Comparison of Three Isomers of Butanol
An alcohol's reactivity is determined based on the attachment of their hydroxyl functional
group. The location of this hydroxyl functional group will impact the molecular structure of the
alcohol, making it either primary (1° ), secondary (2° ), or tertiary (3° ). If the OH is bonded to only one other carbon, it is a primary alcohol (eg. 1-butanol); if bonded to two other carbons, it is a secondary alcohol (eg. 2-butanol); if bonded to three other carbons, it is a tertiary alcohol (eg. 2-methyl-2-propanol). Due to the placement of the hydroxyl functional group in each of the degrees of alcohol, the reactivity of each should be impacted. This means that all three alcohols should have a different level of reactivity.
The hydrogen atom on the hydroxyl functional group can be easily substituted because of its enhanced acidity. This is the basis of a halogenation reaction, and controlled oxidation reaction, which will be observed through this investigation. Purpose:
To observe and compare the reactivity of the isomers of butanol as examples of 1°, 2° , and 3° alcohols. Hypothesis:
The reactivity of primary alcohol will be lower than that of secondary and tertiary alcohol. Also, the reactivity of secondary alcohol will be lower than that of tertiary alcohol. Materials:
Ref: “Comparison of Three Isomers of Butanol” (Textbook pg. 84 and 85 Investigation:1.5.1) Safety:
Ref: “Comparison of Three Isomers of Butanol” (Textbook pg. 84 and 85 Investigation:1.5.1) Procedure:
Ref: “Comparison of Three Isomers of Butanol” (Textbook pg. 84 and 85 Investigation:1.5.1)
Experiment 1: Using HCl(aq)
-White and cloudy
Experiment 2: Using KMnO4 (aq)
-Very thick liquid
-Brown in colour with all purple gone
-Mainly brown with some purple
1. Does each alcohol undergo halogenation and controlled oxidation?
As seen in experiment 1, in each alcohol there is a hyrdrogen atom that is being substituted with a chlorine atom. Therefore each alcohol does undergo halogenation, forming an alkyl halide. However in experiment 2, only 1-butanol and 2-butanol undergo a controlled oxidation reaction to form an aldehyde and a ketone, respectively. The tertiary alcohol 2-methyl-2-propanol, does not undergo a controlled oxidation reaction as there is no hydrogen attached to a carbon atom, for the oxygen to remove. Therefore 2-methyl-2-propanol undergoes a dehydration reaction instead.
2. Evaluate the theory you used to make your predictions.
The theory that I used to make my predictions is, primary and secondary alcohols are the most reactive as they easily undergo oxidation, and tertiary alcohols are resistant to oxidation because they have no hydrogen atoms attached to the oxygen bearing carbon. This theory was accurate because it described what occurred when KMnO4 was added to each of the degrees of alcohol.
3. How can the results of this investigation be accounted for in terms of intermolecular forces? The results of this investigation can be accounted for in terms of intermolecular forces in the first experiment where the alkyl halide and water (products of halogenation) do not have the same intermolecular forces. This is the reason why they will not completely dissolve, leaving the solution cloudy. In this case, the HCl has stronger intermolecular forces as it can break the bond of the OH to form water. The same is true in the second experiment, as the one of the products of the primary and secondary alcohol reaction is water. In the second experiment, 1-butanol and 2-butanol do not have the...