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
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