By: Krishnaa Siva
Chemistry 222
Fall 2012
1
By: Krishnaa Siva Date: November 8, 2012, Thursday 1:30-5:30 PM Lab Group TA: Mathieu Bedard Experiment 6: Azo Dyes 1. Objective: 1 mark What is the purpose of this experiment? The purpose of this experiment is to prepare the azo dye of methyl red by the diazonium coupling reaction of N,N-dimethylaniline with Anthranyl diazonium chloride, which was obtained from the Anthranilic acid. Use the melting point test to check if it was indeed methyl red that is produced. 2. Introduction: 2 marks Brief description of the concept/reaction studied – azo dyes, electrophilic aromatic substitution The basic structure of an azo dyes is Aromatic group-N=N-Aromatic group'. Azo group is the name of the structure that has the nitrogen double bond. Azobenzene derivatives have a conjugated pi-electron system, which cause the absorption of different wavelengths of light and is the reason why we witness a colored compound. Azo dyes are synthesized by the conversion of the primary amine (R-NH2) into a diazonium salt with the help from substances such as sodium nitrite and then have this salt react with another substance that contains an aromatic ring (ex. Dimethylaniline). It is the electron releasing resonance effect that allows for the reaction between the diazonium salt (the electrophile: diazonium cation) and the aryl unit. The regioselectivity in the diazonium coupling reaction is at the para position due to the presence of the large phenyls and alkyl groups. 3. Reaction Equation: 2 marks Include proper structures and data for reactants and products [molar mass, concentration, density, volume, mass, moles, etc – including theoretical yield]
Chemistry 222
Anthranilic acid 0.52 mmol Sol'n of sodium nitrite in water (Values for Sodium Nitrite) 10% Hydrochloric acid (Hydrogen Chloride) N,NDimethylanil ine
Fall 2012
2
Molar mass (g/mol)
Melting Point ( C) ̊
Boiling Point ( C) ̊
Density (g·cm−3)
Volume/ Mass
137.14
146.0-147.0
-
1.40
0.07
Moles 5.104 x 10^(-4)
(69.00)
271.0
-
2.17
(300 µl of solutin)
0.52 x 10^(3) of sodium nitrite
(-36.46)
(-112.0)
(-85.0)
(1.49x10^(3))
550.0 µL
-
2.0
194.0
0.96
89.0 µL=85.0 mg
0.79
Theoretical: 0.1375 g ExperimentalCrude Product: 0.04 g Pure Product: 0.01 g
121.18
Theoretical: 179.0-182.0 Experiment al: 169.0174.0
Methyl Red
269.30
-
0.71 x 10^(3) Theoretical: 5.104 x 10^(-4) Experimental Crude: 1.485 x 10^(-4)
Experiment al Crude: 3.713 x 10^(-5)
4. Procedure: 2 marks Summary of lab procedure – what YOU did, third person, past tense – be concise!
Chemistry 222
Fall 2012
3
To a 3.0 mL conical vial, 0.07 g of anthranilic acid, a magnetic spine vane, and 550 µL of 10% hydrochloric acid was added(for the hydrochloric acid a Pasteur pipette was used). Next, the condenser was added to the vial. Since the solution inside the vial did not seem to be a homogenous solution right away, it was placed it on a hot place for stirring purposes until it became a homogenous solution . Next, to cool the solution in the vial, it was placed in an ice bath, and that ice bath with the vial was placed on top of the hot plate with the stirring still taking place. It was left stirring with the ice bath for 10 minutes. In a beaker, some (greater than 300 µL) 0.52 mmol solution of sodium nitrite was poured and this beaker was also placed in the ice bath to cool. A thermometer was placed in the ice bath, which had the two solutions, in order to monitor the temperature. When both solutions were at approximately 5 C, ̊ 300 µL of the nitrite solution was taken using a Pasteur pipette and it was slowly added to the solution in the anthranilic acid solution in the vial that was being stirred. While doing this step, it was made sure that the temperature remained at approximately 5 C. ̊ After 5 min, the solution of anthranyldiazonium chloride was texted for the presence of excess nitrous acid with the placement of a drop of the solution on a potassium iodide-starch paper. An immediate blue color was not seen, therefore more nitrite solution was added to the vial. The solution was left to stir for a couple of minutes and then this process was repeated and an immediate blue color was observed. Next, the air condenser was removed from the vial, 89 µL of N,N-Dimethylaniline was added quickly to the vial, and then condenser was reattached. The solution was left to stir for 15 minutes, while making sure the temperature was still around 5 C. ̊ 200 µL of 0.83 mmol solution of sodium acetate was then transferred to the vial with the reaction mixture using the Pasteur pipette (without removing the air condenser). The solution was left to stir for an additional 20 minutes, while making sure the temperature was still around 5 ̊ C. The vial was then removed from the ice bath and it was left out (on top of the hotplate?) for 15 min so that the solution. returns to room temperature. Next, 100 µL of 10% aqueous NaOH solution was added to the reaction mixture in the vial using a pipette and then it was left to stand at room temperature for another 30 minutes. The crude Methyl Red was collected through vacuum filtration using a Buchner Funnel. The reaction flask was rinsed with 0.5 mL of water, and is rinse was then used to wash the crystals as well. In addition to this, the crystals were washed with 0.5 mL of 3 M acetic acid, and then again with 0.5 mL of water. The weight and melting point of the crude product were then taken. Following this step, the crude product was dissolve in 2.0-2.5 mL of methanol in a flask (this was placed in a beaker with hot water for faster dissolution). The solution was then cooled in an ice bath, and the pure methyl red crystals were collected with again performing vacuum filtration using a Buchner funnel. The weight, melting point and IR of the product was taken. (This entire experiment was performed in the fume hood.)
Chemistry 222
Fall 2012
5. Results: 5 marks a. Table 1: Yield and Physical Properties Crude Azo Dyes
4
Purified Azo Dyes
Yield (g)
0.04 g
0.01
Yield (%)
29 % Dark Red Solution with some black dots
7% Dark Red Solution (not many black spots) - (not enough pure product to get the melting point)
Appearance Melting Point Recovery (%)
169-174
25 %
6. Discussion: 8 marks (Since almost no pure product was collected, the crude product will be used for most of the analysis purposes) a. Discussion of Yield – crude, purified, % recovery – why less than 100%? The experiment yield of the crude product was only 29% and the experimental yield of the pure product was only 7% due to various experiment errors. The highly inaccurate transfer of liquid using the Pasteur pipette certainly contribute a lot to having a low yield. Another aspect of the experiment, that could have played a role in the yield of the the product is the temperature. It is very important to maintain that low temperature of less than 5 ̊ C throughout the experiment, since the Anthranyl diazonium chloride salt solution is unstable and prone to deterioration at room temperature (E College of Chemistry & Chemical Engineering, 2011). Exposing the Anthranyl diazonium chloride salt solution to a higher temperature for an extended period of time while transferring a solution into the vial could have affected the yield of the experiment. b. Determination Identity of both reactants and product i. Analysis of Melting Point The melting point is unique to every substance. The melting point of the product was measured to be 169.0-174.0 ̊ C and this is very close to the actual melting point of the product of 179-182 C, ̊ considering the possible human error in this experiment. It is important to note that the melting point taken was of the crude product (since there was not enough pure product to take the melting point) which means
Chemistry 222
Fall 2012
5
that the presence small amounts of other substance could have affected the melting point. Nevertheless, since the experimental melting point is very close to the actual melting, it can be said that there is very little "other" substances and for the most part it is methyl red. The melting point values being similar supports the idea that the correct product was formed and that the product is very close to being pure. c.
Mechanism – detailed mechanism for the synthesis of your azo dye
d.
Discussion Questions – i. How does the pH affect the reactivity of the N,N-dimethylaniline? By increasing the pH of the solution, the rate of the azo compound formation (normally a slow reaction) is increased. The higher the pH the lower the concentration of dimethylanilineH+ and that is why 10% aqueous NaOH was added to the vial near the end of the experiment.
Chemistry 222
Fall 2012
6
The lone pair on the N,N dimethylaniline is what that makes it possible to stabilize the positive charge on the nitrogen of anthronyl diazonium chloride, a substance that is in its transition state. It is the electron releasing resonance effect of the N,N dimethylaniline that makes it possible for the stabilization to happen. If the pH of the solution was lower (solution being more acidic), the nitrogen with the lone pair (of the N,N dimethylaniline) would be protonated, making the substance unreactive. Moreover, this would take away N,N dimethylaniline's ability to help stabilize the positive charge on Nitrogen of the anthronyl diazonium chloride (it does not allow for the electrophilic attack from the benzene ring of the demethylaniline). Making the solution more acidic, would certainly increase the concentration dimethylanilineH+ (leaving fewer proton free-dimethylaniline) and as a result the overall production of the Methyl Red would be reduced. ii. Why are azo dyes coloured compounds? Azo dyes absorb different wavelength of light within the visible rang of 400-800nm. The colors that are observed are the wavelengths of light that were not absorbed by the dye ("the leftover wavelengths of light") (Clark, 2004). For example, if it is red light (wavelength: 647700 nm) that is absorbed by the dye the complementary color of green would be seen. The presence of the delocalized electron system made up of alternating double bonds, also known as chromophores, in the dyes, results in the absorption of the energy from the light (Clark, 2004). It is also important to keep in mind that energy is inversely proportional to the wavelength of light: E = hc/λ. With higher amount of double bonds present in the dye, less energy is needed to get the electrons excited and longer wavelengths of light would be absorbed. On the other hand, with less double bonds present, more energy would be needed to get the electrons excited and shorter wavelengths of light would be absorbed. iii. Explain the regioselectivity in the diazonium coupling reaction (para only) The regioselectivity in the diazonium coupling reaction is at the para position only because the diazonium cations are poor electrophiles that are relatively large (Biele, 2010). Since the electron density is at its highest at both the ortho and para positions of the demethylaniline, it may first appear that the regioselectivity in the diazonium coupling reaction can be at both of those positions, but that is not the case. It
Chemistry 222
Fall 2012
7
would not easily bind at the ortho position, due to the strong steric hindrance as there would be a clash between the large phenyl groups and the alkyl groups. With the phenyl groups "needing more space", the para position will be that ideal position for the binding to take place. Furthermore, with the para substitution, the positive charge's stabilization is at its maximum. It is only when the para position is taken, the regioselectivity in the diazonium coupling reaction shifts to the ortho position. iv. What reagents would you need for the synthesis of Methyl Orange (another azo dye)? Give an overall reaction equation (with structures). For the synthesis of Methyl Orange, sulfanilic acid, sodium nitrite and dimethylaniline would be needed (Senese, 2010). Other substances such as sodium carbonate, HCL solution, NaOH (or Sodium Acetate) will help with the reaction.
References Biele, Carsten and Kamp, Oliver. " Diazonium Coupling (Diazo Coupling, Azo Coupling)". CHEMGAROO ChemgaPedia. 2010. Web. 13 Nov. 2012. Clark, Jim. "REACTIONS OF DIAZONIUM SALTS". Chemguide. 2004. Web. 12 Nov. 2012. "Experiment 1: Synthesis of an Azo Dye: 1-(4-hydroxyphenylazo)-2naphthol". E College of Chemistry & Chemical Engineering. 2011. Web. 13 Nov. 2012.
Chemistry 222
Fall 2012
Mayo, Dana W., Ronald M. Pike, and David C. Forbes. "Chapter 6: EXPERIMENT 26 Diazonium Coupling Reaction: Methyl Red." Microscale Organic Laboratory: With Multistep and Multiscale Syntheses. Hoboken, NJ: J. Wiley & Sons, 2011. N. pag. Print. Senese, Fred. " What is methyl orange? How is it made?" General Chemistry Online! 15 Feb. 2010. Web. 12 Nov. 2012.
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Fall 2012
1
By: Krishnaa Siva Date: November 8, 2012, Thursday 1:30-5:30 PM Lab Group TA: Mathieu Bedard Experiment 6: Azo Dyes 1. Objective: 1 mark What is the purpose of this experiment? The purpose of this experiment is to prepare the azo dye of methyl red by the diazonium coupling reaction of N,N-dimethylaniline with Anthranyl diazonium chloride, which was obtained from the Anthranilic acid. Use the melting point test to check if it was indeed methyl red that is produced. 2. Introduction: 2 marks Brief description of the concept/reaction studied – azo dyes, electrophilic aromatic substitution The basic structure of an azo dyes is Aromatic group-N=N-Aromatic group'. Azo group is the name of the structure that has the nitrogen double bond. Azobenzene derivatives have a conjugated pi-electron system, which cause the absorption of different wavelengths of light and is the reason why we witness a colored compound. Azo dyes are synthesized by the conversion of the primary amine (R-NH2) into a diazonium salt with the help from substances such as sodium nitrite and then have this salt react with another substance that contains an aromatic ring (ex. Dimethylaniline). It is the electron releasing resonance effect that allows for the reaction between the diazonium salt (the electrophile: diazonium cation) and the aryl unit. The regioselectivity in the diazonium coupling reaction is at the para position due to the presence of the large phenyls and alkyl groups. 3. Reaction Equation: 2 marks Include proper structures and data for reactants and products [molar mass, concentration, density, volume, mass, moles, etc – including theoretical yield]
Chemistry 222
Anthranilic acid 0.52 mmol Sol'n of sodium nitrite in water (Values for Sodium Nitrite) 10% Hydrochloric acid (Hydrogen Chloride) N,NDimethylanil ine
Fall 2012
2
Molar mass (g/mol)
Melting Point ( C) ̊
Boiling Point ( C) ̊
Density (g·cm−3)
Volume/ Mass
137.14
146.0-147.0
-
1.40
0.07
Moles 5.104 x 10^(-4)
(69.00)
271.0
-
2.17
(300 µl of solutin)
0.52 x 10^(3) of sodium nitrite
(-36.46)
(-112.0)
(-85.0)
(1.49x10^(3))
550.0 µL
-
2.0
194.0
0.96
89.0 µL=85.0 mg
0.79
Theoretical: 0.1375 g ExperimentalCrude Product: 0.04 g Pure Product: 0.01 g
121.18
Theoretical: 179.0-182.0 Experiment al: 169.0174.0
Methyl Red
269.30
-
0.71 x 10^(3) Theoretical: 5.104 x 10^(-4) Experimental Crude: 1.485 x 10^(-4)
Experiment al Crude: 3.713 x 10^(-5)
4. Procedure: 2 marks Summary of lab procedure – what YOU did, third person, past tense – be concise!
Chemistry 222
Fall 2012
3
To a 3.0 mL conical vial, 0.07 g of anthranilic acid, a magnetic spine vane, and 550 µL of 10% hydrochloric acid was added(for the hydrochloric acid a Pasteur pipette was used). Next, the condenser was added to the vial. Since the solution inside the vial did not seem to be a homogenous solution right away, it was placed it on a hot place for stirring purposes until it became a homogenous solution . Next, to cool the solution in the vial, it was placed in an ice bath, and that ice bath with the vial was placed on top of the hot plate with the stirring still taking place. It was left stirring with the ice bath for 10 minutes. In a beaker, some (greater than 300 µL) 0.52 mmol solution of sodium nitrite was poured and this beaker was also placed in the ice bath to cool. A thermometer was placed in the ice bath, which had the two solutions, in order to monitor the temperature. When both solutions were at approximately 5 C, ̊ 300 µL of the nitrite solution was taken using a Pasteur pipette and it was slowly added to the solution in the anthranilic acid solution in the vial that was being stirred. While doing this step, it was made sure that the temperature remained at approximately 5 C. ̊ After 5 min, the solution of anthranyldiazonium chloride was texted for the presence of excess nitrous acid with the placement of a drop of the solution on a potassium iodide-starch paper. An immediate blue color was not seen, therefore more nitrite solution was added to the vial. The solution was left to stir for a couple of minutes and then this process was repeated and an immediate blue color was observed. Next, the air condenser was removed from the vial, 89 µL of N,N-Dimethylaniline was added quickly to the vial, and then condenser was reattached. The solution was left to stir for 15 minutes, while making sure the temperature was still around 5 C. ̊ 200 µL of 0.83 mmol solution of sodium acetate was then transferred to the vial with the reaction mixture using the Pasteur pipette (without removing the air condenser). The solution was left to stir for an additional 20 minutes, while making sure the temperature was still around 5 ̊ C. The vial was then removed from the ice bath and it was left out (on top of the hotplate?) for 15 min so that the solution. returns to room temperature. Next, 100 µL of 10% aqueous NaOH solution was added to the reaction mixture in the vial using a pipette and then it was left to stand at room temperature for another 30 minutes. The crude Methyl Red was collected through vacuum filtration using a Buchner Funnel. The reaction flask was rinsed with 0.5 mL of water, and is rinse was then used to wash the crystals as well. In addition to this, the crystals were washed with 0.5 mL of 3 M acetic acid, and then again with 0.5 mL of water. The weight and melting point of the crude product were then taken. Following this step, the crude product was dissolve in 2.0-2.5 mL of methanol in a flask (this was placed in a beaker with hot water for faster dissolution). The solution was then cooled in an ice bath, and the pure methyl red crystals were collected with again performing vacuum filtration using a Buchner funnel. The weight, melting point and IR of the product was taken. (This entire experiment was performed in the fume hood.)
Chemistry 222
Fall 2012
5. Results: 5 marks a. Table 1: Yield and Physical Properties Crude Azo Dyes
4
Purified Azo Dyes
Yield (g)
0.04 g
0.01
Yield (%)
29 % Dark Red Solution with some black dots
7% Dark Red Solution (not many black spots) - (not enough pure product to get the melting point)
Appearance Melting Point Recovery (%)
169-174
25 %
6. Discussion: 8 marks (Since almost no pure product was collected, the crude product will be used for most of the analysis purposes) a. Discussion of Yield – crude, purified, % recovery – why less than 100%? The experiment yield of the crude product was only 29% and the experimental yield of the pure product was only 7% due to various experiment errors. The highly inaccurate transfer of liquid using the Pasteur pipette certainly contribute a lot to having a low yield. Another aspect of the experiment, that could have played a role in the yield of the the product is the temperature. It is very important to maintain that low temperature of less than 5 ̊ C throughout the experiment, since the Anthranyl diazonium chloride salt solution is unstable and prone to deterioration at room temperature (E College of Chemistry & Chemical Engineering, 2011). Exposing the Anthranyl diazonium chloride salt solution to a higher temperature for an extended period of time while transferring a solution into the vial could have affected the yield of the experiment. b. Determination Identity of both reactants and product i. Analysis of Melting Point The melting point is unique to every substance. The melting point of the product was measured to be 169.0-174.0 ̊ C and this is very close to the actual melting point of the product of 179-182 C, ̊ considering the possible human error in this experiment. It is important to note that the melting point taken was of the crude product (since there was not enough pure product to take the melting point) which means
Chemistry 222
Fall 2012
5
that the presence small amounts of other substance could have affected the melting point. Nevertheless, since the experimental melting point is very close to the actual melting, it can be said that there is very little "other" substances and for the most part it is methyl red. The melting point values being similar supports the idea that the correct product was formed and that the product is very close to being pure. c.
Mechanism – detailed mechanism for the synthesis of your azo dye
d.
Discussion Questions – i. How does the pH affect the reactivity of the N,N-dimethylaniline? By increasing the pH of the solution, the rate of the azo compound formation (normally a slow reaction) is increased. The higher the pH the lower the concentration of dimethylanilineH+ and that is why 10% aqueous NaOH was added to the vial near the end of the experiment.
Chemistry 222
Fall 2012
6
The lone pair on the N,N dimethylaniline is what that makes it possible to stabilize the positive charge on the nitrogen of anthronyl diazonium chloride, a substance that is in its transition state. It is the electron releasing resonance effect of the N,N dimethylaniline that makes it possible for the stabilization to happen. If the pH of the solution was lower (solution being more acidic), the nitrogen with the lone pair (of the N,N dimethylaniline) would be protonated, making the substance unreactive. Moreover, this would take away N,N dimethylaniline's ability to help stabilize the positive charge on Nitrogen of the anthronyl diazonium chloride (it does not allow for the electrophilic attack from the benzene ring of the demethylaniline). Making the solution more acidic, would certainly increase the concentration dimethylanilineH+ (leaving fewer proton free-dimethylaniline) and as a result the overall production of the Methyl Red would be reduced. ii. Why are azo dyes coloured compounds? Azo dyes absorb different wavelength of light within the visible rang of 400-800nm. The colors that are observed are the wavelengths of light that were not absorbed by the dye ("the leftover wavelengths of light") (Clark, 2004). For example, if it is red light (wavelength: 647700 nm) that is absorbed by the dye the complementary color of green would be seen. The presence of the delocalized electron system made up of alternating double bonds, also known as chromophores, in the dyes, results in the absorption of the energy from the light (Clark, 2004). It is also important to keep in mind that energy is inversely proportional to the wavelength of light: E = hc/λ. With higher amount of double bonds present in the dye, less energy is needed to get the electrons excited and longer wavelengths of light would be absorbed. On the other hand, with less double bonds present, more energy would be needed to get the electrons excited and shorter wavelengths of light would be absorbed. iii. Explain the regioselectivity in the diazonium coupling reaction (para only) The regioselectivity in the diazonium coupling reaction is at the para position only because the diazonium cations are poor electrophiles that are relatively large (Biele, 2010). Since the electron density is at its highest at both the ortho and para positions of the demethylaniline, it may first appear that the regioselectivity in the diazonium coupling reaction can be at both of those positions, but that is not the case. It
Chemistry 222
Fall 2012
7
would not easily bind at the ortho position, due to the strong steric hindrance as there would be a clash between the large phenyl groups and the alkyl groups. With the phenyl groups "needing more space", the para position will be that ideal position for the binding to take place. Furthermore, with the para substitution, the positive charge's stabilization is at its maximum. It is only when the para position is taken, the regioselectivity in the diazonium coupling reaction shifts to the ortho position. iv. What reagents would you need for the synthesis of Methyl Orange (another azo dye)? Give an overall reaction equation (with structures). For the synthesis of Methyl Orange, sulfanilic acid, sodium nitrite and dimethylaniline would be needed (Senese, 2010). Other substances such as sodium carbonate, HCL solution, NaOH (or Sodium Acetate) will help with the reaction.
References Biele, Carsten and Kamp, Oliver. " Diazonium Coupling (Diazo Coupling, Azo Coupling)". CHEMGAROO ChemgaPedia. 2010. Web. 13 Nov. 2012. Clark, Jim. "REACTIONS OF DIAZONIUM SALTS". Chemguide. 2004. Web. 12 Nov. 2012. "Experiment 1: Synthesis of an Azo Dye: 1-(4-hydroxyphenylazo)-2naphthol". E College of Chemistry & Chemical Engineering. 2011. Web. 13 Nov. 2012.
Chemistry 222
Fall 2012
Mayo, Dana W., Ronald M. Pike, and David C. Forbes. "Chapter 6: EXPERIMENT 26 Diazonium Coupling Reaction: Methyl Red." Microscale Organic Laboratory: With Multistep and Multiscale Syntheses. Hoboken, NJ: J. Wiley & Sons, 2011. N. pag. Print. Senese, Fred. " What is methyl orange? How is it made?" General Chemistry Online! 15 Feb. 2010. Web. 12 Nov. 2012.
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