Preparation of Alkenes by E1 and E2 Elimination Reactions, Baeyer and Bromine Tests for Unsaturation PDF

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Preparation of Alkenes by E1 and E2 Elimination Reactions, Baeyer and Bromine Tests for Unsaturation.docx...

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Gandhi 1 Preparation of Alkenes by E1 and E2 Elimination Reactions, Baeyer and Bromine Tests for Unsaturation

Partner: Victor, Zahra 03/29/2016 Methods and Background: The objective of this lab is to prepare distribution of alkenes by E1 and E2 reactions. To isolate the products fractional distillation will be used. Further on, to confirm the functional groups in the products, Baeyer and bromine classification tests will be done. Finally, GC analysis of the products would help to calculate the mole percent and percent yield of the final products. There are two types of Elimination reactions which produces alkenes as the products. In dehydrohalogenation reaction (E2), the carbon-halogen bond and adjacent carbon-hydrogen bonds are converted into carbon-carbon double bonds whereas in dehydration reaction (E1), the carbon-hydroxyl bond of a tertiary alcohol is converted into a carbon-carbon double bond.

Figure 1: Dehydrohalogenation Reaction KOH

+ Figure 2: E2 Reaction Performed in this Lab Figure 1 shows the dehydrohalogenation reaction which is also known as E2 reaction where E refers to elimination and 2 refers to the molecularity of the rate determining step. E2 is a bimolecular reaction because both the substrate and the base are involved in the transition state of the rate determining step and its rate of the reaction is k2 [alky halide][B:-]. Due to the partial positive charge on α-carbon atom of an alkyl halide, α-carbon becomes electrophilic and allows nucleophile to attack it which produces a Sn2 reaction rather than an E2 reaction. The elimination process is favored only as the degree of substitution on the α-carbon atom increases. Due to steric hindrance, the nucleophile is enabled to attack the α-carbon, remove the β-proton, and form the corresponding elimination product. Due to the regioselective nature of E2 reaction, it favors the formation of the more highly substituted alkene.

Gandhi 2 The Zaistev product in E2 reaction is the predominant product which is formed from the removal of the β-hydrogen on the carbon that is most substituted; whereas the minor product is called the Hoffman product. Zaitsev’s rule produces a stable elimination product but there are few exceptions. An example would be that if the nucleophile is a bulky base or the non-hydrogen substituents on the β-carbon are large, then it cannot easily access the β-hydrogen on the alkyl halide. So, in these cases, the major product is the Hoffman’s product whereas Zaitsev’s product is the minor product. For second exception has to deal with stereochemistry in which for the E2 reaction, the β-hydrogen and the halide ion which are going to be eliminated must be anticoplanar. In other words, the C-H bond and the C-X bond must lie in the same plane in order for orbital overlap to occur in the transition state, resulting in the formation of the new π bond. So, ultimately the β-hydrogen being removed to form the Zaitsev’s product cannot adopt an anticoplanar geometry relative to the leaving group, and then the β-hydrogen being removed to form the Hoffman product will be the major product.

Figure 3: Dehydration Reaction H2SO4

+

Figure 4: E1 Reaction Performed in this Lab The secondary and tertiary alcohols are converted to alkenes in dehydration reaction. The first step in the mechanism of this reaction is protonation of alcohol which is rapid and produces an oxonium ion. Since water is a good leaving group, it results in the formation of a carbocation which is kinetically a first-order process and it is the rate determining step of the overall reaction. Since, only one species is involved in the process, this reaction is called E1, unimolecular elimination, reaction. The dehydration reaction mechanism is shown in Figure 3 which describes that the carbocation attacked by a nucleophile could result in the formation of a substitution product or an elimination product. In E1 reactions, the intermediate carbocation combines with the conjugate base of the acid catalyst or with the some other nucleophile such as a solvent. If HCl were chosen as a catalyst, chloride ion could trap the carbocation to produces an alkyl chloride. Whereas H2SO4 is used as a catalyst in E1 reactions so its intermediate alkyl

Gandhi 3 bisulfate substitution product readily deionizes to the intermediate carbocation which results in formation of alkenes. The formation of the carbocation is an endothermic process where the transition state looks like the product and is characterized by heterolysis of the carbon-oxygen bond with partial positive charge on the carbon atom. Thus, dehydration reaction follows the order of 3º >2º >>1º with respect to stabilities of carbocation. The intermediate carbocation can undergo rearrangement through a hydride or a methyl shift from a carbon atom adjacent to the cationic center. The rearrangement is more likely to occur if the new carbocation is more stable than the original ion. In E1 reactions, two or more isomeric alkenes may be formed which depends on the stability of the transition state. The lower energy transition state will lead to the more highly substituted and more stable alkene. E1 reaction is reversible, so an alkene may undergo acid-catalyzed hydration to form an alcohol. However, reversal of the dehydration reaction, which is hydration reaction, can be avoided by removing the alkene whose boiling point is lower than the parent alcohol from the reaction mixture by distillation. This shifts the equilibrium to the right and maximizes the yield of alkene. Thus, Le Chatelier principle is illustrated where the equilibrium of a reversible is shifted to the right by removing one of the products and a new equilibrium has been established. In Baeyer tests, the double bond is oxidized. In this reaction, the alkene is reaction with purple color KMnO4 and water to produce cyclic manganite ester which then produces into vicinal diol, a colorless solution and manganese dioxide a brown color residue. Bromine reacts with alkenes to produce dibromoalkanes. So, bromine reaction is the addition reaction across the double bond. The alkene reacts with bromine of orange color to produce cyclic bromonium ion which becomes in to vicinal dihalide a colorless solution. To obtain higher percent yield of alkenes, fractional distillation of the samples will be performed. Fractional distillation is used to separate two or more volatile components present in the mixture. Since fractional distillation contains a packed Hempel column which provides a vertical path through which the vapor must pass from the still pot into the condenser before being collected in the receiver. The uncondensed vapor along with that produced by revaporization of the condensate in the column rises higher and higher in the column and undergoes a repeated series of condensation and revaporization. The repeated cycle of condensation and vaporization is equivalent to performing simple distillation and is termed as theoretical plates in the column. The vapor phase produced in each step of fractional distillation becomes richer in the more volatile component and the condensate that flows down the column becomes richer in the less volatile component. Thus, the number of theoretical plates helps us determine the efficiency and separating power in fractional distillation. Fractional distillation separates the mixture of two volatile compounds better than simple distillation because it contains more theoretical plates which increase the efficiency of separation. Finally, in order to verify the formation of alkene product, bromine and Baeyer tests will be done.

Gandhi 4 Procedure: Part I: Synthesis of 2-Methyl-1-Butene and 2-Methyl-2-Butene E2 Reaction: Calculate the amount of each reagent used in the experiment using the reaction table and equivalents. 2.02 g of KOH and 20 mL of 1-propanol were measured out and transferred to a 100 mL round bottom flask. 1 boiling stone was also added to the flask. Then, the mixture was warmed to completely dissolve KOH. After dissolution of the base, the flask was cooled to room temperature, and then by placing it in an ice bath. Slowly 2-chloro-2-methylbutane was added to the mixture. Then, the reflux apparatus was assembled as shown in Figure 5 and also the joints were greased to avoid freezing from the caustic KOH solution. The Hempel column was filled with Raschig rings. The reaction mixture was heated under reflux for about an hour. The reaction was allowed to cool below its boiling point. The water hoses were removed and water was drained the outside of the Hempel column. The same apparatus was assembled for fractional distillation. Before performing the distillation, the receiving flask was pre-weighed. The distillate was collected below 45ºC were collected in a flask which was kept in an ice-water bath at all times. The final product was weighed and then GC analysis, Baeyer, and bromine tests were performed.

Figure 5: Heating under reflux apparatus

E1 Reaction: In a 25 ml of round bottom flask, 2 mL of 2-methyl-2-butanol and slowly10 mL of 6M of H 2SO4 was added to the reaction mixture. The mixture was distilled using fractional distillation so the Hempel column was filled with Raschig rings. Before performing the distillation, the receiving flask was pre-weighed. The distillate was collected below 45ºC were collected in a flask which

Gandhi 5 was kept in an ice-water bath at all times. The final product was weighed and then GC analysis, Baeyer, and bromine tests were performed Part II: Analysis of Product Distribution by as Chromatography A gas chromatography graphs were obtained of both reaction mixtures. The molar percentages of 2-methyl-1-butene and 2-methyl-2-butene (alkenes) in each fraction were calculated. These molar percentages were used to calculate the percent yield. Par III: Confirmation of Alkene Formation by Classification Tests Baeyer Test: Three test tubes were prepared by adding 1 mL of 95% ethanol to each. In a first test tube, 1-2 drops were added of the product mixture from the E1 reaction. In second test-tube, 1-2 drops of the product mixture were added from the E2 reaction and into the third test tube nothing was added. In all test tubes, solution of 0.1 M KMnO4 was added to three test tubes drop wise. The drops were counted until the purple color persisted. It is a positive test when the number of drops required for the reaction solution exceeds the number of drops required for the blank solution. Bromine Test: Three test tubes were prepared by adding 1 mL of methylene chloride to each. In a first test tube, 1-2 drops were added of the product mixture from the E1 reaction. In second test-tube, 1-2 drops of the product mixture were added from the E2 reaction and into the third test tube nothing was added. In all test tubes, solution of 0.1 M Br2 in dichloromethane was added to three test tubes drop wise. The drops were counted until the light orange color just persisted. It is a positive test when the orange color will rapidly disappear to yield a colorless solution.

Data Acquisition/Calculations: Table 1: Summary of results from different tests Elimination

E1

E1

E2

E2

Product name

2-Methyl-1Butene

2-Methyl-2Butene

2-Methyl-1Butene

2-Methyl-2Butene

Yield

0.0834 g

0.867 g

0.382 g

0.598 g

Bp (oC)

31.2

37.5-38.5

31.2

37.5-38.5

Structure

Gandhi 6

Baeyer Test

POSITIVE

POSITIVE

POSITIVE

POSITIVE

Bromine Test

POSITIVE

POSITIVE

POSITIVE

POSITIVE

(Boiling Point Reference Website: http://pubchem.ncbi.nlm.nih.gov) Table 2: Reaction table of E1 Compound

Molecular weight

d(g/ml) or M (mmol/ml)

Rxn weight or volume (g or ml)

mmol Equivalents

2-methyl-2 butanol

88.15

0.805 g/ml

2.0 ml

18.3

1.0

Sulfuric acid

98.08

6M

10 ml

60

3.28

Table 3: Reaction table of E2 Compound

Molecular weight

d(g/ml) or M (mmol/ml)

Rxn weight or volume (g or ml)

mmol

Equivalents

2-chloro-2 methylbutane

106.59

0.866 g/ml

2 ml

16.25

1.0

KOH

56.11

N/A

2.006 g

35.75

2.2

1-proponal

60.10

0.804 g/ml

20.04 ml

268.11

16.5

Observations: Products

Baeyer Test

Bromine Test

E1 reaction Products

After 3 drops of KMnO4 were added to the test tube, it turned brown in color.

E2 reaction Products

After 2 drops of KMnO4 were added to the test tube, it turned brown in color.

After 3 drops of Br2 in dichloromethane were added to the test tube, it turned orange in color and disappeared to colorless solution. After 1 drop of Br2 in dichloromethane were added to the test tube, it turned orange in color and disappeared to colorless solution.

Gandhi 7 Blank solution in test tube

After 1drop of KMnO4 was added to the test tube, it turned purple in color

After 2 drops of Br2 in dichloromethane was added to the test tube, it turned orange in color and disappeared to colorless solution.

The percent yield will help in determining the success of this experiment. The formula for calculating the percent yield as follows, Percent Yield = [(Actual Yield) / (Theoretical Yield)] x 100 Percent yield of Products in E2 Reaction: Weight of the distillate (Actual Yield) = 0.98 g Theoretical Yield = (16.25 * 10-3 mol 2-chloro-2 methylbutane) * (1 mol of alkene / 1 mol of 2chloro-2 methylbutane) * (70.05 g of alkenes / 1mol alkenes) = 1.14 g alkenes Percent Yield: [(0.98 g) / (1.14 g)] x 100 = 86.1 % Percent yield of Products in E1 Reaction: Weight of the distillate (Actual Yield) = 0.95g Theoretical Yield = (18.3 * 10-3 mol of 2-methyl-2 butanol) * (1 mol of alkene / 1 mol of 2chloro-2 methylbutane) * (70.05 g of alkenes / 1mol alkenes) = 1.28 g alkenes Percent Yield: [(0.95 g) / (1.28 g)] x 100 = 74.2 % GC Analysis: The gas chromatography was performed and the graphs were printed out. The peaks in the graph were analyzed and Mole percentages of the compound were calculated. See attached copies of the three fractions from gas chromatography. Mol % (of compound A) =

Area of compound A [(Area of compound A) + (Area of compound B)]

x 100

E1 Reaction: Fraction one: Peak one: Area --- 104.4 s*mV Retention time --- 39 sec Identity --- 2-methyl-1-butene

Peak two: Area --- 1085 s*mV Retention time --- 43 sec Identity --- 2-methyl-2-butene

Peak one: Mol % = {104.4 s*mV/ [104.4 +1085] s*mV} x 100 = 8.78 % Peak two: Mol % = {1085 s*mV/ [104.4 +1085] s*mV} x 100

= 91.22 %

Gandhi 8 E2 Reaction: Fraction one: Peak one: Area --- 69.27 s*mV Retention time --- 40 sec Identity --- 2-methyl-1-butene

Peak two: Area --- 108.2 s*mV Retention time --- 46 sec Identity --- 2-methyl-2-butene

Peak one: Mol % = {69.27 s*mV/ [69.27 +108.2] s*mV} x 100 = 39.0 % Peak two: Mol % = {108.2 s*mV/ [69.27 +108.2] s*mV} x 100 = 61.0 % Calculating yield of each alkene in E1 and E2: Yield = GC% of actual yield E1 Weight of the distillate (Actual Yield) = 0.95 g 2-methyl-1-butene= 8.78 % of 0.95 g = 0.0834 g 2-methyl-2-butene= 91.22 % of 0.95 g = 0.867 g E2 Weight of the distillate (Actual Yield) = 0.98 g 2-methyl-1-butene= 39.0 % of 0.98 g = 0.382 g 2-methyl-2-butene= 61.0 % of 0.98 g = 0.598 g Conclusion: In this lab, 2-methyl-1-butene and 2-methyl-2-butene was prepared from 2-methyl-2-butanol and H2SO4 by an E1 mechanism and from 2-chloro-2-methylbutane and KOH by an E2 mechanism. This lab was successful in preparing these alkene products by using techniques such as fractional distillation and heating under reflux. The yield of products in E1 reaction was 74.2 % and of E2 reaction is 86.1 %. Even though, the lower percent yield of products from the E1 path, still E1 is a more efficient method for preparing alkenes because E1 is a reversible reaction where it forms alkene via dehydration reaction and forms alcohols via hydration reaction. If alkene is removed, then, according to Le Chatelier principle, the equilibrium of the reaction will shift to the right resulting in formation of alkenes. And in this lab experiment, alkene was removed through fractional distillation. That is why, E1 preferable path for the formation of secondary and tertiary alkenes. Since 2-methyl-1butene’s boiling point is less than 2-methyl-1-butene’s boiling point, it elutes out faster and thus the first peak in gas chromatograph was of 2-methyl-1-butene.

Gandhi 9 Also, the alkenes produced in E1 were in pure form as we analyze the graphs from the GC results E1 reaction, the Mol % of 2-methyl-1-butene was 8.78 % and of 2-methyl-2-butene was 91.22 %. In comparison to E2 reaction, mol % of 2-methyl-1-butene was 39.0 % and of 2methyl-2-butene was 61.0 %. The distribution of products in E2 reaction is different to distribution of products in E1 reaction. In E1 reaction, 2-methyl-1-butene eluted out early but it had a very small area and correspondingly a very small mol %. Whereas in E2 reaction, 2methyl-1-butene eluted out early but had an area mol % larger than that found in E1 reaction. The gas chromatograph of E2 reaction suggests that the products were separated efficiently based on their boiling points. 2-methyl-1-butene eluted out first because of its high volatility and 2methl-2-butene eluted out last because it had a higher boiling point than 2-methl-2-butene. Therefore, with respect to mol % data, one-step synthesis E1 reaction is better than multi-step synthesis involving E2 reaction. The Baeyer test and Bromine test were successful in proving the presence of alkenes in both E1 and E2 products. The results for E1 Baeyer test were after 3 drops of KMnO4 were added to the test tube, it turned brown in color whereas the results for E2 Baeyer test were after 2 drops of KMnO4 were added to the test tube, and it turned brown in color. However; both of the reactions E1 and E2 exhibit positive test because the number of drops required for the reaction solution exceeds the number of drops required for the blank solution. The number of drops needed for the blank solution was only one to persist purple color. For the Bromine test, E1 reaction mixture turned orange in color and disappeared to colorless solution after 3 drops whereas E2 mixture turned orange in color and disappeared to colorless solution after 1 drop. The blank solution turned orange in color and disappeared to colorless solution after 2 drops. This test also shows positive results for E1 and E2 indicating presence of alkenes because a positive test is when the orange color will rapidly disappear to yield a colorless solution. Few errors occurred in this lab were first in E2 procedure, we forgot to out on pipette bulb during the reflux heating process. In about 30 seconds, we realized it and put them back on. We forgot to pre-weigh receiving flask (25 ml flask), but we later on measured another 25 ml flask, so error must be very little if two same flasks weigh different, therefore it was negligible. Overall, the errors were corrected in order to produce reliable and correct data at the end. Reference Gilbert, John C., and Stephen F. Martin. Experimental Organic Chemistry. Cengage Learning, Massachusetts, 2011, 5th Ed, pp. 337-340, 348-352, 82,83, 867-869.

Gandhi 10...