Dehydration Cyclohexanol Lab Report

Dehydration Cyclohexanol Abstract: In this lab, cyclohexene is prepared by dehydrating cyclohexanol. At first part of the experiment, 6.0 mL of cyclohexanol is treated with sulfuric acid and phosphoric acid and a distillation is conducted. The distillate is collected at boiling temperature range of 77°C to 80°C . Finally, the product is transferred to a tare sample vial. From a 6.0 mL of cyclohexanol, 2.29 grams of cyclohexene is produced. Theoretically, in a dehydration reaction, one mole of alcohol produces one mole of alkene in a 1:1 ratio and 1 mole of water (reference 1). From that, 6.0 mL cyclohexanol weighing 5.7 grams, it is expected to give 4.7 grams of cyclohexene, but in this experiment 2.299 grams of cyclohexene is produced and the percentage recovery calculated was 48%. I. INTRODUCTION The purpose of this lab is to prepare cyclohexene by dehydrating cyclohexanol. Dehydration of alcohol removes water (H-OH) group from alcohols to produce alkenes. The first step in this mechanism is protonation of the alcohol group by the acid. The second step is the loss of water to form the carbocation. The final step is removal of a beta hydrogen by water to form the alkene. At first step, the hydroxyl group is protonated by sulfuric acid which then can function as a good leaving group. Hydroxyl group is poor leaving group, sulfuric acid is used , in this way the hydroxyl group pulls off proton from sulfuric acid and a protonated cyclohexanol is obtained. Water which is neutral, stable and a good leaving group, leaves forming a carbocation. Also, in this process, in order to improve the yield of the products, the equilibrium is shifted toward the product. And this is achieved by distilling off the product instead of distilling off water. Cyclohexene has low boiling point and water has higher, as the temperature reaches the boiling range of cyclohexene, it distills up the column and received into the receiver. II. Reaction Scheme: III. RESULTS AND DISCUSSION The lab began by obtaining 6.0 mL of cyclohexanol into a 50-mL round-bottomed flask. Next, to the flask, a boiling chip, 8.0-mL of phosphoric acid and 10 drops of concentrated sulfuric acid were added. The mixture was swirled and meanwhile, a distillation apparatus was set up with a 25-mL round-bottomed flask as receiver and submerged in an ice bath. Next, the flask was placed on a heating mantle set at 7. The mixture was distilled and distillate was collected into the receiver flask. The distillate was allowed to cool for a few minutes. Upon cooling, a small amount of anhydrous calcium chloride was added to the product flask and swirled thoroughly. Finally, the cyclohexene product was transferred from the flask to a pre-weighed sample vial by using a pasturer pippet. The pure product was then weighed. Data: Substance Quantity Density (g/mL) Molecular Weight (g/mol) bp( °C ) cyclohexanol 6.0 mL 0.948 100.16 160 cyclohexene - 82.15 83 phosphoric acid (5%) 8.0 1.685 98.00 sulfuric acid 10 drops 1.840 98.08 - - anhydrous calcium chloride - - 110.99 - - Mass of tared vial - - - - 9.39 grams Mass of tared vial with cyclohexane - - - - 11.68 grams Mass of cyclohexene obtained - - - - 2.29 grams Boiling range of distillate - - - - 0.811 78 to 80°C Calculations: In the experiment, 6.0 mL of cyclohexanol was used. Density of cyclohexanol, as shown in Table = 0.948 g/mL Mass of cyclohexanol (g) = volume of cyclohexanol (mL) x density of cyclohexanol (g/mL) = 6.0mL x 0.958g/mL = 5.7 g From this data, number of moles of cyclohexanol reacted is calculated. = 0.057 mol The reactant to product molar ratio is 1:1, number of moles of cyclohexene = 1 x 0.057 mol = 0.057 mol From this data, the theoretical amount of cyclohexene is calculated. Theoretical mass of cyclohexene (g) = moles of cyclohexene (mol) x molecular weight (g/mol) = 0.057 mol x 82.15 g/mol = 4.7 g Percent yield: = (mass of recovered compound,g /mass of compound in original mixture)*100 =(2.29/4.7g)*100%=48% Although, the percentage recovery of cyclohexene is low, the outcomes of the experiment is achieved because cyclohexene is obtained by dehydrating cyclohexanol. The low percentage recovery could be due to human error during the experiment, also as with all distillations some of the substance is left behind in the flask and in the column, this could make the percentage yield lower. On the other hand, during the transfer of cyclohexanol to the pot, a significant amount could have been lost. IV. Conclusion This experiment shows that one could prepare cyclohexene by dehydrating cyclohexanol. Dehydration of alcohol removes water (H-OH) group from alcohols to produce alkenes. The first step in this mechanism is protonation of the alcohol group by the acid. The second step is the loss of water to form the carbocation. The final step is removal of a beta hydrogen by water to form the alkene. In this experiment from 6.0 mL of cyclohexanol, 2.29 grams of cyclohexene is produced. Theoretically, in a dehydration reaction, one mole of alcohol produces one mole of alkene in a 1:1 ratio and 1 mole of water (reference 1), From that, the percentage recovery calculated for cyclohexene was 48%. I. QUESTIONS 1. Calculate the percentage yield that you obtained from this reaction Percent yield: = (mass of recovered compound,g /mass of compound in original mixture)*100 =(2.29/4.7g)*100%=48% 2. What would be the major product obtained from the E1 dehydration of 2-methylcyclohexanol? The major product obtained from E1 dehydration of 2-methylcyclohexanol is 1-methylenecyclohexene. Although 1-methylcyclohexne could also be produced but due to Saytzeff’s rule, because of higher stability, the products from 3° carbocation are favored over the products of 2° carbocation. Between the 1-methylenecyclohexene and 1-methylcyclohexene, 1-methylcyclohexene is more favored. 3. Outline a mechanism for the dehydration of 1-methyl-1-cyclohexanol. Would you predict this reaction to be faster or slower than the reaction you performed? Mechanism: The above reaction occurs at faster rate as the formation of stable intermediate.There is a formation of a tertiary carbocation. Tertiary carbocations are more stable, therefore the formation of the product is faster.