Question

On the basis of physical properties, an unknown compound is believed to be one of the following: n-pentane (b.p. 36 \(^{\circ}\) ) 1-pentyne (b.p. \(40^{\circ}\) ) 2 -pentene (b.p. 36 \(^{\circ}\) ) methylene chloride (b.p. \(40^{\circ}\) ) Describe how you would go about finding out which of the possibilities the unknown actually is. Where possible, use simple chemical tests; where necessary, use more elaborate chemical methods like quantitative hydrogenation and cleavage. Tell exactly what you would do and see.

Short answer

To identify the unknown compound, perform the following series of tests: 1. Test the solubility of the compound in water: If it is insoluble, it is likely an organic compound (n-pentane, 1-pentyne, or 2-pentene); if soluble, it is likely methylene chloride. 2. React the compound with bromine (Br₂): If the reddish-brown color disappears, the compound has a double or triple bond (1-pentyne or 2-pentene); if it remains, the compound is saturated (n-pentane). 3. If a double or triple bond is present, react the compound with Baeyer's reagent (KMnO₄): If the purple color disappears, the compound has a double bond (2-pentene); if it remains, it has a triple bond (1-pentyne). These tests will help identify the unknown compound among the given possibilities.

Step-by-step solution

Test 1: Solubility in Water

First, you will test the solubility of the unknown compound in water. Add a small amount of the unknown compound to a test tube containing water. Shake well and observe for any change. - If the compound is insoluble in water, it is likely to be an organic compound, which narrows down our choices to n-pentane, 1-pentyne, or 2-pentene. Methylene chloride is soluble in water. - If the compound is soluble in water, it is likely to be methylene chloride.

Test 2: Reaction with Br₂ (Bromine)

Next, test the unknown compound's reactivity towards bromine (Br₂). Add a solution of bromine to a test tube containing the unknown compound. Observe for any change in color. - If the bromine solution's reddish-brown color disappears or reduces, it indicates the presence of a double bond (alkene) or triple bond (alkyne) in the compound, which means it could be 1-pentyne or 2-pentene. - If the bromine solution remains reddish-brown, it suggests that the unknown compound is saturated (no double or triple bonds), which points to n-pentane as the unknown compound.

Test 3: Reaction with Baeyer's Reagent

If the above tests indicate the presence of a double or triple bond, you can further differentiate between an alkene (2-pentene) and an alkyne (1-pentyne) using Baeyer's reagent (aqueous solution of potassium permanganate, KMnO₄). Add the reagent to a test tube containing the unknown compound and observe for any change in color. - If the purple color of potassium permanganate disappears or turns brown, it indicates the presence of a double bond, which means the unknown compound is 2-pentene. - If the purple color remains, it suggests the presence of a triple bond, which means the unknown compound is 1-pentyne. By following these simple chemical tests, the unknown compound can be identified among the given possibilities.

Key concepts

Chemical Tests

Chemical tests are fundamental in identifying organic compounds. They are performed to spot specific functional groups and structural elements in unknown substances. In this exercise, we take advantage of chemical reactions that yield observable changes to pinpoint our unknown compound.

For instance, when testing for alkenes or alkynes, bromine water is a common reagent. A positive test is indicated by a loss of the bromine's reddish-brown color, which signals the presence of either a double or triple carbon-carbon bond. These tests are simple yet revealing, helping to differentiate among compounds that appear similar physically.

Additionally, Baeyer's test using potassium permanganate (KMnO₄) further distinguishes alkenes and alkynes. The decolorization of the purple solution to brown determines the presence of an alkene due to a double bond, whereas no change indicates an alkyne with a triple bond. These straightforward reactions are brilliant tools in the chemist's toolkit.

Physical Properties

Physical properties, such as boiling point and solubility, are intrinsic to a substance. They play a crucial role in identifying unknown compounds. For the unknown in this problem, we use the boiling point as a critical piece of evidence. The compounds considered have very close boiling points, necessitating a careful approach to distinction.

Solubility in water is a quick and effective test. Organic compounds like n-pentane, 1-pentyne, and 2-pentene generally show little solubility in water. Methylene chloride, however, being polar, is soluble. Hence, by observing whether the compound dissolves in water, we can first eliminate some possibilities.

These initial physical property tests are essential as they offer non-reactive separation of options before moving on to more involved chemical tests.

Alkene

Alkenes are hydrocarbons with at least one carbon-carbon double bond, denoted as C=C. This functional group is reactive because of the loosely held \( ext{π}\) bonds. In our exercise, the compound 2-pentene is an alkene, which distinguishes itself from other hydrocarbons through this reactivity.

The presence of a double bond in an alkene causes a reaction with both bromine and Baeyer's reagent. Importantly, for bromine, a successful reaction is marked by a fading of the reddish-brown color. With Baeyer's reagent, the change from purple to brown notes the alkene’s ability to undergo epoxy formation.

Alkenes' reactivity is what makes these tests useful—by interacting with specifically targeted reagents, they make themselves known even in a mix of similar boiling point compounds.

Alkyne

Alkynes are unsaturated hydrocarbons containing one or more carbon-carbon triple bonds. These bonds are represented as R-C≡C-R. The compound 1-pentyne mentioned in this problem is an alkyne.

These triple bonds give alkynes a unique status due to their linear structure and sp hybridization. Alkynes react with bromine similarly to alkenes by decolorizing it, but do not react with Baeyer's reagent. This difference is critical in distinguishing 1-pentyne from 2-pentene when both dissolve the bromine color.

By applying bromine and Baeyer's test, we can ascertain the nature of the unsaturation. This step is crucial for telling apart molecules like 1-pentyne, despite physical similarities to its alkene or alkane counterparts.