See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Reaction of Phenol with CHCl₃ and NaOH (Kolbe's Reaction) **Step 1: Formation of phenoxide ion** In aqueous $NaOH$ at 70°C, phenol is deprotonated: $$\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{O}^- + \text{Na}^+ + \text{H}_2\text{O}$$ **Step 2: Activation of CHCl₃** In basic conditions, $CHCl_3$ loses a hydrogen as $H^-$ (hydride), forming the carbanion/carbenoid species: $$\text{CHCl}_3 + \text{OH}^- \rightarrow \bullet\text{CCl}_3 + \text{H}_2\text{O}$$ The activated species is $\bullet\text{CCl}_3$ (trichloromethyl carbanion/radical), which is equivalent to $\cdot\text{CHCl}_2$ (dichloromethylene electrophile when viewed as $: \text{CCl}_2$ or the resonance form $\dot{C}HCl_2$). **Step 3: Electrophilic attack on the phenoxide ring** The electrophile attacking the aromatic ring is $\dot{C}HCl_2$ (with the dot representing the electron-deficient carbon), which is option **(B)**. This carbon is less substituted than $\bullet\text{CCl}_3$ and acts as the electrophilic center in the aromatic substitution step. **Answer: (B) $\cdot CHCl_2$** ✓