- Created by: teague sheldon
- Created on: 01-01-13 13:17
- Halogenalkanes are made by substituting the -OH group in an alcohol for a halogen atom.
- This can be done using hydrogen halides (HX) or using phosphorus halides.
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- When using hydrogen halides to form halogenalkanes, the general formula is: ROH + HX = RX + H20
- The way in which we get each HX that we use for a specific halide is different, and depends on the halide.
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- Alcohol + hydrochloric acid gives a chloroalkane + water.
- However, this only works well with tertiary alcohols, because the reaction rate is too slow with primary or secondary alcohols (use other method instead).
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- Alcohol + hydrogen bromide gives bromoalkane + water.
- However, we do not use hydrobromic acid to give HBr.
- Instead, we treat the alcohol with a mixture of potassium bromide and sulphuric acid. This produces HBr which reacts with the alcohol.
- The first step does not appear in the overall formation of bromoalkane equation.
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- Alcohol + hydrogen iodide gives iodoalkane + water.
- The alcohol is treated with a mixture of potassium iodide and phosphoric (V) acid, which produces the HI that reacts with the alcohol.
- Phosphoric (V) acid is used instead of conc sulphuric acid because sulphuric acid oxidises the iodide ions and so produces hardly any hygrogen iodide.
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- Alcohol + phosphorus (III) chloride (liquid) gives chloroalkane + H3PO3.
- Alcohol + phosphorus (V) chloride (solid) gives chloroalkane + HCl + POCl3.
- This is a violent reaction producing steamy fumes of HCl, which is very visible.
- This is why we use phosphorus (V) chloride as a test for the -OH group.
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- Bromoalkanes and iodoalkanes are made using the same steps with phosphorus.
- The alcohol is heated with the halogen and red phosphorus under reflux.
- The halogen reacts with the red phosphorus to produce the phosphorus (III) halide.
- The phosphorus (III) halide then reacts with the alcohol to give the halogenalkane and H3PO3.
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- When we make bromoethane (from KBr, H2SO4 and ethanol), there will be impurities such as HBr, Bromine, sulphur dioxide and unreacted ethanol.
- These impurities are removed in 5 stages.
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- Pour the impure bromoethane into a separating funnel, add some water and shake. A bromoethane layer will form on the bottom with an aqeuous layer on top. Discard the aqueous layer, which contains all the soluble impurities.
- Add sodium carbonate/hydrogencarbonate solution to the bromoethane, which will react with any acidic impurites.Separate and remove the bromoethane layer as above.
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- Wash the bromoethane with water in the separating funnel to remove inorganic impurities (excess sodium carbonate). Transfer remaining lower bromoethane layer to a dry test tube.
- Add anydrous calcium chloride to the tube, shake well and leave to stand. This removes any excess water and ethanol.
- Fractionally distil the bromoethane, collecting at 35'C.
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- When a halogenalkane reacts with sodium/potassium hydroxide, a substitution or an elimination reaction will occur.
- A subsitution reaction forms an alcohol + sodium/potassium halide.
- An elimination reaction forms an alkene + sodium/potassium halide + water.
- If the halogenalkane is primary it is more likely to undergo substitution, if it is tertiary, it is far more likely to be an elimination reaction (almost exclusively).
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- The proportion of water to ethanol in the KOH/NaOH effects which reaction will occur.
- Water encourages substitution, ethanol encourages elimination.
- Higher temperatures favour elimination.
- Concentrated KOH/NaOH favours elimination, dilute favours substitution.
- Both elimination and substitution require heat and reflux.
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- Halogenalkane + ammonia gives amine + ammonium halide.
- The ammonia is concentrated and dissolved in ethanol.
- The reactants are heated with pressure, no reflux because the ammonia would escape through the condenser as a gas.
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- The reactivity of a halogenalkane depends on the halogen present, and the number of methyl groups attached to the carbon attached to the halogen (primary, secondary or tertiary).
- The weakest carbon-halogen bond breaks more easily and so is more reactive.
- The further the two atoms are from eachother (atomic radii), the weaker the bond will be. So C-I is the weakest bond.
- Tertiary is the most reactive, because more methyl groups weaken the carbon-halogen bond.
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- Halogenalkanes are generally unreactive, have low boiling points and are non-flammable, so they are used as fire-retardants and refrigerants.
- Chlorofluorocarbons (CFCs) were found to deplete the ozone layer in the atmosphere, allowing harmful UV radiation to reach the earth surface, increasing the risk of skin cancer and eye cateracts.
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