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11 Dec 2019

Ethyl alcohol may be dehydrated to ethylene by action of acatalyst. Since ethylene is a gas, it is not generally collected inthat form, but is instead brominated. Starting with 9.2 g of ethylalcohol and an excess of bromine, calculate the amount ofdibromoethane that would be obtained if the reaction proceeded inthe 75% yield.

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Casey DurganLv2

13 Dec 2019

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Please answer the prelab questions only that are 1,2 and 3. Thankyou

Using models may be helpful in answering these questions. If the brominating of room -still bone proceeded by seen addition what isomer(s) of stolen bromide would be formed? If the brominating of trans-stolen proceeded by anti-addition what isomer(s) of stolen bromide would be formed? Which isomer(s) of stolen bromide would be formed if the reaction followed Mechanism 4? Bromine is very toxic and corrosive. It can cause severe chemical burns on skin contact. Bromine vapor is very harmful. Make all transfers of bromine in the hood and if any is spilled, contact the instructor at once. Carry out the brominating part of this experiment in a hood. There is room to allow 2-3 people to work at each hood. Add about 2 g of trans-stolen to a thoroughly dried 125-ml. Erlenmeyer flask containing a dry magnetic stir bar. Add 20 mL of methylene chloride to the flask and swirl until the solid has dissolved. Clamp your flask to a ring stand over a magnetic stirrer. Adjust the stirring rate so that it is vigorous but without splashing. Obtain about 6 mL of a 2 M bromine solution in methylene chloride (approximately 10% bromine by volume) in a test tube. (Your instructor will provide the prepared solution.) A graduated cylinder will be placed in the hood to let you measure the bromine solution, this will avoid taking bromine-contaminated glassware out of the hood or to your bench. Once you have obtained your bromine solution, place the test tube in a small Kroc mayor flask for support and take it to your set-up in the hood. Carefully add about 1/3 of your bromine solution to your flask and note what color changes occur. Once the characteristic red-brown color of bromine has faded, add another third of your bromine solution to the flask. Again, note what changes occur. After the color has faded, add the remaining bromine solution. This lime, the color of bromine should persist for 3-5 minutes and not fade. If the color persists, the reaction is completed, and the color is due to excess bromine in the solution. If the color still fades, the reaction is not quite complete. If this is the case, add 10 drops of bromine solution and continue stirring until the color persists. A considerable amount of precipitate (the crude product) should be visible by this time. It may be difficult to determine if you have an excess of bromine. Allowing the material to settle for a minute or two should help you determine whether there is excess bromine in the solvent. Once you have a solution containing a small amount of excess bromine, add 1-2 drops of cyclohexane. This should cause the bromine color to disappear quickly. If the color still persists, add cyclohexane one drop at a time until the color fades, taking care not to add too much. Once the bromine color is gone, suspend the flask in an ice bath for 10-15 minutes, then recover the product by vacuum filtration on the Buchner funnel. Rinse the flask and product with two 3-ml. portions of ice-cold methylene chloride. Allow air to pass through your solid for at least 10 minutes to give your product lime to dry. Remove the solid from the Buchner funnel, and allow it to air dry for 10 minutes. Obtain the mass of your isolated product and determine its melting point. The melting point of racemic stilbene dibromide is 112-113 degree C. The melting point of meso-stilbene dibromide is 241-243 degree C. Calculate the yield of your product. You may be asked to obtain an IR spectrum for your product. Submit your product in an appropriately labeled vial.

Total amount of dye obtained (crystal violet) = 6.5mL

The absorbance using UV spectrometer is =1.75

Please help within the following

Using the reported value of the extinction coefficient of 112,000 M^-1 cm^-1, calculate the concentration of diluted sample and (according for dilution ) estimate the percent yield of this reaction by determining the number of moles of the dye formed. You can assume that the dye follows the Beer-Lambert Law and that A=ebc which for a 1cm cell simplifies to Absorbance A= ( 112,000 M^-1 cm^-1)(1.00cm) (concentration of crystal violet).

Please help thank you!

Organic Laborat 2. The Grignard Reaction: Preparation of Crystal Violet Reference: Journal of Chemical Education, 73, 259, (1996) Douglass Taber, Robert Meagley and Danielle Supplee Laboratory Objectives: 1. Prepare a Grignard Reagent from a bromoaromatic compound and magnesium. 2. Perform a water sensitive reaction at a micro-scale. 3. Observe the visible color of a highly conjugated triphenylmethyl dye. 4. Predict reaction products expected from the reaction of Grignard Reagents with carbonyl compounds. Introduction: The Grignard Reaction is a chemical reaction which to form the Grignard reagent would be classified as a reduction. Magnesium reacts with alkyl and aryl halide to form an organometallic reagent which can subsequently react as a carbon nucleophile in a chemical reaction. Grignard reagents can be stored as THF solutions but it is common to make the organometallic compound and then react it without work-up with some suitable compound. This could be carbon dioxide to form a carboxylic acid or water or deuterium oxide to form the reduced hydrocarbon. More commonly carbonyl compounds (organic compounds which contain a C-O group) are used to form the addition product of the organometallic compound and the carbonyl group. This results in an alcohol if aldehydes and ketones are used as the carbonyl group. If the carbonyl group used is an ester (R-COp-R') then the nucleophilic addition to the carbonyl carbon occurs as expected but the carbonyl is reformed by loss of the alkoxide ion. This new carbonyl compound can add an additional mole of organometallic reagent to again give an alcohol but one that contains two of the organic groups from the Grignard reagent. In this reaction you will prepare an aromatic organometallic reagent which will react with diethyl carbonate. Diethyl carbonate will perform the reaction described for esters above but since there are two alkoxide groups that can leave, this results in the addition of not two but three moles of the Grignard reagent. Also the product expected would be a tertiary alcohol in which each of the carbon groups is aromatic. The triphenyl methyl group is commonly called a "trityl" group and since the central carbon is both tertiary and benzylic to each of the rings the alcohol that might be expected will quickly dehydrate to give a stable carbocation (a so-called trityl or triphenylmethyl cation). In the compound that you will prepare called Crystal Violet or Gentian Violet the cation is not only a trityl cation but each of the rings has a dimethylamino group in the para position. These groups being electron donating by resonance will also help to stabilize the cation. This highly conjugated compound will absorb visible radiation and will be highly colored. This compound has been used as a dye in commercial applications including biological staining.

I want another 600 word discussion on 3-methyl-3-pentanol, here is the attached example if you need an idea. I don't want it to be the same.

"Discussion:

In this experiment, the 3-methyl-3-pentanol solution was being used in the dehydration process. The alcohol was a five carbons aliphatic hydrocarbon which one of the hydrogen atoms, H was substituted by one hydroxyl group, OH^-. Due to the low melting point, the pentanol appear in liquid form at room temperature. The dehydration process of alcohol converted pentanol which the hydroxyl group, OH^-was removed to become pentene. Pantene was a five carbons aliphatic hydrocarbon with a single double bond in the molecule. The solution was added with concentrated phosphoric acid in a 5-ml vial and was mixed together by swirling. The phosphoric acid was added as catalyze as such increased the rate of reaction in dehydration without affected the particular chemical reaction. The spin vain added into the solution in order to prevent over boiling of the solution.

The heating of mixture was carried out in a fractional distillation apparatus. As the mixture was heated, the alkene and water were produced as the products in the reaction. Besides, the side product and impurities of the reaction was being produced at the same time. The temperature of mixture when heating was fluctuated. During the temperature 80Â°C, the temperature of mixture dropped suddenly by 2Â°C and the temperature remained at 78Â°C constantly for few seconds. This was because some impurities or side products were being produced in the mixture which might had the boiling point of 78Â°C. The temperature of mixture dropped suddenly because the heat being was absorbed which used to break down the bonding of the side products. When the boiling point was reached, then temperature of mixture remained constant as the state of the side products are converted from liquid form to gaseous form. The temperature of mixture was increased until 107Â°C after all the side products were being converted.The temperature of mixture was reached to maximum at 107Â°C. This temperature was known as the activation temperature which the pentanol started to be dehydrated. In all the distillation process, some of the product was being lost since it washoled up in the apparatus which reduce the product yield. In order to maximize the yield, the mixture was continued to be heated at higher temperature range which more than the boiling point of pentanol. When the mixture was heated at 90Â°C-100Â°C, the water in the mixture pushed over the products into the receiving vial along the condenser. The products produced were collected in the same receiving vial.

Then, the receiving vial containing pentene, water and small amount of the impurities. Two layers liquid were presented in the receiving vial, one drop of distilled water was added into vial in order to determine the location of aqueous layer. Since the water droplet mixed with the lower layer, so the upper layer was determined as cloudy solution while the lower layer was aqueous layer. The upper cloudy layer was pentene with some impurities and water inside. The lower aqueous layer was removed and discarded. But, that is not easy to remove all the water in the receiving vial. So, anhydrous calcium chloride was being added. The purpose of adding of anhydrous calcium chloride was used to remove residual water in the organic solvent. It is known as drying agent in the organic solvent which are not dissolves in the solvent but drying the solvent. The calcium chloride clumped together with the water droplets as it solidified them. In another word, it reacted with water to form hydrates which was their preferred form when water was available. An excess drying agent should be used to ensure that all the water in solvent was removed. If the water remains in the materials collected, it could interfere with the analysis.Water had been successfully removed from the organic compound mixture, so it was very important to not reintroduce water into the mixture. The final distillation of unpurified 3-methyl-3-pentene must be done very carefully to obtain purified products. "

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