Household Bleach Lab Report

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This experiment oxidized an unknown secondary alcohol into a ketone through the use of the oxidizing agent NaOCl (household bleach). The mechanism of this reaction involves the removal of two hydrogens, one from the carbon attached to the oxygen and one from the oxygen. The unknown secondary alcohol was either cyclopentanol, cyclohexanol, 2-heptanol, or 3-heptanol. The unknown compound we oxidized was compound B, and an IR spectrum of this compound was obtained. Initially, acetic acid was combined to the unknown secondary alcohol that way when NaOCl was slowly added to the solution, it would react with the acetic acid and form hypochlorous acid (HOCl). The hypochlorous acid is the reagent that actually oxidizes the alcohol. After, the NaOCl …show more content…

Then, 0.5 mL of sodium bisulfite solution was added to destroy the excess oxidant giving a negative potassium iodide result. The solution was then made basic by adding 6 M NaOH yielding a pH of 6. Next, the solution was extracted with dichloromethane and the organic layer was removed. The layer was then distilled to remove the dichloromethane. Since the dichloromethane distills first, multiple fractions were collected to attain as pure of a fraction as possible. The final fraction was weighed and used to find an IR spectrum. The amount of product collected weighed 0.941 g. The percent yield was calculated to be 53.19% by the formula 0.941 of product ÷ 1.769 of the unknown alcohol × 100. Furthermore, after analyzing the spectrum attained from reactant, the characteristic peaks of an alcohol were identified, including an OH stretch at 3331.32 and a C-O stretch 1112.38. However, since all four unknowns were alcohols consisting of the same groups, it could not be decided the identity of the unknown. After analyzing the IR spectrum and the fingerprint region of the product, my product most closely corresponded to that of 2-heptanone meaning the unknown reactant would be