Benzoquinone Reasearch Lab Report

In order to isolate benzoic acid, benzocaine and 9-fluorenone, each component needed to be separated from one another. All three compounds began together in one culture tube, dissolved in methylene chloride and formed into a homogenous mixture. In this culture tube, two milliliters of aqueous three molar hydrochloric acid was added, which immediately formed two layers, the top acidic aqueous layer was clear in color and contained benzocaine, and the bottom organic formed was yellow and contained benzoic acid and 9-fluorenone. Benzocaine’s amino group is protonated by the aqueous layer hydronium. This protonation forms the conjugate acid of benzocaine, benzocaine hydrochloride. Thus, the conjugate acid, benzocaine hydrochloride is a salt in which is soluble in water and furthermore can be isolated from the organic mixture. When testing out the pH levels in benzocaine, the pH test strip was dark blue in color, indicating a pH level of around 5 to 7. When isolating benzoic acid, two milliliters of aqueous three molar sodium hydroxide was added, which deprotonates the carboxylic group in benzoic acid, forming its conjugate base, sodium benzoate. As with benzocaine hydrochloride, sodium benzoate is a water soluble ionic salt in the aqueous layer that can then be separated from the bottom organic layer containing the 9-fluorenone. The pH test strip was a vibrant red for benzoic acid, indicating a pH of 2. Now the 9-fluorenone is left, deionized water is added to remove any excess

And finally into test tube 3, I pipetted 1.0 ml turnip extract and 4.0 ml of water. The contents of test tube 1 was poured into a spectrometer tube and labeled it “B” for blank. “B” tube was now inserted it into the spectrometer. An adjustment to the control knob was made to zero the absorbance reading on the spectrometer since one cannot continue the experiment if the spectrometer is not zeroed. A combination of two people and a stop watch was now needed to not only record the time of the reaction, but to mix the reagents in a precise and accurate manner. As my partner recorded the time, I quickly poured tube 3 into tube 2. I then poured tube 2 into the experiment spectrometer tube labeled “E” and inserted it into the spectrometer. A partner then recorded the absorbance reading for every 20 seconds for a total of 120 seconds. After the experiment, a brown color in the tube should be observed to indicate the reaction was carried out. Using sterile techniques, any excess liquid left was disposed

Turnips and horse radish roots are rich source of this enzyme. In this experiment, we would carry out a reaction between hydrogen peroxide and guaiacol which is colorless dye, using peroxidase as a catalyst, to produce water and an oxidized form of guaiacol which is brown. The formation of brown color would serve as an indicator that the breakdown of Hydrogen Peroxide took place. The enzyme activity would be directly proportional to the brown color intensity. The color intensity would be measured using a spectrophotometer and standardized to find the corresponding concentration for each absorbance unit.

Catechol, in the presence of oxygen is oxidized by catechol oxidase to form benzoquinone (Harel et al., 1964). Bananas and potatoes contain catechol oxidase that acts on catechol which is initially colorless and converts it to brown (Harel et al., 1964). In this experiment, the effect of pH on the activity of catechol oxidase was conducted using buffers ranging from pH2 to pH10. Two trials were conducted due to the first trial results being altered by an external factor. The results were acquired by taking readings every 2 minutes for 20 minutes from a spectrophotometer and then recorded on to the table. The data collected in the table were then made into graphs to illustrate the influence of pH on the catechol oxidase catalyzed reaction. After analysis, the data revealed that pH did have a significant influence on the enzyme as recorded by absorbance per minute. However, the data was collected was not accurate due to external factors, thus the results are debatable and should be experimented again for validation.

Within the experiment, pure catechol was mixed with different concentrations of catechol oxidase and the rate at which each solution produced benzoquinone was measured. The amount of benzoquinone made throughout the trials was measured by using a colorimeter to measure the level of “brownness” of the liquid. The colorimeter worked by shining a light through the liquid and then measuring that light on the other side to see how much of it was absorbed. In this experiment, absorbance of blue light was measured because blue light is absorbed by the color brown. The amount of blue light absorbance was measured every 15 seconds for five minutes. Because enzymes speed up reactions, more enzymes would cause the reaction to be even faster.1

The role of an enzyme is to catalyse reactions within a cell. The enzyme present in a potato (Solanum Tuberosum) is catechol oxidase. In this experiment, the enzyme activity was tested under different temperature and pH conditions. The objective of this experiment was to determine the ideal conditions under which catechol oxidase catalyses reactions. In order to do this, catechol was catalyzed by catechol oxidase into benzoquinone at diverse temperatures and pH values. The enzyme was exposed to its new environment for 5 minutes before the absorbance of the catechol oxidase was measured at 420 nm using a spectrophotometer. The use of a spectrophotometer was crucial for the collection of data in this experiment. When exposed to hot and cold temperatures, some enzymes were found to denature causing the activity to decrease. Similarly, when the pH was too high or low, then the catechol oxidase enzyme experienced a significant decrease in activity. It can be concluded after completing this experiment that the optimal pH for catechol oxidase is 7 and that the prime temperature is 20º C. Due to the fact that the catechol oxidase was only tested under several different temperatures and pH values, it is always possible to get a more precise result by decreasing the increments between the test values. However, our experiment was able to produce accurate results as to the

The goal of this lab was to find the effect of different substrate concentrations of hydrogen peroxide on the rate of catalase enzyme activity. This experiment was specifically meant to find a correlation between an increasing substrate concentration of hydrogen peroxide on the rate of catalase enzyme activity. Four test tubes were used, one filled with 5 milliliters of water and three with 5 milliliters of different concentration of hydrogen peroxide ranging from 0.5%, 1%, and 1.5%. Four drops of yeast solution were dropped into each test tube using a pipette. Then, a lab quest probe was attached to each of the test tubes. The test tubes were tapped twice every second in order to agitate the solution. The rates of the catalase enzyme activities were recorded after 400 seconds of tapping. After the lab, all the materials were cleaned. This experiment was repeated the next day. After both trials were completed, the average from the two trials was taken. Then, the data from the experiment was graphed, with the different substrate concentrations of hydrogen peroxide recorded on the x-axis and the average rates of catalase enzyme activity in kilopascals per second recorded on the y-axis. The experiment results showed that there was a positive correlation

In this lab, liquid-liquid extraction was performed to isolate a mixture of benzocaine and benzoic acid. 2.0107 grams of the mixture was first weighed out for the trials. When HCl was added to the mixture for the first acid extraction of benzocaine, an emulsion formed during inversion and venting that prevented a defined separation of the two layers. 8 mL of water was therefore added before continuing the extraction. The addition of NaOH then turned the top aqueous layer basic, indicated by the pH strips that turned blue when tested. A vacuum filtration isolated 0.29 grams of benzocaine and a MelTemp apparatus measured the crystal’s melting point ranges to be 85.1C-87.4C. For the base extraction of benzoic acid, the aqueous layers were retrieved

Why I Chose This Topic: I chose this topic because I was interested in investigating the level of citric acid in commercial and freshly squeezed lemon juice in order to compare results and suggest theories and ideas about what increases or decreases the acidity of the lemon juice and why?? (Temperature, adding chemicals/preservatives etc…)

If cucumbers are kept in lemon juice, water, or vinegar, which will preserve the cucumbers best? A question like this could solve simple, everyday issues. Preserving food is a common problem, and many people could benefit from knowing one of the best liquids to preserve a cucumber and its benefits. The expected outcome of this experiment is that vinegar will preserve cucumbers the best because it is an acetic acid that can kill microorganisms and prevents mold-induced softening. To find the answer to this question each of the three liquids were poured into a different labeled cup, and one baby cucumber was placed in each cup. The cucumbers were left in the cups for one week and observed to see which liquid kept the cucumber preserved the

The use of multiple test tubes and Parafilm was used for each experiment. Catechol, potato juice, pH 7 phosphate buffer, and stock potato extract 1:1 will be used to conduct the following experiments: temperature effect on enzyme activity, the effect of pH on enzyme action, the effect of enzyme concentration, and the effect of substrate concentration on enzyme activity. For the temperature effect on enzyme activity, three test tube were filled with three ml of pH 7 phosphate buffer and each test tube was labels 1.5 degrees Celsius, 20 °C, and 60 °C. The first test tube was placed in an ice-water bath, the second test tube was left at room temperature, and the third test tube was placed in approximately 60°C of warm water. After filling the test tubes with three ml of the

There were three test tubes in which the experiment was held. A relatively equal sized portion of raw potato (this contained the enzyme [a biological catalyst] hydrogen peroxidase) was placed in each tube. Then, enough water to cover the potato was added. Proceeding this, each of the test tubes were assigned a temperature; cold, room temperature or warm (this was written on the tag so that they were not confused). The test tube destinated ‘cold’ was placed in a ice bath for five minutes. At the same time, the ‘hot’ test tube was placed in a hot water bath for five minutes. Meanwhile, the room temperature test tube sat at room temperature for five minutes. When the five minutes were over, the test tubes were returned to the rack (so that they were able to be observed). Then, the test tubes were allowed to sit at room temperature for five more minutes. Once that period of time was over, 2 ml of hydrogen peroxide (the substrate) was added to each tube.

Samples of benzophenone, malonic acid, and biphenyl were each tested with water, methyl alcohol, and hexane. Benzophenone was insoluble in water as it is nonpolar while water is highly polar. Benzophenone was soluble in methyl alcohol, dissolving in 15 seconds, because methyl alcohol is intermediately polar as benzophenone is nonpolar. Methyl alcohol is polar but not as much as water. Thus, the nonpolar benzophenone was soluble in methyl alcohol. Benzophenone was partially soluble in hexane because hexane is nonpolar as is benzophenone. Thus, benzophenone was dissolved in hexane. Malonic acid was soluble in water because both malonic acid and water are polar. It took 25 seconds for malonic acid to dissolve in water. Malonic acid was soluble in methyl alcohol because malonic acid is polar and methyl alcohol is intermediately polar, allowing malonic acid to dissolve in the methanol in 15 seconds. Malonic acid was insoluble in hexane because hexane is nonpolar while malonic acid is polar. Biphenyl was insoluble in water as water is highly polar whilst biphenyl is nonpolar. Biphenyl was partially soluble in methanol which is intermediately polar whilst biphenyl is nonpolar, allowing it to dissolve a little. Biphenyl was soluble in hexane because both biphenyl and hexane are nonpolar molecules. Biphenyl dissolved in hexane in 10 seconds.

The aim of my investigation is to see how pH affects the activity of potato tissue catalase, during the decomposition of hydrogen peroxide to produce water and oxygen.

When fruit is outside for a certain period of time it turns brown. Fruits such as apples turn brown due to oxidisation. Enzymes in the fruit such as polyphenol oxidase and catechol oxidase react with oxygen in the air and an iron or copper cofactor in the fruit. A cofactor is a component that is necessary for a certain enzymatic reactions to happen. The fruit starts to oxidize, when electrons are lost to another molecule (in this case the air), and the food turns brown.