The question asks us to identify which compound would readily react with dilute NaOH. Let's analyze each option:
Phenol (\(C_6H_5OH\)) is the compound that will readily react with dilute NaOH, forming phenoxide ion and water. The acidity in phenols arises because of the electron-withdrawing properties of the aromatic ring, making the hydrogen on the hydroxyl group more acidic than in typical alcohols.
The reaction is as follows:
| \(C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O\) |
Therefore, the correct answer is \(C_6H_5OH\) (Phenol).
Phenol (C$_6$H$_5$OH) reacts readily with dilute NaOH because it is more acidic than water. The reaction is as follows:
\[ \text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{O}^- \text{Na}^+ + \text{H}_2\text{O}. \]
The phenoxide ion (C$_6$H$_5\text{O}^-$) formed is stabilized by resonance, making phenol a stronger acid than water.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,