For 2-bromobutane, the reagent that gives mainly the substitution product is

Ahot alcoholic KOH
B $(CH X_{3}) X_{3} COK$ in $(CH X_{3}) X_{3} COH$
C $NaNH X_{2}$ in liquid ammonia
D $NaN X_{3}$ in DMF

Answer: D

Azide is only weakly basic (the conjugate acid has pKa about 4.7) but highly nucleophilic, and DMF leaves it unsolvated, so SN2 prevails. The other three reagents are strong bases and give mainly elimination.

Practice Organic Reaction Mechanisms IV: Elimination More on Substitution vs elimination competition

Previous questionNext question

More questions on Substitution vs elimination competition

In ethanol with no added base, 2-bromo-2-methylpropane gives an ether (SN1) and an alkene (E1) through the same carbocation. On raising the ... JEE Main level
2-Bromopropane is treated with NaI in acetone. The major product is JEE Main level
Which of the following bromides has no beta-hydrogen and therefore cannot undergo a direct one-step E2 dehydrohalogenation (without rearrang... JEE Main level
With NaOEt in ethanol, the order of increasing proportion of elimination (alkene) for CH3CH2Br, (CH3)2CHBr and (CH3)3CBr is JEE Main level
Which of the following changes favour E2 elimination over SN2 for a secondary alkyl halide? (select all that apply) JEE Main level
2-Bromo-3,3-dimethylbutane is (i) treated with NaOEt/ethanol (E2 conditions), and (ii) warmed in aqueous ethanol without base (E1/SN1 condit... JEE Advanced level