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(B) The reaction described is the decarboxylation of a carboxylic acid salt using sodalime $(NaOH + CaO)$.Decarboxylation removes the $-COONa$ group a

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$3-$methylbutanoic acid

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(B) The reaction described is the decarboxylation of a carboxylic acid salt using sodalime $(NaOH + CaO)$.Decarboxylation removes the $-COONa$ group and replaces it with a hydrogen atom,effectively reducing the carbon chain length by one.To produce isobutane $(CH_3-CH(CH_3)-CH_3)$,which has $4$ carbon atoms,the starting acid $X$ must have $5$ carbon atoms.$3-$methylbutanoic acid is $(CH_3)_2CH-CH_2-COOH$.Upon reaction with $NaOH$,it forms sodium $3-$methylbutanoate: $(CH_3)_2CH-CH_2-COONa$.Heating this salt with sodalime results in the loss of $CO_2$ (as $Na_2CO_3$) to yield $(CH_3)_2CH-CH_3$,which is isobutane.

An acid $X$ is neutralized with $NaOH$ and the salt produced is fused with sodalime,resulting in the production of isobutane. $X$ can be:

Similar Questions

The reaction of $RCOOH$ with $R'OH$ takes place in the presence of $H_2SO_4$ to form an ester. The intermediate in the reaction is:

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$RCOOH + NaHCO_3 \to RCOONa + H_2O + CO_2$. In this reaction,the $CO_2$ evolved comes from:

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