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Regioselective Nitration of Acetanilide Essay Sample

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Regioselective Nitration of Acetanilide Essay Sample


Table of reagents
Compound Mol.WT. Amount Density Moles (gm/mol) (gm or mL) (g/mL) (mmol) Acetanilide 135.16 1.06 1.219 7.84 96% Sulfuric acid 98.08 4.4 1.84 Concentrated Nitric acid 63.01 0.7 1.42 5:5 EtOAc:Hexanes 174.16 10 0.682 9:1 EtOAc:Hexanes 174.02 10 0.678 Ice Cubes 18.02 4 cubes 0.9167 Acetone 58.08 10 0.791 Water 18.02 20 1.00 Ethanol 46.07 100 0.789 o-nitroacetanilide 180.16 2 spots 1.42 p-nitroacetanilide 180.16 2 spot 1.34

1. I Placed 1.06 g of acetanilide into a clean 25 mL round bottom flask. 2. I
added 3.1 mL of H2SO4 and stirred gently for two minutes to dissolve the acetanilide. 3. I cooled the acetanilide solution in an ice bath by placing the ice bath container on the stir plate under the flask. 4. In a 50 mL Erlenmeyer flask, I combined concentrated nitric acid (0.7 mL) and concentrated sulfuric acid (1.3 mL) and then I added this mixture to the reaction flask. 5. After 10 minutes, I took a TLC (ether hexanes 5:5) of the reaction to verify that all the reaction is complete; sprinkles of acetanilide and acetone were used as the reference. 6. I placed 4 small ice cubes in a 125 mL Erlenmeyer flask along with ~20 mL of cold water.

The flask was placed on the stir plate for about 10 minutes. 7. I poured the contents of the reaction flask into this 125 mL flask, rinsing with a small volume of water. 8. I collected the solid that formed by suction filtration using a Buchner funnel, and washing with cold water. 10. I did a TLC sample of the crude product by placing a few small crumbs of the crude product in a small vial, and added 10 drops of acetone to dissolve it. 11. I prepared TLC plates that compare the crude product to reference samples of the ortho and para isomers; I eluted the TLC plates using the solvent system ether: hexanes 9:1 for ortho and para isomers.

12. I visualized each plate using the UV chamber and then took pictures with my phone. 13. I dissolved the remaining crude product in 100 mL of boiling ethanol (enough to dissolve the solid) in a 250 mL beaker containing a stir bar. 14. Once dissolved, I removed the flask from the heat and allowed it to cool to room temperature undisturbed until crystals form. 17. Once crystallization was complete, I isolated the crystals by suction filtration. 18. I weighed the pure product (0.099 g) and analyzed the product by TLC (ether: hexanes 9:1) comparing the purified product to the crude product. On a separate TLC plate, I compared the purified product to the mother liquor. I then visualized each plate using the UV chamber and photographed each plate.

– The acetanilide powder was white in colour
– The mixing of sulfuric acid with acetanilide created a gaseous fizz
– reaction flask initially had a black mixture, then orange in the erlenmeyer flask – some product in the erlenmeyer flask condensed and stuck to the sides of the flask
– Final purified crystals was yellow/orange


Ortho: 27.5/ (27.5+40.9) = 40.2%
Para: 40.9/ (27.5+40.9) = 59.8%

Para : 40.9/ (40.9+31.6) = 56.4%
Sample: 31.6/ (40.9+31.6) = 43.6%

Ortho:Para Para:Sample 2:3 3:2.3 Ortho:para:sample = 2:3:2.3


RCP (crude product): 28.64/ (28.64+36.6) = 44%
Pure Product: 36.6/ (28.64+36.6)= 56%
Crude:Product = 1:1.27


RML (mother liquor): 35.3/ (31+35.3) = 53%
Pure Product: 31/ (31+35.3) = 47%
RML:Product= 1:1.13

Yield of Nitroacetanilide Product
1) Number of moles of acetanilide
nacetanilide = macetanilide / MWacetanilide
nacetanilide = [1.06g]/ 135.16 g/mol
nacetanilide = 0.0078 mol
2) Theoretical yield of nitroacetanilide
nnitroacetanilide= acetanilide (1:1 ratio)
mnitroacetanilide= nnitroacetanilide x MWnitroacetanilide
mnitroacetanilide= (0.0078 mol) x (180.16g/mol)
mnitroacetanilide acid = 1.40 g
3) Yield of nitroacetanilide product
yield of nitroacetanilide = (experimental yield / theoretical yield) yield of nitroacetanilide = (0.099 g / 1.40 g)
yield of nitroacetanilide = 0.071 g
n= m/M
n= 0.071/180.16 = 0.00039 mol (or 7.1%)

The mechanism for the nitration is that of an electrophilic aromatic substitution. In this experiment, acetanilide produced ortho and para mono nitroacetanilides; the nitronium ion (formed from the reaction between concentrated nitric acid and sulfuric acid), is directed to the ortho and para positions by the –NHCOCH3 group attached to the benzene ring of acetanilide. This is largely because of the resonance delocalizing the benzene ring by the N: lone pair. Moreover, substitution of the nitro group to the para position is favored over substitution to ortho position due to lack of steric hindrance. This steric hindrance makes ortho substitution less likely than para substitution. Overall, acetanilide mainly gives the p-nitroacetanilide, mixed with smaller proportion of the o-acetanilide. This is evident when observing the TLC plate. The TLC method was used to determine the different isomers produced and the purity of the final product.

As shown in the calculations, the para isomer had a 3:2 ratio, indicating that it would make up the majority of the final product. Additionally, since ortho is more soluble in ethanol, most of it remained in the filtrate during suction filtration while more of the para was filtered. However, in general, the purified product was very minute. The yield, 7.1%, is extremely small because not much solid product was retrieved through suction filtration. This was a common phenomenon in the lab section, so perhaps it was not allowed to cool enough or too much ethanol was used to dissolve the unpurified product. Either way, majority of that yield is composed of para-nitroacetinalide. Now, some key steps had to be taken in this lab, so perhaps a mistake was made along the way, also resultingin this small yield.

In order to prevent di-nitration of the acetanilide, the mixture of the two acids had to be added in small portions to the acetanilide solution, so that the concentration of the nitrating agent is kept at minimum. The addition of nitric acid would be exothermic so the mixture would have got too hot and overpass the maximum temperature range suitable for the nitration this is why less was used and why it was added slowly. In this experiment, cold temperatures were used to slow and cool down the rate of reaction so that over nitration does not occur. An error in these steps could be the culprit as well. Questions:

1. More polar compounds are attracted more strongly to the silica than to the solvent, and so move more slowly up the TLC plate. Polar molecules travel slowly through the silica gel and appear at the bottom of the TLC plate. According to the TLC, the ortho isomer is the more polar compound. Less polar compounds are attracted less strongly to the silica and so move more quickly up the TLC plate. If they move up quickly through the silica gel, the spots will appear near the top of a TLC plate. This is the case with the para isomer (para spot is not visible due to low quality of picture)

2. Ethanol is a polar solvent and since the minor product, ortho-nitroacetanilide, is more polar than para-nitroacetanilide and like-dissolves-like, it is more soluble in ethanol than para nitroacetanilide.

3. After the first nitration, the benzene ring is activated; since nitro groups are electron-withdrawing, the addition of more nitro groups causes each successive product to be less reactive to further nitration. 4. Para isomers are more favoured over ortho isomers due to steric interaction in ortho isomers.



7. C6H6 + HNO3 ——–H2SO4———- C6H5NO2 + H20

a. Benzene: (780g) / (78g/mol) = 10 mol
Nitric Acid: 0.75 L * 16M = 12 mol
Since the ratio is 1:1, the limiting reagent is benzene.

b. Calculate the percent yield of nitrobenzene.
1) Number of moles of benzene
nbenzene = mbenzene / MWbenzene
nbenzene = [780g] / 78 g/mol
nbenzene = 10 mol
2) Theoretical yield of nitrobenzene
nnitrobenzene = nbenzene (1:1 ratio)
mnitrobenzene= nnitrobenzene x Mnitrobenzene
mnitrobenzene= (10 mol) x (123 g/mol)
mnitrobenzene = 1230 g
3) Percent yield of nitrobenzene
Percent yield of nitrobenzene = (experimental yield / theoretical yield) x 100 % Percent yield of nitrobenzene = (1000 / 1230 g) x 100%
Percent yield of nitrobenzene = 81.3 %
c. The side product is most likely water

d. The sulfuric acid behaves as a catalyst, and allows this nitration reaction to proceed more quickly. Mechanism: Sulfuric acid reacts with nitric acid to generate the electrophile- nitronium ion (NO2+)

8. The nitration favors orthonitrofuran over metanitrofuran due to resonance; there are more resonance forms from adding the nitro group onto the ortho position (3 resonance forms) than adding to the meta position (2 resonance forms). The more the resonance, the faster the reaction, meaning the orthonitrofuran product is favored.

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