Relative stability of the contributing structures is :
\((I)>(III)>(II)\)
\((I)>(II)>(III)\)
\((II)>(I)>(III)\)
\((III)>(II)>(I)\)
Explanation:
1. Neutral structures are more stable than charged ones. Therefore, structure (I) is the most stable as it is neutral.
2. Among the charged structures, a positive charge (+) on a less electronegative atom (like carbon) is more stable than a positive charge on a more electronegative atom (like oxygen).
- Hence, structure (II), where C+ is present, is more stable than (III), where O+ is present.
Order: \( I > II > III.\)
Final Answer: Option (2).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32