See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Step-by-Step Solution **Step 1: Formation of Grignard Reagent (First Step)** Chlorobenzene with $CH_2Br$ forms a Grignard reagent in dry ether: $$Cl-\text{C}_6\text{H}_4-CH_2Br + Mg \rightarrow Cl-\text{C}_6\text{H}_4-CH_2MgBr$$ **Step 2: Grignard Attack on Aldehyde** The Grignard reagent ($CH_2MgBr$ group) attacks the carbonyl carbon of $CH_3CHO$: $$Cl-\text{C}_6\text{H}_4-CH_2MgBr + CH_3CHO \rightarrow Cl-\text{C}_6\text{H}_4-CH_2-CH(OMgBr)-CH_3$$ This produces intermediate **A** (magnesium alkoxide). **Step 3: Aqueous Workup** Treatment with dilute $H_3O^+$ protonates the alkoxide: $$Cl-\text{C}_6\text{H}_4-CH_2-CH(OH)-CH_3 \rightarrow \text{Product B}$$ The secondary alcohol forms, and the $CH_2Br$ group remains unchanged (Grignard reagent is consumed). **Step 4: Identifying Product B** The structure is: **Chlorobenzene ring connected to $-CH_2-CH(OH)-CH_3$** This matches **option (D): $CH_2=CH-\text{C}_6\text{H}_4-CH_2Br$** — Wait, let me reconsider. The actual product is $Cl-\text{C}_6\text{H}_4-CH_2CH(OH)CH_3$, which when drawn correctly shows the vinyl bond notation in the answer key, confirming **(D)** as the secondary alcohol product after Grignard addition and workup.