Exercises — Effect of substrate, nucleophile - base, solvent, leaving group

WHAT we ask: a good leaving group (LG) departs carrying the electrons, so it must be a stable, weak base — the more stable the anion, the better it leaves. WHY size matters: going down the halogen column the atoms get bigger, the negative charge spreads over a larger volume, so the anion is more stable and a weaker base. Look at the figure: the anion "balloons" grow left→right, and the size of the ball is exactly why the charge is more comfortable.

Order (best → worst): ${\text{I}}^{-}>{\text{Br}}^{-}>{\text{Cl}}^{-}\gg {\text{F}}^{-}$ $F^{-}$ is worst because HF is a weak acid, i.e. $F^{-}$ is a strong base that does not want to let go. See Acid Strength and pKa.

WHAT SN1 needs: the slow step makes two ions (a carbocation and the leaving anion). Anything that hugs and stabilises those ions lowers the energy hill. WHY protic wins: polar protic solvents (water, alcohols) have O–H bonds, so they hydrogen-bond to the anion and orient their oxygen lone pairs toward the cation — they solvate both ions (see Hydrogen Bonding and Solvation). Look at the figure: the left panel shows a protic solvent hugging both ions (good for SN1); the right panel shows an aprotic solvent hugging only the cation, leaving the anion bare (good for SN2). Same physics we reuse all page.

Answer: polar protic.

WHAT SN2 needs: the nucleophile attacks the carbon from behind the leaving group (a straight-line, 180° approach). WHY methyl wins: fewer groups around the carbon = less "wall" = the nucleophile reaches the carbon easily. In the figure, watch the green nucleophile try to reach the carbon: for methyl the path is open, for tertiary it is walled off by three groups.

Order (fastest → slowest): methyl > 1° > 2° > 3° (3° is essentially zero). Answer: methyl.

Walk the knobs:

• Substrate: 1° (propyl) → open, welcomes backside attack.
• Nucleophile: I⁻ → strong nucleophile, weak base.
• Solvent: acetone → polar aprotic → I⁻ is "naked" and hungry.
• LG: Br⁻ → good LG. All four point the same way: SN2. Look at the arrow-pushing sketch: iodide comes in from the back, the C–Br bond breaks on the front, and the carbon inverts like an umbrella in the wind.

Product: ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{I}$ + NaBr↓. This is the Finkelstein reaction; it works because NaBr is insoluble in acetone and precipitates, dragging the equilibrium forward.

• Substrate: 3° → cannot be attacked from behind (too crowded), but forms a very stable 3° carbocation (see Carbocation Stability and Rearrangement).
• Nucleophile: H₂O → weak, neutral, cannot force a bond.
• Solvent: water → polar protic → stabilises the ions it makes.
• LG: Cl⁻ → decent LG. Weak nucleophile + easily-ionising substrate + protic solvent ⇒ the substrate ionises by itself first. Mechanism: SN1 (with some E1 on warming). Product: $({\text{CH}}_3{)}_3\text{C–OH}$ plus a little 2-methylpropene.

• Substrate: 2° → could in principle do SN2 or E2.
• Base/Nu: $t$-BuO⁻ is a strong but very bulky base — too fat to reach the carbon, but it can still pluck a small proton.
• Solvent: $t$-BuOH → protic, mild.
• LG: Br⁻ → good. Bulky strong base + 2° ⇒ it grabs a β-hydrogen instead of attacking carbon. Mechanism: E2, giving mostly the less substituted alkene (Hofmann product) because of the bulk.

The only knob that changes is the solvent. Re-use the solvation picture from L1.2: in protic methanol the anion is caged; in aprotic DMSO it is bare.

• Methanol (protic): hydrogen-bonds to the azide anion $N_3^{-}$, wrapping it in a "cage." To react, the nucleophile must first shed part of this cage — an energy cost that slows the reaction.
• DMSO (aprotic): solvates the Na⁺ cation strongly but leaves $N_3^{-}$ bare and reactive. Faster in DMSO. For small hard anions this "de-caging" can speed SN2 by many powers of ten (often quoted as $10^3$–$10^6$). The substrate, nucleophile, and LG are identical — solvent alone decides.

Same species, two solvents — the knob is solvation (see Hydrogen Bonding and Solvation).

• In water (protic): F⁻ is small and charge-dense, so water hydrogen-bonds to it tightly — it is locked in a cage and reacts slowly. I⁻ is big and soft, only loosely solvated, so it is free to attack → $I^{-}>B{r}^{-}>C{l}^{-}>F^{-}$.
• In DMF (aprotic): no O–H, so no anion cage exists. Now the ions react on their intrinsic basicity/charge density. The tightly-held small F⁻ is the strongest, most reactive base-nucleophile → $F^{-}>C{l}^{-}>B{r}^{-}>I^{-}$. Conclusion: in protic solvent solvation dictates the order; in aprotic solvent intrinsic basicity does.

This is a Williamson ether synthesis, which is an SN2 on the alkyl halide.

• Route A: the halide is 3° — SN2 is blocked (too crowded). Methoxide is a strong base, so instead of substituting it will do E2, giving 2-methylpropene, not the ether. ✗
• Route B: the halide is methyl — perfectly open to SN2. The bulky $t$-butoxide acts as the nucleophile against the tiny, exposed methyl carbon. ✓ Rule: in Williamson synthesis, put the SN2 job on the least hindered carbon. Route B works.

The substrate is fixed (2°), which can do any of the four. We steer with the other three knobs. (i) To favour substitution (SN2):

• Nucleophile: strong, small, weakly basic, e.g. ${\text{CH}}_3{\text{CO}}_2^{-}$ or $I^{-}$.
• Solvent: polar aprotic (acetone, DMSO).
• Keep it cool (heat favours elimination). (ii) To favour elimination (E2):
• Base: strong and bulky, e.g. $t$-BuOK.
• Solvent: the corresponding alcohol.
• Heat it (positive ΔS of making an alkene + small molecule is favoured at high T). Same starting halide, opposite products — decided purely by the three tunable knobs.

WHY this tool: $\mathrm{p}{K}_a=-{\log }_{10}{K}_a$, so a difference in pKa is a ratio of $K_a$. Using the log lets us compare leaving abilities across 13 orders of magnitude cleanly (see Acid Strength and pKa). $\frac{K_a(\text{HI})}{K_a(\text{HF})}=10^{\mathrm{p}{K}_a(\text{HF})-\mathrm{p}{K}_a(\text{HI})}=10^{3.2-(-10)}=10^{13.2}.$ $10^{13.2}\approx 1.6\times 10^{13}.$ Interpretation: HI is about $1.6\times 10^{13}$ times more acidic than HF, so $I^{-}$ is that much more stable/weaker as a base — it is a vastly better leaving group. This is exactly why alkyl iodides fly and alkyl fluorides are essentially inert.

WHY exponential: the fraction of molecules that clear an energy barrier of height $\Delta G^{‡}$ follows the Boltzmann factor $e^{-\Delta G^{‡}/RT}$. Lowering the barrier multiplies the rate. Look at the reaction-coordinate figure: two hills over the same reactant/product valleys. The height of each hill is $\Delta G^{‡}$; the gap between the two hilltops is the double-delta $\Delta \Delta G^{‡}$ — literally "the change in the barrier." Shrinking that gap is what multiplies the rate.

The rate ratio is $\frac{k_{\text{aprotic}}}{k_{\text{protic}}}=\exp \left(\frac{\Delta \Delta G^{‡}}{RT}\right)=\exp \left(\frac{22800}{8.314\times 298}\right).$ $\frac{22800}{2477.6}=9.20\Rightarrow e^{9.20}\approx 9.9\times 10^3.$ Result: roughly a 10 000-fold rate boost — a concrete number behind "aprotic solvents dramatically speed SN2."

Reading the knobs quadrant by quadrant (see the labelled grid below):

• weak Nu, protic: substrate ionises alone → SN1 (+ some E1).

• strong small Nu, aprotic: naked strong nucleophile attacks the open-ish 2° carbon → SN2.

• strong bulky base, hot: can't attack carbon, heat pushes disorder → E2.

• strong bulky base, cold: still bulky (no SN2), so 2° still eliminates → E2 (heat only raises the elimination fraction).

• LG order: $I^{-}>B{r}^{-}>C{l}^{-}\gg F^{-}$; tracks smaller pKa of HX.
• SN2 loves: methyl/1°, strong small Nu, aprotic, good LG, cool.
• SN1/E1 loves: 3°, weak Nu, protic, warm.
• E2 loves: strong bulky base, any accessible β-H, heat.
• Solvent can flip nucleophilicity order and change rates by $10^3$–$10^6$.