Lab 7 Preparation of Alkenes by E1 and E2 Elimination Reactions; Baeyer and Bromine tests for unsaturation PDF

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Tanvi Kulkarni 10/27/2020 Preparation of Alkenes by E1 and E2 Elimination Reactions; Baeyer and Bromine tests for unsaturation Methods and background: The purpose of this lab was to prepare a distribution of alkenes using E1 and E2 reactions. We prepared 2- methyl-2-butene and 2-methyl-1-butene using the E1 mechanism with sulfuric acid and 2methyl-2-butanol. The E2 mechanism is used to get 2-methyl-1-butene from 2-chloro-2-methylbutane and potassium hydroxide. We will use fractional distillation to isolate the products and use the Baeyer and bromine tests to confirm functional groups in our products. The two different types of elimination reactions used in this experiment are E1 and E2. E2 is the dehydrohalogenation reaction and this is where the carbon-halogen bond and adjacent carbon-hydrogen bonds are converted into carbon-carbon double bonds. The other reaction is E1 which is a dehydration reaction where the carbon-hydroxyl bond of a tertiary alcohol is changed into a carbon-carbon double bond. When analyzing the Elimination reactions, we see that there is a partial positive charge on the alpha carbon atom of an alkyl halide so it becomes electrophilic and will produce a Sn2 reaction instead of an E2 reaction because a nucleophile will attack. The major product in the E2 reaction is a Zaitsev product which occurs when the B-hydrogen is removed from the most substituted carbon. The minor product is a Hoffman product. Dehydration is the loss of water of the oxonium ion in order to produce an alkene. The movement of a proton from a carbon to water is classified as an E1 elimination because it is a first order process. In order to avoid side reactions occurring, H2SO4 is used to effect the dehydration of alcohol because the substitution product becomes ionized to the carbocation and then continues to lose a proton and produce an alkene. The rate of an E1 reaction is only dependent on the concentration of the original substrate. Then we can apply La chatelier's principle to an E1 mechanism reaction. If we follow La Chatelier’s principle, the alkene will be removed from the product and the equilibrium will shift towards the formation of alkenes. When talking about regioselectivity, more than one product can form if the leaving

Tanvi Kulkarni 10/27/2020 group is not symmetrical in the molecule. So, regioselectivity tells us that more than one product can be formed but one is preferred over another. Zaitsev's rule predicts the product and the major product is the one that is more stable. With our lab, 2-methyl-1-butene is less stable than 2-methyl-2-butene which is why 2-methyl-2-butene, is the major product. The exceptions to Zaitsev’s rule can be for steric factors which occur if the steric factors increase the energy with the less substituted alkene. The other exception is due to stereoelectronic factors which can stop the formation of a more substituted product. We used fractional distillation to separate the products from each other. We use this method over simple distillation because the boiling point difference between the compounds is less than 50 degrees celsius. Fractional distillation uses Hempel columns and Raschig rings to increase the surface area and perform multiple vaporizations in a shorter time frame with more efficiency. The apparatus is shown below.

We used two tests to confirm that we synthesized the alkenes, Bromine and Baeyer tests. The Baeyer test involves a reaction between potassium permanganate and Carbon-carbon bond. If the color changes from the purple to yellow-brown, there is a presence of a c-c pi-bond. The Bromine test has the reaction between an alkene and Br2. If the orange color disappears during this test, it is indicative of a positive test. The last thing we did in the experiment was gas chromatography which proved the presence of two alkenes at different concentrations showing a presence of major and minor products of the reactions.

Experimental procedure:

Tanvi Kulkarni 10/27/2020 Before beginning the experiment make sure all safety attire is worn, like goggles and gloves. The first experiment is the E2 mechanism with 2-chloro-2-methylbutane. Potassium hydroxide pellets are added to a round bottom flask. Using a graduated cylinder 1-propanol was measured and added to the flask. A boiling stone is added and clamp the flask to a ring stand. The solution was gently heated to allow the hydroxide to dissolve. When the hydroxide was mostly dissolved, it was removed from the heat and placed on ice. The Sn1 product from the last experiment was added via pipette. The flask is then clamped to a ring stand. A hempel column was filled with raschig rings like in fractional distillation. The water was also attached like a condenser. The joints were greased to prevent them from sticking together. The condenser was attached and the water and heat were turned on. The reaction was refluxed for an hour. After an hour in reflux, the heat was removed and the flask was cooled. Once cooled, the remaining fractional distillation apparatus was assembled with the stillhead attached to the hempel column that was already filled with raschig rings. The condenser was attached to the arm of the stillhead followed by the vacuum adapter. Keg clips held everything together. On the received side, a preweighed round bottom flask is attached. Ice was then placed under the receiving flask and the thermometer adapter was added. The heating mantle was raised and the water was turned on. The distillation was continued until the stillhead temperature was 45 degrees celsius. The flask was removed and massed. The second experiment was an E1 elimination with T-amyl alcohol. To begin 2 mL T-amyl alcohol was measured in a graduated cylinder and put into a round bottom flask. 10 mL of 6 M sulfuric acid was measured and added to the flask. A boiling stone was added and the flask was clamped to the ring stand. The fractional distillation apparatus was assembled over the flask, similar to the previous experiment. The receiving flask was again pre weighed and put on ice. The thermometer adapter was then attached and the water and heat were turned on. The product was collected until the stillhead temperature was 45 degrees celsius. Once done, the flask was removed and massed. In 6 small test tubes about 1 mL of water was added to each. A small amount of E1 product was added to 2 of the test tubes. An equally small amount of E2 product was added to the next 2 test tubes.

Tanvi Kulkarni 10/27/2020 The last 2 test tubes were kept empty as a constant. Using a pipette a small amount of 0.1 M potassium permanganate were added to the E1, E2 and control test tubes. Record the results. A small amount of bromine solution was added to the remaining E1,E2, and control test tubes. Record the results. Using GC, determine the ratio of zaitsev to hoffman products. Using a microsyringe about 2-3 microliters were injected into the GC. Integrate and record the peaks. Repeat for the other products.

Calculations: E1 pre lab calculation: Mmol of 2-methyl-2-butanol: 2 mL x (0.805 g/1 mL) x (1 mol/88.15 g) x (1000 mmol/1 mol)= 18.3 mmol H2SO3 mmol: 10 ml x (6 mmol/ 1mL)= 60 mmol E2 pre lab calculations: Grams 2-chloro-2-methylbutane:673261106/10^8= 6.73 Mmol 2-chloro-2-methylbutane: 6.73 g x (1 mol/106.59 g) x (1000 mmol/ 1 mol) = 63.5 mmol Mmol KOH: 63.5 mmol x 2.2 eq= 139 mmol Mmol 1- propanol: 63.5 x 16.5 eq= 268.11 mmol mL 1-propanol: 268.11 mmol x (1/1000 mmol) x (60.1 g/ 1 mol) x (1 mL/ 0.804 g)= 20.04 mL Equivalents: 2-methyl-2-butanol= 18.3/18.3=1 H2SO4=60/18.3=3.28

Table 1. E2 and E1 calculations Compound

MW

d or M

Rxn weight or

mmol

equivalents

Tanvi Kulkarni 10/27/2020 V 2-methyl-2butanol (1)

88.15

0.805 g/mL

2.00 mL

18.3

1

H2SO4

98.08

6M

10.0 mL

60

3.28

2-chloro-2methylbutane (4)

106.59

0.866 g/mL

6.73 g

63.5

1.00

KOH

56.11

-

7.79 g

139

2.20

1-propanol

60.10

0.804 g/mL

62.9 mL

1047

16.5

E1 reaction percent yield: Weight of distillate= 0.8 g Theoretical yield= (18.3E-3 mol 2-methyl-2-butanol)( 1mol alkene/ 1 mol 2-chloro-2 methylbutane) (70.05 g alkenes/ 1 mol alkenes)= 1.28 g alkene Percent yield of alkene product= 0.8/1.28 *100= 62.5% Percent yield 2-methyl-1-butene= peak percentage x percent yield/ 100= 8.27x62.5/100= 5.17% Percent yield 2-methyl-2-butene= 91.7x 62.5/100= 57.3% E2 percent yield: Weight of distillate=3.6 g Theoretical yield= 63.5*10^-3 mol x( 1mol/1mol) x 70.05 g alkene/ 1mol alkene= 4.45 g alkene percent yield of alkene product= 3.6 g/ 4.45 g*100= 80.9% Percent yield 2-methyl-1-butene= peak percentage x percent yield/ 100= 12.9 x 80.9/100=10.4% Percent yield 2-methyl-2-butene= 28.1 x 80.9/100=22.7% GC analysis mol%= area of compound A/ (Area compound A + area compound B) x100 E1 reaction: Peak 1: area- 82.6 s*mV Retention time- 9.5 s

Tanvi Kulkarni 10/27/2020 2-methyl-1-butene mol%= 82.6/(82.6+916.3)*100=8.27% Peak 2: area-916 s*mV Retention time- 11 s 2-methyl-2-butene mol%= 916/(82.6+916.3)*100=91.7% E2 reaction: Peak 1: area-128.5 s*mV Retention time: 9.7 s 2-methyl-1-butene Mol %= 128.5/ (128.5+280.6)*100=31.4%

Peak 2: area- 280.6 s*mV Retention time: 12 s 2-methyl-2-butene Mol %= 280.6/ (128.5+280.6)*100=68.6% Table 2 % yield, g

8.27%

91.7%

31.4%

68.6%

Bayer test

positive

positive

positive

positive

Bromine test

positive

positive

positive

positive

Table 3: Baeyer and Bromine test observations

Test

E1 product

E2 product

Control

Baeyer test

The color turned from purple to brown so it is positive

The color change was slower than E1 but it changed from purple to brown so its positive

Purple color remained

Bromine Test

The orange color disappeared and left colorless solutionpositive

Orange color disappeared- positive

Orange color remained

Conclusion: The objective of this lab was to analyze 2-methyl-1-butene and 2-methyl-2-butene with

Tanvi Kulkarni 10/27/2020 Elimination reactions. We saw the reactions using a E1 mechanism and E2 mechanism. After we used gas chromatography to find the percent composition of the alkene products. Then we used Baeyer and Bromine tests to confirm if the separation was successful. From the results, the goal was successfully achieved. We found that the percent yield of E1 reaction was 62.5% and the percent yield of the E2 reaction was 80.9%. Both of these percent yields are in the acceptable range for the products. Because of the larger amounts of alkene products that we got, we can assume that the synthesis reactions were successful. When analyzing the gas chromatography graphs, we notice that the first peak had smaller areas for both E1 and E2 mechanisms and it was 2-methyl-1-butene. The larger peak was 2-methyl-2-butene. Usually in GC the liquid with the lower boiling point is the less substituted alkene because it takes a shorter time to evaporate. This means this is the Hofmann product which was 8.27% in E1 and 31.4% in E2. The larger peak was 2-methyl-2-butene or the Zaitsev which produced 91.7% in E1 and 68.6% E2. Reviewing the results of the Baeyer and Bromine tests, it is confirmed that there is a presence of alkenes in the solutions. The Br2 and potassium permanganate react to the presence of a c-c double bond. With the Baeyer test, the test is positive if the purple color changes to a brown-yellow shade. In the bromine test, it is positive if the orange color disappears and the liquid left is colorless. All the test tubes that were tested with the solutions ended up changing colors thus proving the presence of alkenes in each. When comparing the two methods of synthesis, we see that the E2 method was more effective. The reason being that the E2 reaction produces a larger Zaitsev product compared to the E1 reaction. The percent yield for E2 was larger than the E1 reaction. This proves that the E2 reaction is more efficient.

Post lab questions: 1) The reason the excess of a base favors the E2 mechanism is because the alkyl halide in the elimination process is attacked by an abstract hydrogen and the halide is the leaving group. If the strong base is used, the reaction proceeds faster. Increasing the amount of base will increase the

Tanvi Kulkarni 10/27/2020 rate of the E2 reaction but will not affect the E1 reaction. 2)

3)

4) The alcohol will have a higher boiling point because it consists of mainly hydrogen bonds. Alkenes only have van de wall forces. 5) It is kept below 90 degrees celcius to keep the temperature lower than the boiling point of water. 6) Both products have a lower boiling point than water so if water is removed, the product generated will also be removed with the water. The removal of water would not be successful.

References: Gilbert, J.C., and Martin, S.M., Experimental Organic Chemistry, 5th Edition, Cengage Learning, Boston, MA, 2011 Landrie, C.L., McQuade, L.E., Yermolina, M.V., Organic Chemistry: Lab Manual and Course Materials, 8th edition, Hayden McNeil, LLC, Plymouth, MI, 2018

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