E-Stilbene Lab Report
Wittig Reaction and Photoisomerization
Abstract:
For this laboratory experiment stilbene was produced through a Wittig reaction with benzyltriphenyl phosphonium and benzaldehyde producing a form of stilbene (Figure 1). This reaction favored a crude Z-Stilbene crystal product over its E counterpart. When Z-Stilbene underwent photoisomerization with iodine for 1 hour it reconfigured almost exclusively into its more stable counterpart E-Stilbene. The reaction produced very low yield of 6.3% due to the nature of the reaction and the speed at which iodine reacts. The purity of E-Stilbene could have been increased by allowing the reaction to perform longer and to use a faster reactant such as Bromine.
Introduction:
In this experiment a Wittig …show more content…
The limiting reagent was benzyltriphenyl phosphonium and the percent yield was very low due to the nature of the reactions and workup in this experiment. If more benzyltriphenyl phosphonium was added it or purer reactants were used it could have increased the yield of the reaction. The crude product Z-Stilbene was completely favored 100/0 to E-Stilbene as an integration of 10.73 was recorded at 6.6 ppm for Z-Stilbene but no peak was recorded at 7.1 ppm for Z-Stilbene. When Stilbene undergoes photo isomerization it favors almost exclusively E-Stilbene as an integration of 15.14 at 7.1 ppm was observed compared to an integration of 0.89 at 6.6 ppm. This ratio of 94/6 almost completely favors the E product as it is more stable due to its conjugation. In order for proper conjugation a molecule must be flat which Z-Stilbene does not possess due to steric hindrance of its benzene rings being so close together. It causes the molecule to twist. E-Stilbene on the other hand does not have to twist due to hindrance and thus has full conjugation. The recrystallized E-Stilbene product is mostly pure as its NMR matches the literature value for E-Stilbene except for the small peak at 6.6 ppm due to a small amount of leftover Z-Stilbene2. This reaction could be improved by using Bromine instead of Iodine because Iodine is very slow and it is possible that more product could …show more content…
It was then irradiated with stirring for 1 hour. The resulting solution was washed with sodium bisulfate (5 ml) in a seperatory funnel and shaken until de-colorization. The organic layer was washed with sodium chloride (5 ml) and transferred to a flask where sodium sulfate was added for evaporation. The remaining product was roto-vapped to produce crystals. The remaining product was then recrystallized in hot ethanol (12 ml) and cooled to room temperature. E-Stilbene: NMR: 6.1 ppm (0.89, 7.1 ppm
Abstract:
For this laboratory experiment stilbene was produced through a Wittig reaction with benzyltriphenyl phosphonium and benzaldehyde producing a form of stilbene (Figure 1). This reaction favored a crude Z-Stilbene crystal product over its E counterpart. When Z-Stilbene underwent photoisomerization with iodine for 1 hour it reconfigured almost exclusively into its more stable counterpart E-Stilbene. The reaction produced very low yield of 6.3% due to the nature of the reaction and the speed at which iodine reacts. The purity of E-Stilbene could have been increased by allowing the reaction to perform longer and to use a faster reactant such as Bromine.
Introduction:
In this experiment a Wittig …show more content…
The limiting reagent was benzyltriphenyl phosphonium and the percent yield was very low due to the nature of the reactions and workup in this experiment. If more benzyltriphenyl phosphonium was added it or purer reactants were used it could have increased the yield of the reaction. The crude product Z-Stilbene was completely favored 100/0 to E-Stilbene as an integration of 10.73 was recorded at 6.6 ppm for Z-Stilbene but no peak was recorded at 7.1 ppm for Z-Stilbene. When Stilbene undergoes photo isomerization it favors almost exclusively E-Stilbene as an integration of 15.14 at 7.1 ppm was observed compared to an integration of 0.89 at 6.6 ppm. This ratio of 94/6 almost completely favors the E product as it is more stable due to its conjugation. In order for proper conjugation a molecule must be flat which Z-Stilbene does not possess due to steric hindrance of its benzene rings being so close together. It causes the molecule to twist. E-Stilbene on the other hand does not have to twist due to hindrance and thus has full conjugation. The recrystallized E-Stilbene product is mostly pure as its NMR matches the literature value for E-Stilbene except for the small peak at 6.6 ppm due to a small amount of leftover Z-Stilbene2. This reaction could be improved by using Bromine instead of Iodine because Iodine is very slow and it is possible that more product could …show more content…
It was then irradiated with stirring for 1 hour. The resulting solution was washed with sodium bisulfate (5 ml) in a seperatory funnel and shaken until de-colorization. The organic layer was washed with sodium chloride (5 ml) and transferred to a flask where sodium sulfate was added for evaporation. The remaining product was roto-vapped to produce crystals. The remaining product was then recrystallized in hot ethanol (12 ml) and cooled to room temperature. E-Stilbene: NMR: 6.1 ppm (0.89, 7.1 ppm