Question

Conversion of carboxylic acid to hydrocarbon is carried out using

• A. \(\text{KMnO}_4\)
• B. \(\text{LiAlH}_4\) and ether
• C. \(\text{NaOH}\) and \(\text{CaO}\)
• D. \(\text{B}_2\text{H}_6\) (Diborane)

Hint:

To convert \(\text{R-COOH}\) into \(\text{R-H}\) (step-down alkane), always use soda lime (\(\text{NaOH} + \text{CaO}\)).
Do not confuse this with Kolbe's electrolytic decarboxylation, which produces a dimerized alkane (\(\text{R-R}\)).

Answer 1

The Correct Option is C

Concept:
Carboxylic acids lose carbon dioxide to form hydrocarbons containing one fewer carbon atom through decarboxylation.
This chemical transformation is carried out using soda lime, which is a dry mixture of sodium hydroxide (\(\text{NaOH}\)) and calcium oxide (\(\text{CaO}\)).

Step 1: Mechanism of Soda-Lime Decarboxylation:
When the sodium salt of a carboxylic acid is heated strongly with soda lime (\(\text{NaOH} : \text{CaO}\) in a \(3:1\) mass ratio), it undergoes decarboxylation to produce an alkane:
\[ \text{R}-\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \, \Delta} \text{R}-\text{H} + \text{Na}_2\text{CO}_3 \] In this reaction, the carboxyl group is removed as sodium carbonate, yielding a hydrocarbon containing one less carbon atom than the parent carboxylic acid.
Calcium oxide (\(\text{CaO}\)) serves as a porous, hygroscopic support that keeps the reaction mixture dry and raises the fusion temperature, preventing the glass vessel from melting.

Step 2: Evaluating Alternative Reagents:
- \(\text{KMnO}_4\) is a powerful oxidizing agent that oxidizes alkyl benzenes, alcohols, and aldehydes to carboxylic acids, rather than converting carboxylic acids to hydrocarbons.
- \(\text{LiAlH}_4\) in dry ether reduces carboxylic acids to primary alcohols (\(\text{R}-\text{CH}_2\text{OH}\)), not hydrocarbons.
- \(\text{B}_2\text{H}_6\) (diborane) selectively reduces carboxylic acids to primary alcohols and does not cleave the carbon chain to form hydrocarbons.
Final Answer:
The reagent used to convert a carboxylic acid to a hydrocarbon is \(\text{NaOH}\) and \(\text{CaO}\), which corresponds to option (C).