Solvent-Dependent Elimination vs. Substitution in Haloalkanes

Steps:

• Recognize that propene forms from 1-bromopropane via elimination of HBr, requiring a strong base.
• Note that primary alkyl halides like 1-bromopropane undergo substitution with aqueous reagents but elimination with ethanolic ones.
• Evaluate options: aqueous conditions promote SN2 substitution; ethanolic conditions favor E2 elimination.
• Select the ethanolic base for maximum alkene yield.

Why D is correct:

• Ethanolic sodium hydroxide provides a non-aqueous medium that favors the E2 elimination mechanism, yielding propene (CH3CH2CH2Br + OH- → CH3CH=CH2 + HBr + H2O).

Why the others are wrong:

• A: Aqueous ammonia acts as a nucleophile, favoring SN2 substitution to form propylamine.
• B: Aqueous potassium hydroxide promotes SN2 substitution, producing propan-1-ol.
• C: Potassium cyanide undergoes SN2 substitution, forming propanenitrile (CH3CH2CH2CN).

Final answer: D