Carboxylic Acids - Decarboxylation
Sodium ethanoate is heated strongly with soda lime in a decarboxylation reaction, so which hydrocarbon is liberated as the chief gaseous product?
Select the correct option:
Solution
Methane
Decarboxylation removes carbon dioxide from the carboxylate group of a salt, and when a sodium salt of a carboxylic acid is heated with soda lime, the carbon bearing the carboxylate is replaced by hydrogen. This means the alkyl group attached to the COONa fragment is liberated as an alkane with one fewer carbon than the original acid salt. Sodium ethanoate, CH3COONa, has a two-carbon acid, so removing CO2 leaves the methyl group, which picks up hydrogen to give methane as the chief product, making methane the correct answer. Ethane is wrong because it would require the original carbon skeleton to be retained, whereas decarboxylation shortens the chain by one carbon. Ethene cannot form because no elimination to a double bond occurs in this reaction. Ethanal is wrong since decarboxylation does not yield an aldehyde. This laboratory route to alkanes is described in NCERT and is a recurring JEE Advanced product question. Soda lime, a mixture of sodium hydroxide and calcium hydroxide, is preferred over pure sodium hydroxide because it is less corrosive to glass and easier to handle while still supplying the strong base needed to expel carbon dioxide. The reaction is a useful descending step in a homologous series, trimming one carbon at a time from an acid. As a check, an n-carbon acid salt gives an (n-1)-carbon alkane, so ethanoate gives methane.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing oxygen
- Topic
- carboxylic acids - decarboxylation
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
Methane
Decarboxylation removes carbon dioxide from the carboxylate group of a salt, and when a sodium salt of a carboxylic acid is heated with soda lime, the carbon bearing the carboxylate is replaced by hydrogen. This means the alkyl group attached to the COONa fragment is liberated as an alkane with one fewer carbon than the original acid salt. Sodium ethanoate, CH3COONa, has a two-carbon acid, so removing CO2 leaves the methyl group, which picks up hydrogen to give methane as the chief product, making methane the correct answer. Ethane is wrong because it would require the original carbon skeleton to be retained, whereas decarboxylation shortens the chain by one carbon. Ethene cannot form because no elimination to a double bond occurs in this reaction. Ethanal is wrong since decarboxylation does not yield an aldehyde. This laboratory route to alkanes is described in NCERT and is a recurring JEE Advanced product question. Soda lime, a mixture of sodium hydroxide and calcium hydroxide, is preferred over pure sodium hydroxide because it is less corrosive to glass and easier to handle while still supplying the strong base needed to expel carbon dioxide. The reaction is a useful descending step in a homologous series, trimming one carbon at a time from an acid. As a check, an n-carbon acid salt gives an (n-1)-carbon alkane, so ethanoate gives methane.
This hard difficulty chemistry question is from the chapter organic compounds containing oxygen, covering the topic of carboxylic acids - decarboxylation. It appeared in the 2025 exam.
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