This is an experiment to determine relative rates of hydrolysis of the halogenoalkanes.

Aim

The task is to design an experiment that would be suitable for investigating whether the electonegativity of the halogen or the bond polarity and strength affects the rate of hydrolysis.  Therefore, which halogenoalkane reacts faster?

Equipment

• 6 test tubes
• Rubber bands
• Teat Pipettes
• Beaker
• Thermometer
• Measuring cylinder
• Stop clock
• 1 – Chlorobutane
• 1 – Bromobutane
• 1 – Iodobutane
• Ethanol
• Silver nitrate solution

The rate of reaction can be followed by carrying it out in the presence of silver ions.  The halogenoalkanes, being covalently bonded, give no precipitate of silver halide, but as the reaction proceeds and halide ions are produced, a precipitate of silver halide gradually appears. AgCl = white precipitate, AgBr = off-white precipitate, and AgI forms a pale yellow precipitate.  Ethanol will be used as a solvent. But doesn’t interfere with the reaction in any way.

Calculations on how much of each halogenoalkane should be used

1 – Chlorobutane

Molar mass of CH3CH2CH2CH2Cl = (4 x 12) + (9 x 1) +35.5 =92.5gmol-1

Density = 0.890gm -2

Volume = mass

Volume of one mole = 92.5

0.890 = 10.4cm 3

Volume of 0.05 moles = 4.625

0.890 = 5.2cm3

Volume of 0.025 moles = 2.3125

0.890 = 2.6cm3

I think it would be best to use 0.025 moles of each halogenoalkane

1 - Bromobutane

Molar mass of CH3CH2CH2CH2Br = (4 x 12) + (9 x 1) + 80 = 137gmol-1

Density = 1.28gcm-2

2.5% of 137gmol-1 = 3.425gmol –1

Volume of 0.025moles = 3.425

1.28 = 2.7cm3

1 – Iodobutane

Molar mass of CH3CH2CH2CH2I = (4 x 12) + (9 x 1) + 127 = 184gmol-1

Density = 1.62gcm-2

2.5% of 184gmol-1 = 4.6gmol –1

Volume of 0.025moles = 4.6

1.62 = 2.8cm3

Method

1. Measure out 2cm³ of Ethanol and put into a test tube.  Repeat this twice more so that you have three test tubes each with 2cm³ of ethanol in each.
2. Using separate pipettes, measure out the calculated amounts of the halogenoalkanes. Place the 1 – Chlorobutane in the first test tube, 1 – Bromobutane in the second and 1 – Iodobutane in the third.
3. Put a different coloured rubber band around the rim of each of the three test tubes to identify the different halogenoalkanes.
4. Prepare three more test tubes each containing 2cm³ of AgNO3 (silver nitrate solution).
5. Stand all six test tubes in a beaker of water at about 60-65°C.
6. Wait until the contents of all the test tubes have reached approximately 60°C and then add the three test tubes containing the AgNO3  solution to the halogenoalkane and ethanol solution.  Make sure this is done all at the same time (you may need a few extra people to help).  Then start the stop clock
7. Record the time when the precipitate of the silver halide appears: AgCl = white precipitate, AgBr = off-white precipitate, and AgI forms a pale yellow precipitate.

Variables

• A different halogenoalkane in each test tube

Constants

• The volume of the halogenoalkanes
• The volume of ethanol
• The volume of silver nitrate solution
• The temperature of the water bath

Safety/Environmental care

• Take care when using the hot water
• The aqueous residues will contain the heavy metal ions on Ag and the organic residues will contain halogenated hydrocarbons so it is important not to dispose of any waste down the sink.

Fair Testing

• Use same volume for each solution
• Make sure you use different teat pipettes for each halogenoalkane else they will get contaminated and the experiment will not be a fair test.

Prediction

Looking back on the scientific research that I have carried out, I think that the 1 – iodobutane will react fastest due to the weaker bond enthalpy and the longer bond length.  This will overcome the fact that the Chlorine has a higher electronegativity than any of the other halogens and will therefore be the slowest to react.

Sources Used

• Text books: Chemistry 1, Chemistry for you, Laboratory guide, Chains and Rings