GATE 2025 BT Q.26: solution and answer

Official answer: A, B, CSource: Official answer key, GATE 2025 BT (IIT Roorkee)

1. The degree of reduction per C-mole, $\gamma$, of an organic compound whose C-mole formula is ${\text{CH}}_a{\text{O}}_b{\text{N}}_c$ is $\gamma =4+a-2b-3c$. It uses the valences C = 4, H = 1, O = $-2$, N = $-3$, with ${\text{CO}}_2$, ${\text{H}}_2\text{O}$ and ${\text{NH}}_3$ as the reference state.

2. Write each compound per carbon atom by dividing its molecular formula by its number of carbon atoms. None of the four contains nitrogen, so $c=0$.

3. Glucose: ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6\to {\text{CH}}_2\text{O}$, so $\gamma =4+2-2(1)=4$.

4. Lactic acid: ${\text{C}}_3{\text{H}}_6{\text{O}}_3\to {\text{CH}}_2\text{O}$, so $\gamma =4+2-2(1)=4$.

5. Acetic acid: ${\text{C}}_2{\text{H}}_4{\text{O}}_2\to {\text{CH}}_2\text{O}$, so $\gamma =4+2-2(1)=4$.

6. Formic acid: ${\text{CH}}_2{\text{O}}_2$ already has one carbon, so $\gamma =4+2-2(2)=2$.

7. Glucose, lactic acid and acetic acid all have the same C-mole formula, ${\text{CH}}_2\text{O}$, so all three have $\gamma =4$. Formic acid is more oxidised ($\gamma =2$) and does not match them. Correct options: A, B and C.