Problem 40 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

Acetic acid, when reacts with excess of HI in the presence of red phosphorus gives (a) Ethanol (b) Ethane (c) Acetaldehyde (d) Acetone

Step-by-Step Solution

Answer

The final product is ethane (b).

Step 1: Understand the Reaction

Acetic acid \(CH_3COOH\) reacts with hydrogen iodide \(HI\) and red phosphorus. This reaction is known as HI reduction, where the carboxylic acid is reduced.

Step 2: Identify the First Reaction Product

In the presence of red phosphorus, HI provides hydrogen ions which reduce the acetic acid. Initially, \(CH_3COOH\) converts to ethanol \(CH_3CH_2OH\) through reduction.

Step 3: Recognize Further Reduction

Furthermore, excess HI will continue to react with ethanol. The ethanol is further reduced to ethane \(C_2H_6\).

Step 4: Conclusion

Given that acetic acid reacts with excess HI, the final product of the reduction process is ethane \((b) Ethane\).

Key Concepts

Acetic Acid Reduction

Acetic acid, with the chemical formula \(CH_3COOH\), is a simple carboxylic acid. An important reaction involving acetic acid is its reduction, a process where the acid group is converted into a less oxidized form. In the laboratory setting, reductions of carboxylic acids often involve strong reducing agents.When acetic acid is exposed to hydrogen iodide (HI) in the presence of red phosphorus, a complete reduction process occurs. Initially, the acetic acid is reduced to ethanol (\(CH_3CH_2OH\)), which is an alcohol. This is because HI acts as a proton donor in the presence of red phosphorus, leading to the replacement of the carboxylic group with hydrogen atoms.The reaction sequence can be understood in two steps:

• Reduction of acetic acid (\(CH_3COOH\)) to form ethanol (\(CH_3CH_2OH\)).
• Further reduction of ethanol to ethane (\(C_2H_6\)) if excess HI is present.

These transformations highlight the versatility of reduction reactions in organic chemistry, allowing the conversion of an acid into different types of hydrocarbons.

Hydrogen Iodide Reactions

Hydrogen iodide (HI) is a diatomic molecule known for its strong reducing capabilities in organic chemistry. Its reactivity makes it suitable to break down various functional groups, especially in the presence of red phosphorus. When it comes to the reduction of carboxylic acids like acetic acid, HI plays a crucial role. It dissociates into hydrogen and iodide ions. The hydrogen ions act as the reducing agent, while the iodide ions balance the charge. Here's how it contributes to the process:

• Provides a source of hydrogen to facilitate the reduction of carboxylic acids to alcohols.
• Allows further reduction from alcohols to alkanes under excess conditions.

HI, with its dual role as both a reagent and catalyst, enables complex reactions like converting acetic acid to ethane. This demonstrates the effectiveness of hydrogen iodide in organic transformations.

Red Phosphorus in Organic Reactions

Red phosphorus is often employed in organic synthesis, particularly in reactions involving reductions and low-temperature processes. Its presence optimizes conditions that facilitate the reduction reactions like the one involving acetic acid and hydrogen iodide. Here’s how red phosphorus operates in these reactions:

• Acts as a catalyst when combined with HI, enhancing its reducing power.
• Maintains stable reaction conditions by preventing undesirable oxidation or side reactions.

In the HI reduction process, red phosphorus supports the sustained activity of hydrogen iodide, ensuring that the reduction proceeds efficiently from initial acid to the final alkane products. This role in organic reactions makes red phosphorus a valuable additive, especially when precise reductions are necessary to yield specific hydrocarbons such as ethanol and ethane from acetic acid.