Sodium salts of carboxylic acids (RCOONa) when heated with sodalime, yield _____.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(R-H\)
When the sodium salt of a carboxylic acid is heated with sodalime (NaOH and CaO in the ratio 3 : 1), the carboxylate group loses carbon dioxide, which is taken up by the alkali as sodium carbonate. The alkyl group is left with a hydrogen in place of –COONa, so a hydrocarbon with one carbon atom less is formed. This is decarboxylation:
\( \displaystyle \ce{R-COONa + NaOH} \xrightarrow[\Delta]{\ce{CaO}} \ce{R-H + Na2CO3} \)
For example, sodium ethanoate gives methane. So the product is R–H.
Why the other options are incorrect
Option (2) \((R-COO)_2Ca\) — Incorrect. A calcium carboxylate is formed when the acid is neutralised by a calcium base; in sodalime the calcium oxide only keeps the sodium hydroxide dry, and the carboxylate loses \( \ce{CO2} \).
Option (3) \(RCOONa\) — Incorrect. This is the starting salt; heating with sodalime changes it.
Option (4) \(RCHO\) — Incorrect. An aldehyde would need a reduction of an acid derivative, for example Rosenmund reduction of an acyl chloride; decarboxylation removes the carboxyl carbon altogether.
NCERT (2026-27 edition):
Decarboxylation
