See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Solution **Reactants Analysis:** - **p-Hydroxyacetophenone**: $HO-\text{C}_6H_4-COCH_3$ (has $-OH$ and $-COCH_3$ groups) - **Methylmagnesium iodide (Grignard reagent)**: $CH_3MgI$ (strong nucleophile and base) **Reactivity of Functional Groups:** Grignard reagents react with electrophilic groups in order of reactivity: 1. **Carboxylic acids/Phenols** (acidic $-OH$) — most reactive 2. **Carbonyl groups** ($C=O$) — very reactive 3. **Ethers, aromatic rings** — unreactive under normal conditions **Reaction Sequence:** $$CH_3MgI + HO-\text{C}_6H_4-COCH_3 \rightarrow$$ 1. **First**: The acidic phenolic $-OH$ is deprotonated by $CH_3MgI$ (acts as a strong base): $$HO-\text{C}_6H_4-COCH_3 + CH_3MgI \rightarrow CH_3O^-MgI^+ + CH_4 + \text{C}_6H_4-COCH_3$$ 2. **Second**: The ketone carbonyl ($-COCH_3$) is attacked by remaining Grignard reagent, forming a magnesium alkoxide intermediate: $$\text{C}_6H_4-COCH_3 + CH_3MgI \rightarrow \text{C}_6H_4-C(OMgI)(CH_3)_2$$ 3. **Workup** with aqueous acid protonates the alkoxide to give a secondary alcohol: $$\text{C}_6H_4-C(OH)(CH_3)_2$$ **Final Product**: $CH_3O-\text{C}_6H_4-\text{C}(OH)(CH_3)_2$ This matches **option (C)**: a methoxy-substituted aromatic ring with a tertiary alcohol side chain.