How will you prepare carboxylic acids from Olefins ? Explain with example.

Answer

To prepare carboxylic acids from olefins, we can use a specific reaction involving carbon monoxide and water in the presence of a catalyst. Here’s a step-by-step explanation:

### Step 1: Identify the Olefin
We start with an olefin, which is an alkene. For this example, we will use ethene (C2H4), which has the structure:
\[ \text{CH}_2= \text{CH}_2 \]

### Step 2: Set Up the Reaction Conditions
To convert the olefin to a carboxylic acid, we need to treat ethene with carbon monoxide (CO) and water (H2O). This reaction is typically carried out at a temperature of around 573 Kelvin.

### Step 3: Use a Catalyst
We also need a catalyst to facilitate the reaction. In this case, we will use phosphoric acid (H3PO4) as the catalyst.

### Step 4: Conduct the Reaction
When ethene is treated with carbon monoxide and water in the presence of phosphoric acid at the specified temperature, a reaction occurs. The carbon monoxide and water react with the ethene to form a carboxylic acid.

### Step 5: Formation of Carboxylic Acid
The product of this reaction is propanoic acid (C3H6O2), which has the structure:
\[ \text{CH}_3-\text{CH}_2-\text{COOH} \]

### Summary of the Reaction
The overall reaction can be summarized as follows:
\[ \text{CH}_2= \text{CH}_2 + \text{CO} + \text{H}_2\text{O} \xrightarrow{\text{H}_3\text{PO}_4, 573 \text{K}} \text{CH}_3-\text{CH}_2-\text{COOH} \]

Thus, we have successfully prepared propanoic acid from ethene through the reaction with carbon monoxide and water in the presence of a catalyst.

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