See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

To determine whether each reaction proceeds via substitution or elimination, we apply the following principles: 1. Strong, bulky bases (like tert-butoxide, -OC(CH3)3) favor elimination (E2) over substitution, especially with secondary or tertiary substrates. 2. Strong, non-bulky nucleophiles/bases (like NaN3, azide) favor substitution (SN2) with secondary substrates. 3. Weak nucleophiles/weak bases (like H2O) with tertiary substrates favor SN1 substitution. 4. Acidic conditions (H2SO4) with alcohols under heat favor elimination (E1). 5. Alcoholic KOH (alc-KOH) favors elimination (E2). 6. Tertiary substrates with strong bases and heat favor elimination. Analysis of each reaction: (a) 2-chloropropane (secondary alkyl chloride) + K+ -OC(CH3)3 (potassium tert-butoxide, a strong BULKY base) in H2O: - Potassium tert-butoxide is a strong, sterically hindered base. With a secondary substrate, it strongly favors E2 elimination over substitution. - Answer: ELIMINATION (b) tert-Butanol (tertiary alcohol) + H2SO4, heat: - Acid protonates the OH, converting it to a good leaving group (water). With a tertiary carbocation formed, and heat applied, E1 elimination is strongly favored. - Answer: ELIMINATION (c) 2-Chlorobutane (secondary alkyl chloride) + alcoholic KOH (alc-KOH): - Alcoholic KOH provides KOH dissolved in alcohol solvent. This is the classic condition for E2 elimination. Strong base (OH-) in non-aqueous (alcoholic) medium with a secondary substrate favors elimination. - Answer: ELIMINATION (d) 2-Iodopropane (secondary alkyl iodide) + NaN3 (sodium azide): - Azide (N3-) is a strong nucleophile but a weak base. With a secondary substrate, strong nucleophiles that are weak bases favor SN2 substitution over elimination. - Answer: SUBSTITUTION (e) 1-Chloro-1-methylcyclohexane (tertiary alkyl chloride) + EtO- (ethoxide, strong base), heat: - Ethoxide is a strong base. With a tertiary substrate and heat, E2 elimination is strongly favored. Tertiary substrates undergo elimination preferentially with strong bases due to steric hindrance preventing SN2, and the driving force of heat. - Answer: ELIMINATION (f) tert-Butyl chloride (tertiary alkyl chloride) + H2O (weak nucleophile, weak base): - Water is a weak nucleophile and weak base. With a tertiary substrate, SN1 substitution is favored because the tertiary carbocation forms readily and water (though weak) acts as the nucleophile. Elimination (E1) can also occur but SN1 gives the major product under aqueous conditions without heat emphasis. - Answer: SUBSTITUTION Summary: - Substitution: (d) NaN3 is a good nucleophile/weak base → SN2; (f) H2O with tertiary → SN1 - Elimination: (a) bulky strong base → E2; (b) acid + heat + tertiary → E1; (c) alc-KOH → E2; (e) strong base + tertiary + heat → E2 Therefore, the correct answer is {"Substitution": ["D", "F"], "Elimination": ["A", "B", "C", "E"]}.