Match List-I with List-II.
| List-I (Reactant and reagent) | List-II (Organic product) |
|---|---|
| (A) Ethanoic acid heated with P2O5 | (I) Methane |
| (B) Ethanoic acid with NH3, then heated strongly | (II) Ethanoic anhydride |
| (C) Sodium ethanoate heated with NaOH and CaO | (III) Ethane |
| (D) Electrolysis of an aqueous solution of sodium ethanoate | (IV) Ethanamide |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-IV, C-I, D-III
Two of the reactions replace the –OH of the carboxyl group; the other two remove the carboxyl group as carbon dioxide, and they differ in how many carbon atoms the hydrocarbon keeps.
(A) Ethanoic acid heated with P2O5 — This matches (II) Ethanoic anhydride. P2O5 takes one molecule of water from two molecules of the acid:
\( \displaystyle \ce{2CH3COOH} \xrightarrow{\ce{P2O5},\ \Delta} \ce{(CH3CO)2O + H2O} \)
(B) Ethanoic acid with NH3, then heated strongly — This matches (IV) Ethanamide. The acid first gives ammonium ethanoate; at a high temperature this salt loses water to give the amide:
\( \displaystyle \ce{CH3COOH + NH3 <=> CH3COONH4} \xrightarrow[-\ce{H2O}]{\Delta} \ce{CH3CONH2} \)
(C) Sodium ethanoate heated with NaOH and CaO — This matches (I) Methane. Heating with soda lime is decarboxylation: CO2 is lost (as Na2CO3) and the hydrocarbon has one carbon atom less than the acid:
\( \displaystyle \ce{CH3COONa + NaOH} \xrightarrow[\text{heat}]{\ce{CaO}} \ce{CH4 + Na2CO3} \)
(D) Electrolysis of an aqueous solution of sodium ethanoate — This matches (III) Ethane. In Kolbe electrolysis two alkyl groups join, so the hydrocarbon has twice the number of carbon atoms of the alkyl group (CH3– gives CH3–CH3):
\( \displaystyle \ce{2CH3COONa + 2H2O} \xrightarrow{\text{electrolysis}} \ce{CH3-CH3 + 2CO2 + H2 + 2NaOH} \)
NCERT (2026-27 edition):
8.9.2 Reactions Involving Cleavage of C–OH Bond – “1. Formation of anhydride
Carboxylic acids on heating with mineral acids such as H2SO4 or with P2O5 give corresponding anhydride.”
“4. Reaction with ammonia
Carboxylic acids react with ammonia to give ammonium salt which on further heating at high temperature give amides.”
8.9.3 Reactions Involving –COOH Group – “Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaO in the ratio of 3 : 1). The reaction is known as decarboxylation.”
“Alkali metal salts of carboxylic acids also undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid. The reaction is known as Kolbe electrolysis…”