ORGANIC cHEMISTRY - iI

Experiment 3

Aim of the Experiment

Determination of amount of ester by hydrolysis method

Materials required:

1.5 N NaOH solution , 0.1 N HCl solution , phenolphthalein indicator , round bottom flask , water condenser etc…

Procedure

PART 1 BLANK TITRATION

• Take 10 ml of given 1.5 N NaOH solution in a 100 ml volumetric flask and dilute it upto the mark.

• Shake well to make the solution homogeneous .

• Pipette out 10 ml of diluted solution in a conical flask . Add 2-3 drops of phenolphthalein indicator and titrate against 0.1 N HCl solution.

• The end point is marked by colour change from pink to colourless .

• Repeat the titrations till atleast two constant readings are obtained

PART 2 HYDROLYSIS

• Take 10 ml of ethyl acetate in round bottom flask.

• Add 10 ml of 1.5 NaOH solution and then add 2-3 pieces of porcelain and fit it with water condenser .

• Reflux the mixture for about 90 minutes on the sand bath and then cool under tap water and transfer quantitatively to a 100 ml of the volumetric flask and dilute it upto the mark and make it homogeneous .

• Take 10 ml of mixture and titrate against 0.1 N HCl using phenolphthalein as an indicator.

• Repeat till atleast two constant readings are obtained .

Observation(Part 1)

PART 1

BURETTE - 0.1 N HCl

CONICAL FLASK -1.5 N NaOH + phenolphthalein

INDICATOR – phenolphthalein

END POINT – pink to colourless

RESULT

1.__________ ml of 0.1 N HCl is required for complete neutralisation of dilute sodium hydroxide solution

2._______ ml of 0.1 N HCl solution is required for neutralisation of diluted sodium hydroxide solution left unreacted after hydrolysis

3.Sodium hydroxide solution required for hydrolysis of ester in terms of 0.1 N HCl solution =

4.Amount of ethyl acetate in 10ml of diluted solution =

5.Amount of ethyl acetate in 100 ml of original solution =

6. g/L of ethyl acetate =

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