The Preparation of 1-Bromobutane

The Preparation of 1- Bromobutane

Introduction...

Aim : To prepare 1-Bromobutane.

Background : The most common way of preparing alkyl halides, which are very useful intermediates in syntheses, is the replacement of the OH group of an alcohol by a halogen.

This replacement is a nucleophilic substitution reaction, and alcohols do not undergo nucleophilic substitution reactions because hydroxide ison is strongly basic and a poor leaving group. However, alcohols readily undergo nucleophilic substitutions if the hydroxyl group is first activated to produce a better leaving group, so reaction is carried out in the presence of a strong acid. The acid protonates the alcohol to create a suitable leaving group, water, for the SN2 reaction. [IMAGE]

Designing...

In this experiment 1-butanol will be converted to 1-bromobutane by an
SN2 reaction.

CH3CH2CH2CH2OH (aq) + HBr (aq) -----> CH3CH2CH2CH2Br (aq) + H2O (l)

CH3CH2CH2CH2-OH + HBr [IMAGE] CH3CH2CH2CH2-OH2 (+) Br(-) [IMAGE] CH3CH2CH2CH2-Br
+ H2O SN2

Reaction is nucleophilic substitution, and the products are impure, and so various stages of purification are required before a sample of reasonable purity can be obtained.

I'm planning to produce 20 g of 1-Bromobutane as a result of this experiment. Amount of Butan-1-ol needed:

C4H9OH (aq) + HBr (aq) -----> C4H9Br (aq) + H2O (l)

74 : 70 : 137 : 18 ß Relative Molecular Mass.

Proportion is 74 to 137 at 100% yield, so 137 g of Bromobutane will be formed from 74 g of Butan-1-ol.

I will produce 20 g of 1-Bromobutane, my planned yield is 70%.

So;

At 70% yield : (137/100) x 70 = 95.9 g (1 mole)

I want 20 g of C4H9Br so;

( 20 x 74 ) / 95.9 = 15.5 g , so :

Production of 1-Bromobutane will require 15.5 g of Butan-1-ol to obtain the full amount at 70% yield.

Chemicals Needed:

* Butan-1-ol, 7.72 g. (C4H9OH)

* Sodium Bromide (powdered). (NaBr)

* Concentrated