Lab 8

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CHEM 282 Experiment #8: Hypochlorite Oxidation of Methyl Ketones by the Haloform Reaction Sophie Wolkoff (20107258) & Kaelen Partridge (20127197) TA: Billy Deng March 27, 2020 Experimental:

The first step of the procedure for this experiment included an introduction to the haloform reaction executed in the lab. This process involved reacting methyl ketones with a halogen in an alkaline medium, forming two products. The products were a haloform, whose components depended on the halogen employed, and a carboxylic acid that was missing one carbon atom from the methyl ketone starting material. Step two was the hypochlorite oxidation of acetophenone which began with mixing 0.6 mL of acetophenone and 23 mL of 5% (w/v) sodium hypochlorite solution in a flask under the fume hood. Once a Liebig condenser was attached to this system, the mixture was stirred with a magnetic stir bar for 30 minutes. Once completed, 0.15 g of sodium sulphite was added to remove the unreacted bleach and the resulting mixture was washed twice with 15 mL of diethyl ether in a separatory funnel. The collected aqueous layer was then acidified by adding 10% hydrochloric acid dropwise, continuing until the reaction appeared as red on neutral litmus paper instead of blue. At this point, a precipitate of benzoic acid was formed which was collected by vacuum filtration, washed with cold water, and then recrystallized from a minimum amount of hot water. The crystals were collected again and air-dried, then their mass, percent yield, and experimental melting point was determined. Step three, which was completed in the interim while the solution from step two was being mixed, involved performing an iodoform test. This test was performed twice in a test tube, first with 3 drops of pure acetone and then with 3 drops of acetophenone. In the test tubes, 10 drops of 10% aqueous potassium hydroxide solution and 3 drops of KI/I solution were added and swirled, and the mixtures were left to stand for 3 minutes. ₂ The test tubes were observed for the formation of a yellow-white precipitate of iodoform,

and the test tubes were then cooled in an ice-water bath if its appearance was not observed. All results and observations were recorded. Results: Table 1: Recrystallized Product Data Mass of Recrystallized Product (g) Percent Yield (%) Experimental Melting Point (°C) Literature Melting Point (°C) for Benzoic Acid 0.097 16.2% 120.0 - 122.2 122 Calculation of the Mass of Recrystallized Product (Mass of watch glass and product (g) - mass of watch glass(g)) = 48.465 g - 48.368 g = 0.097 g Calculation of Percent Yield Balanced reaction: C 6 H 5 COCH 3 + 3 NaOCl + H + ⟶ C 6 H 5 COOH + 2 NaOH + HCCl 3 Moles of acetophenone Moles of C 6 H 5 COCH 3 = Mass of C 6 H 5 COCH 3 / MW of C 6 H 5 COCH 3 = 0.6 g / 120.15 g/mol = 0.005 mol Moles of sodium hypochlorite Moles of NaOCl = Mass of NaOCl / MW of NaOCl = 23 g / 74.44 g/mol = 0.31 mol

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Calculate the Limiting Reagent 0.31 mol C 6 H 5 COCH 3 x 1 mol C 6 H 5 COCH 3 / 3 mol NaOCl = 0.015 mol C 6 H 5 COCH 3 To use up all of the sodium NaOCl we would need to use 0.103 moles of C 6 H 5 COCH 3 so therefore C 6 H 5 COCH 3 is the limiting reagent. Moles of C 6 H 5 COOH Moles of C 6 H 5 COOH = 0.005 mol C 6 H 4 (OCH 3 ) 2 x 1 mol C 6 H 5 COOH / 1 mol C 6 H 5 COCH 3 = 0.005 mol of C 6 H 5 COOH Mass of C 6 H 5 COOH Mass of C 6 H 5 COOH = Moles of C 6 H 5 COOH / MW of C 6 H 5 COOH = 0.005 mol x (122.12g/mol) = 0.6 g Percent Yield Percent Yield = [Experimental Mass of C 6 H 5 COOH / Theoretical Mass of C 6 H 5 COOH ] x 100% = [0.097 g / 0.6g] x 100% = 16.2% Table 2: Iodoform Tests with Acetone and Acetophenone Compound used Observations Acetone - Liquid was transparent with a slight yellow tint - Grainy, white precipitate formed at the bottom of the test tube

- No need to cool in ice water bath Acetophenone - Liquid became a murky white/yellow colour, that was translucent - Layer of KI/I 2 solution sat on the surface (dark brown, liquid, translucent) - No precipitate was formed - No change occurred when the test tube was cooled in an ice water bath Discussion: Part A The haloform reaction is the process of reacting a methyl ketone with a halogen, chlorine, bromine or iodine, in the presence of hydroxide ions.¹ This reaction generates two products in its alkaline medium. The first is a haloform, dependent on the kind of halogen treated as a starting material, and the second is a carboxylate ion, containing one less carbon atom that the methyl ketone material used.² When the halogen employed in the reaction is iodine, a yellow solid with distinct odour, the haloform reaction can also be used to identify methyl ketones, called the iodoform test.¹ The iodoform test is performed to discern the presence of an acetyl group, or CH CO, in aldehydes or ketones. This can be carried out by use of one of two different ₃ chemically equivalent reagents. The first reagent is a mixture of iodine combined with a small amount of sodium hydroxide with the aldehyde or ketone solution.³ A positive result from this test would appear as a pale yellow precipitate of triiodomethane, known previously as iodoform, with a distinguishable odour. The second reagent that can be used is potassium iodine and sodium chlorate(I) with a small amount of the aldehyde or

ketone solution.³ A positive result is also produced as a pale yellow precipitate. In both cases, it may be necessary to gently warm the solutions if no precipitate form in the cold solution. Due to its known resulting structure, the only aldehyde that can form the precipitate as the positive result is ethanal, or acetaldehyde.³ Many different ketones are able to produce this result, which are all known as methyl ketones since each have a methyl group on one side of their carbon-oxygen bond. Part B According to our experimental results from this lab, we determined that we were reasonably successful at carrying out this experiment. This is due to parts of our data falling within the accurate range of its according theoretical data. In relation to our recrystallized product data from step two, the percent yield we obtained from performing the recrystallization of the product was 16.2%. This was fairly low as our experimental product mass was 0.092g, but its theoretical mass was 0.6 g. This low outcome may have been the result of experimental errors during the crystallization or recrystallization processes, such as possibly losing product through the transfer from one container to another. The experimental melting point data for our product was consistent with theory as we observed a melting point range of 120.0°C to 122.2°C, and the theoretical melting point was 122°C. The theoretical melting point of the product fell within our observed range which deemed our melting point observations as successful. Our observations from part three were not as accurate in regards to our performance of the iodoform test. Our observations revealed that a white grainy precipitate formed in the test tube containing acetone, but no precipitate formed in the

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test tube with acetophenone. In the case of acetone, our performance was successful as acetone is a methyl ketone and does pass this test. However for acetophenone, we detected no precipitate which was incorrect. Acetophenone should pass the iodoform test since it is also a methyl ketone, and therefore should have formed the precipitate. Questions: 1. An experimental error occurred during the separation with the separatory funnel which affected our results. During this process, while pouring the aqueous layer into the funnel, some of the contents spilled which ended up resulting in a lower yield for our final product than what we should have had. Another error could have occurred when transferring our crystals to the watch glass from the funnel. It is possible that we did not properly scrape out all of the crystals onto the watch glass and thus resulted in a lower mass of crystals and a lower final yield than expected. 2. The chlorine comes from household bleach in which one of its active components is sodium hypochlorite. Sodium hypochlorite will react with water and sodium chloride in order to produce molecular chlorine, as well as sodium hydroxide. Redox half reactions: Cl + + 1e - → Cl Cl - → 1e - + Cl Overall balanced net ionic equation: H 2 O + NaOCl + NaCl → 2NaOH + Cl 2 3. Overall Net Equation:

2OCl - + SO 3 2- + 2H + → Cl 2 + SO 4 2- + H 2 O Two redox half reactions: SO 3 2- + H 2 O → SO 4 2- + 2H + + 2e - OCl - + 2H + + 1e - → ½ Cl 2 + H 2 O The oxidation state of the chlorine atom in the hypochlorite ion is 1+ and that of sulfur in the sulfite ion is 4+. Lewis structures: 4. Yes, cyclopentanone would give a positive result in the iodoform test. This is because the compound contains a ketone functional group which is characteristic of the iodoform test being positive. 5.

References: 1. Gunawardena, G. Haloform Reaction. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/ Organic_Chemistry_Glossary/Haloform_Reaction (accessed Mar 17, 2020). 2. Vlahakis, J. Queen’s Chemistry 2020 Laboratory Manual: CHEM 282 General Organic Chemistry II ; Queen’s University: Kingston, ON, 2020; p 53. 3. Clark, J. The Triiodomethane (Iodoform) Reaction. https://chem.libretexts.org/Bookshelves/Organic_Chemistry/ Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/ Reactivity_of_Aldehydes_and_Ketones/ The_Triiodomethane_(Iodoform)_Reaction (accessed Mar 17, 2020).

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