In this experiment, 9-fluorenone, a ketone, was reduced to fluorenol, an alcohol. The product was then identified using melting point and IR data, and percent yield was calculated. Reduction is one of the two processes that occur during a redox reaction, and it involves the gain of an electron by one of the species. The other species in the reaction loses an electron, and is by definition oxidized. In this experiment, fluorenone, the oxidizing agent, was reduced, and sodium borohydride, the reducing agent, was oxidized. In this experiment, hydride reducing agents were used, since hydrides have spare electrons that they can donate to other compounds. Two popular hydride reducing agents, lithium aluminum hydride and sodium borohydride, were considered for this experiment. Since hydride reducing agents were used in this reaction, the reaction would have been extremely sensitive to proton sources, since …show more content…
This would be an excellent yield if it all indeed consisted of fluorenol, but given that -OH peaks were observed in the IR, and that a good yield for this reaction was around 60%, it is possible that this percent is artificially high, and that some of this yield consisted of impurities like water and methanol that had not evaporated away. Of the product that was lost, some was lost due to bad filtration technique, as some of the product was observed to have passed through the filter paper, and into the flask. Some of the product may have also clung to the vial, as the precipitate was difficult to remove from the vial in its entirety. In order to improve this yield, more care could have been taken while removing product from the vial, and during filtration, ensuring that the filter paper was sufficiently wet and no product passed through. As some error most likely occurred due to impurities, inflating the percent yield, the product could have been allowed to dry