H H
According to bond enthalpy the reverse is true. Bond enthalpy shows use that C-I bond is the most reactive and the C-F bond is lest reactive. This is due to the bond energy between the atoms. Bond energy is the average standard molar enthalpy changes for the breaking of a mole of bonds in a gaseous molecule to from gaseous atoms. Bond energies indicate the strength of the forces holding together atoms in a covalent molecule. Bond energy is increased with the number of shared electron pairs. So C-I bond has bond energy of 228 kJ mol-1, which is more reactive because the bond is weak. Compared to C-F, which has bond energy of 467 KJ mol-1, which is a strong bond.
On the basis of bond energies I would predict C-I to be the fastest reaction followed be C-Br, then C-Cl.
E.G.
Strengths of various bonds (values in kJ mol-1).
Halogenoalkanes are classified as tertiary, secondary and primary; this depends on the number of carbon atoms bonded to the carbon atom to which the halogen is bonded.
Primary halogenoalkanes
Examples:
Secondary halogenoalkanes
Examples:
Tertiary halogenoalkanes
Examples:
The reaction of a halogenoalkane is characterised by nucleophilic substitution of the halogen atom in which the electron-deficient carbon- halogen bond is attacked by an electron rich species called nucleophiles. With primary and secondary halogenoalkanes, the reaction is very slow at room temperature, but rapid with tertiary halogenoalkanes. As a nucleophile approaches the electron deficient carbon atom donating a pair of electrons, the halide ion moves away taking with it a pair of electron. A translation state is involved, in which the nucleophile and halide ions are both partially bonded to the same carbon.
Plan
The plan for this experiment is to take four different halogenoalkanes and add a nucleophile to each of the halogenoalkanes record and compare my observations for each halogenoalkene experiment. In the experiment the number of carbons in the halogenoalkene will not vary. The carbon compound I am going to use will be halogen butane. The nucleophile I am going to use is hydroxide ion (:OH-). In this experiment I will not use the halogen fluorine because of the strong bond it from with carbon. According to bond enthalpy the C-F bond is the less reactive of the halogenoalkanes because of how strong the bond energy is between them. The bond energy in the C-F bond is 467 kJ mol-1. So the C-F bond will be too strong to be affect by nucleophiles. Equation for nucleophilic substitution reaction.
C4H9Br + :OH- C4H9OH + Br-
C4H9Cl + :OH- C4H9OH + Cl-
C4H9I + :OH- C4H9OH + I-
Prediction
My prediction for which bond will react most vigorously with the nucleophile will be the C-I halogenoalkane. This is because it has the lowest bond enthalpy of all the halogenoalkanes. The C-CL halogenoalkane will be the leased reactive because it has one of the highest bond enthalpy of the entire bond being tested. The C-BR will be in the middle because it bond enthalpy show that it is more reactive than the C-CL bond but less reactive than the C-I bond.
Procedure
Before undertaking this experiment a hazard analysis will need to be done. The equipment needed to ensure safety would be: goggles, lab coat and gloves. Halogenoalkanes are highly flammable and alcohols are corrosive and can cause irritation to the skin.
Equipment
- 3x test tubes
- Test tube rack
- 2x 10cm measuring cylinder
- Bunsen burner
- Stop watch
- Ethanol
- Halogenoalkanes, which are chlorobutane, bromobutane, iodobutane
- 0.02m of silver nitrate
- Test tube holder and pipette
Method
I will first collect three test tubes and place them in a test tube rack. Then I will place 3cm3 of iodobutane into one test tube, then 3cm3 of bromobutane into a test tube, then 3cm3 of chlorobutane into the last test tube. Then I will add 4-5 drops of hydroxide to each test tube. Then using a stopwatch record how look it would take for each of them to react. I will repeat this experiment 4 times and then take an average of the times for each reaction. I will then do a simple chemical test distinguish between the compounds. The chemical test will involve adding to each test tube silver nitrate solution and then place in a beaker of heated water around 60oC. Then after 10 minutes of precipitate should appear in each test tube. The precipitate indicates which test tube contains which halide ions.
- Arrange three test tubes, into a test tube rack, in a row.
- Using the pipette, add 3 drops of halogenoalkanes in the following sequence: chlorobutane, bromobutane, iodobutane
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Using the measuring cylinder, measure out 2 cm3 of ethanol and add to each test tube. Then measure out another 2 cm3 of 0.02m of silver nitrate and pour into each test tube.
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Then measure out 2 cm3 of chlorobutane, bromobutane, iodobutane
- Using the test tube holder, hold the test tube containing the halogenoalkanes with the ethanol and silver nitrate to the Bunsen burner with a blue flame. Make sure that the solution boils then quickly add the halogenoalkane and time with a stopwatch the speed of the colour change. Then take a note of the colour of the precipitate formed.
- Repeat steps 4 and 5 three times.
Results should be recorded in a table like this
My method would work because of the equipment I used for the experiment. Like the measuring cylinder, which will ensure that, the equal and right amounts are being used in the experiment.
Molar quantities
2g of chlorobutane (Mr=92.5)
Bromobutane (Mr=136.9)
Iodobutane (Mr=184)
No. Of mole of chlorobutane= 2g/92.5=0.02m
No. Of mole of bromobutane=2g/136.9=0.01m
No. Of mole of iodobutane=2g/184=0.01m
Information gathered from: