"Synthesis of Copper Oxide Nanoparticles by using precipitation method"

 

"Synthesis of Copper Oxide Nanoparticles by using precipitation method"

Theory:

Copper oxide nanoparticles can be synthesized by the precipitation method, where copper ions react with a precipitating agent (in this case, sodium hydroxide) to form copper oxide nanoparticles.

Materials Required:

1. Copper sulfate (CuSO₄)
2. Sodium hydroxide (NaOH)
3. Distilled water
4. Volumetric flask (100 mL)
5. pH paper
6. Hot plate/stirrer
7. Centrifuge
8. Oven

Chemicals Required:

• Copper sulfate (CuSO₄)
• Sodium hydroxide (NaOH)
• Distilled water

Chemical Reaction:

Procedure:

1. Weigh 1.596 g of copper sulfate (CuSO₄) and transfer it into a 100 mL volumetric flask.
2. Add a small amount of distilled water to dissolve the copper sulfate. Then, fill the flask up to the mark with distilled water.
3. Place the volumetric flask on a hot plate/stirrer and stir the solution.
4. Heat the solution to a temperature between 50 to 60°C, ensuring complete dissolution of copper sulfate.
5. Add 100 mL of sodium hydroxide (NaOH) solution dropwise into the copper sulfate solution while stirring continuously.
6. Check the pH of the solution using pH paper. Continue adding NaOH solution dropwise until the pH reaches 14, which indicates high alkalinity.
7. Observe the color of the solution turning dark blue after the addition of NaOH. This color change indicates the formation of copper oxide nanoparticles.
8. Maintain the stirring and temperature for two hours to allow for proper nanoparticle formation.
9. Allow the solution to settle overnight, allowing the copper oxide nanoparticles to precipitate.
10. Carefully discard the supernatant solution and perform centrifugation to separate the copper oxide nanoparticles.
11. Dry the separated nanoparticles in an oven at 60°C for two days.
12. Once dried, the copper oxide nanoparticles are ready for further characterization and use.

 

Calculations:

Mass of copper sulfate taken = 1.596 g

Volume of the volumetric flask = 100 mL

Molar mass of CuSO₄ = (63.55 + 32.07 + (4 x 16.00)) = 159.61 g/mol

Moles of CuSO₄ = (mass of CuSO₄) / (molar mass of CuSO₄)

Moles of CuSO₄ = 1.596 g / 159.61 g/mol = 0.01 mol

Calculate the volume of water required to make a 0.1 M solution

Moles = Molarity x Volume (in liters)

0.01 mol = 0.1 M x Volume (in liters)

Volume (in liters) = 0.01 mol / 0.1 M = 0.1 L

Convert the volume to milliliters

Volume (in milliliters) = 0.1 L x 1000 mL/L = 100 mL

Therefore, to prepare a 0.1 M CuSO₄ solution using 1.596 g of CuSO₄, dissolve the given mass of CuSO₄ in distilled water and dilute it to a final volume of 100 mL.

"Preparation of p bromo aniline from p bromo acetanilide"

 

"Preparation of  p bromo aniline from p bromo acetanilide"

Theory:

Para-bromoaniline is synthesized by the conversion of p-bromoacetanilide to p-bromoaniline through a hydrolysis reaction using sodium hydroxide (NaOH). The p-bromoacetanilide is reacted with NaOH, resulting in the replacement of the acetyl group with a hydroxyl group.

Chemicals Required:

1. 5 g p-Bromoacetanilide
2. Ethyl alcohol (as a solvent)
3. 5% Sodium hydroxide (NaOH) solution
4. 7 mL Concentrated hydrochloric acid (HCl)

Apparatus Required:

1. Round-bottom flask
2. Reflux condenser
3. Separatory funnel
4. Buchner funnel
5. Filter paper
6. Glass stirring rod
7. Glassware for measuring and transferring chemicals (graduated cylinders, pipettes, etc.)

Chemical Equation:

Mechanism:

Procedure:

• In a round-bottom flask, I added 5 g of p-bromoacetanilide.
• Then I fitted a reflux condenser to the flask.
• I added 7 mL of concentrated hydrochloric acid (HCl) to the flask.
• Heat the reaction mixture under reflux (boiling) for 1-2 hours. The reflux temperature should be around 70-80°C. During the reaction, the p-bromoacetanilide will react with HCl to form p-bromoacetanilide hydrochloride.
• After the reaction, Icooled the reaction mixture to room temperature.
• I transferred the reaction mixture to a separatory funnel and added water to the separatory funnel. I shaked the separatory funnel gently, allowing the layers to separate.
• I drained the lower aqueous layer (containing impurities) and discard it properly.
• I transferred the upper organic layer (containing p-bromoacetanilide) to a clean beaker.
• To the organic layer I added 5% sodium hydroxide (NaOH) solution dropwise until the pH is around 9-10. This will convert p-bromoacetanilide to p-bromoaniline.
• I stirred the mixture for some time to ensure complete conversion.
• I extracted the p-bromoaniline product by adding a suitable organic solvent (e.g., ethyl acetate or dichloromethane) to the beaker and shaked the mixture gently to ensure good extraction by using the above mentioned procedure of extarction
• I concentrated the organic solution using a rotary evaporator or by heating, once the volume was reduced, I cooled the concentrated solution in an ice bath. And collected the solid p-bromoaniline product by filtration using a Buchner funnel and filter paper.
• I rinsed the solid product with a cold solvent (e.g., cold water or ice-cold ethanol) to remove impurities.

Calculations:

Mass of p-bromoacetanilide = 5 g

Moles of p-bromoacetanilide = Mass / Molar mass

Moles of p-bromoacetanilide = 5 g / 214.1 g/mol

Moles of p-bromoacetanilide = 0.0234 mol

The stoichiometric ratio between p-bromoacetanilide and p-bromoaniline is 1:1. Therefore, the moles of p-bromoaniline formed will be the same as the moles of p-bromoacetanilide

Mass of p-bromoaniline = Moles of p-bromoaniline x Molar mass

Mass of p-bromoaniline = 0.0234 mol x 172.1 g/mol

Mass of p-bromoaniline = 4.026 g

The theoretical yield of p-bromoaniline is approximately 4.026 grams.

To calculate the percentage yield, we need the actual yield of p-bromoaniline.

Let's assume the actual yield of p-bromoaniline is 3.5 grams.

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

Percentage yield = (3.5 g / 4.026 g) x 100 Percentage yield = 86.8%

Therefore, the theoretical yield of p-bromoaniline is approximately 4.026 grams, and the percentage yield is approximately 86.8%.

"Preparation of Schiff’s Base from benzaldehyde"

 

Preparation of Schiff’s Base from benzaldehyde

Theory:

Schiff bases are a class of compounds that contain a carbon-nitrogen double bond with an imine functional group (R-N=CR'). They are typically synthesized by condensation reactions between a primary amine and an aldehyde or ketone. Schiff bases find applications in coordination chemistry, organic synthesis, metal ion detection, biological activities, dye chemistry, optoelectronic devices, corrosion inhibition, and chelation. They are versatile compounds with uses ranging from catalysis and drug development to sensing and materials science. In this case, we will be synthesizing a Schiff base by reacting aniline (primary amine) with benzaldehyde (aldehyde) in the presence of ethanol as a solvent.

Chemicals Required:

1. Aniline (C₆H₅NH₂)
2. Ethanol (C₂H₅OH)
3. Benzaldehyde (C₆H₅CHO)

Apparatus Required:

1. Round-bottom flask
2. Magnetic stirrer
3. Ice bath
4. Heating source (e.g., hot plate)
5. Filtration setup (filter funnel, filter paper)
6. Weighing balance
7. Glassware (beakers, measuring cylinders)

Chemical reaction:

Mechanism:

Procedure:

1. Measure 2 mL of aniline (C₆H₅NH₂) using a measuring cylinder and transfer it to a round-bottom flask.
2. Add 20 mL of ethanol (C₂H₅OH) to the round-bottom flask containing aniline. Stir the mixture until the aniline is completely dissolved.
3. In a separate container, measure 3.5 mL of benzaldehyde (C₆H₅CHO).
4. Place the round-bottom flask with the aniline-ethanol solution on a magnetic stirrer.
5. Add the benzaldehyde dropwise to the aniline-ethanol solution while stirring continuously. The reaction mixture should turn yellow as the Schiff base is formed.
6. After the addition is complete, continue stirring the reaction mixture at room temperature for 10 minutes to ensure complete reaction.
7. Remove the round-bottom flask from the heating source and allow it to cool to room temperature.
8. Set up a filtration setup with a filter funnel and filter paper. Transfer the reaction mixture to a separating funnel. Extract the Schiff base precipitate from it.
9. Dry the filtered Schiff base by placing it in an oven at a suitable temperature (e.g., 60°C) or by placing it in a desiccator until a constant weight is achieved.
10. Weigh the dried Schiff base to calculate the yield and determine its purity.

Observations:

Colour	pale yellow to dark brown
Crystal shape	can be diverse, including needle-like crystals, plate-like crystals, or irregular-shaped crystals
Melting point	2200C
Boiling Point	2480C

 

Calculations:

To calculate the theoretical yield, we need to determine the limiting reagent. The reactants are aniline and benzaldehyde. We will calculate the moles of each reactant and compare them to find the limiting reagent.

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

Molar mass of benzaldehyde (C₆H₅CHO): 106.12 g/mol

Moles of aniline = (volume in mL x density) / molar mass Moles of aniline = (2 mL x 1.021 g/mL) / 93.13 g/mol = 0.0219 mol

Moles of benzaldehyde = (volume in mL x density) / molar mass

Moles of benzaldehyde = (3.5 mL x 1.044 g/mL) / 106.12 g/mol = 0.0345 mol

Now, let's compare the moles of each reactant:

Moles of aniline= 0.0219 mol

Moles of benzaldehyde= 0.0345 mol

From the comparison, we can see that aniline is the limiting reagent because it has fewer moles than benzaldehyde. Therefore, we will use the moles of aniline to calculate the theoretical yield.

The reaction between aniline and benzaldehyde produces 1 mole of the Schiff base.

Molar mass of C₆H₅-CH=N-C₆H₅ = (Molar mass of C₆H₅) + (Molar mass of CH) + (Molar mass of N) + (Molar mass of C₆H₅)) = (12.01 g/mol x 6) + (1.01 g/mol) + (14.01 g/mol) + (12.01 g/mol x 6)

Molar mass of Schiff base = 152.19 g/mol

Theoretical yield = Moles of limiting reagent x molar mass of the Schiff base

Theoretical yield = 0.0219 mol x 152.19 g/mol

Theoretical yield = 3.34 g

Actual yield = 2.8 grams

Theoretical yield = 3.34 grams

Now we can calculate the percentage yield:

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

Percentage yield = (2.8 g / 3.34 g) x 100

Percentage yield = 83.83%

Therefore, the calculated percentage yield of the Schiff base would be approximately 83.83%.

"Synthesis of Ortho-chlorobenzoic acid from anthranilic acid"

 

Step III: Ortho-chlorobenzoic acid from anthranilic acid.

Chemicals Required:

1. Anthranilic acid
2. Sodium nitrate
3. Conc. HCl
4. CuSO₄ (Crystalline)
5. Sodium Chloride
6. Cu-turnings

Theory:

Diazo coupling:

The reaction in which aryl diazonium or diazocation, ArN⁺, may attack highly activated aromating ring. The electrophile is generated of primary reaction of aromatic amine and HNO₂ by the diago Coupling. A diazocation is unstable so it decomposes to lose N₂. In order to present this decomposition, diazotization is done at temperature (0-5⁰C). Primary or Secondary aromatic amines also react with diazocation yielding N-coupled Product which may be re-arranged to more Stable C-coupled product by heating, with mineral acid.

Sandmeyer Reaction:

The Substitution of amine group via preparation at its diazonium salt with nucleophile (Cl, I, CN, RS, OH) under the catalysis of Copper II.

Chemical reaction:

 

Mechanism:

 

Procedure:

1. I dissolved 20 g CuSO₄ and 8.5 g of NaCl in about 40 ml water taken in 500ml round bottom flask.
2. Then I boiled it and added 55ml of HCl and 10 g of copper turnings into it.
3. I heated the flask under reflux until mixture become Cooled.
4. Then I prepared anthranilic acid solution in HCl and water and cooled it to 0⁰C.
5. After that I added NaNO₂ in it slowly with continuous stirring.
6. Then I pouerd this solution to above prepared solution.
7. After that I allowed the solution to stand for about 4 hours.
8. Then, I filtered the crystals.

Calculations:

Mass of anthranilic acid = 3.2g

Molecular mass of athranilic acid = 137g/mol

Moles of anthranilic acid = m/M = 3.2/137 = 0.02335 moles

From chemical equation

1eq of anthranilic acid = 1eq of orthochlorobenzoic acid

0.02335 moles of anthranilic acid = 0.02335 moles of orthochlorobenzoic acid

So,

Moles of orthochlorobenzoic acid = 0.02335 moles

Molecular mass of orhtochlorobenzoic acid = 156.5g/mol

Mass of orthochlorobenzoic acid = n×M = 0.02335 × 156.5 =3.654g

Theoretical yield = 3.654g

Actual yield         = 2.7g

%age yield of o-chlorobenzoic acid = 2.7÷3.654×100 = 73.89%

 

 %age yield of O-chlorobenzoic acid= 73.89%

 

Melting point:

The melting point of this compound is 141⁰C.

"Preparation of p-Bromo acetanilide from Acetanilide"

 

"Preparation of p-Bromo acetanilide from Acetanilide"

Theory:

p-Bromo acetanilide is prepared by the bromination of acetanilide. Acetanilide is an amide compound that reacts with bromine to substitute one of the hydrogen atoms on the aromatic ring with a bromine atom. This reaction occurs under acidic conditions.

Chemicals required:

1. Acetanilide (C₈H₉NO)
2. Bromine (Br₂)
3. Glacial Acetic acid (CH₃COOH)
4. Ice

Material required:

1. Round-bottom flask
2. Filter funnel
3. Ice bath
4. Glass rod
5. Distilled water

Chemical reaction:

Mechanism:

The reaction mechanism involves the bromination of acetanilide in the presence of acetic acid as a solvent and sulfuric acid as a catalyst. The bromine molecule attacks the benzene ring, and one of the hydrogen atoms on the ring is replaced by a bromine atom.

Procedure:

1. I dissolved 4.5 grams of acetanilide in 15 mL of glacial acetic acid in a round-bottom flask.
2. Then, I placed the flask on ice bath.
3. On the other hand, I dissolved 3.5 ml of bromine in 15 mL of glacial acetic acid in another flask.
4. I poured the bromine solution into the round bottom flask containing solution of acetanilide with continuous stirring set in ice bath.  
5. Now, the reaction mixture was stand for about 10 minutes to maintain the temperature of reaction for about 20-25^oC.
6. After that, the mixture was poured over the beaker containing crushed ice.
7. The white shiny crystals of p-bromo acetanilide was obtained.
8. The filter funnel was used to separate the crystals from water. Then, I dried them.
9. I weighed the dried product to determine the yield.
10. I measured the melting point of the p-bromoacetanilide to confirm its purity.
    
    
    

Calculations:

Molecular weight of acetanilide (C₈H₉NO) = 135.17 g/mol

Molecular weight of p-bromoacetanilide (C₈H₈BrNO) = 214.08 g/mol

Now, let's calculate the theoretical yield and percentage yield;

Theoretical yield:

The molar ratio between acetanilide and p-bromoacetanilide is 1:1, which means

1 mole of acetanilide = 1 mole of p-bromoacetanilide.

Moles of acetanilide used = mass of acetanilide / molecular weight of acetanilide = 4.5 g / 135.17 g/mol = 0.0333 mol

Therefore, the theoretical yield of p-bromoacetanilide is also 0.0333 mol.

Mass of p-bromoacetanilide = moles of p-bromoacetanilide x molecular weight of p-bromoacetanilide = 0.0333 mol x 214.08 g/mol = 7.13 g

Actual yield:

The mass of the obtained p-bromoacetanilide crystals.

Assuming you obtained 6.5 g of p-bromoacetanilide crystals as the actual yield.

Percentage yield = (Actual yield / Theoretical yield) x 100 = (6.5 g / 7.13 g) x 100 = 91.2%

The percentage yield in this case is 91.2%.

Melting Point:

The melting point of p-bromoacetanilide is approximately 164-166°C.

"Synthesis of Anthranilic Acid from Pthalamide"

 

Step II

"Synthesis of Anthranilic Acid  from Pthalamide"

Chemicals Required:

1. Phthalamide       
2. Bromine
3. Sodium hydroxide
4. Conc. HCI
5. Glacial acetic acid

Theory:

The most Common migration rearrangement reactions involve migration of atom or group of atoms from one atom or group of atoms i.e. 1-2 Shift.

A group may migrate to an e deficient carbon, nitrogen, oxygen, to e^- rich Carbon.

The Species which migrate may be an atom or a group a sigma bond.

Classification of Rearrangement:

Depending on electronic nature of migrating unit.

A rearrangement may be classified as follows,

(a) Nucleophillic or Aniontropic Rearrangment:

Aniontropic Rearrangment is where the migrating group migrate along with the electrons pair to electron deficient center.

 

(b) Electrophillic or cationtropic Rearrangement:

Cationtropic Rearrangement is where migrating group migrate without electron pairs to rich migarting center.

 

(C) Free Radical rearrangement:

Free Radical rearrangement is where the migrating groups changes its Position just with one electron.

 

(d) Sigma tropic Rearrangement:

Sigma tropic rearrangement is where the Cleavage or breakage of Sigma bonds occur simultaneously through cyclic transitions.

 

(a) Nucleophillic Rearrangement to an é^- deficient Nitrogen:

The electron deficient nitrogen is first generated by loss of alpha leaving group from nitrogen atom.

 

Hoffmann Re-arrangement:

There is a group related closely rearrangement, which converts N-Substituted amides to anthranilic acid. These migration involves migration of alkyl or aryl group from adjacent carbon to e-deficient Nitrogen.

Chemical reaction:

Mechanism:

Procedure:

I prepared sodium hypobromite by dissolving 7.5g of sodium hydroxide in about 40ml water in a conical flask. Then, I cooled the solution to 0^oC in ice bath and then I added 2.1 ml of bromine to it with stirring. Then I added 6g of phthalamide in cold solution with stirring and then poured the solution in 5.5g "NaOH" in 20ml water. Upon this addition flask becomes hot. Then, I warmed the reaction mixture in water bath at 80^oC for about 5 minutes. Then, I filtered it off. After that, I cooled the filterate in 500 ml flask containing crushed ice. Then, I added 15 ml HCL in filterate slowly with stirring untill the solution became neutral. Then, I added 5-6ml of glacial acetic acid acid. Finally, anthranilic acid was prepared.

Calculations:

Mass of pthalimide used = 6g

Molecular mass of pthalimide = 147g/mol

No. of moles of pthalimide = xm/M

                                                 6/147 = 0.04081 moles

Moles of pthalimide = 0.04081 moles

Amount of Br₂ used = V = 2.1ml

Density of Br₂ = 3.119 g/cm³

Density = mass/volume = m/v

d = m/v   =›   m = dV =› 2.1×3.11

                                   = 6.531g

Molecular mass of Br₂ = 70g/mol

No. of moles of Br₂ = m/M   = 6.531/70 = 0.0933 moles

Moles of Br₂ = 0.0933moles

Mass of NaOH used = 5.5g

Molecular mass of NaOH = 40g/mol

No. of moles of NaOH = m/M   =   5.5/50 = 0.1375 moles

Moles of NaOH = 0.1375 moles

1eq of pthalimide = 1eq of Br₂ = 1eq of NaOH = 1eq of anthranilic acid

0.04081moles of pthalimide=0.04081moles of Br₂=0.04081moles of NaOH= 0.04081 moles of anthranilic acid

So,

Moles of anthranilic acid = 0.04081moles

Molecular mass of antranilic acid = 137g/mol

Mass of anthranilic acid = n×M = 0.04081×137 = 5.59g

Theoretical yield of anthranilic acid = 5.59g

Actual yield of anthranilic acid = 3.2g

%age yield of anthranilic acid = 3.2/5.59×100 = 57.24%

Melting Point:

The melting point of anthranilic acid is 145^oC

Yield:

The Calculated amount of product is theoretical yield. The amount of product actually obtained is actual yield and when actual yield is divided by theoretical give obtained yield. It would give expected product yield.