Question

What does sodium phthalate give on reaction with sodalime?

• A. Benzene
• B. Toluene
• C. Phthalic acid
• D. Phenol

Hint:

Sodalime decarboxylation is a common method to convert sodium salts of carboxylic acids into hydrocarbons. For dicarboxylic acids or their salts, if both carboxyl groups are removed, the resulting hydrocarbon will have two fewer carbons than the original acid if it were aliphatic, or result in the base aromatic ring if it was an aromatic dicarboxylic acid like phthalic acid.

Answer 1

The Correct Option is A

Step 1: Understanding the reaction of a carboxylate with sodalime.
The reaction of a sodium salt of a carboxylic acid with sodalime (NaOH + CaO) is a decarboxylation reaction. In this reaction, a carboxyl group (-COOH) is removed from the molecule as carbon dioxide, typically leading to the formation of an alkane or an aromatic hydrocarbon with one less carbon atom in the main chain or ring for each carboxyl group removed. Calcium oxide (CaO) helps to keep the mixture dry and acts as a flux.
Step 2: Identifying Sodium Phthalate and its structure.
Sodium phthalate is the disodium salt of phthalic acid. Phthalic acid is an aromatic dicarboxylic acid with two carboxyl groups attached to adjacent carbon atoms on a benzene ring.
The structure of sodium phthalate is a benzene ring with two -COONa groups attached to adjacent carbons.
Step 3: Predicting the product of decarboxylation.
When sodium phthalate undergoes decarboxylation with sodalime, both -COONa groups are removed as Na₂CO₃. Since there are two carboxyl groups, two molecules of CO₂ (in the form of Na₂CO₃) are removed, leaving behind the benzene ring.
The reaction can be represented as:
C₆H₄(COONa)₂ + 2(NaOH + CaO) → C₆H₆ + 2Na₂CO₃
Sodium Phthalate → Benzene + Sodium Carbonate
Step 4: Evaluate the options.

• Option (1): Benzene: This is the correct product, as the two carboxylate groups are removed from the benzene ring, leaving benzene.
• Option (2): Toluene: Toluene has a methyl group attached to a benzene ring (C₆H₅CH₃). Decarboxylation of sodium phthalate does not introduce a methyl group.
• Option (3): Phthalic acid: Phthalic acid is the starting material (or its precursor) and not the product of decarboxylation.
• Option (4): Phenol: Phenol has a hydroxyl group attached to a benzene ring (C₆H₅OH). Decarboxylation does not introduce a hydroxyl group.

Step 5: Conclusion.
The reaction of sodium phthalate with sodalime results in the formation of benzene.
\[ \boxed{\text{Benzene}} \]