Questions: Competition Between Substitution and Elimination Pathways

Question 1

A tertiary alkyl bromide is treated with sodium tert-butoxide (NaOtBu) in a polar aprotic solvent. Which product predominates, and why?

• A. SN1 substitution product, because tertiary substrates readily form stable carbocations
• B. E2 elimination product, because tert-butoxide is a strong, bulky base that cannot access the crowded tertiary carbon but can abstract a β-hydrogen
• C. SN2 substitution product, because tert-butoxide is a strong nucleophile
• D. E1 elimination product, because tertiary substrates always ionize in polar protic solvents

Answer: (B)

Tertiary substrates cannot undergo SN2 — backside attack is blocked by steric hindrance. SN1 requires a weak nucleophile/base in a polar protic solvent, not a strong base like NaOtBu. Tert-butoxide is too bulky to attack the tertiary carbon but is an effective base for abstracting a β-hydrogen in an E2 mechanism. The result is E2 elimination. The common mistake is defaulting to SN1 for tertiary substrates without considering the strength and character of the reagent.

Question 2

A secondary alkyl bromide is treated with NaCN in DMSO. Which product predominates?

• A. E2 elimination product, because secondary substrates preferentially undergo elimination
• B. SN1 substitution product, because polar solvents stabilize carbocation intermediates
• C. SN2 substitution product, because CN⁻ is a strong nucleophile and DMSO enhances nucleophilicity without solvating the anion
• D. No reaction, because CN⁻ is too weak to displace a bromide leaving group

Answer: (C)

CN⁻ is a strong nucleophile but a weak base — it favors substitution over elimination. DMSO is a polar aprotic solvent: it does not hydrogen-bond with the nucleophile, leaving CN⁻ unsolvated and highly reactive, strongly favoring SN2. Secondary substrates are accessible to SN2 when the nucleophile is strong and the substrate is not overly hindered. SN1 requires a polar protic solvent and weak nucleophile/base, neither of which applies here.

Question 3

Increasing reaction temperature generally favors elimination over substitution because elimination produces more molecules from one substrate, giving a larger positive entropy change.

• T. True
• F. False

Answer: True

Elimination reactions produce two molecules (alkene + HX or conjugate acid) from one substrate, yielding a positive ΔS. Because ΔG = ΔH − TΔS, larger positive entropy terms become increasingly favorable at higher temperatures. This is a consistent principle: when temperature is raised in an S_N vs. E competition, the E product fraction typically increases. Enthalpy differences also play a role, but the entropy advantage of elimination is the dominant thermodynamic driver of this temperature dependence.

Question 4

A primary alkyl halide treated with a strong base like NaOEt will primarily undergo SN1 because primary substrates ionize readily to form primary carbocations stabilized by the alkoxide.

• T. True
• F. False

Answer: False

Primary carbocations are extremely unstable — they essentially cannot form under normal conditions. This rules out both SN1 and E1 for primary substrates, since both require prior ionization to a carbocation. With NaOEt (a strong base but not excessively bulky), the primary substrate undergoes either SN2 (backside attack is unhindered) or E2 (proton abstraction by ethoxide). The statement is wrong in predicting SN1 and in invoking carbocation stabilization by alkoxide, which is not a real stabilization pathway.

Question 5

What is the first step when predicting which mechanism dominates in a substitution/elimination problem, and why is it performed before analyzing the reagent?

Model answer: Identify the substrate class — primary, secondary, or tertiary — because substrate structure immediately eliminates impossible mechanisms. Primary substrates cannot undergo SN1 or E1 (primary carbocations are too unstable). Tertiary substrates cannot undergo SN2 (steric hindrance blocks backside attack). This filtering step narrows the field to feasible mechanisms before reagent character, solvent, or temperature are evaluated.

Starting with substrate structure is decisive because it imposes hard constraints, not just tendencies. Once impossible mechanisms are eliminated, the remaining candidates are assessed using: nucleophile/base character (strong nucleophile-weak base favors S_N; strong base-poor nucleophile favors E; both properties → look at substrate and solvent), solvent polarity (polar protic → SN1/E1; polar aprotic → SN2), and temperature (higher → more elimination). Skipping the substrate analysis step leads to predicting impossible mechanisms and wrong products.