See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

**Analysis of the Elimination Reaction:** The starting material is a branched butane with deuterium (D) at C-2 and a methyl group at C-3. The reaction with $Br_2$ under conditions that favor elimination (free radical mechanism) produces an alkene 'X' and HBr. **Step 1: Identify the mechanism** This is a free radical bromination followed by elimination. The $Br_2$ abstracts a hydrogen from the molecule, creating a radical intermediate that undergoes $\beta$-elimination to form a C=C double bond. **Step 2: Determine which H is abstracted** The molecule has two $\beta$-hydrogens available for abstraction: - One on C-1 (primary position) - One on C-4 (primary position) Free radical bromination is not highly selective between equivalent primary hydrogens, but **elimination typically follows Zaitsev's rule** — the major product is the more substituted alkene. **Step 3: Apply Zaitsev's rule** Removing the hydrogen from C-4 (or C-1, by symmetry) would give a **monosubstituted alkene** (terminal). Removing the hydrogen from C-3 creates a **disubstituted alkene** (more stable), giving: $CH_3-\dot{C}H-C(D)(CH_3)-CH_3$, which forms the double bond between C-2 and C-3. **Step 4: Identify product X** This produces **option (B)**: $H_3C-\dot{C}H-\dot{C}-CH_3$ with D and $CH_3$ substituents. The double bond forms between the two central carbons, creating a more stable, disubstituted (or tetrasubstituted) alkene following Zaitsev's rule. **Answer: (B)**