DECARBOXYLATION refers to the loss of $\mathrm{CO_2}$ from a carboxylic acid (or its salt). When sodium acetate is HEATED with soda lime ($\mathrm{NaOH}$ + $\mathrm{CaO}$), the major organic product is

1. NCERT §8.6 (Carboxylic Acids — chemical reactions): heating the sodium SALT of a carboxylic acid with SODA LIME removes CO$_2$. 2. Generic equation: $\mathrm{R{-}COONa} + \mathrm{NaOH} \xrightarrow{\text{CaO},\,\Delta} \mathrm{R{-}H} + \mathrm{Na_2CO_3}$. 3. For sodium acetate, $\mathrm{R} = \mathrm{CH_3}$. So the alkane formed is $\mathrm{CH_3{-}H} = \mathrm{CH_4}$ — METHANE. 4. Equation: $\mathrm{CH_3COONa} + \mathrm{NaOH} \xrightarrow{\text{CaO},\,\Delta} \mathrm{CH_4} + \mathrm{Na_2CO_3}$. 5. Option A is the WURTZ-style coupling product (two methyl radicals joining), which is NOT what soda-lime decarboxylation produces. Option C ignores the reaction. Option D is the partial oxidation product, not what soda lime gives. 6. This is one of the classic laboratory preps of methane. _Source: NCERT Class 12 Chemistry Part 2, Ch 8, §8.6 (Decarboxylation of carboxylic acid salts), p. 19–20._