Decarboxylation of all six possible forms of diaminobenzoic acids $$C_6H_3(NH_2)_2COOH$$ yields three products A, B and C. Three acids give a product 'A', two acids give a product 'B' and one acid give a product 'C'. The melting point of product 'C' is

We need to determine how many of the six diaminobenzoic acid isomers give each diaminobenzene product upon decarboxylation, and find the melting point of product C (obtained from only one acid).

Identify all six isomers of diaminobenzoic acid.

Diaminobenzoic acid has the formula $$C_6H_3(NH_2)_2COOH$$. On a benzene ring with a COOH group at position 1, the two $$NH_2$$ groups can be at:

1. Positions 2,3

2. Positions 2,4

3. Positions 2,5

4. Positions 2,6

5. Positions 3,4

6. Positions 3,5

Determine the decarboxylation product for each isomer.

Decarboxylation removes the COOH group. The relative positions of the two $$NH_2$$ groups determine the product:

1. (2,3)-diaminobenzoic acid $$\rightarrow$$ 1,2-diaminobenzene (o-phenylenediamine)

2. (2,4)-diaminobenzoic acid $$\rightarrow$$ 1,3-diaminobenzene (m-phenylenediamine)

3. (2,5)-diaminobenzoic acid $$\rightarrow$$ 1,4-diaminobenzene (p-phenylenediamine)

4. (2,6)-diaminobenzoic acid $$\rightarrow$$ 1,3-diaminobenzene (m-phenylenediamine) [positions 2,6 relative to removed COOH become 1,3]

5. (3,4)-diaminobenzoic acid $$\rightarrow$$ 1,2-diaminobenzene (o-phenylenediamine)

6. (3,5)-diaminobenzoic acid $$\rightarrow$$ 1,3-diaminobenzene (m-phenylenediamine)

Count the products.

Product A (from 3 acids): 1,3-diaminobenzene (m-phenylenediamine) — from isomers 2, 4, and 6. Melting point = 63°C.

Product B (from 2 acids): 1,2-diaminobenzene (o-phenylenediamine) — from isomers 1 and 5. Melting point = 104°C.

Product C (from 1 acid): 1,4-diaminobenzene (p-phenylenediamine) — from isomer 3 only. Melting point = 142°C.

The melting point of product C is 142°C.

The correct answer is Option D.