Primary haloalkanes favor substitution over elimination with ethanolic NaOH Steps:

• Verify molecular formulas: C4H9Br requires 4 carbons; A, B, C describe C5 structures (2-methylbutane and pentane bases), yielding C5H11Br.
• Only D (1-bromo-2-methylpropane, (CH3)2CHCH2Br) matches C4H9Br.
• Classify halide types: primary halides like D undergo SN2 substitution; secondary/tertiary undergo E2 elimination to alkenes.
• Thus, D does not form an alkene, fitting the condition.

Why D is correct:

• As a primary halide (Br on CH2 group), it follows the rule that primaries resist E2 elimination, preferring SN2 with ethanolic NaOH.

Why the others are wrong:

• A: Molecular formula is C5H11Br, not C4H9Br.
• B: Molecular formula is C5H11Br; tertiary halide would undergo E2 to form alkene.
• C: Molecular formula is C5H11Br; secondary halide would undergo E2 to form alkene.

Final answer: D