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Problems · Example 9.6

Q.Sodium salt of which acid will be needed for the preparation of propane ? Write chemical equation for the reaction.

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Sodalime decarboxylation removes exactly one carbon from a carboxylic acid's sodium salt. Propane has 3 carbons, so the salt must come from the 4-carbon acid — butanoic acid: CHX3CHX2CHX2COONa+NaOH→CaO,ΔCHX3CHX2CHX3+NaX2COX3\ce{CH3CH2CH2COONa + NaOH ->[\text{CaO}, \Delta] CH3CH2CH3 + Na2CO3}.

The concept: decarboxylation

Section 9.2.2 gives a standard laboratory route from carboxylic acids to alkanes: heat the sodium salt of the acid with sodalime — a mixture of sodium hydroxide and calcium oxide, written NaOH\ce{NaOH} (CaO). The carboxylate group is eliminated as carbonate, a process called decarboxylation. The essential bookkeeping is that the product alkane always contains one carbon atom fewer than the parent acid, because the carboxyl carbon is the one that leaves.

Step-by-step reasoning

  1. Count the carbons the product needs. Propane is CHX3CHX2CHX3\ce{CH3CH2CH3} — three carbons.

  2. Work backwards to the acid. Since decarboxylation removes one carbon, the acid must have four: CHX3CHX2CHX2COOH\ce{CH3CH2CH2COOH}, butanoic acid. The salt actually heated is its sodium salt, sodium butanoate, CHX3CHX2CHX2COOX−NaX+\ce{CH3CH2CH2COO^-Na^+}.

  3. Write the reaction. The CaO\ce{CaO} does not appear in the equation — it keeps the mixture dry and porous and acts as a heat-transfer medium, which is why it is written over the arrow:

CHX3CHX2CHX2COONa+NaOH→CaO,ΔCHX3CHX2CHX3+NaX2COX3\ce{CH3CH2CH2COONa + NaOH ->[\text{CaO}, \Delta] CH3CH2CH3 + Na2CO3} …

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