Beilstein Test Lab Report

Purpose: To help students identify the functional groups in a mixture of two compounds by interpreting the IR spectroscopy, measuring the boiling point, comparing the physical and chemical properties of an unknown with a known substance. In addition, students will learn how impurities affect the boiling point and solubility of a substance.

Procedure: The student will read the IR spectroscopy of the unknown first. Then perform various tests and measure the boiling point.

Beilstein Test

The Beilstein test determines if there is a haloalkane (R-Cl, R-Br, or R-I) in the mixture. A copper wire is heated to remove any residue. Then it is dipped into the sample and heated again to observe the presence of a green flame. The green flame confirms

the presence of a haloalkane. This is due to the CuO reacting with a haloalkane (R-X) to produce CuX2 with excited Cu ions that produce the green color flame. This reaction only works with R-Cl, R-Br, or R-I because they are exothermic and R-F is too endothermic. If there is no positive result, then there might be a R-F or a no haloalkane.

Reaction: 2CuO +R-X + Heat 2CuX2 + Heat Green flame

Baeyer Test
This test requires ethanol, KMnO4, and the unknown sample to be mixed in a vial to observe if there are any alkenes or alkynes in the mixture. If the orange color from the KMnO4 becomes colorless, then there is an alkene or an alkyne because the strong oxidizing agent, KMnO4, oxidizes the alkene to an alcohol, thus producing a brown color. KMnO4 is not a good oxidizing agent since it also produces other functional groups.

Reaction: Alkene + KMnO4 (purple)+ H2O diols + MnO2 (brown)

Chromic Acid Test
For this test, the unknown, acetone, and the chromic acid reagent are mixed to observe if there are any alcohols or aldehydes in the mixture. For a primary or secondary alcohol, the orange color from H2CrO4 will transform into a green color in Cr2(SO4)3 as it is oxidized to a ketone or a carboxylic acid. However, a tertiary alcohol or no alcohol will have a negative test result which is observed by a persistent pink color after it is mixed.

Reaction: H2CrO4 (orange) + H2SO4 + R-CH2OH R-COOH + Cr2(SO4)3 (Green) 4H2O

Solubility Test
For this test, the unknown and water is mixed to determine the solubility (insoluble, slightly soluble, or soluble). If the mixture is soluble then it contains functional groups that have polar bonds.

Results: Unknown C

Discussion: The IR contains a broad peak at 3400-3650 cm-1 for an alcohol and a peak around 800 cm-1 for a haloalkane.

My mixture had an average boiling point of 99.5 degrees Celsius. My unknown C was positive for the Beilstein test, meaning that there is either a bromoalkane, chloroalkane or iodoalkane. The IR spectroscopy confirms this because there is a peak around 800 cm-1 which is the haloalkane range. It was slightly soluble; therefore, there might be a functional group that is polar such as an alcohol. Once again, the IR spectroscopy confirms this since the broad peak is located at around 3400 cm-1. Unknown C was negative for the Chromic acid test so there might be a tertiary alcohol or no alcohol. Since the IR has an alcohol I assume that there might be a tertiary alcohol. After completing the test, I can minimize the options that correspond to the IR and the tests to draw a conclusion for the functional groups found in the mixture. For my unknown C, I narrowed down the options to 2-methyl-2-butanol and 1-bromobutane because these compounds have a boiling point of 102 degrees Celsius which is close to my readings of 99 and 100 degrees Celsius. 2-Methyl-2-butanol has a tertiary alcohol, which prevents it from undergoing a chromic acid test and the 2-methyl-2-butanol has a bromoalkane which was identified by the Beilstein test and the

IR.

Conclusion: I learned how to identify two functional groups found in a mixture by completing various tests such as a chromic acid test. I also learned how and why the boiling point may differ due to impurities. Besides the boiling point being affected, I also learned that we should perform tests with a known substance and compare it to the unknown substance to have a control group and confirm that there are no false positive results. In addition, I confirmed some possible answers with the peaks from the IR. With the help of all previous tests and readings, I can draw a conclusion as to which functional groups are in my mixture.

Possible Alcohol: 2-methyl-2-butanol
Possible Haloalkane: 1-bromo-butane