Conc. HNO\(_3\)
When phenol reacts with concentrated nitric acid (HNO₃), it undergoes nitration primarily at the ortho and para positions relative to the hydroxyl group (-OH). The electron-donating effect of the hydroxyl group activates the aromatic ring for electrophilic substitution, making the ring more reactive towards nitration.
The nitration of phenol with concentrated nitric acid produces **2,4,6-Trinitrophenol (picric acid)**, which is a highly reactive compound. The reaction can be represented as: \[ \text{C}_6\text{H}_5\text{OH} + 3\text{HNO}_3 \rightarrow \text{C}_6\text{H}_2(\text{NO}_2)_3\text{OH} + 3\text{H}_2\text{O} \]
The hydroxyl group on phenol is an electron-donating group that activates the benzene ring, making it more susceptible to electrophilic attack. During nitration, the nitronium ion (\( \text{NO}_2^+ \)) attacks the ortho and para positions of the phenol ring, resulting in the formation of **2,4,6-Trinitrophenol**, also known as **picric acid**.
The nitration of phenol with nitric acid results in the formation of **2,4,6-Trinitrophenol (picric acid)**, which is a highly reactive compound due to the electron-donating effect of the hydroxyl group. The major product is formed at the ortho and para positions relative to the hydroxyl group.

Identify the major product (G) in the following reaction (Bromination with \( Br_2/FeBr_3 \)). 


A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).