Identify the product \(‘X’\) in the following reaction:
\(C_2H_5COONa \underset{Heat}{\overset{NaOH + CaO}{\longrightarrow}} X + Na_2CO_3\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(C_2H_6\)
In decarboxylation, the sodium salt of a carboxylic acid is heated with sodalime (NaOH and CaO in the ratio 3 : 1). The –COONa group is removed as \( \ce{Na2CO3} \) and replaced by H, so the hydrocarbon formed has one carbon atom less than the acid:
\( \ce{R-COONa} \xrightarrow[\Delta]{\ce{NaOH + CaO}} \ce{R-H + Na2CO3} \)
Sodium propanoate has R = \( \ce{C2H5} \), so X is ethane:
\( \ce{C2H5COONa + NaOH} \xrightarrow[\Delta]{\ce{CaO}} \ce{C2H6 + Na2CO3} \)
Why the other options are incorrect
Option (1) C2H4 — Incorrect. Ethene is an alkene; decarboxylation puts an H in place of –COONa and gives a saturated hydrocarbon, not a double bond.
Option (3) C3H8 — Incorrect. Propane has three carbons, the same as propanoate; it would need the carboxyl carbon to stay. It is formed from sodium butanoate.
Option (4) C3H6 — Incorrect. Propene has both the wrong number of carbons and a double bond.
NCERT (2026-27 edition):
Decarboxylation
