Simple Distillation, Gas Chromatography: Preparation of Synthetic Banana Oil

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Simple Distillation, Gas Chromatography:
Preparation of Synthetic Banana Oil

This experiment gave us the opportunity to work with a variety of new procedures as well as practice procedures that are new to us within the past few weeks of labs.
Within the context of the scenario, we find that a distilling company that markets a popular banana liqueur is having problems. The banana plantation that they use to create their banana extract was hit hard by a hurricane and their reserves of extract are running low. It is our job to formulate a synthetic banana flavoring and determine whether it will work as a substitute for natural extract. The problem that the distilling company foresees is that the process used to create this synthetic extract tends to leave behind considerable amounts of starting material in the product, which is okay in small amounts, but we need to test the product to see if it is within their acceptable ranges to be used.

We will be synthesizing synthetic banana oil, or otherwise known as isopentyl acetate. We will synthesize thie oil by combining isopentyl alcohol with acetic acid and sulfuric acid, and heating under reflux for one hour. After this time, we will separate and purify our product via washing with sodium bicarbonate (that separates compounds via solubility properties) and distilling (that purifies via different boiling point properties of the compounds). The reaction that occurs is modeled below:

After the product has been synthesized, it will be tested using a gas chromatograph to measure the percentages of different compounds in the product. This measurement will tell us if the product that we have synthesized could be used as an effective substitute to natural banana extract.

My hypothesis for this experiment was twofold. As far as the experiment and procedures itself, I was a little concerned for how well my yield would turn out. Heating under reflux as well as the new piece of distillation gave made me a little uncertain that I would end with a high yield, thus my hypothesis for the yield bit was that I would end with a significantly smaller yield than theoretically possible.

My hypothesis for the outcome of the experiment, however, was a bit more positive. I know that the process we use to synthesize the compound usually results in high percentage of the starting material, but we were introduced to Le Châtelier's principle that helps shift the equilibrium toward the products. This gave me the hypothesis that we would be within, or at least close to, the requirements given to us by the distilling company.


Theoretical yield: Using stoichiometric properties based on the above reaction equation, we know that there is a 1:1 ratio between the original isopentyl alcohol and the synthesized isopentyl acetate. * Grams isopentyl alcohol: 1.775g

Molecular weight isopentyl alcohol: 88.1g/mol (therefore .0201 mol or 20.1 mmol) Molecular weight isopentyl acetate: 130.2g/mol
* .0201mol X 130.2g/mol = theoretical yield isopentyl acetate = 2.61g

* Actual Yield isopentyl acetate:
* Distillation yielded 2.42g, 84.92% of which was isopentyl acetate. * 2.05g isopentyl acetate
moles isopentyl acetate: .0157mol (or 15.74 mmol)

Percent Yield = mass actual/mass Theoretical yield X 100%
% Yield = 2.05g/2.61g X 100%
% Yield = 78.5%

Before You Begin Exercises

20.0 mmol isopentyl alcohol X 1 mol X 88.1g = 1.76g 1000 mmol1 mol

20.0 mmol isopentyl alcohol X 1 mol X 88.1g X 1mL =2.16mL
1000 mmol 1 mol .815


There were several major sections to this experiment: Reaction, purification, and analysis. The results for each section is given in the tables below:

Compound| Mass/Vol|
initial isopentyl alcohol| 1.775g|
acetic acid| 2.3 mL|
sulfuric acid| 5 drops|...
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