Making a Halogenoalkane

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Making a Halogenoalkane

Aim

The aim of this activity is to prepare a sample of 1-bromobutane, an example of a halogenoalkane. This activity will involve learning how to use new techniques and equipment, firstly to carry out the reaction and form the product, and secondly to separate and purify the halogenoalkane.

I will attempt to obtain the maximum yield possible of 1-bromobutane. I shall examine the product that I have obtained and comment on the amount. I shall also evaluate any practical difficulties I experience in obtaining the halogenoalkane.

Background Information

The overall process involved in making a halogenoalkane is simply to replace the -OH group in an alcohol by a halogen atom.

The actual process to produce a halogenoalkane is slightly more complex, involving more than one single reaction. Butan-1-ol is heated in a mixture of concentrated sulphuric acid and sodium bromide. There are two main reactions. Firstly, the sulphuric acid reacts with the sodium bromide to form hydrogen bromide. Secondly, the hydrogen bromide reacts with the butan-1-ol to form 1-bromobutane and water.

. Making HBr:

The reaction for formation of HBr can be summarised by the following equation:

NaBr + H2SO4 ? NaHSO4 + HBr

There is also an unwanted side reaction that takes place at this stage. That is the oxidation of the product we want, the HBr into water and bromine:

2HBr + [O] ? H2O + Br2

(It does serve some purpose in colouring the halogenoalkane, which later on will be useful).

The oxidising agent is the sulphuric acid, which is reduced to sulphur dioxide amongst other things.

As I will explain later, the conditions of the first reaction will have to be configured so that production of HBr is maximised, and so that the oxidation of HBr is minimised.

2. Production of 1-bromobutane:

The formation of 1-bromobutane can be summarised by the following equation:

CH3CH2CH2CH2OH + HBr ? CH3CH2CH2CH2Br + H2O

butan-1-ol 1-bromobutane

I will now describe the exact mechanism involved in the reaction.

The butan-1-ol picks up a H+ ion (from the HBr), it is protonated:

This leaves the butan-1-ol positively charged. The Br- which is left, is negatively charged, carrying four lone pairs of electrons. Such species are called nucleophiles. They are deficient in positive charge and will attack anything positively charged, such as the protonated butan-1-ol.

This is an example of a nucleophilic attack. One of the lone pairs on the bromide ion begins to make a bond to the carbon atom, and at the same time the C-+OH2 bond begins to weaken.

A transition state is reached in which the C atom is partially bonded to both

-+OH2 AND Br-. This state only exists for a moment, before it is rapidly converted into the products.
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The transition state is made from two molecules. The reaction is said to have a molecularity of 2. We also know that it involves a substitution, and a nucleophilic attack. The mechanism is described as an SN2 reaction.

So far I have explained the reaction for making 1-bromobutane. There are however, many impurities in the liquid product (I shall explain in more detail about unwanted substances in my evaluation). To obtain a pure yield of the halogenoalkane, it is necessary to put the product through a few stages of purification.

The four main experimental stages are ...

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