Chem2210-Lab Report 2 Part 2 PDF

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Patrick Rukundo, 7775181 Section B05, Wednesday-PM Parker 290 Lab Report 2: Separation and Purification Techniques Due Date and Time: 08/11/2017 - 2:30pm

1. Identity and Purity Assessment Solid A – 9-Fluorenone The identity of solid A was deduced using the melting point determination and the IR spectrum. After the extraction and crystallization of unknown A, we were left with a yellow solid substance. We first ran the solid through the IR spectrum machine, and we got a distinct peak of interest at 1711.24cm-1. The peaks are used to determine what type of functional groups are observed in the compound. The IR spectrum we obtained indicated that a carbonyl(C=O) functional group was present. After the IR spectrum, we then determined the solid’s melting point range, and the value we got was 80.9˚C-81.5˚C. So, when we looked at all this data, we able to determine what our solids identity was, starting first with its colour. Among the six possible compounds, the only compound that was listed to have a yellow like colour was 9-flourenone, the rest were said to be only white in colour. We then decided to look at 9-fluorenone’s chemical structure, and in its structure we were able to observe the functional groups that corresponded with our solid’s IR spectrum. Furthermore to finalize our deduction, we compared the literature melting point of 9-fluorenone, which is 80.2˚C, to the experimental melting point of our solid A, which is 80.9˚C-81.5˚C, and we observed that the literature melting point falls in our solid’s melting point range. Thus, from all this data we concluded that our solid A must be 9-fluorenone. When determining purity what we look for is whether the compound is contaminated with some other different compound, and the best way to spot contamination is through the use of the IR spectrum, the melting point determination, and most easily the TLC. For the IR spectrum of 9-fluorenone what we expect to observe is the peak for the functional group carbonyl, and if any other functional group’s peaks are present it will indicate that they may be a contamination. However, since w only observed carbonyl in our solid’s IR, we can say that it did not contain any contamination. As for the

melting point determination, we can identify a compounds purity if it has a reasonably sharp melting point, or a range that does not span more than 2˚C, such a compound will be considered pure [3]. Knowing this we can say that our solid was pure, because its melting point range had a lesser span than 2˚C. Lastly the TLC in my opinion was the simplest in terms of spotting any contamination. After conducting a TLC each column of the plotted spots should contain only one spot. If multiple spots appear in a column, this is an indication that there is more than one type of compound present. Our TLC, had only one spot present in the solid A column, which indicates that there is only one type of compound present (i.e. no contaminants). Thus, looking at all the results I can conclude that solid A had no contaminants, meaning that it is a pure compound. Solid B – Ethyl 4-aminobenzoate The identity of solid B was deduced using the same methods as for solid A, starting off with the IR spectrum. Solid B IR spectrum had distinct peaks at 3342.27cm-1 - 3421.33cm-1 and 1680.68cm-1, these peaks indicated the presence of a primary amine group (NH2), and a carbonyl group (C=O), respectively. We then put solid B into the melting point apparatus, and we able to determine its melting point to be 89.5˚C-90.5˚C. After obtaining this information, what we able to conclude was that out of the six possible substances, our solid had to be ethyl 4-aminobenzoate. Ethyl 4-amino benzoate has a literature melting point of 89.0˚C-92.0˚C, and our solid’s melting point falls within this range. When looking at the chemical structure of ethyl 4-amino benzoate, we observe the functional groups NH2 and a carbonyl (C=O), which are functional groups that correspond with our solid’s IR spectrum. Furthermore ethyl 4-aminobenzoate is a white substance, same as our solid B, and from all this we can conclude that our solid B is indeed ethyl 4-aminobenzoate. The purity of solid B was also determined using the same procedures as for solid A. Our IR spectrum for solid B showed no signs of contaminants, there

was no peak observed that did not correspond to the structure of ethyl 4aminobenzoate. Our melting point range (89.5˚C-90.5˚C) had a span less than 2˚C, which indicates that it is relatively pure. Furthermore, the solid B column on the TLC had only one spot present, an indication of no contaminants. Thus, I conclude that solid B is pure. 2.

Rf=X/Y x=distance of spot from the origin y=distance of solvent front from the origin A = 9-Fluorenone Rf= 4.0/5.2 = 0.77 X=4.0c

Y=5.2c

Co-spot = 0.69, 0.77

X=3.6c

A

Co-

B = Ethyl 4-aminobenzoate Rf= 3.6/5.2 = 0.69

B

As we can see from the figure above the Rf values differ. The Rf values for solid A and solid B are 0.77 and 0.69, respectively. This difference can be explained by the polarity each of these solids display. As described in question 1, solid A was determined to be 9-fluorenone, and solid B was determined to be ethyl 4-aminobenzoate. Ethyl 4-aminobenzoate is known to have a relatively higher polarity than 9-flourenone. Due to this higher polarity, ethyl 4-aminobenzoate will stick to the plate much stronger than 9fluorenone because the plate itself is made of a very polar substance (silica gel), and as we know polar attracts polar. So as the eluent (mobile phase) is pulling the two substances up the plate (stationary phase) the more polar substance will stick much stronger to the plate and will thus exhibit less movement up the plate. 3. Tables Melting Point Compound 9-fluorenone Ethyl 4aminobenzoate

Experimental Melting Point (˚C) 80.9 – 81.5

Literature Melting Point (˚C) 83.5

Referenc es 1

89.5 – 90.0

89.0 – 92.0

2

IR Spectrum – 9-Fluorenone Wavenumber (cm-1)

Strength of Band

Assignment

1711.24

Strong

C=O of Carbonyl

3012.15, 3060.07

Weak

Alkanes

IR Spectrum – Ethyl 4-aminibenzoate Wavenumber (cm-1)

Strength of Band

Assignment

3342.27, 3421.33

Medium

Primary Amine (NH2)

1680.68

Strong

C=O of Ester

TLC Compound A – 9-fluorenone

Rf 0.77

Conditions Eluent: Ethyl acetate: Hexanes

Co-Spot

0.69, 0.77

(3:1) Plate: Silica gel

B – Ethyl 4-aminobenzoate

0.69

Visualization: UV

4. Scifinder Search Results for Organic Neutral Compound; 9Fluoronone Screenshot of 9-Fluorenone – HNMR Spectrum Results

Screenshot of 9-Fluorenone – IR Spectrum Results

5. If we were to use ethanol instead of ethyl acetate for the separation of the two compounds, I believe the results would be much worse. Ethanol is much more polar than ethyl acetate, so it will tend to interact more with polar substance. In our experiment, we used 1M HCl and ethyl acetate as our solvents for the extraction, and these two liquids are immiscible due to the difference in their polarity. This tendency to be immiscible when mixed together is what gives us the separation of our unknowns because while one unknown prefers the non-polar liquid, the other unknown will prefer the polar liquid, and from this arises the separation of our two unknowns. So, if we were to instead use 1M HCl and ethanol as our solvents for the extraction,

we would not have this separation because HCl is very polar and so is ethanol, and as we know polar likes to mix with polar (i.e. they are not immiscible). What we would observe is that HCl and ethanol would actually mix more with each other than HCl and ethyl acetate, giving us essentially only one liquid layer, so our two unknowns would in essence be mixed in with each other as well, giving us little to no separation.

6. We were assigned unknown solution 3. The identities of the components in the unknown solution were the known solvent and internal standard, which is respectively acetone and butanone, and lastly the unknown analyte which in our case was determined to be 2-pentanone. We deduced the identities, more specifically the analyte, using the Gas Chromatography (GC) graphs of the given standards. We did this by first graphing standard 2 and standard 5 in the GC, each graph contained three peaks, the first two peaks were the acetone and butanone which we knew will be observed in all the GC graphs, so we didn’t concern ourselves with them in terms of deducing the unknown. However, the third peak of standard 2 and 5 differed significantly, and what we were interested in was the time that they peaked at, also known as their retention time because this will help us in determining the unknown analyte of solution 3. So, we then graphed the unknown using the GC and observed the three peaks, the first two were the same as both standard 3 and 5; that is that they peaked at about the same time. However, the third peak of unknown solution 3 resembled that of standard 2 only; the retention time of the unknown was 3.260min and the retention time of standard 2 was 2.845min. The components of standard 2 were acetone, butanone and 2pentenone, and since our unknown resembled the retention times of standard 2, specifically the analyte 2-pentanone, we concluded that our unknown had the same components. The molarity of the unknown ketone’s concentration is 1.31M. Calculation; Step 1: Calculate % volume - from linear equation: y=0.2528x – 1.6792

y= integration ratio of 2-pentanone/butanone (this is derived from the unknown GC) =integrated area of 2-pentanone / integrated area of butanone = 98.59 / 50.51 = 1.952 x= 2-pentanone concentration (%) =? (We’ll use the linear equation here) = (y + 1.6792) / 0.2528 = (1.952 + 1.6792) / 0.2528 = 14.36% volume Step 2: Convert % volume to ml 14.36% >> 0.1436 x 10.0ml(this is the total volume of the solution) = 1.40ml Step 3: Convert ml to g – we’ll use its density (0.809g/ml) 1.40ml x (0.809g/1.00ml) = 1.13g Step 4: Convert g to mol – we’ll use its molecular weight (86.134g/mol) 1.13g x (1.00mol/86.134g) = 0.0131mol Step 5: Calculate Molarity Molarity = mol of solute/ L of solution = 0.0131 / 0.01 = 1.31M

7. The internal standard is used as a base comparison for the analyte. If we were to inject the same amount of a certain sample into the GC we will not always get the same reading; that is that the integrated areas will differ in each graph. The internal standard is used to accommodate for this variability, and it does so by responding to any disturbance or external stimuli in the same manner that the analyte will respond [4]. So, if the

disturbance causes the integrated area of the analyte to be halved then the internal standard integrated area will also be halved.

8. Standard Curve of 2-Pentanone with Butanone as the Internal Standard: Analyte= 2-Pentanone, Internal Standard= Butanone

Integration Ratio of 2-Pentanone/butanone

Standard Curve of 2-Pentanone with Butanone as Internal Standard 12 f(x) = 0.25 x − 1.68 R² = 1

10

Ra tio

8 6 4 2 0 5

10

15

20

25

30

35

40

45

50

55

2-Pentanone Concentration (%)

9. (Next Page)

10. The IR spectra for both my solids came out just fine. The peaks that I expected to observe were present in the graphs. It is also mentioned in the lab manual that a bad quality IR spectrum would appear “hairy”, and since both my spectra do not appear so, I can confidently say that my spectra are of good quality. So, in conclusion I don’t think there is any need for improvements

9. IR Spectrum for Solid A – 9-Fuorenone – Patrick & Uzair

IR Spectrum for Solid B – Ethyl 4-aminobenzoate – Patrick & Uzair

References (1) Science Lab: Chemicals and Laboratory Equipment. 9-Fluoronone MSDS https://www.sciencelab.com/msds.php?msdsId=9924071 (November 02, 2017) (2) Material Safety Data Sheet. Ethyl 4-aminobemzoate 98% https://fscimage.fishersci.com/msds/01438.htm (November 02, 2017) (3) Loung, H. CHEM2210 Introductory Organic Chemistry: Structure and Function. 2017, 33-35. (4) LCGC: Solutions for Separation Scientist. When Should an Internal Standard be Used? http://www.chromatographyonline.com/when-should-internal-standard-beused-0 (November 02, 2017)...