Question

In a fit of frustration or inspiration you decide to boil (reflux) your unanticipated \(\mathrm{C}_{10} \mathrm{H}_{12}\), and, in order to do that, you attach a long condenser to a receiving vessel that is kept very cold. Your \(\mathrm{C}_{10} \mathrm{H}_{12}\) boils at about \(165^{\circ} \mathrm{C}\), but you collect a product of the formula \(\mathrm{C}_{5} \mathrm{H}_{6}\) boiling at \(40^{\circ} \mathrm{C}\). Moreover, although this \(\mathrm{C}_{5} \mathrm{H}_{6}\) product is stable at low temperature for a long time, at room temperature it rapidly reverts to your starting \(\mathrm{C}_{10} \mathrm{H}_{12}\). Explain.

Short answer

\text{C}_{10}\text{H}_{12} dissociates into \text{C}_{5}\text{H}_{6} on heating. At room temperature, \text{C}_{5}\text{H}_{6} reverts back to \text{C}_{10}\text{H}_{12}.

Step-by-step solution

- Understand the Chemical Reaction

First, recognize the key details: the starting compound is \(\text{C}_{10}\text{H}_{12}\), and the product formed during reflux is \(\text{C}_{5}\text{H}_{6}\). The reaction also indicates the boil points are different: \(\text{C}_{10}\text{H}_{12}\) boils at \(165^{\text{\degree} \text{C}}\) while \(\text{C}_{5}\text{H}_{6}\) boils at \(40^{\text{\degree} \text{C}}\).

- Analyze Molecular Formulas

Notice that the molecular formula \(\text{C}_{10}\text{H}_{12}\) is exactly double that of \(\text{C}_{5}\text{H}_{6}\). This suggests a possible dissociation reaction where one \(\text{C}_{10}\text{H}_{12}\) molecule splits into two \(\text{C}_{5}\text{H}_{6}\) molecules.

- Check Stability and Reversion

Given that \(\text{C}_{5}\text{H}_{6}\) product is stable at low temperatures but reverts to \(\text{C}_{10}\text{H}_{12}\) at room temperature, infer that \(\text{C}_{10}\text{H}_{12}\) is the favored, more stable form under standard conditions. This reversible reaction dynamic is a clue to thermodynamic principles in play.

- Identify Reaction Type

Determine that this situation is likely a dynamic equilibrium between dimer and monomer forms. At high temperatures, energy favors dissociation into \(\text{C}_{5}\text{H}_{6}\), while at room temperature, the system shifts back to the more stable dimer \(\text{C}_{10}\text{H}_{12}\).

Key concepts

Molecular Formula

The molecular formula of a compound tells you the number and type of atoms present. For instance, in the exercise, you start with a compound that has the molecular formula \(\text{C}_{10}\text{H}_{12}\). This means the compound contains 10 carbon atoms and 12 hydrogen atoms. When you boil this compound, it produces a new compound with the formula \(\text{C}_{5}\text{H}_{6}\), indicating the new product has 5 carbon atoms and 6 hydrogen atoms. The new compound's formula is exactly half of the original. This suggests a dissociation or splitting reaction where one molecule of \(\text{C}_{10}\text{H}_{12}\) breaks down into two molecules of \(\text{C}_{5}\text{H}_{6}\). Knowing the molecular formula helps us understand how the compounds relate to each other and what the reaction likely involves.

To summarize:

• \(\text{C}_{10}\text{H}_{12}\) is the original compound.
• \(\text{C}_{5}\text{H}_{6}\) is the dissociated product.
• The product's formula implies a splitting or dissociation reaction.

Boiling Point

The boiling point is the temperature at which a liquid turns into vapor. It's a critical physical property that helps us identify and separate compounds. In this exercise, the initial compound \(\text{C}_{10}\text{H}_{12}\) boils at \(165^{\text{\textdegree} \text{C}}\), indicating it needs significant heat to vaporize. Upon boiling, a new product, \(\text{C}_{5}\text{H}_{6}\), is collected, which boils at \(40^{\text{\textdegree} \text{C}}\). The much lower boiling point of \(\text{C}_{5}\text{H}_{6}\) compared to \(\text{C}_{10}\text{H}_{12}\) shows it requires less energy to vaporize. This difference is likely due to the smaller size and mass of the \(\text{C}_{5}\text{H}_{6}\) molecules, making them less stable and easier to vaporize.

Key points:

• \(\text{C}_{10}\text{H}_{12}\) boils at \(165^{\text{\textdegree} \text{C}}\).
• \(\text{C}_{5}\text{H}_{6}\) boils at \(40^{\text{\textdegree} \text{C}}\).
• The boiling point reflects molecular size and stability.

Dynamic Equilibrium

Dynamic equilibrium occurs in reversible reactions where the forward and reverse reactions happen at the same rate. This means that the concentration of reactants and products remains constant over time. In the given exercise, \(\text{C}_{10}\text{H}_{12}\) can dissociate into \(\text{C}_{5}\text{H}_{6}\) at high temperatures but reverts back at room temperature. When both compounds are present and can transform into each other, they form a dynamic equilibrium. This state is temperature-dependent, meaning high temperatures favor dissociation into the monomers \(\text{C}_{5}\text{H}_{6}\), while at room temperature, the system shifts back to the more stable dimer form \(\text{C}_{10}\text{H}_{12}\).

Summary of points:

• Dynamic equilibrium means constant concentrations of reactants and products.
• In this reaction, \(\text{C}_{10}\text{H}_{12}\) ⇌ 2 \(\text{C}_{5}\text{H}_{6}\).
• Temperature affects which direction the equilibrium favors.
• At high temperatures, dissociation is favored; at room temperature, recombination is favored.

Stability

Stability in chemistry refers to how likely a compound is to remain in its current form without changing. The compound \(\text{C}_{5}\text{H}_{6}\) from the exercise is stable at low temperatures, meaning it doesn’t easily revert to \(\text{C}_{10}\text{H}_{12}\) under those conditions. However, at room temperature, it rapidly reverts back to \(\text{C}_{10}\text{H}_{12}\), indicating that \(\text{C}_{10}\text{H}_{12}\) is the more stable form at those temperatures. This suggests that the dimer form is favored because it is energetically more stable under standard conditions. The reversion implies that while \(\text{C}_{5}\text{H}_{6}\) can exist temporarily, \(\text{C}_{10}\text{H}_{12}\) is the preferred stable state at typical temperatures.

Important takeaways:

• \(\text{C}_{5}\text{H}_{6}\) is stable at low temperatures.
• \(\text{C}_{10}\text{H}_{12}\) is more stable at room temperature.
• Stability depends on environmental conditions like temperature.

Thermodynamics

Thermodynamics involves the study of energy changes, including heat and work, during chemical reactions. In this context, when \(\text{C}_{10}\text{H}_{12}\) boils and dissociates into \(\text{C}_{5}\text{H}_{6}\), the process involves an energy input (heat). At high temperatures, the energy provided is enough to break the bonds in \(\text{C}_{10}\text{H}_{12}\), leading to its dissociation. However, as the temperature drops to room temperature, the thermodynamic equilibrium shifts. The energy conditions favor reforming the stable \(\text{C}_{10}\text{H}_{12}\) dimer due to lower energy and entropy changes. Thus, the thermodynamic principles help explain why \(\text{C}_{10}\text{H}_{12}\) is the stable product under standard conditions while at higher temperatures, dissociation occurs.

Key insights:

• Thermodynamics involves the study of energy changes.
• At high temperatures, bond energies favor dissociation to \(\text{C}_{5}\text{H}_{6}\).
• At room temperature, equilibrium shifts to the more stable \(\text{C}_{10}\text{H}_{12}\).