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.
Consider the following reaction of benzene. the percentage of oxygen is _______ %. (Nearest integer) 





MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
