"Extraction of Nicotine from Tobacco Leaves"

 

"Extraction of Nicotine from Tobacco Leaves"

Theory:

Nicotine is an alkaloid primarily found in tobacco (Solanaceae Family), as well as in smaller quantities in tomato plants and green peppers. It is also present in the Coca plant. Nicotine was first isolated from tobacco plants in 1828 by German chemists Posselt and Reimann. It is a potent neurotoxin and is used in various insecticides.

Apparatus Required:

• 250 ml beaker
• Separatory funnel
• Measuring cylinder
• Hot plate
• Magnetic stirrer
• Weighing balance
• Iron stand
• Pipette

Chemicals Required:

• 5% NaOH solution
• Diethyl ether
• Tobacco leaves
• Distilled water

Structure:

Procedure:

1. Clean and dry all the apparatus carefully.
2. Take 10g of tobacco leaves, wash them thoroughly, and ensure they are completely dried.
3. Grind the leaves and add them to a solution of 100g 5% NaOH.
4. Stir the mixture thoroughly and filter the solution.
5. Dilute the filtrate with 30ml of distilled water to remove impurities.
6. Heat the solution on a hot plate until it boils and a strong smell of nicotine is obtained.
7. Transfer the solution into a separatory funnel and extract it by adding 25ml of diethyl ether. Repeat the extraction process three times.
8. Evaporate the ether using a hot plate placed in an ice bath.
9. Separate the upper brown layer of nicotine oil.

Result: The yield of nicotine oil obtained was 3.8 ml.

Calculations:

Initial amount of tobacco leaves = 10 grams

Concentration of nicotine in the extracted oil = 5%

Step 1:

Calculate the amount of nicotine in the extracted oil: Amount of nicotine = Volume of nicotine oil * Concentration of nicotine Amount of nicotine = 3.8 ml * 5% (0.05) Amount of nicotine = 0.19 grams

Step 2:

Calculate the percentage yield:

Percentage yield = (Amount of nicotine / Initial amount of tobacco leaves) * 100

Percentage yield = (0.19 g / 10 g) * 100

Percentage yield = 1.9%

Therefore, with the new assumptions, the percentage yield of nicotine oil would also be approximately 1.9%.

Observations:

Molecular Formula: C₁₀H₁₄N₂

Molar Mass: 162.23 g/mol

Color: Nicotine is a colorless to pale yellow liquid in its pure form.

Melting Point: Nicotine has a melting point of -79°C (-110°F).

Boiling Point: Nicotine has a boiling point of 247°C (477°F) at normal atmospheric pressure.

Precautions:

• Always wear gloves while conducting the experiment.
• Use a mask to protect against the pungent smell of nicotine oil.  
• Ensure thorough cleaning and drying of apparatus before use.
• Handle nicotine with caution due to its neurotoxic properties.
• Follow proper disposal protocols for chemical waste generated during the experiment.

“Lab Report For The Synthesis of Chalcone”

 

“Lab Report For The Synthesis of Chalcone”

Theory:

Chalcone is synthesized through the Claisen-Schmidt condensation reaction between an aromatic aldehyde and an acetophenone or ketone. The reaction is catalyzed by a base, usually a strong base like sodium hydroxide, and proceeds through the formation of an enolate intermediate. The enolate reacts with the aldehyde to form the chalcone product.

Chemicals Used:

1. Sodium hydroxide (NaOH) - 5.5 grams
2. Water - 50 ml
3. Rectified spirit - 30 ml
4. Ice - 13 grams
5. Acetophenone - 14 ml
6. Benzaldehyde - 11 ml

Apparatus Used:

• Magnetic stirrer
• Glass beaker 
• Stir bar
• Weighing scale
• Filtration setup
• Drying apparatus

Chemical Equation:

Mechanism:

Procedure:

1. Dissolve 5.5 grams of sodium hydroxide (NaOH) in 50 ml of water to prepare a sodium hydroxide solution.
2. Add 30 ml of rectified spirit (ethanol) to the sodium hydroxide solution.
3. Add 13 grams of ice to the solution and stir until it is dissolved.
4. Add 14 ml of acetophenone to the mixture and shake well.
5. Add 11 ml of benzaldehyde to the solution and observe the formation of a milky white mixture.
6. Maintain the temperature below 20°C, preferably by placing the reaction vessel in an ice bath or a cold-water bath.
7. Place the reaction vessel on a magnetic stirrer and stir the solution for 2 hours with the vessel covered.
8. After 2 hours, the solution should become thick and change color from white to yellow.
9. The yellow precipitate formed is chalcone.
10. Filter the chalcone precipitate using a filtration setup (e.g., Buchner funnel) to separate it from the liquid portion.
11. Wash the chalcone precipitate with a small amount of cold solvent (e.g., rectified spirit) to remove impurities.
12. Dry the chalcone product using a suitable drying apparatus (e.g., desiccator) to remove any residual solvent.

“Synthesis of dibenzalacetone (DBA)”

 

“Synthesis of dibenzalacetone (DBA)”

Theory:

Dibenzalacetone can be synthesized via a crossed aldol condensation reaction between benzaldehyde and acetone. The reaction involves the nucleophilic addition of the α-carbon of the carbonyl group of acetone to the carbonyl carbon of benzaldehyde, followed by dehydration to form dibenzalacetone.

Materials Required:

• Weigh Balance
• Ice bath
• Beaker
• Stirrer
• Filter paper

Chemicals Required:

1. Benzaldehyde (C₇H₆O) (20 mL)
2. Acetone (C₃H₆O) (10 mL)
3. Sodium hydroxide (NaOH) (catalytic amount)
4. Hydrochloric acid (HCl) (few drops)
5. Water (for washing)

Chemical Equation:

Mechanism:

Procedure:

1. Set up an ice bath by placing a large beaker or bowl filled with ice and water.
2. In a separate reaction flask, add 20 mL of benzaldehyde and 10 mL of acetone.
3. Place the reaction flask in the ice bath to maintain a low temperature.
4. Slowly add a catalytic amount of a base, such as sodium hydroxide (NaOH), to the reaction mixture while stirring continuously.
5. Continue stirring the reaction mixture in the ice bath for about 30 minutes.
6. After the reaction time has elapsed, remove the flask from the ice bath and add a few drops of concentrated hydrochloric acid (HCl) to acidify the mixture.
7. Allow the reaction mixture to settle, and collect the yellow precipitate formed, which is dibenzalacetone.
8. Wash the collected dibenzalacetone with water to remove any impurities, and then dry it. 

Calculations:

To calculate the theoretical yield and percentage yield, we need the actual yield value from the experiment. Let's assume the actual yield of dibenzalacetone obtained is 8.5 grams.

Theoretical Yield:

In the synthesis of dibenzalacetone, the molar ratio between benzaldehyde and dibenzalacetone is 1:1.

To calculate the moles, we can use the formula:

Moles = Mass / Molar mass

Moles of Benzaldehyde= Molar mass of benzaldehyde (C₇H₆O) = 106.12 g/mol

Moles of benzaldehyde = 20 g / 106.12 g/mol

Calculating the above expression gives:

Moles of benzaldehyde = 0.188 moles

Moles of Benzaldehyde: Since the molar ratio between benzaldehyde and dibenzalacetone is 1:1, the moles of benzaldehyde will be the same as the moles of dibenzalacetone.

Moles of dibenzalacetone = 0.188 moles

The molar mass of dibenzalacetone (C₁₇H₁₄O) = 234.29 g/mol

The molar mass of benzaldehyde (C₇H₆O) = 106.12 g/mol

Since the molar ratio is 1:1, the theoretical yield can be calculated as follows:

Theoretical yield = (Actual yield) × (Molar mass of dibenzalacetone) / (Molar mass of benzaldehyde)

Theoretical yield = (8.5 g) × (234.29 g/mol) / (106.12 g/mol)

Calculating the above expression gives:

Theoretical yield = 18.73 g

Percentage Yield:

Percentage yield = (Actual yield / Theoretical yield) × 100

Plugging in the values:

Percentage yield = (8.5 g / 18.73 g) × 100

Calculating the above expression gives:

Percentage yield = 45.4%

Therefore, if the actual yield of dibenzalacetone obtained in the experiment is 8.5 grams, the theoretical yield would be approximately 18.73 grams, and the percentage yield would be approximately 45.4%.

"Preparation of iron oxide nanoparticles (NPs)"

 

Preparation of iron oxide nanoparticles (NPs)

Theory:

The co-precipitation method involves the precipitation of iron hydroxide from iron salts in the presence of a base. In this case, the FeCl₃ and FeSO₄ salts are used to provide Fe⁺² ions for the reaction. By gradually adding NH₄OH as the base, the pH of the solution increases, leading to the formation of iron hydroxide. The iron hydroxide precipitate can then be further processed to obtain iron oxide nanoparticles.

Chemicals Required:

• Iron (III) chloride (FeCl₃) solution, 0.2 M
• Iron (II) sulfate (FeSO₄) solution, 0.1 M
• Ammonium hydroxide (NH₄OH) solution
• Solvent: Water or appropriate organic solvent

Materials Required:

1. Reaction vessel or flask
2. Magnetic stirrer or mechanical stirrer
3. pH meter
4. Centrifuge
5. Drying equipment (e.g., vacuum oven or desiccator)

Chemical Equation:

Procedure:

• Prepare a reaction vessel and add the FeCl₃ solution (50 ml, 0.2 M) and the FeSO₄ solution (50 ml, 0.1 M) to it.
• Start stirring the mixture using a magnetic or mechanical stirrer for 15 minutes to ensure proper mixing.
• Gradually add NH₄OH dropwise to the reaction mixture while monitoring the pH using a pH meter.
• Continue stirring the mixture for an additional 20 minutes or until the pH reaches 11. The pH adjustment to 11 promotes the formation of iron hydroxide precipitates.
• Once the reaction time is complete, stop the stirring and allow the precipitate to settle.
• Separate the precipitate from the solution using a centrifuge. Discard the supernatant.
• Wash the obtained precipitate several times with water or a suitable solvent to remove any impurities or unreacted reagents.
• Dry the washed precipitate using a vacuum oven or desiccator to remove any residual moisture.
• The resulting dried material will likely consist of iron oxide nanoparticles, which can be characterized using various techniques such as X-ray diffraction (XRD), transmission electron microscopy (TEM), or Fourier-transform infrared spectroscopy (FTIR).

Alternating method for synthesis of Iron Oxide Nanoparticles by using ammonia:

Theory:

Iron oxide nanoparticles are commonly synthesized through a precipitation method using iron chloride and iron sulfate solutions. The addition of ammonia helps in the formation of iron hydroxide, which is subsequently transformed into iron oxide nanoparticles through heat treatment or aging.

Formation of iron hydroxide:

FeCl₂ + 2NaOH → Fe(OH)₂ + 2NaCl

FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄

Conversion of iron hydroxide to iron oxide:

2Fe(OH) ₂ → Fe₂O₃ + 2H₂O

Overall reaction:

2FeCl₂ + 4NaOH → Fe₂O₃ + 4NaCl + 2H₂O

2FeSO₄ + 4NaOH → Fe₂O₃ + 2Na₂SO₄  + 2H₂O

Procedure:

1. Prepare a 0.2 M solution of iron chloride (FeCl₂) by dissolving the appropriate amount of iron chloride in 50 ml of distilled water.
2. Prepare a 0.1 M solution of iron sulfate (FeSO₄) by dissolving the appropriate amount of iron sulfate in 50 ml of distilled water.
3. Mix both solutions (0.2 M FeCl₂ and 0.1 M FeSO₄) in a glass beaker.
4. Stir the mixture for approximately 15 minutes to ensure thorough mixing. Apparatus required: Magnetic stirrer, stir bar.
5. Add 40 ml of ammonia (NH₃) to the mixture while continuing to stir.
6. Continue stirring the mixture until the formation of iron oxide nanoparticles.

Note:

The stirring time required for nanoparticle formation may vary depending on the desired size and properties of the nanoparticles. Further heat treatment or aging may be necessary to complete the transformation.

Calculations:

Molar mass of FeCl₃ = 55.845 g/mol + 106.359 g/mol = 162.204 g/mol

To prepare a 0.2 M solution of FeCl₃ in 50 ml water:

Molarity (M) = moles of solute / volume of solution (L)

0.2 M = moles of FeCl₃ / 0.05 L (50 ml = 0.05 L)

Moles of FeCl₃ = 0.2 M x 0.05 L = 0.01 moles

Mass of FeCl₃ = Moles of FeCl₃ x Molar mass of FeCl₃

Mass of FeCl₃ = 0.01 moles x 162.204 g/mol

Therefore, the mass of FeCl₃ required is 1.62204 grams.

Similarly, for FeSO₄:

Molar mass of FeSO₄ = 55.845 g/mol + 32.06 g/mol + 63.996 g/mol = 151.901 g/mol

To prepare a 0.1 M solution of FeSO₄ in 50 ml water:

Molarity (M) = moles of solute / volume of solution (L)

0.1 M = moles of FeSO₄ / 0.05 L (50 ml = 0.05 L)

Moles of FeSO₄ = 0.1 M x 0.05 L = 0.005 moles

Safety Precautions:

• Wear appropriate personal protective equipment, such as gloves and safety goggles, when handling chemicals.
• Work in a well-ventilated area or under a fume hood to avoid inhalation of fumes.
• Handle ammonia with care, as it is a corrosive substance. Avoid direct contact with skin and eyes.
• Follow proper waste disposal protocols for chemicals used in the synthesis.

"Preparation of p-aminoazobenzene (4-phenylazoaniline) from Aniline"

 

Preparation of p-aminoazobenzene (4-phenylazoaniline) from Aniline

The preparation of p-aminoazobenzene (4-phenylazoaniline) involves two main steps: the synthesis of diazoamino benzene (4-diazoanilinobenzene) and its subsequent conversion to p-aminoazobenzene. Here's a step-by-step outline of the process

Step 1:

Synthesis of Diazoamino Benzene (4-diazoanilinobenzene)

Theory: The synthesis of diazoamino benzene involves the conversion of aniline to its diazonium salt, followed by the reaction of the diazonium salt with an excess of aniline to form the diazoamino compound.

Chemicals Required:

• Aniline (C₆H₅NH₂)
• Concentrated hydrochloric acid (HCl)
• Sodium nitrite (NaNO₂)
• Ice-cold water

Materials Required:

1. Reaction flask
2. Ice bath
3. Stirring apparatus
4. Filtration setup
5. Drying oven

Chemical Reaction:

Mechanism:

Procedure:

• Start by setting up an ice bath and placing a reaction flask in it to keep it cold throughout the process.
• In the reaction flask, add 10 mL of concentrated hydrochloric acid (HCl) and cool the solution in the ice bath.
• In a separate container, dissolve 6.8 g of sodium nitrite (NaNO₂) in 30 mL of water to prepare a sodium nitrite solution.
• Slowly add the sodium nitrite solution dropwise to the cold hydrochloric acid solution in the reaction flask, while stirring continuously. Maintain the temperature below 5°C.
• In another container, prepare a solution of 7.2 g of aniline in 30 mL of water.
• Slowly add the aniline solution to the reaction flask containing the diazonium salt solution, while stirring vigorously.
• Continue stirring the reaction mixture for about 30 minutes to ensure complete reaction.
• Pour the reaction mixture into a large volume of ice-cold water to precipitate the diazoamino benzene.
• Collect the precipitate by filtration and wash it with water to remove impurities.
• Dry the precipitate in a drying oven at a suitable temperature.
• Weigh the dried diazoamino benzene product and record its mass for further calculations.

Calculations:

Aniline (C₆H₅NH₂) = 7.2 g

Sodium nitrite (NaNO₂) = 6.8 g

First, we need to determine the limiting reagent by comparing the moles of aniline and sodium nitrite used:

Molar mass of aniline (C₆H₅NH₂) = 93.13 g/mol

Molar mass of sodium nitrite (NaNO₂) = 69.01 g/mol

Number of moles of aniline: mass / molar mass

Moles = 7.2 g / 93.13 g/mol = 0.0774 mol

Number of moles of sodium nitrite = 6.8 g / 69.01 g/mol = 0.0984 mol

The stoichiometry of the reaction between aniline and sodium nitrite is 1:1. Since the number of moles of aniline is lower than the number of moles of sodium nitrite, aniline is the limiting reagent.

To calculate the theoretical yield of diazoamino benzene, we'll use the moles of aniline and the molar mass of diazoamino benzene:

Molar mass of diazoamino benzene = 181.2 g/mol

Theoretical yield = Moles of limiting reagent × Molar mass of diazoamino benzene

Theoretical yield = 0.0774 mol × 181.2 g/mol

Theoretical yield = 14.04 g

Therefore, the theoretical yield of diazoamino benzene, based on the given quantities, is approximately 14.04 grams.

 

Step 2: Conversion of Diazoamino Benzene to p-Aminoazobenzene

Theory:

The conversion of diazoamino benzene to p-aminoazobenzene involves the reaction of diazoamino benzene with aniline under specific conditions to form the desired product.

Chemicals Required:

1. Diazoamino benzene (from Step 1)
2. Aniline (C₆H₅NH₂)
3. Sodium hydroxide (NaOH)
4. Ethanol (C₂H₅OH)

Materials Required:

1. Reaction flask
2. Ice bath
3. Stirring apparatus
4. Filtration setup
5. Drying oven

Chemical Reaction:

Mechanism:

Procedure:

• Start by setting up an ice bath and placing a reaction flask in it to keep it cold throughout the process.
• In the reaction flask, add the diazoamino benzene obtained from Step 1.
• In a separate container, prepare a solution of 4 g of aniline in 30 mL of ethanol.
• Slowly add the aniline solution to the reaction flask containing diazoamino benzene, while stirring continuously.
• Add a solution of 5 g of sodium hydroxide (NaOH) in 30 mL of water dropwise to the reaction mixture while maintaining the temperature below 10°C.
• Continue stirring the reaction mixture for about 30 minutes.
• Pour the reaction mixture into ice-cold water to precipitate the p-aminoazobenzene.
• Collect the precipitate by filtration and wash it with water to remove impurities.
• Dry the precipitate in a drying oven at a suitable temperature.
• Weigh the dried p-aminoazobenzene product and record its mass for further calculations

Calculations:

Theoretical yield of diazoamino benzene = 14.04 g

The stoichiometry of the reaction between diazoamino benzene and aniline to form p-aminoazobenzene is 1:1. Therefore, the moles of p-aminoazobenzene formed will be equal to the moles of diazoamino benzene used.

Molar mass of p-aminoazobenzene = 197.23 g/mol

Theoretical yield of p-aminoazobenzene = Moles of diazoamino benzene × Molar mass of p-aminoazobenzene

Moles of diazoamino benzene = Theoretical yield of diazoamino benzene / Molar mass of diazoamino benzene Moles of diazoamino benzene

Moles of diazoamino benzene = 14.04 g / 181.2 g/mol = 0.0774 mol

Theoretical yield of p-aminoazobenzene = 0.0774 mol × 197.23 g/mol

Theoretical yield of p-aminoazobenzene = 15.26 g

Therefore, the theoretical yield of p-aminoazobenzene, based on the given quantities, is approximately 15.26 grams.

"Preparation of Acetanilide"

 

"Preparation of Acetanilide"

Theory:

Acetanilide is an organic compound that can be synthesized by acetylation of aniline. Acetylation refers to the process of introducing an acetyl group (-COCH3) to a compound. In this case, acetic anhydride will be used as the acetylating agent to react with aniline, resulting in the formation of acetanilide.

Materials Required:

• Acetic anhydride (C₄H₆O₃): Acetylating agent used to introduce the acetyl group.
• Aniline (C₆H₇N): Amine compound that undergoes acetylation to form acetanilide.
• Zinc dust (Zn): Catalyst used to promote the reaction.
• Glacial acetic acid (CH₃COOH): Solvent and reaction medium.
• Distilled water (H₂O): Used for washing and purification steps.

Chemicals Required:

1. Acetic anhydride
2. Aniline
3. Zinc dust
4. Glacial acetic acid
5. Distilled water

Chemical Equation:

Mechanism:

Procedure:

• Set up a round-bottom flask equipped with a reflux condenser and a magnetic stirrer.
• Add 10 mL of aniline to the flask.
• Slowly add 20 mL of acetic anhydride to the flask while stirring.
• Add a small amount (a few grams) of zinc dust to the reaction mixture as a catalyst.
• Heat the mixture gently under reflux for about 30 minutes. The reaction is exothermic, so monitor the temperature and adjust the heat accordingly to maintain refluxing conditions.
• After 30 minutes, remove the heat source and allow the reaction mixture to cool to room temperature.
• Carefully add the reaction mixture to a beaker containing ice-cold water while stirring vigorously. This will cause the acetanilide to precipitate out of the solution.
• Filter the precipitated acetanilide using a Buchner funnel or a simple gravity filtration setup.
• Wash the collected solid with cold distilled water to remove impurities. Repeat this washing process a few times.
• Finally, dry the acetanilide by spreading it out on a watch glass or filter paper in a well-ventilated area until it is completely dry.
• Weigh and record the yield of acetanilide obtained.
  
  
  

Calculations:

Amount of Aniline = 10ml

Molar mass of Aniline = 93 g/mol

Density of Aniline = 1.022 g/mol

Mass of Aniline = 10 x 1.02 = 9.784g

Moles of Aniline =  m/M

                    9.784÷93g/mol = 0.1052 moles

Moles of Aniline = 0.1052 moles

Amount of Acetic anhydride = 20ml

Molar mass of Acetic anhydride = 102 g/mol

Density of Acetic anhydride = 1.08 g/mol

Mass of Acetic anhydride = 20 x 1.08 = 18.51g

Moles of Acetic anhydride = m/M

                       18.51g÷102g/mol = 0.1815 moles

Moles of Acetic anhydride = 0.1815 moles

From chemical equation;

1eq of Aniline = 1eq of Acetic anhydride = 1eq of Acetanilide

so,

Molecular mass of Acetanilide = 135 g/moles

Moles of Acetanilide = 0.105 moles

Mass of Acetanilide = n×M

                                (0.105mol)(135 g/mol) = 7.77g

Theoretical yield = 7.77 g

Actual Yield = 7 g

Percentage Yield = (7 g / 7.77 g) × 100

 Percentage Yield = 90.01%

Precautions:

The reaction should be performed with caution as acetic anhydride is a corrosive and toxic compound. It is also important to wear appropriate safety equipment, such as gloves and goggles, and work in a well-ventilated area.