The purpose of this experiment was to perform elimination reactions with two different bases: potassium hydroxide and potassium t-butoxide. We explored the principles of dehydrohalogenation, Zaitsev’s rule, Hofmann’s rule, and elimination. Using these strong bases eliminated problems that could arise due to by-products produced by substitution reactions, and also allows a study of effects on the product ratios because one base is stronger than the other and has different steric requirements and molecular structure. In comparing the ratios of the different products produced by potassium hydroxide and potassium t-butoxide, we learned how different solvents and reagents affect the course and outcome of a chemical reaction. Reaction Scheme Theory Reaction Mechanism An elimination reaction is a reaction in organic chemistry with which two substituents are …show more content…
The resulting products were 2-methyl-2-butene and 2-methyl-1-butene. For the first part of the experiment we used 0.945 moles of KOH and 0.02 moles of 2-methyl-2-bromobutane, making 2-methyl-2-bromobutane the limiting reagent. To calculate theoretical yield, I multiplied the moles of the limiting reagent and the molecular weight of the product. The major product for the first part of the experiment was 2-methyl-2-butene, so: (0.02 moles)(70.13 g/mol) = 1.4 grams. The weight of 2-methyl-2-butene recovered was 1.056 grams, so in finding percent yield I divided the two: (1.056)/(1.4) x 100 = 75.4%. For the second part of the experiment we used .20 moles of KOt-Bu and 0.02 moles of 2-methyl-2-bromobutane, again making 2-bromo-2-methylbutane the limiting reagent. In calculating the theoretical yield: (0.02 moles)(70.13 g/mol) = 1.4 grams. The weight of the major product (2-methyl-1-butene) recovered was 1.299 grams, making the percent yield: (1.299)/(1.4) x 100 =
The theoretically yield for this experiment, based off of the mass of aluminum pieces originally weighed out, was 19.0144 g, while our actual yield was 12.7222 g. This is a 66.91% yield. Ideally, of course is a 100% yield. Factors that may have caused this in our own experiment include, but may not be limited to: 1) There was barely any hydrogen gas being produced in the first step when we took it off the heat, but enough to infer that the reaction may not have been completely over, possibly affecting the amount of potassium aluminum hydroxide produced 2) Having to filter the reaction mixture in the first step more than once, because we neglected to turn on the vacuum the first time, we may have lost some of the mixture in the process. 3)
Aims of experiment • Determine the rate constants for hydrolysis of (CH3)3CCl in solvent mixtures of different composition (50/50 V/V isopropanol/water and 40/60 V/V isopropanol/water) • Examine the effect of solvent mixture composition on the rate of hydrolysis of (CH3)3CCl Introduction With t-butyl chloride, (CH3)3CCl, being a tertiary halogenoalkane, it is predicted that (CH3)3CCl reacts with water in a nucleophilic substitution reaction (SN1 mechanism), where Step 1 is the rate-determining step. The reaction proceeds in a manner as shown
The melting point ranges supported that the purified product could either be sample D (4-chlorobenzyl phenyl ether) or sample E (4-Tert-butylbenzyl phenyl ether) because both share a similar melting point range. However, the tiny layer chromatography data supported only sample D (4-chlorobenzyl phenyl ether). Although the experiment was conducted to avoid mistakes, the yield does not really make sense because it is higher than 100%. The yield could be high because the sodium hydroxide (NaOH) did not completely dissolve or because the product did not fully dry. A way to test these proposed hypotheses is to repeat the experiment and reflux or dry for a longer time.
In the presence of a nucleophile and good leaving group, an organic reactant in specific conditions is likely to undergo a chemical reaction, namely, nucleophilic substitution. Nucleophilic substitution consists of two different mechanisms, SN1 and SN2. In this experiment, SN2 is the mechanism tested. SN2 is a relatively fast, one-step mechanism in which the nucleophile attacks the organic reactant and the leaving group diverges from the reactant to become a weak base (Fig. 1). The overall speed of the reaction can increase based on the size or basicity of the nucleophile, or the bulkiness of the carbon group with the leaving group.
The initial amount was 2.003 grams and the end product weighed 1.468 grams. The results show that the crude sample that was made from the lab had around the same purity with that of the known sample, thus, the experiment was
Then dimensional analysis was used to find the theoretical yield for the solid substance of each given equation. Starting with 2.0 gram of sodium bicarbonate, the molar ratio from each equation was used to convert from moles of sodium bicarbonate to moles of the solid product in that equation and then molar mass was used to convert the answer back into grams. Once this step was repeated with all four equations the theoretical yield of each was compared with the actual yield found by decomposing the
In this experiment, (mass of isopentyl acetate) was formed directly by esterification of acetate acid with isopentyl alcohol. Sulfuric acid was used as a catalyst. The percentage yield of isopentyl acetate was approximately 20% with a theoretical of 2.082g. This might due to the reason that I lost some of my product in the distillation process. It also might be the fact that my TF told me to add a little drop of sodium chloride in the solution because it was cloudy right after I added 5ml of water. Another error might include using unclean glassware.
As seen in table 1, the theoretical yield was .712 g of C_17 H_19 NO_3. The % yield of this experiment was 7.51 % of C_17 H_19 NO_3. . This low yield can be explained from a poor recrystallization technique combined with potential contamination. Throughout the experiment, the mixture changed color from green, orange, to yellowish lime, and eventually clear.
The third step is identifying the limiting reagents for each test. It can easily be classified by the smallest number. The higher number will be the excess. 4. After identifying reagents and finding the theoretic yield, it’s possible to find the excess reagent mass and number of moles for each test.
The purpose of this experiment was to learn about metal hydride reduction reactions. Therefore, the sodium borohydride reduction of the ketone, 9-fluorenone was performed to yield the secondary alcohol, 9-fluorenol. Reduction of an organic molecule usually corresponds to decreasing its oxygen content or increasing its hydrogen content. In order to achieve such a chemical change, sodium borohydride (NaBH4) is used as a reducing agent. There are other metal hydrides used in the reduction of carbonyl groups such as lithium aluminum hydride (LiAlH4).
(150.22g/mol)(3.5 x 10^-3 mol of nucleophile) = 0.525 g Actual yield = 0.441 g, Percent Yield = (0.441g/0.525g) x 100% = 84% 10. Percent recovery from recrystallization = (0.172g/0.441g) x 100% = 38% 11.
The percent yield was 22.33%. In most cases, that means that a lot of possible product was lost. However, in this case, that was not true. When Benzaldehyde reacted with the Wittig reagent, it produced two products: E-Stilbene and Z-Stilbene. The Z product was a liquid, while the E product was a solid.
The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different
The reaction adds two electrons to the contaminating molecule, thus reducing the contaminant. In order for the reductive dehalogenation to proceed, a substance other than the halogenated contaminant must be present to serve as an electron donor. Possible electron donors are hydrogen and organic compounds of low molecular weight (lactate, acetate, methanol or