Lab 8 Preparation of Alkenes by E1 and E2 Substitution Reactions, Baeyer and Bromine Test Used for Saturation PDF

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Lab 8: Preparation of Alkenes by E1 and E2 Substitution Reactions, Baeyer and Bromine Test Used for Saturation Katja Gonzalez Lab Partner: Kyle Leonida, Juan and Michael 2018/04/09 Methods and Background The two main purposes of this lab were to prepare 2-methyl-2-butene and 2-methyl-1-butene. From 2-methyl-2-butanol and sulfuric acid you are able to obtain 2-methyl-2-butene from an E1 mechanism. On the other side of this lab 2-methyl-1-butene is obtained from 2-chloro-2methylbutane and potassium hydroxide by an E2 mechanism. Before perming Bromine or Baeyer tests, the approximate relative amounts of the created alkene product need to be calculated and determined by using GC analysis. Once this is complete the Bromine and Baeyer tests are completed to identify if there is presence of the correct alkenes. There are many variants between both the E1 and E2 tests. These variances include the mechanism, rates or reactivity, reactions that are in competition and the regioselectivity. When looking at the steps of the reactions, E1 is known for have 2 separate ones while E2 has a singular step. Starting with the E1 steps, the first step includes the loss of the leaving group and the second step shows the deprotonation of a carbocation that is equal to the product. The singular E2 reaction step is a combination of the steps present in E1 reactions however they are just shrunk to fit into one step. One thing to keep in mind is the removed hydrogen has to be “anti” against the leaving group.

Figure 1: E1 reaction(left) and E2 reaction(right)

For the E1 reaction the rate is completely dependent on the substrate. This is due to the limiting step containing the part where there is the formation of carbocation. With this more stable carbocation the reaction will be able to proceed faster. In the starting material the hydrogen is able to be positioned into any orientation against the leaving group as there is no particular requirement for the stereochemistry. The E2 reaction rate is dependent on two things, both the substrate and the base. This is because the step used to determine the rate is bimolecular. When looking at the stereochemistry, the hydrogen that is removed needs to be anti in regard to the leaving group.

In order to increase the percent yield of the outcome Le Chatelier's principle is utilized which manipulates the concentration used in the procedure. Le Chatelier’s principle states that this should shift the equilibrium towards the right and maximize the actual yield of the alkene. The preference of one side of the reaction to another is known as regioselectivity. A reaction can be considered as regioselective if the reaction is able to happen in more than one direction with a clear direction that is favored over the other. Zaitsev’s rule looks for the most stable product with the least amount of hydrogen and one that is highly substitutable for the alkene. This rule has two exceptions that should be looked out for. The first rule is that steric hindrance will inhibit the deprotonation of hydrogen will induce a duct not considered as substitutable as it should be. The second rule deals with the stereo electronic factors. This is when the halogen and hydrogen are not the anti-coplanar relationship which means the reaction is unable to proceed in the correct direction and instead produces a Hoffman product. The E1 dehydration of alcohols and the E2 dehydrohalogenation of alkyl halides can produce alkenes. To create alkenes by the dehydration reactions the alcohols are heated with the presence of a strong acid. H+ on the acid reagent obtain two electrons from the -OH group located on the alcohol. This will form an alkyloxonium ion that will act as a leaving group that forms a carbocation. The recently deprotonated acid will form a double bond with the hydrogen that is adjacent to the carbocation that has just formed. To create alkenes by dehydrohalogenation of the alkyl halides the base will remove the hydrogen of the C-H bond during the formation of the double bond of the alkene. This will cause the halogen group to separate itself. The next portion of the lab utilizing fractional distillation. The fractional method of distillation is used instead of simple because it has a higher accuracy. When using fractional distillation, the separation of liquids occurs with lower boiling points than with what we see in simple distillation. With fractional distillation you can raise the theoretical plate count which will in turn produce a greater yield that can be considered to be more accurate. Another form of distillation that is used is Reflux which gives heat to the total reaction. With this increase in heat it raises the rate of reaction while maintaining around a constant temperature. To analyze the product distribution gas chromatography(GC) is used. GC should show the fluctuations of each substances gas and liquid phases. To obtain GC a small portion of sample is placed into the chromatographer which can instantly vaporize the sample. Lastly unsaturation is tested by using both Baeyer and Bromine tests. A positive test is seen by a clear color change due to the reactiveness of the alkene by having no carbon-carbon double bonds. During a Bromine test, the presence of bromine will combine with the carbon-carbon double bond of alkenes and create dibromo alkanes. After this reaction occurs the bromine molecule is consumed which will produce the dark-red brown color. A positive test will be seen if there is a quick disappearance of bromines color. Using the Baeyer's test is dependent upon the ability of potassium permanganate to oxidize the carbon-carbon double bonds which will give you the alkane diols. When the permanganate is obliterated a brown precipitate will be left from the production of the MnO2.

Figure 2: Reflux distillation(left) and Fractional distillation(right)

Experimental Procedure E2 Reaction: The first step of the E2 reaction is to calculate the amount of KOH and 1-propanol needed for this experiment. 3.63 g of KOH and 36.6 mL of 1-propanol is used for a 3.00 mL sample of 2chloro-2-methylbutane. The calculated amount of KOH and 1-propanol were then weighed/measured and placed into a round bottom flask. The KOH is in solid form, so the flask needs to be heated slowly until it dissolved. During this process the flask was covered with a watch glass in order to prevent any 1-propanol from evaporating out. Once the KOH is dissolved the flask it cooled to room temperature. To this cooled mixture the 3.00 mL of 2-chloro-2methylbutane is added. The flask is then attached to a reflux apparatus and is heated for one hour. The hour does not start until there is a presence of boiling occurring. It is recommended that the E1 reaction procedure should be completed during this wait. Once the hour has come to an end the flask will be cooled to below boiling point while still being attached to a reflux condenser. Fractional distillation apparatus is set-up during this cooling process. As the flask reaches room temperature it is then placed on the fractional distillation apparatus. All distillate was collected that was boiled off below 45° C into a pre-weighed flask. The final flask and product are then weighed together. The total mass of the product and percent yield are calculated at this time. To assure no product evaporates out a stopper is placed a top it and placed into an ice bath. E1 Reaction 2.00 mL of 2-methyl-2-butanol was measured and placed into a 25 mL round bottom flask. 10 mL of 6.00 M sulfuric acid is then slowly added into this flask. This mixture is then heated on a and distilled on a fractional distillation apparatus. Distillate is collected under the 45° C boiling point into a pre-weighed flask which is positioned in an ice bath. The mass of the product is determined by the re-weighed the flask where percent yield is then calculated. The flask is stoppered in order not to lose any product.

Gas Chromatography Both products obtained from the E1 and E2 reaction were used for gas chromatography. A miniscule portion of each sample was obtain using a syringe and inserted into the gas chromatographer. The computer ran the designated programs and graphs were printed with highlighted peaks. You should have two different graphs when GC is completed. Using the peaks on the graph mole percent is calculated.

Baeyer Test Three test tubes are prepared with 1.00 mL of H2O in each. From the E1 reaction, 1-2 drops of the product are added into one test tube. From the E2 reaction 1-2 drops of the product is added into a separate test tube. The final tube should have no additional factors. Lastly several drops of 0.10 M of KMnO4 was added into each test tube. When looking for a positive test the solution should turn into a brownish color.

Bromine Test Three test tubes are prepped with 1.00 mL of methylene chloride added to each. From the E1 reaction 1-2 drops is added to one test tube and from the E2 reaction 1-2 drops is placed into the second tube. Again, the final tube is left alone. The final step is the addition of several drops of 0.10 M of Br2 into each of the tubes. A positive test is indicated by the solution changing to a clear color.

Data Acquisition I. Reaction Tables Compound Mol. Weight 2-Chloro-2106.59 g methylbutane KOH 56.11 g 1-propanol 60.10 g II.

mmol

Equivalents

0.866 g/mL

Rxn Weight/Vol. 3.00 mL

24.37

1.0

n/a 0.804 g/mL

3.00 g 30.06 mL

53.62 402.17

2.2 16.5

Relevant Equations [𝐴] Molar Percent = [ ] × 100% Percent Yield =

III.

Density

𝐴 +[𝐵] 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑

𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑

× 100%

Calculations Percent yield of E1 Molar Percent 2-methyl-1-butene = Molar Percent 2-methyl-2-butene = 4 mL 2-chloro-2-methylbutene × 4 mL 2-chloro-2-methylbutene ×

1309 𝑚𝑉𝑠 1309 𝑚𝑉𝑠+2575 𝑚𝑉𝑠 2575 𝑚𝑉𝑠 1309 𝑚𝑉𝑠+2575 𝑚𝑉𝑠

0.865 𝑔 𝑚𝐿 0.865 𝑔 𝑚𝐿

× ×

1 𝑚𝑜𝑙 106.59 𝑔 1 𝑚𝑜𝑙 106.59 𝑔

× ×

× 100% = 33.7% × 100% = 66.3%

1 𝑚𝑜𝑙 1 𝑚𝑜𝑙 1 𝑚𝑜𝑙 1 𝑚𝑜𝑙

Mass of product = 0.12 g 0.12 g × 0.337 = 0.040 g 2-methyl-1-butene produced

× ×

70.135 𝑔 𝑚𝑜𝑙 70.135 𝑔 𝑚𝑜𝑙

= 2.28 g 2-methyl-1-butene = 2.28 g 2-methyl-2-butene

0.12 g × 0.663 = 0.080 g 2-methyl-2-butene produced Percent yield of 2-methyl-1-butene = Percent yield of 2-methyl-2-butene =

0.040 𝑔 2.28 𝑔 0.080 𝑔 2.28 𝑔

× 100% = 1.8% × 100% = 3.5%

Percent yield of E2 (Using Michael and Juan’s Data) 3710 𝑚𝑉𝑠

Molar Percent 2-methyl-1-butene = Molar Percent 2-methyl-2-butene

× 100% = 66.0%

1912 𝑚𝑉𝑠+3710 𝑚𝑉𝑠 1912 𝑚𝑉𝑠 = 1912 𝑚𝑉𝑠+3710 𝑚𝑉𝑠

× 100% = 34.0%

Mass of product = 1.6 g 1.6 g × 0.660 = 1.06 g 1.6 g × 0.340 = 0.54 g Percent yield of 2-methyl-1-butene = Percent yield of 2-methyl-2-butene

1.06 𝑔

× 100% = 46.3%

2.28 𝑔 0.54 𝑔 = 2.28 𝑔 × 100% =

23.9%

IV. Table of Products Reaction mechanism E1 E2 Mass of distillate (g) 0.12 1.6 2-methyl-1-butene 2-methyl-2-butene 2-methyl-1-butene 2-methyl-2-butene Products Mass of product (g) 0.040 0.080 1.06 0.54 Percent yield 1.80% 3.50% 46.3% 23.9% V. Test Tube Results

Baeyer Test and Bromine Test Baeyer Test 1 (E1)

2 (E2)

Positive

Positive

3 (Nothing)

Negative

Bromine Test 1 (E1)

2 (E2)

Positive

Positive

3 (Nothing)

Negative

Conclusions The two purposes of this lab were to create 2-methyl-2-butene and 2-methyl-1-butene. This was done by using 2-methyl-2-butanol with H2SO4 and 2-chloro-2-methylbutane with potassium hydroxide using both E1 and E2 elimination reactions respectfully. GC’s were taken at the end of the reaction procedure and the products were then tested by Baeyer and Bromine test to look at unsaturation. The compounds were distilled to isolate the alkene from starting compound. Our obtain percent yield for the E1 reaction was 33.70% of 2-methyl-1-butene and 66.30% of 2-methyl-2-butene. The percent

yields achieved for the E2 reaction were 66.0% of 2-methyl-1-butene and 34.0% of 2-methyl-2butene. Factors that may have altered our yields to show them as not as ideal may include partial evaporation of product due to stopper not placed correctly or quickly enough atop the flask. 2methyl-2-butene has three alkyl groups present on the carbon-carbon double bond which provides a larger stable ground making this the majority product. With this greater stability it makes 2-methyl-2-butene favorable for biomolecular elimination. This favorability is also seen

when viewing the ideal percent composition calculated from the GC graphs. E1 graphs had the presence of two peaks, 33.70% of 2-methyl-1-butene and 66.30% of 2-methyl-2-butene. The larger peak of 2-methyl-2-butene demonstrates is was a highly substituted product. The E2 graphs showed different results with percentages of 66.0% of 2-methyl-1-butene and 34.0% of 2methyl-2-butene. This shows the 2-methyl-1-butene was actually the more highly substitutable product. For both the Baeyer and Bromine test six test tubes were prepared. This test was used to determine the presence of alkenes in the final product. Two positive results were obtained from the tests. For the Baeyer test there was a distinct change from purple to a red brownish color while for the Bromine test there was a quick change from orange to clear. Excluding the control group all tests reacted positively and demonstrated a color change. References Gilbert, J.C., and Martin, S.M., Experimental Organic Chemistry, 5th Edition, Cengage Learning, Boston, MA, 2011. Landrie, C.L., and McQuade, L.E., Organic Chemistry: Lab Manual and Course Materials, 3rd Edition, Hayden-McNeil, LLC, Plymouth, MI, 2013....