logo

Examination of Unknown an Organic Compound

Examination of Unknown an Organic Compound

Written by Parul

Purpose

The purpose of this experiment is to be able to successfully execute a series of classification tests on an unknown compound, ultimately resulting in the deduction of the functional groups present. By performing this experiment, I hoped to be able to utilize and materialize knowledge of stereo-specific, chemical and physical properties of organic compounds.

Procedure

  1. View given infrared spectrum and identify possible functional groups.
  2. Complete a preliminary classification.
  3. Determine boiling point using the microscale boiling point determination method (PLK&E p. 586-589).
  4. Perform a “solubility in water” test (PLK&E p. 432-436). After addition of water to unknown, if concentration gradient lines are present, partial solubility has taken place. This reveals whether the compound is a strong base, a weak acid, a strong acid or a neutral substance.
  5. Perform a Beilstein Test for halogens (PLK&E p. 439). If test result is positive, it is confirmed a halogen is present in the unknown mixture. This presence is represented by a flame color change from yellow to blue or green.
  6. Perform a Potassium Permanganate Baeyer Test for unsaturation (PLK&E 446). If this test is positive, the purple MnO4 ion turns into brown MnO2 precipitate. If this occurs, a double or triple bond is present.

VII. Perform a Chromic Acid Test (PLK&E 463-464). This test is based on the reduction of chromium(VI), which is orange, to chromium(III), which is green when alcohols are oxidized by the reagent. A color shift from orange to green represents a positive test and confirms presence of alcohol.

 

Results

Unknown B

Infrared Spectrum -OH group present
Preliminary Examination Liquid is clear and pungent, giving off smell of rubbing alcohol.
Solubility in water When unknown mixture is added to water, bubbles form and an oily film forms on the top. Mixture is water insoluble.
Beilstein Test for Halogens When a drop of mixture is added onto a piece of copper wire and inserted into a flame, the flame turns a green/turquoise color.
KMnO4 Test Solution turns purple and brown precipitate forms after 30-40 seconds.
Chromic Acid Test Solution turns sage green at the addition of chromic acid reagent.
Boiling Point 93-101° C

 

Discussion

From the Infrared spectrum, we know that there is an alcohol group present. This is affirmed through the Chromic acid test, since the solution turned green in the presence of chromic acid reagent. The solubility test shows us that the mixture is insoluble in water. Since alcohols are generally soluble in water, we now know that there must be some other group present creating the insolubility. Although our KMnO4 test seems to be positive, as the brown precipitate formed, after confirming with another group that was analyzing the same Unknown mixture and re-performing the examination, we came to the conclusion that some type of contamination must have occurred during the first test. The second time we did this, no precipitate formed. Thus, no carbon-carbon double or triple bond is present in the mixture. Thus, we deduce that an alkene or alkyne is not present in the mixture. This is affirmed with the Beilstein test that was performed. The flame became a turquoise color when a drop of the mixture was introduced via a copper wire. Thus, we knew that a halogen is present in the mixture. So, there is an alcohol group and a halogen group present in the unknown mixture. To deduce the structure, we use the boiling point we found. Regarding the alcohol group, we may have 1-propanol, 2-butanol or 2-Methyl-2-butanol which range from 97 to 102° C. However, we know that the mixture is insoluble. Insoluble mixtures tend to contain 5 or more carbons. Thus, 2-Methyl-2butanol seems to be the best fit. Regarding the alkyl bromide group, n-butyl seems to be the best fit, at a boiling point of about 102° C. The other options of alkyl bromides occur at temperatures that are too low relative to the boiling point we found.

Conclusion

We believe the following components are present in the unknown mixture we assessed:

Alkyl Bromide: n-Butyl

Alcohol: 2-Methyl-1-butanol

From this lab, we were able to better understand and materialize the processes to assess the presence of certain types of bonds and functional groups in mixtures. Through the series of tests that were performed, deduction functional groups became simpler after every step. Basically, I found that if one test is positive or negative, many possibilities of present functional groups can be disregarded. Through this careful, step-wise experiment, I was undoubtedly able to better understand the nature of deducibility of unknown organic compounds.

  • Share
0 found this helpful