Step 1: Work backwards:
The product has a symmetrical skeleton with two $(\text{CH}_3)_3\text{C}-\text{CH}_2-$ halves joined by one new C-C bond.
Step 2: Identify the halide:
Each half must come from $(\text{CH}_3)_3\text{C}-\text{CH}_2-\text{Cl}$ with the Cl replaced by the new bond.
The isobutyl chloride in A would join two isobutyl groups, and C and D carry the chlorine on a secondary carbon, so they would join through that carbon and give a different carbon skeleton.
Final Answer:
The alkyl chloride is $(\text{CH}_3)_3\text{C}-\text{CH}_2-\text{Cl}$, option (B).
\[ \boxed{(\text{CH}_3)_3\text{C}-\text{CH}_2-\text{Cl}} \]