Ochem Lab
Lab Experiment 8: Oxidation Puzzle
Post Lab Report for 2-ethyl-1,3-hexandiol
Calculations
Theoretical yield: 1.857g
Product Yield: 1.055g ----> Percent yield = (1.055g/1.87g) x 100% = 56.41%
Starting amount of diol: 1.184g ----> Percent Yield = (1.184g/1.87g) x 100% = 63.32%
Spectroscopy
O-H (Stretch, H-bonded)
C-H (Stretch)
C-H
(2720-2820 cm-1)
Carbonyl
C-O (Stretch)
Product wavelength cm-1
3422
Strong, Broad
2877,2936,2964
Strong, Medium
None Present
1705
Strong
1043
Strong
Structure
The Structure in my compound is: C8H18O2 + 2 ClO- → C8H14O2 + 2 Cl- + 2 H2O producing 2-ethyl-3-keto-1-hexanal.
The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols. This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present. Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol. The textbook states that aldehydes show 2 characteristic absorption’s in the range of 2720-2820 wavenumbers. No such absorption’s were present in my sample. Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketone.
In this experiment, the secondary alcohol is selected over the primary alcohol. In many cases the primary alcohol can be oxidized all the way to a carboxylic acid. In order to achieve selectivity, sodium hypochlorite is used. It is reacted with acetic acid to form HOCl.
Post Lab Report for 2-ethyl-1,3-hexandiol
Calculations
Theoretical yield: 1.857g
Product Yield: 1.055g ----> Percent yield = (1.055g/1.87g) x 100% = 56.41%
Starting amount of diol: 1.184g ----> Percent Yield = (1.184g/1.87g) x 100% = 63.32%
Spectroscopy
O-H (Stretch, H-bonded)
C-H (Stretch)
C-H
(2720-2820 cm-1)
Carbonyl
C-O (Stretch)
Product wavelength cm-1
3422
Strong, Broad
2877,2936,2964
Strong, Medium
None Present
1705
Strong
1043
Strong
Structure
The Structure in my compound is: C8H18O2 + 2 ClO- → C8H14O2 + 2 Cl- + 2 H2O producing 2-ethyl-3-keto-1-hexanal.
The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols. This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present. Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol. The textbook states that aldehydes show 2 characteristic absorption’s in the range of 2720-2820 wavenumbers. No such absorption’s were present in my sample. Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketone.
In this experiment, the secondary alcohol is selected over the primary alcohol. In many cases the primary alcohol can be oxidized all the way to a carboxylic acid. In order to achieve selectivity, sodium hypochlorite is used. It is reacted with acetic acid to form HOCl.