ABSTRACT Isopropyltoluene isomers were synthesized through the friedel-crafts alkylation of toluene with 2-chloropropane in the presence of aluminum chloride as a catalyst. Through extraction of the distillate with water and 5% NaHCO3, followed by a fractional vacuum distillation, the desired aromatic organic compound was isolated. Infrared Spectroscopy and Gas Chromatography were used to analyze the composition, purity and to confirm the identity of the prepared product. The sample weighed approximately 2.66g, which reflected a low yield at 47%. However, from the GC the area % (also known as percent purity) was 91.8%, which indicates that a very pure product was obtained.
| Observations of Chemicals | Zinc Sulfate | Powder of a white solid | Barium Iodide | Powder of a white solid. | Deionized water | Liquid, transparent. | Trial # | BaI2 | ZnSO4 | Theoretical Yield of ZnI2 | Actual Yield | Percent Yield | 1 | .67g | .45g | .499820g | .52g | 104% | 2 | .67g | .45g | .499820g | .52g | 104% | 3 | .66g | .46g | .493117g | .48g | 97% | Calculations for Cost | Double Replacement | Synthesis | 0.48 grams of Zinc Sulfate - $0.02 | 1.00 gram Granular Zinc - $62.50 | 0.67grams of Barium Iodine Dihydrate - $0.886 | 2.00 gram Iodine - ($74.90 × 2) - $149.80 | 0.52 grams of Zinc Iodide - $0.906 | 1.00 gram zinc - $0.212 | 1000 grams of Zinc Iodide = $1,923.00 | 1000 grams of Zinc Ioidide = $212.30 | Focus Question Should chemists prepare Zinc Iodide, from its Elements or from a Double Replacement Reaction between Barium Iodide and Zinc Sulfate?
With the use of this technique we placed chlorine, bromine, and iodine into solutions containing chloride, bromide, and iodide. In the reaction the free halogen (X2) oxidizes the other halide ion (Y-) and gets reduced by gaining electron(s). In table 3, chlorine was the strongest oxidizing agent and iodine was the weakest oxidizing agent. Since chlorine was the strongest oxidizing agent it will react more and the weak agent will react less. This explanation can be demonstrated in table 3 also because the results of the reactions demonstrates that chloride reacted more by the color of the product compared to the color of chloride in the mineral oil.
The NMR spectrum does contain impurities including methanol (4.80 ppm), methyl oleate (5.40 ppm), CHCL3 (7.20 ppm), acetone (2.00 ppm), and water (1.60 ppm). The methyl oleate may be present due to incomplete reaction with hydrogen and the CHCL3 may be present due to the contamination of CDCL3 NMR solvent. Acetone was used as a cleaning agent on the apparatus before starting the experiment. The experimental NMR spectrum does match that of an authentic spectrum except the experimental spectrum contains a greater number of peaks due to
2-propanol (bp=82 degrees C) 3. tetrahydofuran (bp=65 degrees C) 4. 1-butanol (bp=118 degrees C) 5. butanone (bp=80 degrees C) Give a better separation for the mixture to be distilled tetrahydofuran (bp=65 degrees C) because it is farthest from 100 degrees C Which alkyl halide would react fastest in a nucleophilic substitution using silver nitrate in ethanol (weak nucleophile, protic solvent)? 3-bromo-3-methylpentane (most
2 marks 4 Draw the structural formula of Compound G. 1 mark 5 Using the chemical shift correlation for 13C NMR, predict the number of peaks for Compound G and draw in the position of the peaks on the blank spectrum below, annotating each peak with its corresponding structure. (2 marks) 6 Draw the structural formula for 2-chloro but-2-ene. Below this draw a structural formula of an isomer of 2-chloro but-2-ene and name this substance.
CHEM 2124 Unit 2 Worksheet 2 1) Why does benzene undergo a substitution reaction with Br2 while cyclohexene undergoes an addition reaction? Addition of bromine to benzene is highly unfavorable because it would result in a non-aromatic product. 2) Which is more stable, cyclobutadiene or 1,3-butadiene? Explain 1,3-butadiene is more stable. Cyclobutadiene is antiaromatic and antiaromatic systems are less stable than their open chain counterparts.
I had based my hypothesis off of the theoretical yield data we collected after doing a few stoichiometry equations. The actual tests we performed proved that NaHCO3 produced the most CO2 by heating and CaCO3 produced the least. Percent yield is a measure of how much of a product you actually obtained in a reaction compared to the amount that you could have obtained according to calculations. Since our percent yield for NaCO3 is so high, some parts of our experiment may have gone wrong. An error we could have made was assuming all the test tubes weighed the same, and next time we should properly weigh them.
Synthesis of 1-Bromobutane I. Conclusion In this experiment we prepare 1-bromobutane from 1-butanol by using Sn2 reaction. By heating the primary alcohol with two reagents: NaBr and H2SO4, an aqueous solution containing the alko-halide and water will be produced. The overall equation for the reaction: H2SO4 + NaBr + CH3CH2CH2CH2OH -------> CH3CH2CH2CH2Br + H2O + NaHSO4 Questions: 3- By changing the source of the halide from NaBr to NaCl: H2SO4 + NaCl + CH3CH2CH2CH2OH -------> CH3CH2CH2CH2Cl + H2O + NaHSO4 2- In refluxing you gently heating the mixture without losing product to evaporation. If the mixture were boiled some of the solvent that contains some products will be gone with evaporation.
of moles of CuO = 4g63.55+16=0.0503 moles Hence, no. of moles of CuSO4.5H2O = 0.0503 moles Mr. of CuSO4.5H2O =63.55+32.07+416+ 518 = 249.62g/mol Mass of CuSO4.5H2O formed = 0.0503 ×249.62=12.56g Theoretical yield = 12.56g (d) Calculation of percent yield Percent yield =12.0012.56 ×100=95.54% (e) Some CuSO4 solution was removed to use for the reaction in Part B. This has reduced the percent yield by a very small fraction only. This is because only around 1,2 ml have been used to carry out the analysis. B.