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Plan for rate of reaction for halogenoalkanes

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Introduction

Harry Denis Candidate No. 9038 Centre No. 10534 Plan for rate of reaction for halogenoalkanes Aim Aim of the experiment is to find out and compare the rate of displacement of the halide ion and how is varies with respect to the carbon halogen bond. This will occur through a nucleophilic attack. General equation: general formula: R-X +: Nu- ? R-Nu + X- CnH2n+1X (where X is the halogen) Background information halogenoalkanes In the reaction of halogenoalkanes bond polarity show us that C-F would be the most reactive then C-CL, C-Br and C-I would be the lest reactive. This is due to the electronegativity of the halogen atoms. Electronegativity is the measure of how strongly an atom in a compound attaches electron in a bond. The greater the difference in the electronegativity of two atoms, the more polar is the covalent bond between the two atoms. So in a carbon halogen bond the halogen is more electronegative than carbon. Consequently the bond between them is polarized so the halogen atom is slightly negative. Electronegativity values E.G. C 2.5 F 4.0 Cl 3.0 Br 2.8 I 2.5 H :Nu- H R C+ X - R C Nu +:X- H H According to bond enthalpy the reverse is true. ...read more.

Middle

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. ...read more.

Conclusion

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. 4. Then measure out 2 cm3 of chlorobutane, bromobutane, iodobutane 5. 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. 6. Repeat steps 4 and 5 three times. Results should be recorded in a table like this Halogenoalkanes Colour change of precipitate Time taken Chlorobutane Bromobutane Iodobutane 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: * www.chemguide.co.uk/mechanisms/nucsub/water.html * www.scienceatstmarys.net/chemistry/AS2.2biii.htm * www.rjclarkson.demon.co.uk/candr/haloalkanes.htm Harry denis candidate No. 9038 Centre No. 10534 ...read more.

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