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Thin layer chromatography conclusion
Thin layer chromatography conclusion
Thin layer chromatography conclusion
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Hello student! In order to solve your problem, it is important to first understand how thin layer chromatography works. Thin Layer Chromatography is a way in which we can separate components of mixtures and identify qualities about their chemical nature. The TLC plate is made of a silica gel which plays an integral role in the distance the different samples will travel, which we will discuss later. For now, it is important to know that Silica gel is very polar. The solvent, which is what you are placing the TLC plate in, is also very important in determining the distance the spots will travel. The different properties of the mixtures, especially polarity, determine how far the spot will travel once placed in the solvent and this is what allows …show more content…
The only difference between the two is that 4-phenylphenol has an OH bonded to one of the carbon rings. This makes 4-phenylphenol the more polar compound due to the electronegativity of the oxygen. Intermolecular forces such as hydrogen bonds and dipole–dipole interaction can aid in explaining the polarity difference. The dipole-dipole interaction that occurs happens due to the electronegativity difference between Oxygen and Hydrogen and creates partial charges on the atoms, aiding in attracting electrons. Hydrogen bonds can occur when a hydrogen atom interacts with a Nitrogen, Oxygen, or Fluorine of another molecule, and this interaction is one of the strongest intermolecular forces. Because these hydrogen bonds can occur the 4-phenylphenol will form intermolecular hydrogen bonds with the silica gel and therefore not move as far on the TLC plate. When looking at compound C we know understand that biphenyl would travel further than the more polar 4 –phenyl phenol. Understanding this, we would predict that compound B is the 4-phenylphenol and that the lower spot in C is 4-phenylphenol and the higher spot is biphenyl, since it will travel
In this experiment, column chromatography and thin layer chromatography were used to separate a mixture of fluorene and 9-fluorenone. These two methods were then compared, and the results were analyzed. In column chromatography, 0.1010 g of mixture was separated. During the separation, fluorene eluted first. This compound was white in color once dried with the rotary evaporator. A percent yield of 93.47% was calculated for fluorene. The product that eluted first was confirmed to be fluorene by the IR spectrum obtained and the experimental melting point. The IR spectrum RM-02-CC1 was the spectrum obtained for this compound. Aromatic carbon- hydrogen bonds, carbon-carbon double bonds and hydrogens attached to sp2 carbons were shown by peaks 3038
TLC allows identification of compounds based on polarity. Nonpolar compounds move higher up on the TLC plate than polar compounds because polar compounds are more attracted t...
The gas chromatography calculations offer the area values for under the peak curves. Those area values are directly correlated with the relative concentrations of each alkene product in the mixture. When the ratio of the two values is calculated, 3-methylcyclohexene being the 2nd peak area value and 1-methylcyclohexene being the 3rd peak area (represented on table 1), (11152:1283) = 8.7 1-methylcyclohexene molecules: 1 3-methylcyclohexene molecule.
The hypothesis expected that aniline and phenol would be tri-substituted, and anisole an acetanilide would be mono-substituted and have a bromine in the para position, because the para position is favored over the ortho position because of sterics. Results after the experiment were: phenol > aniline > acetanilide. Anisole was not applicable because the experiment was not ran during class and that data was not provided. The results for aniline were one group claiming 2,6 dibromoaniline and two groups claiming 2,4 dibromoanaline. 2,4 dibromoanaline seems more probable because it is preferred to have substituents spaced out on an aromatic ring, for less tension between substrates. The conclusion with phenol was that it was trisubstituted 2,4,6 tribromophenol. Although there are two melting points, it is the same product that was being analyzed, because only one group could correctly brominate phenol. The temperature difference is attributed to two different mel-temps being used, causing two different temperatures. The results for acetanilide, were that all three groups concluded that 4-bromoacetanailde was the product. This is predictable because acetanailde is a bulky substrate. Compounds favor that bulky substituents only attach in the para
The HPLC method used was 20 μL of sample at 25°C through an RI detector for 15 minutes. The mobile phase was .0001M Sulfuric acid at .6ml/min with a column temperature of 60°C.
Chromatography has been developed over the past century and has an important contribution in many areas of modern science. However the main original work of M.S.Tswett was published in a book Chromatographic Adsorption Analysis.
To begin we made sure that the spectrophotometer was warmed up and was set to 605 nm while the chloroplast suspension was being made by the instructor. 605 nm was used because this is the wavelength that is the easiest to see transmittance of DPIP in the chlorophyll. The chloroplast suspension is created using spinach leaves that are deveined because the veins contain less chloroplasts. Once the leaves had been deveined they were placed under a light to begin the photosynthesis process. Then the instructor then placed the leaves in a chilled blender. He used a chilled blender because the motor of the blender when in use would generate heat and the heat would in turn cause the leaves to denature. Next, he added .5 M sucrose to the blender which
This mixture was very good at separating the mixtures because its dielectric constant is 3.832. This relatively high value for the dielectric constant gives a strong effect towards moving the compounds up the TLC plate. The way that I visualized the spots on the TLC plate is first I placed the plate under a UV light. This showed most but not all the spots. The next way was to dip the TLC plate into bromocresol green. The best mixture was mixture 5 at separating the compounds due to the greatest dielectric constant. The worst mixture being the first mixture this is due to the very low dielectric constant. The general values between the ibuprofen and the aspirin are almost the same most of the time while the naproxen is very low in
The purpose and goal of this experiment is to separate a mixture of fluorene and 9-fluorenone through the method of column chromatography, and then determine the purity of the isolated compounds through thin-layer chromatography (TLC), as well as melting point analysis. Through column chromatography, gravity is utilized to move the solvent and compounds down the column, where the changes in the solvent polarity assists in eluting the desired compounds to separate fractions. Each fraction solvents can then be evaporated to obtain the compounds of interest. Through TLC, a thin layer of polar and hydrophilic silica gel on an inert sheet is used to spot the sample on the bottom of the sheet and is then developed in a jar of eluent, where through
Set up tubes; labels 2. Add 1 mL of each sample to be tested. Make sure you stir the solution before pipetting it into your tube. 3. Add 2 mL of Benedict`s reagent to each tube 4.
Start to prepare the sample in a 10mL (10000µm) place 0.4mL (400µm) of the stock Ferritin solution 0.5g/mL in 0.15M sodium acetate (AcONa). Afterwards add 0.8mL (800µm) at 2M of sulphuric acid (H₂SO₄) and 0.8mL (800µm) at 5mM of dihydroxyfumarate (C₄H₄O₆). Wait for 30 minutes then complement the solution with 1.6mL (1600µm) at 2.5M of sodium acetate (NaOAc) followed by 0.8mL (800µm) at 12.5mM ferrozine (C20H12N4Na2O6S2). Lastly wait another 30 minutes afterwards add distilled water using a stopper to complete the solution to 10mL (10000µm) then put the rubber stopper on the volumetric flask and mix till you are sure that all the contents are mixed. Again, set up the spectrometer to 562nm and using the 2 samples that were made take one from each to measure the absorbance plus a third one from either one of the solution.
Two unknown solutions were given to conduct a lab, to determine the macromolecules within them by conducting a series of macromolecule tests. The unknown solution which will be discussed is Solution #1 as shown in Table #2; this solution was known as Unknown #2. The solution was of a watery consistency; it was transparent yet had a pinkish hue to it. There were no visible grains in the solution, unlike other solutions that could not fully dissolve certain products. The Fats test, Protein test, Starch test and Glucose test were conducted to identify the types of macromolecules in the solution. The Fats test, Protein test, and Starch test all came out negative. This was concluded by following the negative observations of these tests from Table #1. The only test that came
Analyze each fraction by spotting 10 times with capillary tubes on a TLC plate, which is exposed to iodine vapor for 15 minutes.
LAB REPORT 1st Experiment done in class Introduction: Agarose gel electrophoresis separates molecules by their size, shape, and charge. Biomolecules such as DNA, RNA and proteins, are some examples. Buffered samples such as glycerol and glucose are loaded into a gel. An electrical current is placed across the gel.
Set up a ring stand in order to support the separatory funnel and place 100ml of beverage into the separatory funnel.