Questions: SN1 vs SN2 Selectivity: Factors and Competition

Question 1

Sodium cyanide (CN⁻, a strong nucleophile) is added to (CH₃)₃CBr in DMSO at room temperature. A student predicts SN2 because CN⁻ is a strong nucleophile. What will actually be the major pathway and why?

• A. SN2 — strong nucleophiles always displace leaving groups regardless of substrate structure
• B. SN1 — the tertiary substrate forms a stabilized carbocation, and steric crowding physically blocks back-side attack even though CN⁻ is strong
• C. No reaction — polar aprotic solvents destabilize carbocations and prevent SN1 on tertiary substrates
• D. E2 elimination — CN⁻ preferentially acts as a base on all tertiary substrates

Answer: (B)

Substrate structure is the most important factor. Tertiary substrates have three alkyl groups shielding the electrophilic carbon, physically preventing the back-side attack that SN2 requires — regardless of how strong the nucleophile is. The tertiary carbocation from (CH₃)₃C⁺ is also well-stabilized by hyperconjugation. The student's error is thinking nucleophile strength alone determines mechanism; it cannot overcome the steric barrier. The combination of tertiary substrate + DMSO (which stabilizes ions) means SN1 dominates.

Question 2

2-Bromopropane (a secondary substrate) is treated with methanol (a weak nucleophile) in water. Which prediction is best, and what factors lead to it?

• A. SN2 — secondary substrates have moderate steric hindrance and will always go SN2 with any nucleophile
• B. SN1 — water and methanol are polar protic solvents that stabilize the secondary carbocation intermediate, and the weak nucleophile cannot drive SN2
• C. No reaction — secondary substrates require a strong nucleophile to react by either mechanism
• D. E2 elimination — methanol is basic enough to deprotonate and favor elimination on secondary substrates

Answer: (B)

Secondary substrates are the borderline case — you must look beyond substrate structure. Here, two factors point to SN1: the solvent (water/methanol) is polar protic, which stabilizes the secondary carbocation through solvation and weakens nucleophilicity through hydrogen bonding; and methanol is a weak nucleophile that cannot drive SN2 by actively attacking the substrate. Weak nucleophiles favor SN1 because they passively trap a carbocation once it forms rather than forcing a direct displacement.

Question 3

A secondary substrate treated with a strong nucleophile (e.g., CN⁻) in a polar aprotic solvent (e.g., DMSO) will predominantly undergo SN1, because secondary carbocations are moderately stable.

• T. True
• F. False

Answer: False

This is a common error. For secondary substrates, the solvent and nucleophile strength matter enormously. A polar aprotic solvent leaves the nucleophile bare and highly reactive (no hydrogen bonding to solvate and weaken it), and a strong nucleophile can drive direct back-side attack. Together, these conditions favor SN2. SN1 on secondary substrates requires polar protic solvents to stabilize the carbocation and a weak nucleophile that cannot force SN2. The substrate alone does not determine the outcome — the combination of all four factors must be evaluated.

Question 4

The selectivity between SN1 and SN2 for a given substrate can be shifted by changing the solvent, even without changing the nucleophile or substrate.

• T. True
• F. False

Answer: True

Solvent is a genuine mechanistic lever. Switching from a polar protic solvent (water, methanol) to a polar aprotic solvent (DMSO, DMF, acetone) can shift a secondary substrate from SN1 toward SN2. Polar protic solvents stabilize carbocation intermediates through solvation, favoring SN1; they also hydrogen-bond to nucleophiles, weakening them. Polar aprotic solvents do neither — they leave nucleophiles reactive. So the same substrate and nucleophile can show dramatically different selectivity depending solely on solvent choice.

Question 5

Why does tertiary substrate structure favor SN1 over SN2? Provide both a steric and an electronic argument.

Model answer: Sterically: three alkyl groups surround the electrophilic carbon in a tertiary substrate, creating a steric barrier that prevents a nucleophile from approaching the back face for the SN2 transition state. Even strong nucleophiles cannot physically reach the carbon for back-side attack. Electronically: when the tertiary substrate ionizes to form a carbocation, the three alkyl groups stabilize the positive charge through hyperconjugation (overlap of adjacent C-H sigma bonds with the empty p orbital) and inductive electron donation. This lowers the activation energy for carbocation formation, making SN1 energetically favorable.

Both effects reinforce each other: the steric crowding blocks SN2, and the electronic stabilization promotes SN1. This is why tertiary substrates almost always go SN1 regardless of nucleophile strength or solvent — the two factors combine to make the SN1 pathway strongly preferred.