Match List-I with List-II.
| List-I (Conversion) | List-II (Reagent and condition) |
|---|---|
| (A) Sodium benzoate → Benzene | (I) Electrolysis of the aqueous solution |
| (B) Sodium propanoate → Butane | (II) Br2 and a small amount of red phosphorus, then H2O |
| (C) Benzoic acid → Phenylmethanol | (III) NaOH and CaO, heat |
| (D) Propanoic acid → 2-Bromopropanoic acid | (IV) LiAlH4 in ether, then H3O+ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-IV, D-II
Look at what happens to the carbon chain in each conversion: losing CO2 with one carbon less, joining two alkyl groups, reducing –COOH to –CH2OH, or putting a halogen on the α-carbon.
(A) Sodium benzoate → Benzene — This matches (III) NaOH and CaO, heat. Heating the sodium salt with soda lime removes CO2; the C6H5 group takes a hydrogen:
\( \displaystyle \ce{C6H5COONa + NaOH} \xrightarrow[\text{heat}]{\ce{CaO}} \ce{C6H6 + Na2CO3} \)
(B) Sodium propanoate → Butane — This matches (I) Electrolysis of the aqueous solution. In Kolbe electrolysis two ethyl groups join, giving a hydrocarbon with twice the carbon atoms of the alkyl group:
\( \displaystyle \ce{2CH3CH2COONa + 2H2O} \xrightarrow{\text{electrolysis}} \ce{CH3CH2CH2CH3 + 2CO2 + H2 + 2NaOH} \)
(C) Benzoic acid → Phenylmethanol — This matches (IV) LiAlH4 in ether, then H3O+. The –COOH group is reduced to a primary alcohol, –CH2OH:
\( \displaystyle \ce{C6H5COOH} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{LiAlH4}\text{/ether}} \ce{C6H5CH2OH} \)
(D) Propanoic acid → 2-Bromopropanoic acid — This matches (II) Br2 and a small amount of red phosphorus, then H2O. This is the Hell-Volhard-Zelinsky reaction, which puts bromine on the α-carbon:
\( \displaystyle \ce{CH3CH2COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{Br2}\text{/red phosphorus}} \ce{CH3CH(Br)COOH} \)
NCERT (2026-27 edition):
8.9.3 Reactions Involving –COOH Group – “Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride or better with diborane.”
“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…”
8.9.4 Substitution Reactions in the Hydrocarbon Part – “Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids. The reaction is known as Hell-Volhard-Zelinsky reaction.”