Experiment 7
Electrophilic Aromatic Substitution Of Salicyamide
Name: Lidia Santiana Palha
Student number: s3333523
Email-address: lidiapalha@gmail.com
Name of demonstrator: H.Helbert
Reaction Equation Summary
Salicylamide and sodium iodide are dissolved in ethanol, and stirred and cooled to 0.
After that household bleach was added while stirring vigorously, solution changed from colorless to pale yellow. Sodium thiosulphate and hydrochloric acid were added aswell.
Product collected by vacuum filtration and recrystallized from 96% ethanol. IR- and H-
NMR-spectrum were taken to determine the location of the I+. I had a yield of 0,7 g
(2,66mmol, 18,2%). The iodide’s position is opposite of the –OH group.
Salicylamide …show more content…
Sodium thiosulphate and hydrochloric acid were added aswell.
Product collected by vacuum filtration and recrystallized from 96% ethanol. IR- and H-
NMR-spectrum were taken to determine the location of the I+. I had a yield of 0,7 g
(2,66mmol, 18,2%). The iodide’s position is opposite of the –OH group.
Salicylamide and sodium iodide are dissolved in ethanol, and stirred and cooled to 0.
After that household bleach was added while stirring vigorously, solution changed from colorless to pale yellow. Sodium thiosulphate and hydrochloric acid were added aswell.
Product collected by vacuum filtration and recrystallized from 96% ethanol. IR- and H-
NMR-spectrum were taken to determine the location of the I+. I had a yield of 0,7 g
(2,66mmol, 18,2%). The iodide’s position is opposite of the –OH group.
Salicylamide and NaI were dissolved in ethanol and cooled in a water bath. Then household bleach was added while stirring vigorously. Then NaOCl and HCl were added as well. Product was collected by vacuum filtration and then recrystallized from 96% ethanol and obtained a yield of 2,07g (54%). IR- and H-NMR-spectrum were taken. The Iodide is in the para position (also known as 5-iodosalicylamide), this is confirmed by the IR and the …show more content…
The second part is the reformation of the double bond. Since iodide is a large atom, it can’t be just placed anywhere, according to the resonance structures, the most stable is C (in Fig.2).
The most stable is C, because the OH and the NH2 give steric hindrance. Also because –OH is a strong activating group (electron donating substituent) and the NH2 is a deactivating group (withdraws electrons from the ring). Therefore –OH is ortho, para directing and -NH2 is meta deactivating. With all this in mind, positions 3 and 5 are more likely (on carbon 2 and 4 on the NMR). 5 is more likely and not 3 because of steric hindrance.
Experimental
In a 100ml round bottomed flask, salicylamide (2.0g, 14.6mmol, 1eqiuv) and sodium iodide (2.4g, 16mmol, 1.1equiv) were dissolved in absolute ethanol (40ml). The solution was stirred and cooled to 0°C using a water bath. The flask was removed from the ice bath then household beach (4%, 272.2ml, 1.2equiv) was added to the solution while stirring vigorously. Solution changed from colourless to dark red-brown to pale yellow. After a short time, sodium thiosulfate (10%, 10ml) was added. HCL (10%, 10ml) was added to acidify the reaction. The precipitate was collected by vacuum filtration and recrystallized from ethanol (96%). Both starting and product
Electrophilic Aromatic Substitution Of Salicyamide
Name: Lidia Santiana Palha
Student number: s3333523
Email-address: lidiapalha@gmail.com
Name of demonstrator: H.Helbert
Reaction Equation Summary
Salicylamide and sodium iodide are dissolved in ethanol, and stirred and cooled to 0.
After that household bleach was added while stirring vigorously, solution changed from colorless to pale yellow. Sodium thiosulphate and hydrochloric acid were added aswell.
Product collected by vacuum filtration and recrystallized from 96% ethanol. IR- and H-
NMR-spectrum were taken to determine the location of the I+. I had a yield of 0,7 g
(2,66mmol, 18,2%). The iodide’s position is opposite of the –OH group.
Salicylamide …show more content…
Sodium thiosulphate and hydrochloric acid were added aswell.
Product collected by vacuum filtration and recrystallized from 96% ethanol. IR- and H-
NMR-spectrum were taken to determine the location of the I+. I had a yield of 0,7 g
(2,66mmol, 18,2%). The iodide’s position is opposite of the –OH group.
Salicylamide and sodium iodide are dissolved in ethanol, and stirred and cooled to 0.
After that household bleach was added while stirring vigorously, solution changed from colorless to pale yellow. Sodium thiosulphate and hydrochloric acid were added aswell.
Product collected by vacuum filtration and recrystallized from 96% ethanol. IR- and H-
NMR-spectrum were taken to determine the location of the I+. I had a yield of 0,7 g
(2,66mmol, 18,2%). The iodide’s position is opposite of the –OH group.
Salicylamide and NaI were dissolved in ethanol and cooled in a water bath. Then household bleach was added while stirring vigorously. Then NaOCl and HCl were added as well. Product was collected by vacuum filtration and then recrystallized from 96% ethanol and obtained a yield of 2,07g (54%). IR- and H-NMR-spectrum were taken. The Iodide is in the para position (also known as 5-iodosalicylamide), this is confirmed by the IR and the …show more content…
The second part is the reformation of the double bond. Since iodide is a large atom, it can’t be just placed anywhere, according to the resonance structures, the most stable is C (in Fig.2).
The most stable is C, because the OH and the NH2 give steric hindrance. Also because –OH is a strong activating group (electron donating substituent) and the NH2 is a deactivating group (withdraws electrons from the ring). Therefore –OH is ortho, para directing and -NH2 is meta deactivating. With all this in mind, positions 3 and 5 are more likely (on carbon 2 and 4 on the NMR). 5 is more likely and not 3 because of steric hindrance.
Experimental
In a 100ml round bottomed flask, salicylamide (2.0g, 14.6mmol, 1eqiuv) and sodium iodide (2.4g, 16mmol, 1.1equiv) were dissolved in absolute ethanol (40ml). The solution was stirred and cooled to 0°C using a water bath. The flask was removed from the ice bath then household beach (4%, 272.2ml, 1.2equiv) was added to the solution while stirring vigorously. Solution changed from colourless to dark red-brown to pale yellow. After a short time, sodium thiosulfate (10%, 10ml) was added. HCL (10%, 10ml) was added to acidify the reaction. The precipitate was collected by vacuum filtration and recrystallized from ethanol (96%). Both starting and product