Alkyl halides — preparation, SN1 vs SN2 (mechanism, kinetics, stereochemistry), E1 vs E2 (mechanism, Zaitsev - Hofmann): Visual walkthrough

This is the visual companion to the parent topic. Prerequisites we lean on: Carbocations — stability, hyperconjugation, rearrangements, Reaction kinetics — molecularity vs order, Stereochemistry — R/S, optical activity, racemisation, and Nucleophilicity vs Basicity.

Step 1 — What a "reaction rate" even means

WHAT. Before any mechanism, define the thing we are deriving. The rate of a reaction is how fast the product concentration climbs — how many molecules turn over per second in a litre.

WHY this first. You cannot derive a "rate law" until you agree what "rate" is. Everything else is built on this brick.

PICTURE. Look at the two beakers. Each dot is a reacting molecule. In one second, more collisions that "stick" means a steeper climb in the product line on the right.

Figure — Alkyl halides — preparation, SN1 vs SN2 (mechanism, kinetics, stereochemistry), E1 vs E2 (mechanism, Zaitsev - Hofmann)

The single idea we will use over and over:

Step 2 — Turning "a meeting" into a formula

WHAT. Convert "molecules must meet" into algebra. If a step needs species $A$ to meet species $B$ , the chance of finding an $A$ and a $B$ together is proportional to how crowded each is: $[A]$ times $[B]$ .

WHY multiply, not add? Because meetings are an AND, not an OR. Double the $A$ 's → twice as many meetings. Also double the $B$ 's → twice as many again → four times total. Independent chances multiply. (Same reason two coin flips both landing heads is $\frac{1}{2} \times \frac{1}{2}$ , not $\frac{1}{2} + \frac{1}{2}$ .)

PICTURE. The grid shows every possible $A$ – $B$ pairing. Add one column of $A$ : a whole new row of meetings lights up. Add one row of $B$ : another column lights. The lit squares (= meetings) grow as the product of the counts.

Step 3 — SN2 has exactly ONE step, and both partners are in it

WHAT. Draw the SN2 event. The nucleophile $Nu^{-}$ (a lone-pair carrier, negatively rich) drives into the carbon from the side directly opposite the leaving group $X$ . Bond making (Nu→C) and bond breaking (C→X) happen together, passing through one 5-connected transition state.

WHY one step matters. If bond forming and bond breaking are the same event, then both $R - X$ and $Nu^{-}$ must be present at that instant — the step's meeting is " $R - X$ AND $Nu^{-}$ ".

PICTURE. Follow the red $Nu^{-}$ arrow entering from the left, the crowded transition state in the middle (five things touching carbon), and $X^{-}$ leaving on the right. Notice the three grey bonds beginning to flip through the flat plane — hold that thought for Step 5.

Apply Step 2's formula with $A = R - X$ and $B = Nu^{-}$ :

Step 4 — SN1 splits into TWO steps, and the nucleophile misses the slow one

WHAT. Draw the SN1 event. Step (1): $R - X$ ionises all by itself — $X^{-}$ walks off with the bonding electrons, leaving a carbocation $R^{+}$ . This is slow (hard: you are ripping a bond with no help). Step (2): the nucleophile swoops onto the now-hungry $R^{+}$ — fast.

WHY the slow step rules the rate. A two-step road is only as fast as its slowest gate. If Step (1) is a narrow slow door and Step (2) is wide open, the whole traffic speed is set by Step (1). We call it the rate-determining step.

PICTURE. The left "energy hill" is tall (slow ionisation, rate-determining). The right hill is a tiny bump (fast capture). The overall pace is stuck behind the tall hill — and only $R - X$ stands on that hill.

Now use Step 2's formula on the slow step only. Its meeting is " $R - X$ falling apart" — a single molecule, no partner:

Step 5 — The pictures also predict the STEREOCHEMISTRY

WHAT. Same two pictures, now watch the shape.

SN2: attack comes from behind $X$ , so the other three groups snap through the flat plane — like an umbrella flipping inside-out in wind. A stereocentre labelled $R$ becomes $S$ : complete inversion (Walden inversion).
SN1: the intermediate $R^{+}$ is flat ( $s p^{2}$ , trigonal planar). The nucleophile can land on the top face or the bottom face with equal ease, so you get roughly 50/50 of each mirror image → racemisation.

WHY the shapes force this. SN2's single-file backside approach has only one geometric outcome (inversion). SN1's flat cation has lost the memory of which side $X$ used to be on — both faces are open, so both products form. (See Stereochemistry — R/S, optical activity, racemisation.)

PICTURE. Left: the umbrella flip, $R \to S$ , one product. Right: the flat cation with two green attack arrows (top and bottom) making the two mirror images in equal amounts.

Recall Why kinetics and stereochemistry agree

Number of steps (kinetics) ::: one step SN2, two steps SN1 What shape the reacting carbon holds ::: SN2 keeps a rigid 4-bond carbon (backside → inversion); SN1 passes through a flat 3-bond cation (both faces → racemic) So the same mechanism picture explains both the rate law and the stereochemistry — they are not two facts, they are one fact seen twice.

Step 6 — Edge & degenerate cases the pictures must also cover

A derivation is only trustworthy if it survives the extremes. Read each off the two mechanism pictures.

Case A — Tertiary substrate ( $3^{\circ}$ ). In the SN2 picture, three bulky groups guard the backside; the $Nu^{-}$ arrow cannot reach carbon (steric wall). So SN2 rate → essentially zero. But the SN1 picture loves it: a $3^{\circ}$ carbocation is the most stable one (Carbocations — stability, hyperconjugation, rearrangements), so the slow ionisation hill is lower → SN1 fast.

Case B — Methyl / primary substrate ( $1^{\circ}$ ). SN2 backside is wide open → fast. SN1 would need a wildly unstable $1^{\circ}$ /methyl cation → the ionisation hill is too tall → SN1 essentially never happens.

Case C — Nucleophile concentration pushed to zero. SN2 rate $= k [R - X] [Nu^{-}] \to 0$ : no attacker, no reaction. SN1 rate $= k [R - X]$ is untouched until Step (2) — but with no nucleophile at all, the cation just waits (or grabs solvent). This is exactly why SN1 tolerates weak, dilute nucleophiles and SN2 does not.

PICTURE. The two-panel chart: as we walk methyl → $1^{\circ}$ → $2^{\circ}$ → $3^{\circ}$ , the SN2 line sinks (steric wall grows) while the SN1 line rises (cation stabilises). They cross around secondary — the borderline where a $2^{\circ}$ halide can do either, depending on conditions.

Recall The crossover in one line

Where does SN2 win, where does SN1 win? ::: SN2 wins at methyl/1° (open backside, unstable cation); SN1 wins at 3° (blocked backside, stable cation); 2° is the tug-of-war decided by solvent, nucleophile strength, and temperature.

The one-picture summary

Everything above compressed: count the steps → read the rate law → read the shape.

Recall Feynman retelling — the whole walkthrough in plain words

Reactions happen when molecules bump. If a step needs two molecules to bump, then having more of either makes more bumps, and having more of both multiplies the bumps — that is why the rate is "amount of one times amount of the other".

SN2 is one clean shove: a nucleophile runs in from behind, kicks the halogen out the front, all in a single motion. Both the carbon-molecule and the nucleophile are in that shove, so the rate multiplies both concentrations — second order. And because the shove comes from behind, the molecule flips inside-out like an umbrella — clean inversion.

SN1 is lazier and comes in two acts. First the halogen falls off on its own, leaving a flat, positively-charged carbon — this is slow and hard, the bottleneck. Then a nucleophile grabs the flat carbon — easy. Since the bottleneck act has only the substrate in it, pouring in more nucleophile changes nothing: first order. And because the flat carbon can be attacked from either face equally, you get a 50/50 mirror-image mixture — racemisation.

Bulky (tertiary) carbons block the SN2 shove but make the SN1 flat cation extra stable — so big carbons go SN1, small carbons go SN2, and the secondary ones in the middle can be talked into either.