CHEM
Experiment #7: Dehydration of 2-MethylcyclohexanolExperiment #7: Dehydration of 2-Methylcyclohexanol

Experiment #7: Dehydration of 2-MethylcyclohexanolExperiment #7: Dehydration of 2-Methylcyclohexanol

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Brett Thompson 10-23-2011 CHEM 2411 John Hatch Experiment #7: Dehydration of 2-Methylcyclohexanol The dehydration of 2-methylcyclohexanol is catalyzed by acid. For this experiment, we use phosphoric acid. This acid is chosen over sulfuric acid, which is a much stronger acid that easily protonates alkenes to form carbocations, which then react with water to form alcohols. This is the reverse of the alkene formation we desire. It is also preferred over HCL because the chloride anion is a strong nucleophile that can react with carbocation. Phospohric acid loses a hydrogen through proton transfer to the hydroxyl group on the 20methylcyclohexanol, forming a water molecule attached to the cyclohexene. The formation of this intermediate carbocation is the slow, rate-determining step. The phosphoric acid acts as a catalyst because the protonation of the hydroxyl group on the 2-methylcyclohexanol makes it a better leaving group due to the increased stability of the H 2 O compared to the original hydroxyl group that was attached. A better leaving group requires less energy to separate from the original molecule, and because of this, the phosphoric acid successfully lowered the activation energy of the reaction. After the loss of the leaving group, a unimolecular elimination reaction occurs, removing a hydrogen atom from the methylcyclohexanol. This results in the formation of two different methylcyclohexenes of different structure depending on which hydrogen is abstracted. In this experiment, we use gas chromatography and steam distillation to aid us in determining which of these methylcyclohexenes is the major product; in other words, the actual course of the reaction.
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Whenever multiple products can be formed, the ratios in the product mixture depend on whether the reaction is kinetically or thermodynamically controlled. When thermodynamically controlled, a high temperature guarantees the achievement of the activation energies for each product, and a long reaction time allows equilibrium to be reached. The product with the lower activation energy will be the major product. Under kinetic control, however, the product with the lower energy transition state will be the major product due to the fact that the reverse reaction
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