) * (137.03 g bromobutane / 1 moe bromobutane) = 1.77 g of 1-bromobutane The result shows that the 1-butanol is the limiting reagent so the theoretical yield is 1.497 g of 1-bromobutane. % yield = (0.17 / 1.497)* 100 =

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