Master Decision Framework

Updated Apr 17, 2026

This section is the single highest-yield MCAT organic chemistry skill in this book. Given any substrate, reagent, and conditions, you can predict the major product via SN1, SN2, E1, or E2 using a four-step flowchart. Get this right and you will handle almost every MCAT substitution/elimination question on test day.

Use the interactive predictor below to stress-test your intuition. Change one variable at a time — substrate, reagent, solvent, or temperature — and watch how the winning mechanism shifts. The rationale line explains why each combination lands where it does.

Choose a substrate and conditions. The predictor applies the master decision framework (substrate → reagent → solvent → temperature) to identify the dominant mechanism and explain why.

Substrate Nucleophile / base Solvent Temperature

SN2 (minor: E2) mixed

2° substrate with a strong non-bulky Nu in polar aprotic solvent: SN2 wins with a minor E2 byproduct. Inversion at the reacting carbon.

Stereochemistry: inversion

Tip: change one variable at a time and watch how the mechanism shifts. Every combination (252 total) follows an explicit rule.

The Flowchart

Ask these four questions in order. The first yes/no that decides the mechanism gives you the answer.

Step 1: What is the substrate?

Methyl or 1° (unhindered) → only SN2 or E2 possible (cation too unstable for SN1/E1).
3° (or 1° neopentyl-like) → only SN1 or E1 possible (SN2 blocked by steric bulk; 3° carbocation readily forms).
2° → all four mechanisms possible.
Allyl / benzyl → all four possible; both SN1 and SN2 are fast because the cation is stabilized AND the substrate is not hindered.

Step 2: What is the reagent - strong nucleophile, strong base, or both/neither?

Strong nucleophile, non-bulky (CN⁻, N₃⁻, HS⁻, I⁻, RS⁻, HC≡C⁻) → SN2 preferred.
Strong base, bulky (tBuO⁻, LDA, DBU) → E2 only; SN2 is blocked by steric bulk.
Strong base AND nucleophile, non-bulky (RO⁻, OH⁻, NH₂⁻) → SN2 vs. E2 depends on substrate.
Weak nucleophile (H₂O, ROH, RCOOH, halide in protic solvent) → SN1 / E1 if substrate allows.

Step 3: What is the solvent?

Polar aprotic (DMSO, DMF, acetone) → favors SN2 (nucleophile is “naked” and reactive).
Polar protic (H₂O, MeOH, EtOH) → favors SN1 / E1 (stabilizes ionic intermediates).
Nonpolar → generally slow for all mechanisms; not common on MCAT.

Step 4: What is the temperature?

Low temperature → favors substitution over elimination (within either SN1-vs-E1 or SN2-vs-E2).
High temperature → favors elimination over substitution. This is why acid-catalyzed dehydration requires heat - you are pushing the cation toward E1.

The Master Table

| Substrate | Reagent | Solvent | Temp | Mechanism |
|-----------|---------|---------|------|-----------|
| Methyl | strong Nu | polar aprotic | any | SN2 |
| 1° | strong Nu | polar aprotic | low | SN2 |
| 1° | bulky strong base | any | moderate/high | E2 |
| 1° | weak Nu (H₂O, ROH) | polar protic | any | Very slow - limited reaction |
| 2° | strong Nu, small | polar aprotic | low | SN2 |
| 2° | strong base, small | polar aprotic | moderate | mix SN2 + E2 |
| 2° | bulky strong base | any | any | E2 |
| 2° | weak Nu/base | polar protic | low | SN1 + E1 mix |
| 2° | weak Nu/base | polar protic | high | E1 (Zaitsev) |
| 3° | strong Nu or base | any | low | E2 (no SN2 due to steric) |
| 3° | weak Nu/base | polar protic | low | SN1 |
| 3° | weak Nu/base | polar protic | high | E1 (Zaitsev) |
| Allyl/benzyl | strong Nu | polar aprotic | any | SN2 (or SN1 if weak Nu) |

Common MCAT Scenarios

Scenario 1: Methyl iodide + NaCN in DMSO.

Substrate: methyl → SN2 only.
Nucleophile: strong, non-bulky.
Solvent: polar aprotic (favors SN2).
Answer: SN2. Product: methyl cyanide (acetonitrile).

Scenario 2: 2-bromopropane + NaOEt in ethanol.

Substrate: 2° (all four possible).
Reagent: strong small nucleophile and base.
Solvent: polar protic.
Answer: mix of SN2 and E2, with E2 dominating slightly due to branching. Actual experimental result: ~80% E2, ~20% SN2.

Scenario 3: tert-butyl bromide + methanol (solvent and nucleophile).

Substrate: 3° → only SN1 or E1 possible.
Nucleophile: methanol is weak.
Solvent: polar protic (favors SN1/E1).
Answer: SN1 at room temperature, E1 at high temperature.

Scenario 4: (R)-2-bromopentane + NaOtBu in t-butanol.

Substrate: 2°.
Base: bulky → E2 only.
Answer: E2, Hofmann product (terminal alkene, 1-pentene), because tBuO⁻ cannot reach the interior beta-H for Zaitsev.

Scenario 5: 1° alkyl chloride + NaN₃ in DMF.

Substrate: 1°.
Reagent: strong nucleophile, weakly basic (azide is a very good nucleophile but a poor base).
Solvent: polar aprotic.
Answer: SN2. Product: 1° azide.

Putting It on Paper

For each MCAT question about SN/E mechanism:

Label the substrate (methyl / 1° / 2° / 3° / allyl / benzyl).
Label the nucleophile/base (strong Nu? strong base? bulky base? weak?).
Label the solvent (polar aprotic? polar protic? nonpolar?).
Check the temperature (low? high?).
Use the flowchart. Decide on the mechanism. Draw the product with the appropriate stereochemistry (inversion for SN2, racemization for SN1, Zaitsev or Hofmann for E1/E2).

This routine takes 30-60 seconds once practiced. It answers a huge portion of MCAT organic chemistry in that time.

Predict the major product when (R)-2-chloro-2-methylpentane is treated with (a) NaOH in cold water, and (b) KOtBu in hot t-butanol.

Click to reveal answer

(a) With NaOH cold H₂O: 3° substrate, strong base/moderate nucleophile, polar protic, low T. Substrate is 3° so no SN2; OH is a decent base; cold water favors substitution. Mechanism: SN1. Product: 2-methyl-2-pentanol (racemic, but since the C attached to OH is already 3° and not chiral, no stereo issue). Some minor E1 product may appear.(b) With KOtBu hot tBuOH: 3° substrate, bulky strong base, polar protic, high T. Bulky base rules out SN2 and SN1. High temp favors elimination. Mechanism: E2 with Hofmann regiochemistry. Product: 2-methyl-1-pentene (terminal alkene, the Hofmann product, preferred because tBuO⁻ can only reach the peripheral terminal methyl Hs).