Effect of substrate, nucleophile - base, solvent, leaving group: Worked examples

You have the four knobs from the parent note: substrate, nucleophile/base, solvent, leaving group. This page drills them until every combination is a reflex. We first lay out a matrix of every case class, then work examples that hit every single cell.

The scenario matrix

Think of a reaction as a coordinate: (substrate class, nucleophile/base type, solvent type, leaving group quality, temperature). Just like an angle has four quadrants, a halide reaction has a handful of "quadrants" — and each one has a winning mechanism. If we miss a cell, you'll meet a problem we never showed you. So here is the complete grid.

Cell	Substrate	Nu / Base	Solvent	Extra knob	Winner
C1	Methyl / 1°	strong Nu, weak base	polar aprotic	good LG	SN2
C2	3°	weak Nu (H₂O/ROH)	polar protic	—	SN1 (+E1)
C3	2°/3°	strong bulky base	any	heat	E2
C4	2°/1°	strong small base (EtO⁻)	—	heat	E2 vs SN2 mix
C5	any	same Nu, vary LG (I vs F)	fixed	LG is the variable	rate contrast
C6	any	same everything, vary solvent	protic vs aprotic	solvent is the variable	rate contrast
C7 (degenerate)	3°	strong Nu (CN⁻)	—	steric block	no SN2 → E2/SN1
C8 (zero case)	1°	F⁻ leaving group	aprotic	worst LG	essentially inert
C9 (real-world)	1° alkyl chloride	I⁻ (recycled)	acetone	Finkelstein	SN2
C10 (exam twist)	2° allylic/benzylic	weak Nu	protic	resonance cation	SN1 (fast)

We now clear every cell.

Worked example Example 1 — Cell C1 (clean SN2)

Statement: $CH_{3} CH_{2} Br + NaI$ in DMSO. Predict the mechanism and the product.

Forecast (guess first!): one substituent on the carbon, a fat happy iodide ion, no O–H in the solvent. Which of the four?

Substrate = 1°. Why this step? One CH₃ group leaves the C–Br carbon almost open — the backside (the face opposite Br) is reachable. That green-lights SN2, red-lights SN1 (a 1° cation is too unstable).
Nucleophile = I⁻, strong. Why? A strong nucleophile lets the slow step involve it → bimolecular. See SN2 Reaction Mechanism.
Solvent = DMSO, polar aprotic. Why? No O–H means the iodide is not caged in hydrogen bonds — it stays "naked and hungry", maximising SN2 speed. See Hydrogen Bonding and Solvation.
Product: $CH_{3} CH_{2} I + NaBr$ .

Verify → SN2. Sanity check: 1° + strong Nu + aprotic + good LG — every knob points the same way. No conflict, so the answer is unambiguous.

Worked example Example 2 — Cell C2 (SN1 with E1 tag-along)

Statement: $(CH_{3})_{3} CCl$ (tert-butyl chloride) warmed in water. Mechanism? What two products?

Forecast: three methyls crowd the carbon, water is a shy nucleophile. Can water ever "push"?

Substrate = 3°. Why? Three methyls form a wall — no backside attack possible, so SN2 is dead. But the same three methyls donate electrons and stabilise the $(CH_{3})_{3} C^{+}$ cation via hyperconjugation → ionisation is easy. See Carbocation Stability and Rearrangement.
Nucleophile = H₂O, weak/neutral. Why? It can't force a bond, so it just waits for the substrate to ionise on its own. That is the signature of a unimolecular path → SN1. See SN1 Reaction Mechanism.
Solvent = water, polar protic. Why? Its O–H groups hydrogen-bond around and stabilise both the forming cation and the leaving Cl⁻, lowering the ionisation barrier.
Products: substitution gives $(CH_{3})_{3} C-OH$ ; the same cation can also lose a β-proton → $(CH_{3})_{2} C=CH_{2}$ (E1). Why both? Once the cation exists, water can attack C (SN1) or grab an adjacent H (E1) — they share the slow step. See E1 and E2 Elimination.

Verify → SN1 (major) + E1 (minor). Sanity check: 3° + weak Nu + protic — again all knobs agree on unimolecular.

Worked example Example 3 — Cell C3 (bulky base forces E2)

Statement: 2-bromobutane $+ t -BuOK$ in $t$ -BuOH, warm. Mechanism and major alkene?

Forecast: the base is strong — but it's enormous. Where can something that big actually reach?

Substrate = 2°. Why? A 2° carbon is borderline: it could do SN2 with a small nucleophile. So the base's size becomes the deciding knob.
Base = $t$ -BuO⁻, strong but bulky. Why? It's too fat to squeeze behind the carbon for substitution, but a β-hydrogen sticks out on the edge of the molecule — easy to pluck. Strong base + bulk ⇒ E2.
Regiochemistry: a bulky base favours the less substituted (Hofmann) alkene, 1-butene, because it grabs the more accessible terminal H. See E1 and E2 Elimination.
Product: 1-butene (major), 2-butene (minor).

Verify → E2, Hofmann. Sanity check: strong bulky base is the textbook E2 flag; heat further tilts to elimination ( $Δ S > 0$ ).

Worked example Example 4 — Cell C4 (small strong base: SN2/E2 competition)

Statement: 1-bromobutane $+$ NaOEt in ethanol, moderate heat. What's the mix?

Forecast: ethoxide is strong and small. Small enough to reach the carbon — so does it substitute or eliminate?

Substrate = 1°. Why? A 1° carbon is wide open → substitution is very feasible.
Base = EtO⁻, strong and small. Why? Being small, it can reach the carbon (SN2) and it's basic enough to grab a β-H (E2). Both channels are live.
Tie-breaker = geometry + heat. Why? For 1° substrates SN2 usually wins (the carbon is accessible and there's no bulk penalty). Elimination is the minor product unless heat is high.
Products: butyl ethyl ether (SN2, major) + 1-butene (E2, minor).

Verify → SN2 major, E2 minor. Sanity check: contrast with Example 3 — same 2°/1° region, but a small base flips the winner back to substitution. The knob that changed was base bulk.

Worked example Example 5 — Cell C5 & C8 (leaving-group contrast, including the zero case)

Statement: Rank the SN2 rates of $CH_{3} CH_{2} I$ , $CH_{3} CH_{2} Br$ , $CH_{3} CH_{2} Cl$ , $CH_{3} CH_{2} F$ with NaCN in DMSO. Everything is identical except the leaving group.

Forecast: which halide leaves most willingly, and which basically refuses?

Freeze all other knobs. Why? All are 1°, all with the same strong Nu (CN⁻) in the same aprotic solvent → all SN2. The only variable is the C–X bond that breaks.
Leaving-group rule. Why? The best LG departs as the most stable / weakest base. Its conjugate acid HX tells us: stronger acid ⇒ weaker base X⁻ ⇒ better LG. Roughly (p $K_{a}$ of HX): HI $\approx - 10$ , HBr $\approx - 9$ , HCl $\approx - 7$ , HF $\approx + 3$ . See Acid Strength and pKa.
Order: $I^{-} > Br^{-} > Cl^{-} ≫ F^{-}$ .
The zero case (C8): F⁻ is a strong base (HF is a weak acid) → a terrible LG → ethyl fluoride is essentially inert. This is the degenerate cell: change nothing but the LG to F and the reaction dies.

Verify → I > Br > Cl ≫ F. Numeric check in the appendix: HF has positive p $K_{a}$ while HI/HBr/HCl are all negative, confirming F⁻ is the outlier.

Worked example Example 6 — Cell C6 (solvent contrast, same reactants)

Statement: $CH_{3} Br + NaCl$ : compare the SN2 rate in methanol (protic) vs DMF (aprotic).

Forecast: same substrate, same nucleophile, same LG — only the solvent changes. Which room lets Cl⁻ attack faster?

Freeze all knobs but solvent. Why? Isolate the solvent's effect so we can name it cleanly.
Methanol = polar protic. Why slower? Its O–H groups build a hydrogen-bond cage around Cl⁻, so the nucleophile must strip off that shell before it can attack — an energy tax. See Hydrogen Bonding and Solvation.
DMF = polar aprotic. Why faster? Same high polarity, but no O–H. DMF wraps the Na⁺ cation and leaves Cl⁻ bare — a naked, high-energy nucleophile that attacks much faster.
Result: SN2 is dramatically faster in DMF.

Verify → aprotic ≫ protic for SN2. Sanity check: this is the exact opposite of the SN1 solvent preference (protic wins there) — proving polarity alone is not the story; the H-bond cage is.

Worked example Example 7 — Cell C7 (degenerate: strong Nu meets 3°)

Statement: $(CH_{3})_{3} CBr + NaCN$ in DMSO. A strong nucleophile in a good SN2 solvent — so it must be SN2, right?

Forecast: the knobs seem to shout "SN2". What quietly vetoes it?

Substrate = 3°. Why decisive? The three methyls are a solid wall — no backside path exists. SN2 is physically impossible regardless of how strong CN⁻ is. This is the trap of the cell.
So SN2 is off the table. Why? Accessibility, not Nu strength, gates SN2 (recall the parent's steel-man mistake).
What's left? CN⁻ is also a base; with a 3° substrate and heat, E2 dominates → 2-methylpropene. In cold, weakly, some SN1 solvolysis can occur but the strong base pushes elimination.
Product: $(CH_{3})_{2} C=CH_{2}$ (E2 major).

Verify → E2, not SN2. Sanity check: the "obvious" SN2 answer is wrong — the degenerate substrate overrides the nucleophile. Always check substrate accessibility first.

Worked example Example 8 — Cell C10 (exam twist: resonance-stabilised cation)

Statement: benzyl bromide $C_{6} H_{5} CH_{2} Br$ + water. It's only "primary-like" — surely SN1 is impossible for a 1° carbon?

Forecast: normally 1° never does SN1. What's special here?

Look at the cation. Why? On ionising, the positive charge sits on a carbon next to a benzene ring. Resonance spreads that charge into the ring → the benzylic cation is unusually stable, comparable to a 3° cation. See Carbocation Stability and Rearrangement.
Nucleophile = H₂O, weak; solvent = protic. Why? Weak Nu can't push, but the stable cation forms readily and protic water stabilises it → SN1 becomes viable even for a formally 1° carbon.
Product: benzyl alcohol $C_{6} H_{5} CH_{2} OH$ .

Verify → SN1 (fast). Sanity check: the exam twist is that cation stability, not substitution count, is the real SN1 driver — allylic and benzylic 1° halides break the "1° never SN1" rule.

Coverage check

Recall Did we hit every cell of the matrix?

C1 (SN2) ::: Example 1 C2 (SN1+E1) ::: Example 2 C3 (bulky E2) ::: Example 3 C4 (SN2/E2 small base) ::: Example 4 C5 & C8 (LG contrast + zero case) ::: Example 5 C6 (solvent contrast) ::: Example 6 C7 (degenerate 3° + strong Nu) ::: Example 7 C9 (Finkelstein real-world) ::: covered by Example 1 pattern → see Finkelstein and Williamson Syntheses C10 (resonance cation twist) ::: Example 8

One picture to hold it all

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