6AL Lab 5 Worksheet

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6AL

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Chemistry

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Jan 9, 2024

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1. In this experiment, the aqueous phase from the extraction with 3 M NaOH(aq) was treated with 6 M HCl(aq) until the cloudiness persisted. What was this cloudy substance and why did we want it to persist before continuing? Upon addition of hydrochloric acid to the basic aqueous phase, a solid white precipitate forms. This is the 2-naphthol precipitating out of solution, as it becomes protonated once again. The addition of the sodium hydroxide deprotonates 2-naphthol molecules which instills a negative charge onto the hydroxyl group and allows it to enter into the aqueous phase via increasing its ability to form hydrogen bonds with water molecules. Then when a strong acid is added, it reprotonates the hydroxyl group, thus reduces its solubility in water, and forms a solid precipitate. With the base still being present in solution, a greater amount of acid is required to both reprotonate the hydroxyl group and neutralize the hydroxide ions. Therefore, a good indication of complete base neutralization is the persistence of the precipitate. 2. How and why was the 3 M HCl(aq) solution used in experiment 5a? In your answer, you should explain on a molecular level what the solution is doing. Use structures in your answer. 3 M HCl (aq) was added to the separatory funnel containing the diethyl ether so as to remove the organic solute containing a basic functional group. In unknown sample #5, this compound was 3-nitroaniline, which had its amine group protonated (-NH 3 + ) giving the molecule a net positive charge, that allowed it to hydrogen bond with water, thus facilitating transfer into the aqueous layer. This leaves the neutral organic compound in the non-polar diethyl ether solvent. 3. What is the purpose of brine, as used in Experiment 5a? Since extraction via a seperatory funnel relies on the experimenter to determine the layer boundary, there is a higher likelihood of an error in completely extracting each layer. Additionally, certain solution can form emulsions, which further reduce accuracy of determining the exact boundary between the two layers. No solvent is completely immiscible with water, and so emulsions should be expected. Once the aqueous layer is removed, brine is added to dry the solution, removing any remaining water from the ether layer. Any remaining water will more strongly associate with the chloride ions than the non-polar ether solvent.
4. Sodium sulfate was used multiple times in Experiment 5a. What was the purpose of this compound and how does it work? When do you know you have used enough sodium sulfate? The purpose of adding sodium sulfate, or any anhydrous inorganic salt to a solution is to remove any remaining water that was not removed with the brine wash. Two of the most common anhydrous salts used are sodium sulfate and magnesium sulfate. They work by forming hydrates with water molecules. When adding an anhydrous salt such as sodium sulfate, the crystals will begin to clump up, indicating they are forming the desired hydrates and removing excess water from the solution. The process is complete once the addition of more sodium sulfate produces free flowing crystals, indicating all the water has been removed. 5. A student isolated an unknown compound that is believe to be either an alcohol or a carboxylic acid. How can IR spectroscopy be used to distinguish these functional groups? While both the alcohol and carboxylic acid have hydroxyl groups that will produce strong broad peaks around 3400 cm -1 , a compound containing a carboxylic acid will also produce a strong sharp band around 1700 cm -1 due to the presence of the C=O carbonyl group. The alcohol will not have this peak, thus the identity of the unknown compound can be narrowed down. 6. In experiment 5a, what was the percent recovery of each compound? Please show your work. Assuming that the unknown compound was an equal mixture by mass of each of the three organic compounds, the percent recovery for 2-Naphthol, 3-Nitroaniline, and naphthalene were 88.8%, 58.8%, and 33.12% respectively. The original unknown sample had a mass of 0.750 grams, so it can be expected that each compound will have a mass of 0.250 grams. So to determine the percent recovery, you simply divide the mass of the product recovered (actual yield) by the the expected yield of 0.250 grams. Finally multiply by 100 to convert the ratio into a percentage.
7. A student isolated an unknown organic compound through extraction. The student measures the melting point and finds it to be 115-122 °C. The student looks through the list of possible structures, provided below, and decides the compound is most likely 2-napthol. Do you believe this choice is correct? Justify your reasoning. Possible Compound Melting Point Range (°C) 3-Nitroaniline 111-113 Acetylsalicylic Acid 134-135 2-Napthol 120-122 Yes, I do believe that the student has chosen the correct compound. The melting point ranges listed for each compound are for pure compounds. However, extractions are not completely pure with some small amount of impurity always being present. These impurities would depress and broaden the melting point range due to a disruption in the crystalline latice formed by a pure compound. The areas of disrupted crystalline structure would melt at a lower temperature due to decreased intermolecular forces holding the molecules together. The top end of the melting point range is unaffected likely due to pure regions of the sample being unaffected by the impurities present, thus can be treated as pure compounds and can be expected to display similar properties as the reference compound.
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8. Fill in the diagram below to show how, using the lab manual procedure, you can purify a mixture of 2-chlorobenozic acid, 9-fluorenone, and 4-nitroaniline that is dissolved in diethyl ether. In the box by each arrow, indicate the solvent or solution added. The other boxes represent different phases; in these boxes indicate the identity of the solvent for that phase ( i.e. ether, 3 M HCl(aq), etc.) and draw the structure of the molecule(s) present.
9. A student performed an extraction on a mixture of 3 compounds: a carboxylic acid, an amine, and a ketone, and obtained an IR spectrum of each. Unfortunately, the student mixed up the IR spectra and doesnâ t know which is which. For each IR spectrum below, indicate which compound it belonged to. Justify your answer by indicating the absorption bands you used and the chemical bonds responsible for those absorption bands. The carboxylic acid, amine, and ketone can be distinguished by looking at the characteristic peaks they each produce. The carboxylic acid and ketone will produce a sharp strong peak at around 1700 cm -1 , however the carboxylic acid will also have a broad strong peak around 3100 cm -1 . To distinguish between the amine and carboxylic acid, one can look for the presence of a double fanged peak around 3400 cm -1 , which will be absent from the carboxylic acid sample. Additionally, the peak created by the carboxylic acid will be much broader than the amine. The peaks around 3000 cm -1 represents a C-H stretching vibration by sp 3 and sp 2 carbons. That is to say that the IR spectra shown represent the carboxylic acid, the amine, and the ketone in order from top to bottom. 10. Identify your 3 unknown compounds. Do so by comparing the data you acquired to the data posted on Gauchospace. Attach your 3 IR spectra here and clearly state, using IR and melting point data, why you believe your assignments are correct. Based off of the IR data and the melting point ranges determined, a reasonable conclusion on the identity of the three compounds could be 2 Naphthol for the compound isolated from part A, 3 nitroaniline from the compound isolated in part B, and Naphthalene from the compound isolated from part C. These conclusions are supported by the melting point ranges falling closely in line with the accepted values presented on the reference sheet. In order of isolation (A,B,C), the determined melting ranges (in degrees Celcius ) were 120 to 124, 103 to 108, and 76 to 78. We knew from the lab procedure that there would be three different organic compounds present. One with an acidic functional group, another with a basic functional group, and a third neutral compound. Because part A of the experiment required precipitation via addition of acid, it can be assumed that the compound contained an acidic functional group. This narrows the possible compounds to 5. From these 5 compounds, there are two likely candidates; those being benzoic acid and 2 naphthol. The distinction between the two will require examination of the IR spectra discussed later. Similar reasoning for compound B narrow the possibilities to 4, with 3 nitro aniline most closely resembling the presented melting range. Compound C can then be assumed to be the
neutral compound both by process of elimination and because it is present in the non polar solvent. Based on the melting range, the reference value for naphthalene most closely resembles the data. When looking at the IR data, compound A most resembles the reference spectra for 2 naphthol. Benzoic acid is a potential candidate, however can be ruled out because it produces a less smooth and broad peak in the ~3400 cm 1 range. Compound B produced a double fanged peak around 3400 cm 1 likely indicating a primary amine, however this data does not narrow down the possibiltles, so other areas must be examined to narrow down the identity of the compound. The identity can be narrowed down to two of the compounds containing nitro groups due to the absence of a strong sharp peak around 1680 cm 1 which would be indicative of a carbonyl group. Determining whether compound B is 4 nitroaniline vs 3 nitroaniline comes down to the breadth of peak produced by the amino group. The reference spectra for 3 nitro aniline most closely resembles the data generated by the IR spec. For compound C, the three available neutral compounds can be ruled out on the basis of absence of certain peaks. Naphthalene only has a peak at ~3050 cm 1 , and lacks any other peaks indicating an absence of other functional groups. 9 fluorenone has a carbonyl group that produces a distinct strong sharp peak around 1700 cm 1 which is absent from the generated IR spectra. 2 methoxy naphthalene is ruled out because it contains a CO bond which can be seen around 1250 and is lacking for naphthalene. Additionally, only naphthalene produced the sharp peak around 3050.
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