Question

Conversion of carboxylic acid to hydrocarbon is carried out using

• A. \(KMnO_4\)
• B. \(LiAlH_4\) and ether
• C. NaOH and CaO
• D. \(B_2H_6\) (Diborane)

Hint:

Think of decarboxylation of the sodium salt with soda lime.

Answer 1

The Correct Option is C

Step 1: Idea: look at what must leave
A hydrocarbon has no oxygen. So the whole $COOH$ group must leave. The only reagent set that takes it out as a carbonate is soda lime.

Step 2: Test each reagent by the product
$KMnO_4$ is an oxidant and would not change a carboxyl group. $LiAlH_4$ adds hydrogen and gives $RCH_2OH$. $B_2H_6$ also gives $RCH_2OH$ from the acid. None of the three reach $R-H$.

Step 3: Soda lime route
The acid is first neutralised to $RCOONa$. Heating with solid NaOH and CaO breaks the C-C bond next to the carboxylate. The product is $RH$ and sodium carbonate, so the carbon of the carboxyl goes out as carbonate.

Step 4: Example
Sodium ethanoate on heating with soda lime gives methane: \[ CH_3COONa + NaOH \xrightarrow{CaO} CH_4 + Na_2CO_3 \] This shows an acid becoming a hydrocarbon.

Step 5: Pick the option
The reagent pair is NaOH and CaO, which is option 3.

Final Answer:
Option 3, NaOH and CaO (soda lime), converts the acid to a hydrocarbon. \[ \boxed{\text{NaOH and CaO}} \]