The reduction of 9-fluorenone experiments in period 8 is a reduction reaction of double bonds, specifically a reduction of a carbonyl compound to prepare an alcohol, 9-flurenol is this experiment. Just to reiterate what a carbonyl is, a carbonyl is a compound that has a carbon double bonded to an oxygen. Since the main reaction behind this week’s experiment is a reduction reaction, it is important to learn and understand how and what takes place in it. In general terms, a reduction is a gain of electrons, opposite of oxidation, and an increase in carbon to hydrogen bonds caused by adding hydrogen atoms across double and triple bonds. This is not necessarily true in organic chemistry because the above process usually takes place on neutral atom or an ion but in this case carbon does not gain the electrons because a covalent bond is made. One way to tell if the reaction that has taken place is a reduction or not is by checking if the oxidation number of the carbon has decreased.
As mentioned above, in a reduction reaction hydrogen atoms are added to the multiple, double and triple, bonds of a compound. So the starting products, or reactants,
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In this experiment, the chemical method using metal hydrides is used to reduce carbonyl compounds into alcohols. As mentioned above, it is important to carefully select the metal hydride to use because of their different properties. For this reduction, sodium borohydride, NaBH4 will be used. NaBH4 is used over LiAlH4 because the LiAlH4 is more likely to react with the protic solvent that the carbonyl, while the NaBH4 does the opposite and selectively reacts with the carbonyl first. So in the reduction of 9-fluorenone to 9-fluorenol, the sodium borohydride will transfer its four hydrogens to the carbonyl to make a intermediate product which then be reacted with acid and water to give the final product of the alcohol,