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# Find out how the rate of hydrolysis of an organic halogen compound depends on the identity of the halogen atom, and the nature of the carbon-hydrogen 'skeleton'.

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Introduction

PLANNING AIM: The purpose of this experiment is to find out how the rate of hydrolysis of an organic halogen compound depends on the identity of the halogen atom, and the nature of the carbon-hydrogen 'skeleton'. I will be comparing the rates of hydrolysis of the primary substances 1-chlorobutane, 1-bromobutane, 1-iodobutane, and will investigate the rate of hydrolysis of the tertiary substance 2-bromo-2-methylpropane. From the results I will then go on to deduce a rate expression/equation and a possible mechanism for the reaction. Calculations that are to be carried out include that of gradients and rates of the graph. This is needed because the gradient of the graph gives us the rate of reaction. A rate for each tangent taken needs to be worked out so that a rate graph can be constructed, which will give the order of the hydrolysis of the haloalkane. Example: Figure 1 (I) Rate = Gradient Gradient = Y2 - Y1 X2 - X1 = [reactants at Y2] - [reactants atY1] (where [ ] refers to the concentration) t2 - t1 (where t refers to the time) Rate = ?[reactants] (where ? refers to the change) ?t The initial rate will be taken, therefore the gradient line will start at zero, and thus Y1 and X1 will equal 0. From this it can be shown that, Rate = Y2 - 0 X2 - 0 = Y2 X2 The rate of reaction at the start is the initial rate, which is what we are interested in. We can find the initial rate by drawing a tangent to the curve at the point t = 0, and measuring the gradient of this tangent. In this example, Initial gradient = Y2 X2 = 5.1 (cm3) 10 (s) = 0.51 cm3 Therefore the rate is 0.51 cm3 of O2/ s-1. This rate will then be drawn on a separate graph along with the other rate values calculated from the other concentrations of the haloalkane, and from that the order of the reaction can be concluded and thus a mechanism deduced. ...read more.

Middle

attached to a hydrocarbon chain. They do not occur in nature, but they are useful for all sorts of human purposes. As with all functional groups, the halogen atom modifies the properties of the relatively unreactive hydrocarbon chain. The simplest examples are the halogenoalkanes (sometimes referred to as haloalkanes), with the halogen atom attached to an alkane chain. Physical properties: Figure 21 Looking at Figure 21 above, the carbon-hydrogen bond is polar, but not polar enough to make a big difference to the physical properties of the compound. For example, all haloalkanes are immiscible with water. Their boiling points depend on the size and number of halogen atoms present: the bigger the halogen atom and the more halogen atoms there are, the higher the boiling point, as can be seen from Table 4 below. Table 4 The influence of halogen atoms on the boiling is important when it comes to designing halogen atoms for particular purposes. If you want a compound with a high boiling point, you have to include a larger halogen atom such as Cl or Br rather than a smaller one such as F: but it is these larger atoms that can cause the greater environmental damage. Chemical reactions of haloalkanes: Reactions of haloalkanes involve breaking the C ? Hal bond (Hal stands for any halogen atom). The bond can break homolytically or heterolytically. Homolytic fission: Homolytic fission forms radicals. One way this can occur is when radiation of the right frequency (visible or ultraviolet) is absorbed by the haloalkane. For example, with chloromethane: H H ? ? H ? C ? Cl + hv H ? C* + Cl* ? ? H H chloromethane methyl chlorine radical radical this is shortened to: CH3 ? Cl + hv CH3* + Cl* (XI) This kind of reaction occurs when haloalkanes reach the stratosphere, where they are exposed to intense ultraviolet radiation. This is how the chlorine radicals that cause so much trouble for ozone in the stratosphere have formed. ...read more.

Conclusion

This error was rectified during the investigation and was not repeated - this ensured that accurate results are obtained and that any error in the experiments, if any, is not due to procedural error. This particular experiment was repeated and the new method, stirring the conductance meter in the solution, was taken into account. The graphs obtained showed to be the same as those predicted and those of what previous experimenters found. Precision error: Percentage error can be calculated as follows: Apparatus Uncertainty value Volumetric flask - 250 cm3 0.2 cm3 Pipette - 25 cm3 0.02 cm3 Therefore, for the volumetric flask, Percentage error = 0.20 cm3 x 100 50 cm3 = 0.4 % For the pipette, Percentage error = 0.02 cm3 x 100 50.00 cm3 = 0.04 % Therefore, the total percentage error that occurred in experiment 2 was 0.44%. Since this is largely below 5%, and can therefore be discarded. Once again, atmospheric conditions may also have had an effect on the rate of reaction. During the day at which the experiments were carried out, the temperature of the room had reached around 30 ?C. This could affect the rate of reaction, as it is known that the temperature has an affect on the rate, as was described in detail on page 11. Ways of improving the experiment includes the use of controlled atmospheric conditions, as changes in the conditions of the atmosphere could affect the collisions of particles and therefore the rate of reaction. This was taken into account at the start of the investigation, but this kind of facility was not available to me. There were many parts of my experiment, which were vital in ensuring that the evidence collected was accurate and reproducible, such as the chosen apparatus. For example, a 250 cm3 volumetric flask was used other than a 100 cm3 volumetric flask, as the larger the size of the apparatus, the more accurate it is. A 250 cm3 volumetric flask has a less percentage error than a 100 cm3. ...read more.

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