ipl-logo

Experiment 3 Resolution Of Phenylethylamine

335 Words2 Pages

Experiment # 3 – Resolution Of ( )- -Phenylethylamine Name: Krishna Binu Class: Chem 2020 Due Date: October 23, 2015 Table 1 Shows the experimental results. Name Result Mass of Amine (g) 3.32 Mass of Salt (g) 1.26 Observed Rotation of Isolated Amine (Degrees) -36.33 1) Draw the chemical structures of the two diastereomeric salts produced in Part A using a proper perspective (line-angle) representation. Please note: (+)-tartaric acid has the (R,R) configuration. Pay special attention to the geometry of the stereocentres in each ion. Clearly label each stereocentre as (R) or (S). Drawing in the Hard copy 2) Calculate the yields of collected salt and isolated amine to the nearest %. Percent yield of isolated Amine …show more content…

Observed rotation = -36.33 (-36.33)/(0.94g/mlx1) = -38.64 Percent yield = (-38.64/40.4) x100 = 95.67% 4) What would happen if you did not add the 2 M NaOH solution to the amine tartrate salt prior to extracting the amine with dichloromethane? If the 2M NAOH were not added to the amine tartrate salt during the extraction method then the separation of amine wouldn’t have been possible. When the NAOH is added to the mixture, it covert the salt back to amine. 5) If you used (-)-tartaric acid in your experiment, what salt would you obtain in the precipitate? Which isomer of α-phenylethylamine would you get from this? The salts will have a (-)(-) and (+)(-) combination Salt that would obtain will be (+) amine (-) Tartrate

Show More
Related
993 Words | 4 Pages

Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029

Read More
801 Words | 4 Pages

Table 1 Results DDA Concentration Initial Mass(g) Time Interval Recovered Mass Cumulative Mass (g) Cumulative Recovery (%) Ln[(Rinf -R)/ Rinf] R=Rinf(1-e-kt) (M) (g) 10^(-5) 160 0 0

Read More
779 Words | 4 Pages

Benzyne Formation and the Diels-Alder Reaction Preparation of 1,2,3,4 Tetraphenylnaphthalene Aubree Edwards Purpose: 1,2,3,4-tetraphenylnaphthalene is prepared by first producing benzyne via the unstable diazonium salt. Then tetraphenylcyclopentadienone and benzyne undergo a diels-alder reaction to create 1,2,3,4-tetraphenylnaphthalene. Reactions: Procedure: The reaction mixture was created. Tetraphenylcyclopentadienone (0.1197g, 0.3113 mmol) a black solid powder, anthranilic acid ( 0.0482g, 0.3516 mmol) a yellowish sand, and 1,2-dimethoxyethane (1.2 ml) was added to a 5-ml conical vial.

Read More
704 Words | 3 Pages

Lab Report 5: Acetylsalicylic Acid (Aspirin) Synthesis Name: Divya Mehta Student #: 139006548 Date Conducted: November 19th 2014 Date Submitted: November 26th 2014 Partner’s Name: Kirsten Matthews Lab Section: Wednesday 2:30 L9 IAs Name: Brittany Doerr Procedure: For the procedure, see lab manual (CH110 Lab Manual, Fall 2014) pages 96-98. Wilfrid Laurier University Chemistry Department. Fall 2014. Acetylsalicylic Acid (Aspirin) Synthesis.

Read More
840 Words | 4 Pages

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.

Read More
795 Words | 4 Pages

Exercise 1 1. Suppose a household product label says it contains sodium hydrogen carbonate (sodium bicarbonate). Using your results from Data Table 1 as a guide, how would you test this material for the presence of sodium bicarbonate? B BoldI ItalicsU Underline Bulleted list Numbered list Superscript Subscript33 Words

Read More
1039 Words | 5 Pages

(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.

Read More
563 Words | 3 Pages

3. In this experiment, the percent yield was 90%. This number implies that there was little error in this experiment. However, this result could have been caused by certain external factors.

Read More
1270 Words | 6 Pages

The last goal was to determine the percent yield of a product formed during a reaction with the unknown compound. Experimental Design The first day of lab consisted of various preliminary tests that helped identify the unknown compound.

Read More
471 Words | 2 Pages

The percent recovery of the copper was calculated using the equation, percent recovery = (the mass of the copper recovered after all the chemical reactions/the initial mass of the copper) x 100. The amount of copper that was recovered was 0.32 grams and the initial mass of the copper was 0.46 grams. Using the equation, (0.32 grams/0.46 grams) x 100 equaled 69.56%. The amount of copper recovered was slightly over two-thirds of the initial amount.

Read More
1125 Words | 5 Pages

The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.

Read More
1237 Words | 5 Pages

(0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube

Read More
1167 Words | 5 Pages

5 water bath were set up each to10 °C. (5 were used do the experiment faster) 5 cm3 of starch solution were added into the 5 test tubes that were labeled test tubes. Then 5 cm3 of amylase enzyme was added into the other 5 test tubes that were labeled. Put one of the starch solution test tube (preferably the one labeled 1) and one of the test tube containing amylase into the water bath (10 °C).

Read More
1052 Words | 5 Pages

The calculated value was 1.6 x 10^-5. Conclusions The resulting Ka of the acetic acid from this experiment’s calculations was consistent with the experimental results. The experimental percent of CH3COOH was calculated at 1.6 x 10^-5, while the actual value was 1.8 x 10^-5.

Read More
767 Words | 4 Pages

Abstract The unknown concentration of benzoic acid used when titrated with standardized 0.1031M NaOH and the solubility was calculated at two different temperatures (20◦C and 30◦C). With the aid of the Van’t Hoff equation, the enthalpy of solution of benzoic acid at those temperatures was determined as 10.82 KJ. This compares well with the value of 10.27KJ found in the literature.

Read More

More about Experiment 3 Resolution Of Phenylethylamine

Open Document