Predicting SN1, SN2, E1, and E2 Products

1. The Decision Framework: Substrate First

Classify the carbon bearing the leaving group:

• Methyl and primary: SN2 is fast (minimal steric hindrance); E2 only with a strong, bulky base.
• Tertiary: no SN2 (too hindered); SN1/E1 with weak nucleophiles in protic solvents, or E2 with a strong base.
• Secondary: the crossroads — strong nucleophile and weak base gives SN2; strong bulky base gives E2; weak nucleophile in a protic solvent gives SN1/E1.
• Allylic and benzylic: these react like a "one-substitution-level lower" substrate for SN2 (resonance-delocalized transition states) but also do SN1/E1 readily because their carbocations are resonance-stabilized.

2. Reagent, Solvent, and Leaving-Group Effects

• Nucleophile vs base: a strong nucleophile that is a weak base (for example, I⁻, Br⁻, RS⁻, CN⁻, N₃⁻) favors SN2; a strong base (OH⁻, OR⁻, NH₂⁻) favors E2, especially if bulky (tert-butoxide, LDA → Hofmann elimination).
• Solvent: polar aprotic favors SN2/E2 (reactive anions); polar protic favors SN1/E1 (stabilizes cations).
• Leaving group: a good leaving group (I⁻ > Br⁻ > Cl⁻) is needed for all four; poor leaving groups (OH⁻, F⁻) usually require protonation (SN1/E1) or fail.
• Heat: favors elimination, so hot conditions push E1 over SN1 and can push E2 over SN2.

3. Predicting the Final Product: Rearrangement, Regio, and Stereo

After choosing the pathway, apply the mechanism's rules: check for carbocation rearrangements (SN1/E1 only), pick the Zaitsev alkene for E1 and for E2 with non-bulky bases (Hofmann with bulky bases), and apply anti-periplanar geometry for E2 and inversion for SN2. Predict the major product, and expect minor products from competing pathways.

How it works

1. Classify the substrate's leaving-group carbon (methyl/1°/2°/3°/allylic/benzylic).
2. Determine whether the reagent is a strong or weak nucleophile and a strong, bulky, or weak base.
3. Combine substrate plus reagent to pick the dominant pathway (SN2, SN1, E2, or E1).
4. Refine with solvent (aprotic vs protic) and temperature (heat favors elimination).
5. Apply mechanism-specific rules: rearrangements (SN1/E1), Zaitsev/Hofmann (E2/E1), anti-periplanar (E2), inversion (SN2).
6. Write the major product and note likely minor products from competing pathways.

Common confusions

Memory aids

"Substrate, then Reagent, then Solvent, then Heat" (S-R-S-H) — classify the Substrate, judge the Reagent, note the Solvent, and check for Heat before predicting the product.

Quick review

Topic Recap

Predicting products of substitution and elimination means classifying the substrate first, then weighing nucleophile strength, base strength and bulk, leaving-group ability, solvent, and heat. Methyl/primary substrates favor SN2; tertiary substrates favor SN1/E1 or E2; secondary substrates depend on the reagent. Apply rearrangements, Zaitsev/Hofmann, anti-periplanar, and inversion rules to finish the prediction, and expect competing minor products.

Knowledge Check

1. What is the major product of 1-bromobutane with NaCN in DMSO?
2. Why can a tertiary substrate not undergo SN2?
3. What reagent and solvent combination most favors SN1 for a tertiary substrate?
4. Which alkene forms when 2-bromo-2-methylbutane reacts with bulky potassium tert-butoxide?
5. Under what conditions does E1 compete with SN1, and which does heat favor?

Answers and Rationales

1. Pentanenitrile (1-cyanobutane) by SN2 — cyanide is a strong nucleophile, the substrate is primary, and DMSO is polar aprotic.
2. Three alkyl groups block backside approach, so the nucleophile cannot reach the carbon.
3. A weak nucleophile (for example, water or an alcohol) in a polar protic solvent — these stabilize the carbocation and leave the nucleophile weak.
4. The Hofmann (less substituted) alkene, 2-methylbut-1-ene — the bulky base removes the most accessible β hydrogen.
5. E1 and SN1 share a carbocation and always compete when the substrate is 2° or 3° with a weak nucleophile in a protic solvent; heat favors the E1 elimination product.