Question

Write the equation for the oxidation of propene by \(\mathrm{KMnO}_{4}\) solution at \(\mathrm{pH}=7.0 .\) Show any intermediates. Assume cold and dilute conditions.

Short answer

The balanced equation for the oxidation of propene by \(\mathrm{KMnO}_{4}\) at \(\mathrm{pH}=7.0\) under cold and dilute conditions is: 3C\(_3\)H\(_6\) (g) + 3OH\(^-\) (aq) + \(\mathrm{KMnO}_4\) (aq) + 2H\(_2\)O (l) → 3C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq).

Step-by-step solution

Write the unbalanced equation

Write the chemical formulas of the reactants and products. Reactants: Propene (C\(_3\)H\(_6\)) and potassium permanganate (\(\mathrm{KMnO}_{4}\)) Products: Propanol (C\(_3\)H\(_7\)OH) and manganese dioxide (MnO\(_2\)) Unbalanced equation: C\(_3\)H\(_6\) (g) + \(\mathrm{KMnO}_{4}\) (aq) → C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq)

Balance the equation

Balance the equation by ensuring that the number of atoms for each element is the same on both sides of the equation. Balanced equation: C\(_3\)H\(_6\) (g) + \(\mathrm{KMnO}_{4}\) (aq) + H\(_2\)O (l) → C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq)

Show the intermediates

In cold, dilute conditions, potassium permanganate is reduced to form manganese dioxide, while the propene is oxidized to form propanol. This can be divided into two half-reactions: one for the oxidation of propene and another for the reduction of potassium permanganate. Oxidation half-reaction: C\(_3\)H\(_6\) (g) + OH\(^-\) (aq) → C\(_3\)H\(_7\)OH (l) + e\(^-\) Reduction half-reaction: \(\mathrm{MnO}_{4}^-\) (aq) + 2H\(_2\)O (l) + 3e\(^-\) → MnO\(_2\) (s) + 4OH\(^-\) (aq)

Combine the half-reactions to form the final equation

Multiply the half-reactions by the appropriate factors to equalize the number of electrons transferred. Then, add the two half-reactions together, canceling out any species that appear on both sides. Oxidation: 3(C\(_3\)H\(_6\) (g) + OH\(^-\) (aq) → C\(_3\)H\(_7\)OH (l) + e\(^-\)) Reduction: (MnO\(_4^-\) (aq) + 2H\(_2\)O (l) + 3e\(^-\) → MnO\(_2\) (s) + 4OH\(^-\) (aq)) Final balanced equation: 3C\(_3\)H\(_6\) (g) + 3OH\(^-\) (aq) + \(\mathrm{MnO}_4^-\) (aq) + 2H\(_2\)O (l) → 3C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + 4OH\(^-\) (aq) + KOH (aq). Thus, the balanced equation for the oxidation of propene by \(\mathrm{KMnO}_{4}\) at \(\mathrm{pH}=7.0\) is: 3C\(_3\)H\(_6\) (g) + 3OH\(^-\) (aq) + \(\mathrm{KMnO}_4\) (aq) + 2H\(_2\)O (l) → 3C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq).

Key concepts

Oxidation Reactions

Oxidation reactions are a fundamental part of organic chemistry. The process of oxidation involves the loss of electrons or an increase in the oxidation state of a molecule. In simpler terms, oxidation is often the gain of oxygen or the loss of hydrogen in organic compounds. This is a key reaction in the synthesis of various organic compounds, allowing for the modification of molecular structures for desired chemical transformations.

In the context of this exercise, the oxidation of propene ( C_3H_6 ) by potassium permanganate ( KMnO_4 ) is highlighted. A characteristic feature of oxidation in organic molecules is the introduction of an additional oxygen atom to the carbon structure. As a result, propene is converted into propanol ( C_3H_7OH ), an alcohol. Note that the addition of oxygen elevates the molecule's oxidation state.

This kind of reaction is crucial in industry and biological systems alike. For example, alcohols, such as propanol, are important intermediates and end products in the production of many chemical goods.

Potassium Permanganate

Potassium permanganate (KMnO_4) serves as a vigorous oxidizing agent in many organic reactions. It's known for its versatility and effectiveness in promoting the oxidation of a wide variety of chemical compounds. Its distinct deep purple color also makes it quite recognizable.

In reactions like the oxidation of propene, potassium permanganate undergoes a chemical change itself. Under the specific conditions mentioned—cold, dilute, and at \(pH = 7\)—KMnO_4does not fully break down. Instead, it is reduced to manganese dioxide (MnO_2). This is a stabilizing factor, ensuring the reaction does not excessively proceed, which might form unwanted by-products. Here's how it breaks down:

• As an oxidizing agent, it drives the transfer of oxygen to propene.
• The change in manganese's oxidation state from +7 in KMnO_4 to +4 in MnO_2 releases energy required for the oxidation of propene.

This makes potassium permanganate an effective choice for controlled oxidations in laboratory settings, especially in reactions requiring precision, such as synthesis and analytical chemistry.

Chemical Equation Balancing

Chemical equation balancing is essential for accurately representing chemical reactions. The purpose is to ensure that the same number of each type of atom appears on both sides of the equation, satisfying the law of conservation of mass. This process can sometimes seem daunting, but by breaking it down into steps, it becomes manageable.

To start, first write the unbalanced equation. Identify all reactants and products involved as seen in this exercise with propene and potassium permanganate converting to propanol and manganese dioxide.

Next, balance one element at a time by adjusting the coefficients (the numbers before molecules), rather than altering the actual chemical formulas. In a complex reaction like the one shown, consider using half-reactions. This method involves splitting the reaction into two parts:

• The oxidation half-reaction where electrons are lost.
• The reduction half-reaction where electrons are gained.

Each half-reaction is balanced separately first for atoms, then for charge, before combining them to formulate the complete and balanced chemical equation.

Once practiced, equation balancing becomes an invaluable skill to accurately communicate chemical processes.