2-Methylcyclohexanol Reaction Lab Report

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The goal of this experiment was to convert 2-methylcyclohexanol into 1-methyl-1-cyclohexene, 3-methyl-1-cyclohexene, methylenecyclohexane, and water through the addition of phosphoric acid and sulfuric acid. This was done through distillation where a mixture of 2-methylcyclohexanol, phosphoric acid, and sulfuric acid was distilled for 30 minutes in a reflux apparatus. Sulfuric acid being a strong acid acts as a catalyst in this reaction. Phosphoric acid also acts as a catalyst in this reaction. The distillate was then added to a centrifuge tube along with 1-2 ml of saturated sodium chloride. The bottom layer in the centrifuge tube was then extracted and dried using anhydrous sodium sulfate. This bottom layer was then distilled again in …show more content…

The product that is formed the most is 1-methyl-1-cyclohexexene because it is the most highly substituted and thus the most stable, while 3-methyl-1-cyclohexene and methylenecyclohexane are produced less because they are less highly substituted and thus less stable. This reaction proceeded through an E1 pathway. In the mechanism the sulfuric acid provides a proton which protonates the hydroxyl group on the 2-methylcyclohexanol. This forms a good leaving group on the 2-methylcyclohexanol which leaves the compound as water. A carbocation results and H2PO4^- deprotonates a hydrogen on a carbon atom next to carbon atom with a positive charge resulting in alkenes with the major and minor products. One major technique used in this experiment was distillation. The reason distillation works is because different organic compounds have different boiling points. Usually a mixture containing two compounds is placed in the round bottom flask in the distillation apparatus. When the distillation apparatus is turned on and heat is applied, the vaporization of the compound in the mixture with the lower boiling point occurs. This compound, then condenses in the condenser and is received by the receiving flask at the end of the distillation …show more content…

Based on the data it appears that 1-methyl-1-cyclohexene had the highest retention time while 3-methyl-1-cyclohexene had the lowest retention time. The area of the 1-methyl-1-cyclohexene is the greatest at 46.75314%, followed by the area of 3-methyl-1-cyclohexene at 16.59539%. The area of the methylenecyclohexane is the lowest at 1.99052%. The results seem to be accurate since the major products formed in this reaction are 3-methyl-1-cyclohexene and 1-methyl-1-cyclohexene. The GC shows that the areas for these two compounds are also the greatest. The minor product formed was methylenecylohexane and its area is the smallest based on the GC. Since it is a minor product, it is logical that it has the smallest observed area of all the products formed. The experiment was started with 6.678 grams of 2-methyl cyclohexanol. As such, based on the theoretical yield calculation, 5.590 grams of the final product should be obtained. In the experiment 3.810 grams of the final product were obtained giving a theoretical yield of 68.15%. Some explanations as to why the percent yield was not ideal could have been the reaction not going to completion during the reflux step, or some quantity of the product being lost when it was transferred from the receiving flask in the distillation apparatus, to the flask

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