A control was used to compare the reactant and the product, showing a clear appearance that would indicate an alkene being present. Bromine was added to the reactant 4-methylcyclohexanol, and a reddish-brown color appeared, indicating that no reaction took place. Bromine was then added to the product 4-methylcyclohexene, and the clear appearance of the product remained, concluding that an alkene is indeed present. Discussion Given the results obtained post-experiment, the percent yield was calculated to be about 35.5%, which may suggest that either product was lost at some point during the experiment, or side product was formed. Regarding the unsaturation test, the bromine test was performed and produced the expected result since the 4-methylcyclohexene had a clear appearance after 5 drops of bromine were added to it, which revealed that the alcohol was eliminated and a carbon-carbon double bond …show more content…
Based on the result from the bromine test conducted, the clear appearance of the 4-methylcyclonexene demonstrated that the bromine had all been used up, and there is an unsaturated compound present. Infrared spectroscopy was another method of identifying an alkene. Figures 1 and 2 show the IR spectra for the reactant (4-methylcyclohexanol) and the product (4-methylcyclohexene), respectively; each were compared and analyzed. The IR spectrum for 4-methylcyclohexanol revealed a broad peak at 3328 cm-1, signifying an OH group and therefore an alcohol. In contrast, the IR spectrum for 4-methylcyclohexene no longer showed that broad band, but instead displayed a sharp peak at 1651 cm-1, representing a double bond and therefore an alkene. With this, the reaction of 4-methylcyclohexanol with a strong acid (H3PO4) acting as a catalyst was successful, and the product 4-methylcyclohexene was obtained as