For Lassaigne’s test:
• Organic compounds must contain nitrogen or halogens for detection.
• Both nitrogen and halogen must be part of the same compound to test positive for both.
CH3NH2.HCl
NH2OH .HCl
NH4Cl
N2H4 . HCl
1. Lassaigne’s Test: This test detects nitrogen, sulfur, and halogens in organic compounds.
- The compound is fused with sodium, converting nitrogen into NaCN and halogen into NaX (\(X = \text{Cl, Br, I}\)).
2. Compound Analysis:
\(\text{N}_2\text{H}_4\cdot\text{HCl}\): Contains nitrogen but no halogen.
\(\text{CH}_3\text{NH}_2\cdot\text{HCl}\): Contains both nitrogen (from the \(-\text{NH}_2\) group) and halogen (\(\text{Cl}\) from \(\text{HCl}\)).
\(\text{NH}_4\text{Cl}\): Contains nitrogen and halogen but is not an organic compound.
\(\text{NH}_2\text{OH}\cdot\text{HCl}\): Contains nitrogen and halogen but does not form the required products in Lassaigne’s test.
3. Conclusion: Only \(\text{CH}_3\text{NH}_2\cdot\text{HCl}\) gives a positive Lassaigne’s test for both nitrogen and halogen.
Final Answer: (2) \(\text{CH}_3\text{NH}_2\cdot\text{HCl}\).




For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: