Questions: Substitution vs Elimination Competition

Question 1

2-Bromopropane (a secondary substrate) is treated with sodium tert-butoxide (NaOtBu) in DMSO. What is the major product?

• A. 2-propanol via SN2 — polar aprotic solvent strongly favors substitution
• B. Propene via E2 — the bulky base cannot access the carbon backside but can abstract a proton
• C. 2-propanol via SN1 — tertiary carbocation forms readily from secondary substrates
• D. Equal SN2 and E1 products, because secondary substrates always split between pathways

Answer: (B)

Even though DMSO (polar aprotic) normally favors SN2, NaOtBu is an exceptionally bulky base — its three methyl groups physically block backside attack on the carbon. Proton abstraction is far less sterically demanding, so E2 wins. The misconception is treating 'polar aprotic = SN2' as absolute; steric bulk of the base can override the solvent effect, especially on secondary substrates.

Question 2

A tertiary alkyl bromide is placed in a water/ethanol mixture at 50°C. What outcome is most likely?

• A. E2 elimination exclusively, because tertiary substrates always eliminate
• B. SN2 substitution, because the polar protic solvent activates the nucleophile
• C. A mixture of SN1 and E1 products, because both pathways share a common carbocation intermediate
• D. No reaction, because tertiary substrates are too hindered for any mechanism

Answer: (C)

Tertiary substrates cannot do SN2 — the three alkyl groups completely block backside attack. Water/ethanol is a polar protic solvent that stabilizes the carbocation and contains only a weak nucleophile, strongly favoring the unimolecular pathways. SN1 and E1 always accompany each other because they arise from the same carbocation intermediate — ionization produces a carbocation that can be captured by the nucleophile (SN1) or lose a proton (E1). You never get pure SN1 without some E1 byproduct.

Question 3

SN1 and E1 reactions always produce product mixtures because they share a common carbocation intermediate.

• T. True
• F. False

Answer: True

Correct. Both SN1 and E1 begin with ionization of the substrate to form a carbocation. That intermediate can then be captured by a nucleophile (giving the substitution product) or lose an adjacent proton (giving the elimination product). Because both pathways originate from the same intermediate, they compete simultaneously whenever conditions favor ionization. A reaction described as 'SN1' is more precisely described as 'predominantly SN1, with SN1/E1 competition.'

Question 4

A strong, small base like hydroxide (HO⁻) generally gives elimination as the major product on primary substrates, because base strength is the key factor in E2 selectivity.

• T. True
• F. False

Answer: False

Base strength is necessary but not sufficient for E2 selectivity. On a primary substrate, the carbon backside is unhindered, so even hydroxide — which is both a strong base and a good nucleophile — performs SN2 faster than E2. Elimination becomes dominant over substitution when the base is both strong AND sterically bulky (like tert-butoxide), making proton abstraction preferred over backside attack. On tertiary substrates SN2 is blocked regardless, but on primary substrates the substrate accessibility matters more than base strength alone.

Question 5

Why is substrate class the first factor to assess when predicting which of SN1, SN2, E1, or E2 will dominate?

Model answer: Substrate class can completely eliminate certain mechanisms before other variables are considered. Tertiary substrates cannot undergo SN2 — the three alkyl groups block backside attack regardless of nucleophile, solvent, or temperature. Methyl and primary substrates cannot form stable carbocations, so SN1 and E1 are negligible. These structural constraints prune the decision tree immediately, so the remaining factors (base/nucleophile identity, solvent, temperature) only need to adjudicate among the pathways that remain possible for that substrate class.

The decision framework is hierarchical: substrate class eliminates mechanisms structurally, then reagent character (nucleophilicity vs. basicity, steric bulk) selects among the surviving options, then solvent and temperature fine-tune the outcome. Starting with substrate class prevents applying irrelevant rules — asking whether hydroxide does SN2 or E2 on a tertiary substrate is a moot question, because neither SN2 nor E2 is efficient there.