ORGANIC cHEMISTRY - Ii

Experiment 2

Aim of the Experiment

Estimation of acetamide by hydrolysis method

Materials required:

acetamide solution , 0.15 N HCl solution , phenolphthalein 1.5 N NaOH solution

Procedure:

Part 1

• Take 10 ml of given acetamide solution in a borosil conical flask.

• Add 10 ml of 1.5 N sodium hydroxide solution and two test tubes of distilled water.

• Add 2 to 3 porcelain pieces.

• Keep a funnel on mouth of the flask to prevent evaporation and heat it on the sand bath for about 50 minutes till there is no smell of ammonia.

• Put off the burner and cool the solution to the room temperature for 5 minutes and then under tap water. Then transfer the contents quantitatively to a 100 ml of volumetric flask avoiding porcelain pieces. Make up the volume and shake well.

• Pipette out 10 ml of this stock solution and titrate against 0.15 N HCl using

• phenolphthalein as an indicator. Colour change will be from pink to colourless.

• Repeat the titration till atleast two constant readings are obtained
  

Part 2

• Pipette out 10 ml of 0.15 N sodium hydroxide solution in a 100 ml of volumetric flask and dilute it with distilled water upto the mark .

• Shake well to make the solution homogeneous .

• Pipette out 10ml of diluted solution into conical flask.

• Add 2-3 drops of phenolphthalein indicator and titrate against 0.15 N hydrochloric acid solution.

• The end point is marked by colour change from pink to colourless. Repeat the titration till atleast two constant readings are obtained.

Observation (Part 1 )

BURETTE- 0.15 N HCl solution

CONICAL FLASK - 10 ml diluted hydrolysed mixture

INDICATOR- phenolphthalein

END POINT – pink to colourless

Observation (Part 2 )

BURETTE – 0.15 N HCl solution

CONICAL FLASK – 10 ml of diluted sodium hydroxide solution

INDICATOR - phenolphthalein

END POINT –pink to colourless

CALCULATIONS

1 mole of CH3CONH2 = 1 mole of NaOH= 1 mole of HCl

59 g = 40 g = 36.5 g

1 mole of HCl = 1 mole of acetamide

1000 ml of 1 N HCl = 59g of acetamide

1 ml of 0.15 N HCl =

_______________ g of CH3CONH2 present in 10 ml of diluted solution

Amount of CH3CONH2 present in original solution =

RESULT

1. ___________ ml of 0.15 N HCl solution is required for the complete

neutralisation of 10 ml of diluted sodium hydroxide solution

2._____________ ml of 0.15 N HCl solution is required for the complete

neutralization of dilute sodium hydroxide solution left unreacted after hydrolysis

3.Sodium hydroxide solution required for hydrolysis of amide in terms of 0.15N

HCl solution =

4.The amount of acetamide in 10 ml diluted solution=

5.The amount of acetamide in original solution =

6. g/L of acetamide =

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