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 : \(A_2 \rightleftharpoons 2A\)