Methyl Stearate

Theoretical Yield

Methyl Stearate

(0.003848 mol)(298.504g/mol product)

= 1.1486 g methyl stearate

Percent Yield

Given the 1:1 stoichiometry of the reaction:

100% x (0.2551g product / 1.1486g theoretical yield)

= 22.2%

Conclusion

A 22.2% yield of synthesized methyl stearate was obtained via the catalytic hydrogenation of methyl oleate in the presence of the catalyst 10% palladium on carbon. The product was centrifuged to remove the catalyst from the mixture and then filtrated and recrystallized to purify the methyl stearate product. The percent yield of the white crystal product may have been low due to incomplete hydrogenation of the methyl oleate, lost product in the filtration/ recrystallization process, and incomplete transfer between the centrifuge tube to a beaker.

The NMR spectrum of the purified product suggests that its identity is methyl stearate. The one peak at 3.482 ppm indicates the –OCH3 group, because the hydrogen atoms (HA) are adjacent to the highly electronegative oxygen atom. The –CH3 hydrogen atoms observe no splitting because they have no neighboring protons (N=0) therefore it is a singlet. The three peaks at around 0.90 ppm represent the –CH3 group on the opposite end of the molecule, because the three hydrogen atoms (HB) are furthest from the electronegative oxygen atoms and because the protons have two neighboring protons (N=2) indicating a triplet. The three peaks at around 2.40 ppm represent the –CH3 group adjacent to the C=O bond, because the hydrogen atoms (HC) have two neighboring protons (N=2) indicating a triplet and it is next to the electronegative double bond.