In Experiment 9, distillation was used to carry out the dehydration of 2-methylcyclohexanol using 4:1 mixture of phosphoric and sulfuric acid. Gas Chromatography was then used to measure the products. Two products were formed. The theoretical yield of the alkene mixture was 3. 914 g. The actual yield of the mixture was 0. 73 g. The percent yield for the reaction was 18. 65%. Peak A had an area of 0. 3 mm2 and a percentage of 4. 6 %. Peak B had an area of 6. 21 mm2 and a percentage of 95. 39%. Introduction The acid used speeds up the dehydration of the alcohols.
This is an elimination reaction. Water is eliminated from the reaction and the products formed are alkenes. The reaction follows Zaitsev’s rule because it is an elimination reaction. Zaitsev’s rule says that the double bond that is more highly substituted or more stable will be dominant in the products. In the experiment the dehydration of 2-methylcyclohexanol takes place. 2 methyl-cyclohexanol has a very poor leaving group which is the hydroxyl group (OH-).
When a strong acid such as H+ protonates the oxygen in the hydroxyl group, it becomes a good leaving group in the form of water.
The leaving group leaves as water, and a carbocation intermediate is formed because of experiment conditions. Since there is no other nucleophile present besides water, the intermediate must undergo elimination to reach stability. Each alpha carbon loses a proton and produces the alkenes 3-methylcyclohexene, and 1-methylcyclohexene. According to Zaitsev’s rule, 1-methylcyclohexane is more substituted and will therefore be the major product. In the first half of the experiment, the alkenes produced were distilled to prevent the reverse reaction from taking place, and to obtain the greatest possible yield.
In part two, Gas Chromatography was used to show the amount of major products formed. Gas Chromatography was also used to determine how much of each alkene was present in the final product, and to determine if those results correlated with the results predicted by Zaitsev’s rule. Reaction Scheme 2-methyl-cyclohexanol 3-Methylcyclohexene 1-Methylcyclohexene C7H14OC7H12C7H12 114. 1996. 1796. 17 Curly Arrow Mechanism Table of Physical Constants*. Experimental Procedure Part 1 A simple distillation apparatus was set up using a 25 mL round bottom flask.
5 mL of 2-methylcyclohexanol and 1mL of 4:1 phosphoric acid: sulfuric acid solution was added to the round bottom flask. The flask was briefly swirled to mix reagents. The flask was heated and a temperature was maintained below 100 oC. The distillation continued until approximately 3 mL of distillate was collected. Sodium sulfate was then added to a beaker and weighed. The distillate was then added to the beaker. The beaker was covered and left for a week. Part 2 A vial was obtained and weighed. The distillate was taken out of the beaker with the sodium sulfate with a pipette and placed in the new vial.
The vial was weighed. 10 micro liters of distillate were obtained with a syringe and placed in the Gas Chromatographer. The distillate was injected at the same time the start button was pushed. Graph results were obtained from the computer. Calculations. Part 1 m = d x v m = (0. 9304 g/mL C7H14O)(5 mL C7H14O) = 4. 652 g C7H14O moles of C7H14O = mass of C7H14O / molar mass of C7H14O = 4. 652 g C7H14O / 114. 19 g/mol C7H14O = 0. 0407 mol C7H14O Theoretical Yield (g) = moles of limiting reagent x molar mass of product = moles of C7H14O x molar mass of C7H12 =0. 0407 mol C7H14O x 96. 17 g/mol C7H12 = 3. 914 g.
Actual Yield of Alkene Mixture C7H12 = (liquid + vial) – vial = 9. 65 g – 8. 92 g = 0. 73 g C7H12 Percent Yield = (Actual Yield (g) / Theorectical Yield (g)) x 100 = (0. 73 g C7H12 / 3. 914 g C7H12) x 100 = 18. 65% Part 2 Area of a triangle= ? base x height Area of Peak A= ? (1. 2mm) x 0. 5mm= 0. 3 mm2 Area of Peak B= ? (2. 3mm) x 5. 4mm= 6. 21 mm2 Percentage of Product A = (Area of Peak A) / (Area of Peak A + Area of Peak B) = (0. 3 mm2) / (0. 3 mm2+ 6. 21 mm2)= 0. 46 x 100= 4. 6% Percentage of Product B = (Area of Peak B) / (Area of Peak A + Area of Peak B) = (6. 21 mm2) / (0. 3 mm2+ 6. 21 mm2)= 0. 9539 x 100= 95.
39% Results and Discussion The objective of the experiment was to carry out the dehydration of the alcohol 2-methylcyclohexanol with an acid as a catalyst. Particularly, a 4:1 mixture of sulfuric acid and phosphoric acid were used in the experiment. The products formed were to be compared to Zaitsev’s rule and analyzed using Gas Chromatography. The graph produced from the Gas Chromatography showed that a reaction occurred because there are peaks on the graph. Each peak represents a major product formed. Since there are two peaks on the graph 2 major products formed. The theoretical yield of the alkene mixture was 3.
914 g; however, the actual yield of the mixture was 0. 73 g. The percent yield for the reaction was 18. 65%. Considering the percent yield, the reaction was not efficient. The difference in the theoretical yield and the percent yield is most likely due to the reaction incomplete reaction. Instead of distilling the reaction to completion, it was stopped after 3 mL of product was obtained. To improve the percent yield the reaction could be taken to closer to completion by continuing distillation past 3 mL of product formed. The low percent yield could also be a result of the extraction.
When removing the liquid from the crystals product could have remained behind. Using the Gas Chromatography data, peak A had an area of 0. 3 mm2 and a percentage of 4. 6 %. Peak B had an area of 6. 21 mm2 and a percentage of 95. 39%. The results are consistent with Zaitsev’s rule. According to Zaitsev’s rule, 1-methylcyclohexane should be the major product. The little peak comes of first which represents the product with the lower boiling point. That product was 3-methylcyclohexene. The larger peak had a longer retention time which represents the product with the higher boiling point, and a greater percentage.
In this case that product is 1-methylcyclohexene. The results are consistent with Zaitsev’s rule. Name_______________________________________________Lab Day and Time__________ Post Lab Quiz. To be completed individually and submitted separately from the lab report 1. Indicate which is dehydrated fastest, second fastest and slowest out of the following alcohols in an acid catalyzed dehydration reaction under reaction conditions similar to that given in the experiment above. Explain your answers and show structures of each compound. (10) Ethanol 1-methyl cyclohexanol 4-methyl cyclohexanol 2.
Where would starting material, 2-methyl cyclohexanol, come out on a gas chromatogram if it was present in our product mixture relative to the methyl cyclohexene peaks? Explain. (10). 2-Methylcyclohexanol boils at 1630C. 3. Propose a curly arrow mechanism for the dehydration of cyclohexanol, catalyzed by H2SO4. How many products will form? (10) 4. A student dehydrated 4-methylcyclohexanol with sulfuric acid as outlined in the above experiment. He found that the product mixture included three alkenes, below. Explain his results. Provide a curly arrow mechanism for the formation of all three products. (20)Hint: carbocation’s can rearrange!