Calculate theoretical yield for the experiment(pictures provided). EXPERIMENTAL

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Calculate theoretical yield for the experiment(pictures provided). EXPERIMENTAL PROCEDURE Read All Notes Pertaining to this Experimental Procedures Before Conti Macro/Miniscale Synthesis nuing Ina 100 mL round bottom flask are added 14g of sodium bromide, 15 mL of water and 10mL of 1-butanol The flask with its contents is then immersed in an ice-water bath (Note 1). Slowly and with swirlin 12mL of sulfuric acid are added Notes 2& 3). After the acid addition is complete, the round bottom flask is placed on a heating mantle over a stir plate and clamped securely. A magnetic stir bar (boiling g, stones can be used in place of a stir bar) is placed in the flask followed by the reflux condenser 4). Gently reflux the mixture for 45 minutes (Notes 5,6 &7). At the completion of the reflux reflux column is removed and the apparatus is set for simple distillation (Note (Notes 8). The mixture should be distilled until the head temperature reads approximately 115 C (Notes 9&10). Once the distillation is complete, the distillate is poured into a separatory funnel containing 15mL of water. The resulting biphasic mixture is shaken and allowed to separate. The organic layer (nearly all 1-bromobutane and some byproducts) is removed and the aqueous layer is discarded Note 11). If the organic layer contains pinkish coloration, add 5mg of sodium bisulfite to discolor before retuning to the separatory funnel Note 12). The organic layer is returned to the separatory funnel and extracted with 10mL of ice cold oncentrated sulfuric acid. The mixture is left to separate for a least 5 minutes and the acid layer should

Explanation / Answer

CH3-CH2-CH2-CH2-OH + NaBr + H2SO4 -----> CH3-CH2-CH2-CH2-Br + H2O + NaHSO4

Mol. wt of 1-butanol = 74 g/mol

Mol. wt of 1-bromobutane = 137 g/mol

From the equation, 1 mole 1-butanol = 1mole 1-bromobutane i.e., 74 g of 1-butanol gives 137 g of 1-bromobutane

Amount(in ml) of 1-butanol taken in the experiment = 10mL

Amount(in g) of 1-butanol taken in the experiment = Vol. x density = 10mL * 0.81 g/mL = 8.1 g

Now, amount of 1-bromobutane that will be formed from 8.1g of 1-butanol = 8.1*137/74 = 15.0 gms

Theoritical yield of product = 15.0 grams

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