The decarboxylation of ethanoic acid will produce carbon (IV)oxide and

1. Analyze the question

The question asks for the product formed alongside carbon(IV) oxide ($CO_2$) when ethanoic acid undergoes decarboxylation.

2. Explain the concept

Decarboxylation is a chemical reaction that involves the removal of a carboxyl group ($-COOH$) from a compound, usually in the form of carbon(IV) oxide ($CO_2$).

In the laboratory, the decarboxylation of ethanoic acid is typically achieved by heating its sodium salt (sodium ethanoate) with sodalime (a mixture of sodium hydroxide, $NaOH$, and calcium oxide, $CaO$). This is a standard method for preparing alkanes in organic chemistry.

The general equation for the decarboxylation of a sodium salt of a carboxylic acid is:

3. Show the solution

• Step 1: Identify the reactant. Ethanoic acid is $CH_3COOH$. Its sodium salt is sodium ethanoate ($CH_3COONa$).
• Step 2: Apply the decarboxylation process. When sodium ethanoate is heated with sodalime, the $-COONa$ group is removed as $Na_2CO_3$ (or $CO_2$ is evolved in the process).
• Step 3: Determine the remaining hydrocarbon. The methyl group ($CH_3$) from the ethanoate picks up a hydrogen atom from the $NaOH$ to form methane ($CH_4$).

Reaction:

(Note: In the presence of sodalime, the $CO_2$ reacts with the base to form sodium carbonate, but the hydrocarbon product remains methane).

4. Identify the answer

Comparing the result ($CH_4$) to the given options: A. methane B. ethane C. propane D. butane

The correct product is methane.

5. Tip

Exam Strategy: Remember the "n-1" rule for decarboxylation. If you start with a carboxylic acid containing $n$ carbon atoms, the resulting alkane will always have $n-1$ carbon atoms. Ethanoic acid has 2 carbons; therefore, the product must have $2 - 1 = 1$ carbon (methane).