Question

homework assignment: substitution-elimination: discerning the differences
review • ends on december 13, 2025
for the given reactions, identify the mechanism that would produce the major product?
1
2
3
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reaction 1

Answer:

To solve this, we analyze each reaction based on the reagent, solvent, and substrate to determine if it's \( S_N2 \), \( S_N1 \), \( E2 \), or \( E1 \):

Reaction 1:

Substrate: 1 - bromopentane (primary alkyl halide).
Reagent: \( KOC(CH_3)_3 \) (t - BuOK, a strong, sterically hindered base).
Solvent: Acetone (polar aprotic).

Sterically hindered bases favor \( E2 \) (elimination) over \( S_N2 \) (substitution) for primary substrates (due to steric hindrance for \( S_N2 \) attack). The strong base abstracts a \( \beta \) - hydrogen, leading to elimination.

Reaction 2:

Substrate: 1 - bromopentane (primary alkyl halide).
Reagent: \( NaOCH_3 \) (sodium methoxide, a strong base/nucleophile).
Solvent: \( HOCH_3 \) (methanol, polar protic).

For primary alkyl halides, strong nucleophiles/bases in polar protic solvents can undergo either \( S_N2 \) or \( E2 \), but \( S_N2 \) is favored here (since the substrate is primary, and \( NaOCH_3 \) is a good nucleophile). However, if we consider the conditions, \( NaOCH_3 \) in \( HOCH_3 \) can also do \( E2 \), but for primary, \( S_N2 \) is more common? Wait, no—wait, \( NaOCH_3 \) is a strong base. Wait, primary alkyl halides with strong bases: actually, \( E2 \) is possible, but for primary, \( S_N2 \) is favored if the base is also a good nucleophile. Wait, \( OCH_3^- \) is a good nucleophile and base. But in polar protic solvent, for primary, \( S_N2 \) is more likely? Wait, no—let’s correct:

Wait, \( NaOCH_3 \) in \( HOCH_3 \): the solvent is polar protic, which stabilizes the transition state for \( S_N2 \) (by solvating the cation, \( Na^+ \)). For primary alkyl halides, \( S_N2 \) is favored over \( E2 \) when the nucleophile is good ( \( OCH_3^- \) is a good nucleophile). So reaction 2 is \( \boldsymbol{S_N2} \)? Wait, no—wait, \( OCH_3^- \) is a strong base. Wait, maybe I made a mistake. Let's re - evaluate:

Primary alkyl halide + strong base/nucleophile: \( S_N2 \) (substitution) is favored over \( E2 \) (elimination) because the nucleophile can easily attack the primary carbon (less steric hindrance). So reaction 2: \( S_N2 \)? Wait, no—wait, \( NaOCH_3 \) is a strong base. Wait, maybe the answer is \( E2 \)? No, for primary, \( S_N2 \) is more common with strong nucleophiles. Wait, perhaps the intended answer is \( S_N2 \) for reaction 2? Wait, no, let's check again.

Wait, the substrate is 1 - bromopentane (primary). \( NaOCH_3 \) is a strong base and good nucleophile. In polar protic solvent (methanol), \( S_N2 \) is favored for primary alkyl halides (since the nucleophile can attack the less - hindered primary carbon). So reaction 2: \( \boldsymbol{S_N2} \)? Wait, no, maybe \( E2 \)? I think I confused. Let's recall:

• \( S_N2 \): primary > secondary >> tertiary (steric hindrance).
• \( E2 \): tertiary > secondary > primary (steric hindrance for \( S_N2 \), so elimination is favored for tertiary with strong bases).

For primary, \( S_N2 \) is favored over \( E2 \) when the base is also a good nucleophile. \( OCH_3^- \) is a good nucleophile, so \( S_N2 \) (substitution: \( CH_3O^- \) attacks the primary carbon, displacing \( Br^- \)).

Reaction 3:

Substrate: 1 - bromopentane (primary alkyl halide).
Reagent: \( NaN_3 \) (sodium azide, a strong nucleophile).
Solvent: Acetone (polar aprotic).

Polar aprotic solvents favor \( S_N2 \) (substitution) because they solvate cations ( \( Na^+ \)) but not nucleophiles ( \( N_3^- \)), making the nucleophile more reactive. Primary alkyl halides with strong nucleophiles in polar aprotic solvents undergo \( S_N2 \) (the azide ion \( N_3^- \) attacks the primary carbon, displacing \( Br^- \)).

Final Mechanisms:

1. Reaction 1: \( \boldsymbol{E2} \) (elimination, due to sterically hindered base).
2. Reaction 2: \( \boldsymbol{S_N2} \) (substitution, good nucleophile/base on primary substrate).
3. Reaction 3: \( \boldsymbol{S_N2} \) (substitution, strong nucleophile in polar aprotic solvent).

(Note: If the question expects drag - and - drop, match each reaction to its mechanism:

• Reaction 1 → \( E2 \)
• Reaction 2 → \( S_N2 \)
• Reaction 3 → \( S_N2 \))