Sodalime decarboxylation of sodium propionate produces

Options

A. propane
B. ethane
C. methane
D. butane

Explanation

To solve the question regarding the product of sodalime decarboxylation of sodium propionate, we will follow these steps:

### Step 1: Understand the Reactant
Sodium propionate is represented by the chemical formula CH3CH2COONa. It consists of a propionic acid (propanoic acid) part and a sodium ion.

### Step 2: Identify the Decarboxylation Process
Decarboxylation is a chemical reaction that removes a carboxyl group (-COOH) and releases carbon dioxide (CO2). In the case of sodium propionate, the carboxyl group will be removed, leading to the formation of a hydrocarbon.

### Step 3: Write the Reaction
When sodium propionate undergoes decarboxylation in the presence of soda lime (a mixture of sodium hydroxide and calcium oxide), the reaction can be summarized as follows:
\[ \text{CH}_3\text{CH}_2\text{COONa} \xrightarrow{\text{soda lime}} \text{C}_3\text{H}_8 + \text{NaOH} + \text{CO}_2 \]
Here, C3H8 represents propane.

### Step 4: Determine the Product
From the reaction, we can see that the product formed is propane (C3H8). Therefore, the answer to the question is propane.

### Final Answer
The sodalime decarboxylation of sodium propionate produces **propane**.

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