Hydrolysis of halogenoalkanes.

* Then measure out 4cm3 of Silver nitrate to add to the haloalkanes.

- The silver nitrate is added to the haloalkanes because when the two are added together the halide ions that are produced form a silver halide precipitate. The precipitate tells us about the rate of hydrolysis in haloalkanes.

* After the addition of Silver nitrate, Ethanol must be added to the solution. About 2cm3 would be enough. And at this point the stop clock must be started.

- The reason why ethanol is added is because the halogenoalkanes are immiscible with just silver nitrate, however ethanol makes the solution of silver nitrate and the halogenoalkanes miscible. And doesn't interfere with the reaction in anyway, it is a helping spectator.

* Monitor the test tubes under room temperature until a precipitate is seen. In which case a note must be made of the time.

- The formation of a precipitate shows the rate of hydrolysis of the haloalkanes.

Calculations on how much of each haloalkanes should be used:

* 1-chlorobutane

H H H H

H-C-C-C-C-Cl

H H H H

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

Density = 0.890 gcm-2

Volume = mass

density

Volume of 1mole = 92.5

0.890 = 103.9cm3

Volume of 0.10moles = 9.25

0.890 = 10.4cm3

Volume of 0.05moles = 4.625

0.890 = 5.2cm3

Volume of 0.025moles = 2.3125

0.890 = 2.6cm3

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

* 1-bromobutane

H H H H

H-C-C-C-C-Br

H H H H

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

Density = 1.28 gcm-2

2.5% of 137gmol-1 = 3.425gmol-1

Volume of 0.025moles = 3.425

1.28 = 2.7cm3

* 1-iodobutane

H H H H

H-C-C-C-C-I

H H H H

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

Density = 1.62 gcm-2

2.5% of 184gmol-1 = 4.6gmol-1

Volume of 0.025moles = 4.6

1.62 = 2.8cm3

Observations

In this experiment I hope to observe the formation of precipitates in each of the test tubes, to compare the rate of reaction. The halides from the halogenoalkanes will form a silver halide precipitate that will be whitish in colour. In the case of 1-bromobutane this should be what is happening in the test tube:

C4H8Br + H2O C4H8OH + H+ (aq) + Br - (aq)

The general formula would be:

C4H8X + H2O C4H8OH + H+ (aq) + X- (aq) (where X is the halogen atom)

At the end of the experiment I should have 3 different times which represent the rate of reaction for each of the haloalkanes I set out to investigate.