Title:

Synthesis of Benzocaine from P-Aminobenzoic acid

Apparatus:

Ø  Quick fit round bottom flask,

Ø  Condenser,

Ø  Water Bath,

Ø  Beaker,

Ø  PH Meter

Chemicals:

Ø  Amino Benzoic acid

Ø  Conc. H₂SO₄

Ø  Ethanol

Ø  Distilled water

Ø  10% Na₂CO₃ Soln.

Procedure:

First of all we will take round bottom flask and add in it 1.2g Para amino benzoic acid with 12ml ethanol and shake the flask so that all the solid dissolve in ethanol. Then we will add 1ml Conc. H₂SO₄ in it. After that we will attach the round bottom flask with condenser and reflux this reaction mixture at 80^oC up to one hour. In the next step we will separate the flask from condenser and transfer this material into the beaker and add 30ml distilled water in that beaker and allow this beaker to cool down. Then we will add 10% NaCO₃ in it to maintain the PH of this solution mixture upto8 or 8.5 after that we will wash the filtrate ppt with distilled water and in this last recrystallize it with ethanol.

Chemical Equation:

P-Aminobenzoic acid                                                Benzocaine

Mechanism:

Observation and calculations:   

Find Percentage Yield:

Molar mass of P-amino benzoic acid    = 137.14 g/mol

Molar mass of Ethanol is = 98.08

Mass of HNO₃ = d= m/v       d= 0.79

m=0.79 X 12 = 9.48g

No of moles of Ethanol used = mass/ molar mass

                                              = 9.48/98.08

                                              = 0.09665 g/mol     

No of moles of Paraamino benzoic acid used = mass/ molar mass

                                                =1.2/ 137.14

                                                = 0.008750 g/mol     

Because there is more No. of moles of ethanol than P-aminobenzoic acid, P-aminobenzoic acid is used up first as the reaction proceeds. We can conclude that ethanol is in excess and P-amino benzoic acid is the limiting reactant.

Theoretical yield:

Theoretical yield = No. of moles of P-aminobenzoic acid X molar mass of P-aminobenzoic acid

                         Theoretical yield = 0.0087 X 137

                          Theoretical yield = 1.1919g

                          Actual yield which we obtained was 0.79g

% Yield:

% Yield = Actual yield / Theoretical yield X 100

                                       % yield =  0.79g / 1.1919g X 100

                                       % yield =  66%

Results:

Mass of obtained benzil = 0.94g

The M.P of the product which we measured is ----^oC

The percentage Yield was = 66%

The Rf value is     2/3=0.67

Discussion:

The product which we prepared in laboratory was Benzocaine from P-Aminobenzoic acid                                          The percentage yield which we obtained of this product was 66% which is satisfactory since the yield may also be reduce due to climate change and it is also possible that there were different impurities present in a resulting product.

                                                                                                                            

Experiment no 1                                                26/08/2013

Preparation of Tetra amine Copper(II) Sulphate and [Cu(NH₃)₄]SO₄ also calculate the percentage yield_.

Apparatus:

                   Following apparatus was used in the preparation of the complex salts.

·        Test tubes

·        Test tube holder

·        Test tube stand

·        Copper Sulphate

·        Ammonia Solution

·        Ethyl Alcohol.

Reaction:

CuSO4.5H2O + 4NH3 → [Cu(NH₃)₄(H₂O)₅]SO₄

Procedure:

                  First of all 3 gm of CuSO4.5H2O was dissolved in distilled water in the 250 ml beaker to make a concentrated solution. Than ammonia solution was added in it drop vise with a constant stirring. Blue precipitates were formed and were dissolved in excess of ammonia. After that 25 ml of ethyl alcohol was added in this solution and that solution was allowed to stand for 10-15 minutes. Blue crystals of complex Tetra amine Copper (II) sulphate were formed and these were separated by filtration. These crystals were allowed to dry and percentage yield was calculated and other reactions were performed.

Observations and Calculations:

Theoretical yield:

249.5 gm of CuSO.5H2O produces Tetra  aminecopper (II) Sulphate=317.5gm

1 gm of CuSO.5H2O produces  Tetra amine  copper (II) Sulphate=317.5gm/249.5gm

3gm of CuSO.5H2O produces  Tetra amine  copper (II) Sulphate=317.5gm/249.5*3gm

= 3.8

Actual Yield:

Weight of Petri dish = 43.51gm

Weight of petridish + complex = 45.31 gm

Wight of complex = 45.31-43.51

= 1.80

Percentage yield:

 percentage yield = Actual yield / Theoretical yield * 100

% yield = 1.8/ 3.8 *100

% yield = 47%

water molecules in it:

Weight of crucible = M1= 16.18 gm

Wt.of Compound+ crucible = M2 = 16.81 gm

Wt. of hydrated complex = M3 = M2-M1= 16.81-16.18 = 0.63gm

Constant wt.of crucible + salt = M4 = 16.73 gn

Mass of removed water = M5 = M2-M4 = 16.81-16.73 =  0.08 gm

Mass of anhydrous salt = M4 – M1 = 16.73 – 16.18 =  0.55gm

0.55 gm of [Cu(NH3)4]SO4 have water = 0.08 gm

1 gm of [Cu(NH3)4]SO4 have water = 0.08/ 0.55 gm

2276.76 gm of [Cu(NH3)4]SO4 have water = 0.08/0.55*227.76=33.12gm

Moles of water = 33.12/18 =  1.84 m

 Test for Sulphate ion: 

Solution of [Cu(NH3)4] SO4.5H2O + pb(CH3COO)2 → 2CH3COO + white PPT of pbSO4

Test for CU:

Solution of [Cu(NH3)4] SO4.5H2O + BaCl2 → White PPT   

Results:

The percentage yield for that experiment was 47%.

Sulphate ions when react with the lead acetate they form white ppt..

Cu reacted with the BaCl2 and form the White ppt.

Experiment no 2

Preparation of Nickle dimethyl glyoxime complex and also calculate the percentage yield.

Apparatus:

·        Test tubes

·        Test tube stand

·        Test tube holder

·        250ml Beaker

·        Flask

·        Petri dish

·        Filter paper

·        Tripod stand

·        Chemicals

Chemicals:         

·        Nickle Sulphate

·        Ammonia Solution

·        Alcholic solution of dimethyl glyoxime

Reaction:

NiSO4.6H2O + 4NH4OH + C8H14O4N4  → (NH4)SO4 + 2H2O + red colour complex

  

Procedure:

First of all in the 250 ml beaker, 0.3 gm of NiSO4.6H2O was dissolved in the 100 ml of distilled water containing 5 ml 1:1 HCL solution. Than 10% ammonia solution was added in it dropwise until it gave smell of ammonia. After that 1% solution of dimethyl glyoxime was added drop wise until the completion of the precipitation. The solution was allowed to stand for 30-45 minute, than the solution was filtered and crystals were weighted after drying .Than the percentage yield was calculated.

Calculations and observation:

Theoratical yield:

262.7 gm of NiSO4.6H2O produces [Ni(C4H7O2N2)2] = 288.7 gm

1 gm of NiSO4.6H2O produces [Ni(C4H7O2N2)2] = 288.7/262.7 gm

0.3 gm of NiSO4.6H2O produces [Ni(C4H7O2N2)2] = 288.7/262.7*0.3 gm

= 0.329gm

Actual yield:

Wt.of filter paper = 1.26 gm

Wt.of filterpaper + complex = 1.54gm

Wt,.of complex = 1.54-1.26 = 0.28 gm

10 ml of NiSO4.6H2O produces [Ni(C4H7O2N2)2] = 0.28m

1 ml of NiSO4.6H2O produces [Ni(C4H7O2N2)2] = 0.28gm

100ml of NiSO4.6H2O produces [Ni(C4H7O2N2)2] = 0.28100 =  2.8gm

 % yield = 0.28/0.329*100 = 85 %