Visual walkthrough — Effect of substrate, nucleophile - base, solvent, leaving group
Step 1 — What a reaction looks like as a picture
WHAT. Draw the reaction as a journey. The horizontal axis is the reaction coordinate — a made-up "distance travelled" that goes from starting materials on the left to products on the right. The vertical axis is potential energy — how much stored energy the molecules have. Higher = less stable = harder to sit there.
WHY this picture and not a table? A table tells you the answer; a graph of energy vs progress tells you why. Reaction rate is decided entirely by ONE number on this graph — the height of the tallest hill you must climb. Every one of the four knobs is just "does this raise or lower a hill / a valley?" So if we get this one picture right, everything else is reading heights off it.
PICTURE. Look at figure s01. The molecule starts in a valley (a stable state), climbs to a peak (the transition state — the least stable, most stretched arrangement), then drops into the product valley.
Step 2 — SN2 as one hill, and where the substrate knob pushes
WHAT. SN2 is a single smooth hump: nucleophile comes in the back door as the leaving group swings out the front, all in one motion. One valley → one peak → one valley.
WHY one hump? Because nothing gets made in the middle — there is no resting intermediate. So there is exactly one hill, and its height is set by how crowded the peak is. The peak has five things touching the carbon at once (the incoming nucleophile, the three original groups, the leaving group). Crowding = high peak.
PICTURE. In figure s02 the two curves share a reactant valley but the tertiary substrate (violet) pushes the peak much higher than the methyl substrate (magenta), because three bulky groups crowd the already-crowded peak.
Step 3 — SN1 as two hills, and why the substrate knob flips direction
WHAT. SN1 is a two-hump journey. First hump: the C–X bond breaks all by itself, dumping the molecule into a middle valley — the carbocation (a carbon missing an electron pair, so it is positive). Second, tiny hump: the nucleophile drops onto that cation.
WHY two humps? Because the cation is a real, resting-for-an-instant species — a valley between two peaks. The first peak is the tall one (the slow, rate-deciding step), so its height is all that matters for the rate.
PICTURE. Figure s03 stacks a tertiary curve (magenta) and a methyl curve (violet). Notice the flip: the same extra alkyl groups that raised the SN2 peak now lower the first SN1 peak, because they stabilise the middle cation valley — and a lower valley pulls the peak just before it down too.
Recall Why can't methyl do SN1?
A methyl cation has no alkyl groups to stabilise it — its middle valley is a mountaintop, not a valley. So the first SN1 peak is astronomically high. Methyl simply never ionises. ::: It has no way to stabilise the carbocation, so the ionisation hill is impossibly tall.
Step 4 — The nucleophile / base knob: which hill it touches
WHAT. A strong nucleophile appears in the rate-deciding peak of SN2/E2, so making it stronger lowers that peak. A weak nucleophile (water, alcohol) is a bystander in the slow step of SN1 — it can't lower the ionisation hill at all.
WHY does strength only matter sometimes? Because a knob only moves a hill if it is present at that hill. In SN2 the nucleophile is in the transition state → strengthen it, drop the hill. In SN1 the slow step is the halide ionising alone → the nucleophile isn't there yet → strengthening it does nothing to the rate.
PICTURE. Figure s04 shows two SN2 curves: a strong nucleophile (orange) sits under a lower peak than a weak one (violet). Beside them, the SN1 first-peak is drawn flat under both nucleophiles — proof it doesn't care.
Step 5 — The solvent knob: it moves the valley, not the peak
WHAT. Solvent wraps around charges and lowers their energy (solvation). The trick: it can lower the starting valley OR the peak, and that changes in opposite ways.
WHY this is subtle. = peak minus valley. If solvent lowers the peak more, the hill shrinks (faster). If it lowers the starting valley more, the hill grows (slower). Same solvent, opposite effect — depending on where the charge is.
PICTURE. Figure s05 has two panels.
- Left (SN1): a polar protic solvent (magenta) drops the ionising peak — where the charge is being born — so SN1 speeds up.
- Right (SN2): a polar protic solvent instead cages the naked nucleophile in the reactant valley, dropping the valley and thus raising the hill → SN2 slows. A polar aprotic solvent (orange) leaves that valley high, so the SN2 hill stays short → faster.
Step 6 — The leaving group knob: it lowers every peak
WHAT. A good leaving group is one that is happy to walk off carrying the negative charge — i.e. a weak base (a stable anion). Because the C–X bond stretches in the peak of all four mechanisms, an easier-leaving group lowers every peak on this page.
WHY it helps all four. Look back: SN2 (Step 2), SN1 first hump (Step 3), and both elimination hills all stretch the C–X bond at their peak. Anything that makes that stretch cheaper drops that peak. So the leaving group is the one knob that never picks a mechanism — it just speeds up whichever one you already have.
PICTURE. Figure s06 overlays the same reaction with I⁻ (orange, low peak), Br⁻, Cl⁻, and F⁻ (violet, tall peak) leaving. The peak height climbs as the leaving group gets worse.
Recall Why must an alcohol be protonated first?
Its leaving group would be — a strong base, so a terrible leaving group (impossibly tall peak). Protonating it makes , a weak base and excellent leaving group — the peak drops to a climbable height. ::: OH⁻ is a strong base (bad LG); protonation converts it to water, a weak base and great LG.
Step 7 — The degenerate cases, drawn
WHAT. Every rule above has an edge where a hill vanishes or becomes infinite. We draw all of them so the reader never meets an un-shown case.
PICTURE. Figure s07 is a four-tile grid of "broken" energy curves.
The one-picture summary
Figure s08 puts all four knobs on one energy diagram: the substrate and nucleophile knobs move peaks, the solvent knob moves valleys, and the leaving group knob lowers every peak at once. Reading a reaction is just: which knobs move which landmark, and does the tallest hill go up or down?
Recall Feynman retelling — the walkthrough in plain words
Picture a skateboarder crossing a hilly park; the park is the reaction and the tallest hill decides how long the trip takes. Step 1–2: SN2 is one hill. Pile more friends around the skater at the top of the hill (a bigger, crowded substrate) and the hill gets taller — the trip slows to a stop for a tertiary carbon. Step 3: SN1 is two hills with a rest-stop (the cation) in between. Those same crowding friends now hold up the rest-stop, so the first hill actually gets shorter — the exact opposite of SN2. That flip is why crowded carbons switch from SN2 to SN1. Step 4: A pushy newcomer (strong nucleophile) is only helpful if it's on the hill with you — it lowers the SN2 hill but ignores the SN1 ionisation hill it isn't part of. A fat pushy one can't squeeze onto the crowded carbon, so it grabs a nearby hydrogen instead — that's E2. Step 5: The solvent is the weather. Protic weather is a warm blanket — great for the newborn charge in SN1, but it smothers the SN2 attacker down in the starting valley, deepening the valley and raising the hill. Aprotic weather leaves the attacker naked and eager. Step 6: The leaving group is the backpack the skater drops. A light, happy-to-leave backpack (I⁻, a weak base) makes every hill in the park shorter. A clingy one (F⁻) makes them all taller. Step 7: At the edges the hills go to infinity (methyl can't do SN1, tertiary can't do SN2) or the weather tilts things — and heating the whole park favours the route that makes more pieces, which is elimination. One picture, four knobs, one rule: watch the tallest hill.
Back to the parent topic · related: SN1 Reaction Mechanism · SN2 Reaction Mechanism · E1 and E2 Elimination.