Step 1: Read the reaction described. The reaction involves an organic halide (a vinylic bromide) treated first with magnesium and then with heavy water $D_2O$. We reason out the major product even though the structures are shown as images. Step 2: Form the Grignard reagent. The carbon halogen compound reacts with magnesium in dry ether to form the corresponding Grignard reagent, where magnesium inserts between carbon and bromine. \[ R\text{-}Br + Mg \rightarrow R\text{-}MgBr \] Step 3: Note the key feature of Grignard reagents. In $R\text{-}MgBr$ the carbon bonded to magnesium carries a strong partial negative charge, behaving like a carbanion. This makes it very reactive towards any source of protons or deuterons. Step 4: React with heavy water. Heavy water $D_2O$ supplies a deuteron $D^+$. The carbanion like carbon grabs this deuteron, so the $-MgBr$ group is replaced by $-D$ at the same carbon. \[ R\text{-}MgBr + D_2O \rightarrow R\text{-}D + Mg(OD)Br \] Step 5: Identify the product position. The deuterium ends up exactly where the bromine originally was, giving the specifically deuterated product. This corresponds to the structure shown in option $4$. Step 6: State the answer. The major product is the compound with deuterium at the original halogen carbon, option $4$, matching the key. \[ \boxed{\text{Option } 4 \;(R\text{-}D)} \]
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