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Methyl Benzoate

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Submitted By jackiefalcon64
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The purpose of this experiment was for the student to be able to perform the dehydrobromination of meso-stilbene dibromide in order to produce an alkyne, diphenylacetalyene. The dehydrobromination of meso-stilbene dibromide requires removal of two hydrogen bromides—in other words, a double elimination. In order for the elimination of the halogen to happen, the hydrogen being attacked and the leaving group need to be on the same plane. More specifically, the molecule has to be in an anti-periplanar orientation in order to have the elimination reaction happen more efficiently. Even though a molecule in a syn-periplanar orientation also has the hydrogen and leaving group in the same plane, the fact that both groups are on the same side makes the molecule sterically hindered. The staggered conformation of the meso-stilbene dibromide is orbitally more favorable than the eclipsed conformation making the anti-periplanar orientation the more favorable orientation for an elimination reaction. Fortunately, the fact that meso-stilbene dibromide had a single bond allowed for this molecule to develop the anti-periplanar orientation by having rotated the bond until the hydrogen and the leaving group were on opposite sides of the molecule. When producing alkynes, there are two steps involved in the reaction: the formation of an intermediate alkene with a vinyl halide followed by the elimination of the halide to result in an alkyne. For the creation of the intermediate vicinal bromide in this experiment, a base-induced E2 mechanism was performed using potassium hydroxide where the base removed the hydrogen on the carbon atom beta to the leaving bromide. This reaction proceeded swiftly due to the anti-periplanar orientation of the meso-stilbene dibromide. However, the intermediate vicinal bromide, an alkene, was locked in the syn-periplanar orientation due to the double bond not

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