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Cases 41: 4-Methylcyclohexanol

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Cases 41: 4-Methylcyclohexanol
Experiment 41: 4-Methylcyclohexene
Introduction:
This experiment was a study of protonating a cyclo-alcohol to become a cyclo-ene. In this case 4-methylcyclohexanol was protonated using phosphoric acid to become 4-methylcyclohexene. This demonstrates the loss of an alcohol group by protonation, the loss of a proton from the cyclohexane to form an alkene through elimination. This also demonstrates Le Chatelier's principle, by using distillation to remove the product as it is formed the equilibrium of the reaction will constantly be in favor of the product.
Reaction:

Side Reaction: Due to a 1,3-hydride shift dehydration of 4-methycyclohexanol will produce a small amount of 1-methylcyclohexene. According to research performed by Chris
…show more content…

The experiment began with 1.603g of 4-methycyclohexanol and resulted in a yield of 0.920g of product. Since the theoretical yield is 1.35g this leads to a percent yield by mass of 68.1% (see calculations). The boiling point of the product obtained was found to be 104.4°C which is higher than the expected 101-102°C (see table 1); this could also be due to the side reaction product whose boiling point is 108-112 °C (see table 1) causing the boiling point found experimentally to be a combination of the two products. The side reaction, however, should not affect the percent yield since both products have the same …show more content…

Being a volatile solution keeping the solutions contained whenever possible was an important consideration.
The results obtained in this experiment deviated only slightly from what was expected, however since results from similar experiments could not be found the percent yield could not be compared to other’s results. The boiling point found experimentally was only 2.4°C higher than the accepted value of the product expected (4-methylcyclohexene). However, as stated above, this error could be due to the fact that some of the product was 1-methylcyclohexene but this would not explain the yield since the MW of both products are the same. Error in the yield could be due to the aforementioned volatility of the product, especially since the solution required multiple transfers exposing it to air and allowing evaporation whenever it was uncapped. Further error would be due to having accidentally left the cap off of the vial while washing the Hickman head with saturated sodium chloride allowing more evaporation than would be expected in this


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