hydrolysis of halogenoalkanes

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Hydrolysis of Halogenoalkanes

In this piece of coursework I aim to investigate the reactivity of three halogenoalkanes. By doing this it allows me to see which halogenoalkane is the most reactive and which is the least reactive.

Halogenoalkanes have an alkane skeleton with one or more halogen atoms in place of hydrogen atoms. The general formula of a halogenoalkane with a single halogen atom is Cn + H2n+1X, where X is the halogen.

The three halogenoalkanes that I will be testing are:

  1. chlorobutane
  1. bromobutane
  1. iodobutane

As you can see above all the halogens are primary halogens indicated by the number 1. This is where the halogen is at the end of the chain, by keeping all the halogenoalkanes as primary ones it becomes a fairer a test as the halogen is situated in the same place in each. This is because reactions may vary due to the halogen not being in the same place therefore results from the experiments taken out may not necessarily be those which are relevant to the objective of the experiment.

One of the main ways which halogenoalkanes react is by nucleophilic substitution. This type of reaction takes place when a halogenoalkane reacts with aqueous Sodium hydroxide solution. A nucleophile is defined as electron pair donors. In this reaction the sodium hydroxide (NaOH) will provide the nucleophile, which is a hydroxide ion (OHֿ).

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The equation for the reaction that will take place for each of the halogenoalkanes:

Halogenoalkane + Hydroxide ion                           Alcohol    + Halide ion

C4H9X       +          OHֿ                 C4H9OH        +        Xֿ

As you can see above the number of carbons and hydrogen’s remains the same this is because it has an alkane skeleton of butane and it is only X which is the halogen which is changed. The oxygen atom is highly electronegative compared to a hydrogen atom and ...

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