Lab8 - Lecture notes 8.6 PDF

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Marlin Amy Halder 8 April 2014 Partner: Maurice Dixon TA: Wiktoria Pecak, Tues 12pm Preparation of Alkenes by E1 and E2 Elimination reactions. Baeyer and Bromine tests for Unsaturation Methods and Background The goal of this experiment is to prepare 2-methyl-1-butene and 2-methyl-2-butene by performing E1 and E2 reactions. For E1 reaction, 2-methyl-2-butanol and H2SO4 was utilized and for E2 reaction, 2-chloro-2-methylbutane and KOH was utilized. Below are the mechanisms and detailed explanation of what elimination reactions are. After the products were obtained, two tests: Baeyer and Bromine, were done to confirm that alkenes were produced, and Gas Chromatography method was used to aid the percent yield calculations. General Mechanism for E1 reaction

General Mechanism for E2 reaction

The E1 mechanism is a first order, unimolecular elimination, and it proceeds through a carbocation intermediate. The reaction rate only depends on the concentration of the substrate, and does not involve the base. The rate determining step depends on the carbocation stability, and goes from tertiary>secondary>>primary. E2, however is a second order is a bimolecular elimination where the nucleophile deprotonates an adjacent carbon to the carbon bearing the leaving group and the leaving group is kicked out simultaneously to produce an alkene. The rate law is dependent on concentrations of both substrate and the base. The combination of strong, bulky bases, high temperature and steric hindrance around the alpha carbon favors bimolecular elimination. When nucleophiles tend to favor elimination it is because the hydrogen is more sterically accessible than the carbon nearing the leaving group. One of the factors is polarizability, which is a measure of how easy the electron cloud of a Lewis Base is distorted by a nearby center bearing a partial or fully positive charge. A nucleophile that is a weak base and highly polarizable (such as I-, Br-and H2O) favor substitution whereas one that is a strong base and only slightly polarizable (H-, OH-, and H2N-) favor elimination. Zaitsev’s rule also comes into play as most of the time, the major product of the elimination reaction is the most substituted alkene. Le Châtelier’s Principle shows why different conditions alter the products of the reaction as change in concentration shifts the equilibrium and yields more of the product wanted.

Fractional distillation was used in this experiment because both of the solutions were volatile, and when comparing simple distillation to fractional, fractional distillation works better in separating two similar volatile liquids with similar boiling points. Bromine Test Bromine adds to the carbon-carbon double bond of alkenes to yield dibromoalkanes and reacts with alkynes to produce tetrabromoalkanes. While the reaction is in place, molecular bromine is consumed, and its dark red-orange color disappears if bromine is not added in excess.

Baeyer Test Baeyer test depends on the ability of potassium permanganate to oxidize the carbon-carbon double bond to give alkanediols or the carbon-carbon triple bond to yield carboxylic acids. The permanganate is destroyed in the reaction and a brown precipitate of MnO2 is produced. The disappearance of the characteristic purple color of the permanganate ion is a positive test for unsaturation.

As for the results of the experiment, something went terribly wrong with either our methods, calculations or something out of our control, and therefore we did not get any distillate after the fractional distillation. The data collection is taken from another lab group. The ideal results would be having positive tests for both Baeyer and Bromine and E1 percent yield should be higher than E2 because they always follow Zaitsev’s rule. Experimental Procedure Synthesis of 2-methyl-1-butene and 2-methyl-2butene (E2) Before starting the actual experiment, the amount of the reagent should be calculated based on how much product was obtained from previous week’s lab. The calculations are shown in the Data Acquisition section. The amounts of potassium hydroxide and 1-propanol is added to 25mL round-bottom flask with 1-2 boiling stones. The mixture needs to be warmed up so that all the KOH is melted and then cooled

down to room temperature, then is placed in an ice bath. 2-chloro-2-methylbutane (the product from Lab 7) is added to the chilled mixture and is placed on the reflux apparatus. All the joints needs to be greased since the solution is very volatile and can escape. Keep the reaction in the reflux for an hour and then cool it. Prepare for fractional distillation (Hempel column was filled with Raschig rings) and remove the water hoses. Begin fractional distillation and have the receiving flask pre-weighed and placed in an ice bath to prevent evaporation. Pictures of reflux and fractional distillation is shown on the picture in the right. Only the distillate that boiled below 45°C should be collected. Afterwards, analyze the distillate using Gas Chromatography and prepare for the classification of alkene tests. E1 Procedures Place 2mL of 2-methyl-2-butanol and 10mL of 6M sulfuric acid, and place it in a 25mL round bottom flask. Attach the flask to the fractional distillation apparatus that is similar to prepared before. Fill the Hempel column with rings and start the distillation with a pre-weighed flask placed on ice bath on the receiving end. Once again, grease the joints to prevent evaporation and collect distillate below 45°C. Measure the end product and prepare for GC analysis Analysis of product by Gas Chromatography Have the products on ice bath and obtain gas chromatograph for both mixtures. Each graph should have two integrals - with 2-methyl-1-butene as Integral A and 2-methyl-2-butene as Integral B. The molar percentages are calculated without the correction factor because both substances have same molar mass and same polarities. Baeyer Test 1 mL of 95% ethanol was added to three test tubes. Then, 1-2 drops of the product from E1 was added to one of the test tubes, 1-2 drops of the product from E2 was added to another test tube, and nothing was added to the third test tube. After that, add KMnO4 (0.1M) to the test tubes until a purple color appears and remains. The test can be called positive if it takes more of KMnO4 for the E1 and E2 products than the blank one. Bromine Test 1 mL of methylene chloride was added to three test tubes. 1-2 drops of the product from E1 was added to one of the test tubes, and 1-2 drops of the product from E2 was added to another test tube. Similar to previously, nothing was added to the third test tube. Br2 is added to the mixtures until an orange color starts to appear. The test can be called positive when the orange disappears quickly, producing a clear solution.

Data Acquisition Below are the calculations on determining how much material to use to start the reaction. These were done using stoichiometry and given densities and molar masses and matched to the amount of product obtained from lab 7.

2.00 mL 2-Methyl-2-Butanol x

10mL H2SO4 x

x

x

x

x

= 18.26mmol

=60mmol

Calculations for Amounts of Reagents for E1 Reaction

Compound

Molecular Weight

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

Reaction Weight (g or mL)

2-Methyl-2Butanol

88.15

0.805 g/mL

2.74 mL

18.26

1

Sulfuric Acid

98.08

6M

10 mL

60

3.28

25mmol 2-chloro-2-methylbutane x

= 55mmol x

mmo l

Equivalent s

x

x

x

= 3.1g KOH

25mmol 2-chloro-2-methylbutane x =30.8mL 1-propanol

= 412.5mmol x

x

x

Calculations for Amounts of Reagents for E2 Reaction Compound

Molecular Weight

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

Reaction Weight (g or mL)

mmol

Equivalents

2-chloro-2methylbutane

106.59

0.866 g/mL

2.74g

91.3

1

Potassium Hydroxide

56.11

n/a

3.1g

55

2.2

1-propanol

60.10

0.804 g/mL

30.8 mL

412.5

16.5

Below are calculations for finding percent yields and distribution of products for each reaction. Because our experiment did not work, all the measurements are used from a different lab group, and their product from Lab 7 obtained was 4.55g of 2-chloro-2-methylbutane Percent Yield Compound

Mass or Volume of Reactant

Actual Mass of Product

Theoretical Mass of Product

Percent Yield

E1 2-Methyl-1Butene

2.0 mL

0.040 g

1.28 g

3%

E1 2-Methyl-2Butene

2.0 mL

0.40 g

1.28 g

31.25%

E2 2-Methyl-1Butene

4.55 g

0.29 g

3.00 g

9.67%

E2 2-Methyl-2Butene

4.55 g

0.47 g

3.00 g

15.67%

Equation used: Percent (%) Yield =

actual amount obtained theoretical amount

x 100

E1: Actual Mass of 2-Methyl-1-Butene = [Integral A/(Integral A + Integral B)][from GC](product mass)

=

x 0.44 = .040g

Theoretical Mass of 2-Methyl-1-Butene = [(reactant volume)(reactant density)/(molar mass reactant)] * (molar mass product)

[2.0mL 2-methyl 2-butanol x % yield: (.040g/1.28)x100 = 3%

]/

x

= 1.28g

E2: Actual Mass of 2-Methyl-1-Butene = [Integral A/(Integral A + Integral B)][from GC](product mass)

x 0.76g = 0.29g Theoretical Mass of 2-Methyl-1-Butene = [(mass reactant)/(molar mass reactant)] * (molar mass product)

[4.55g 2-chloro-2-methylbutane /

]/x

= 3.0g

% yield = (0.29/3)*100 = 9.67% Data for 2-Methyl-1-Butene and 2-Methyl-2-Butene

E1

E2

Product Name

2-methyl-1butene

2-methyl-2butene

2-methyl-1butene

Product distribution

9.2% * 0.44 g = 0.040 g

Yield

3%

31.25%

9.67%

15.67%

Boiling Point

31.6°C

39°C

31.6°C

39°C

Baeyer Test (+ or -)

+

+

+

+

Bromine Test (+ or -)

+

+

+

+

90.8% * 0.44 g = 37.5% * 0.76 g = 0.40 g 0.29 g

2-methyl-2butene 62.4% * 0.76 g = 0.47 g

This list of unified calculations give an idea of how the product distribution differed as well as the differences in percent yields. When performing the classification tests, for the Bayer test, when KMnO4 was added, both the E1 and E2 products turned brown, telling there was al an alkene present. For the Bromine test, compared to the blank test time, when Br2 was added, the color changed from light orange to colorless, once again proving that there is an alkene present.

Conclusion The purpose of this experiment was to make 2-methyl-1-butene and 2-methyl-2-butene through E1 and E2. The reactants used for that are 2-methyl-2-butanol and 2-chloro-2-methylbutane. And after the products were made, two tests: Bayer and Bromine, were done to make sure alkenes were made. Gas chromatography was also used to aide in determining the percent composition and product distribution as well as percent yields for each alkene in each reaction. Unfortunately, our experiment did not yield any sort of distillate during both E1 and E2 collection. At first, we assumed it was because the distillation material has not been greased properly, but when we did it again, it produced no distillation for E1. In terms of E2, the starting material from lab 7 may not have been enough or gotten evaporated, but I went ahead and performed the experiment with what was left, but it yielded no distillation either. Below are the conclusions made from using another lab’s data. The percent yield for both products were much lower than expected. From the E1 product, 2-methyl-1-butene yielded 3% and 2-methyl-2-butene yielded 31.25%. From the E2 product, 2-methyl-1-butene yielded 9.67% and 2-methyl-2-butene yielded 15.67%. There may have been multiple reasons why the yield was so low. Both the products were volatile and a lot of it could have been evaporated. Also the lack of glassware to put products it having residual material left in it may have tampered with the results. But when the GC analysis as well as the classification tests were done, it was verified that whatever amount of products were produced, it was indeed a mixture of 2-methyl-1-butene and 2-methyl-2-butene. The gas chromatography allowed for calculating the distribution of products. The E1 product consisted of 9.2% 2-methyl-1-butene and 90.8% 2-methyl-2-butene while the E2 product produced of 37.5% 2-methyl-1-butene and 62.4% 2-methyl-2-butene. This shows that E1 was more selective and preferred than E2 mainly because E1 is more likely to produce Zaitsev product instead of the Hoffman product. Also, it is seen from percent yields of E1 that overall, it’s better to perform the direct, one-step synthesis of the alkenes over the two-step synthesis (labs 7 and 8). That is because twostep synthesis has more chance of error than one-step synthesis. In this instance, the direct reaction was high yielding and selective, therefore, it is more beneficial to do one-step synthesis. Reference Landrie, Chad and Lindsey McQuade. Organic Chemistry Laboratory I: Lab Manual and Course Materials. 2013. Hayden-McNeil, LLC...