4.3.2 · D5Halides and Oxygenated Derivatives
Question bank — Effect of substrate, nucleophile - base, solvent, leaving group
A rapid-fire trap-clearing sheet for the parent topic. Each line is a question you can cover; the reveal after
:::gives the reasoning, not just the verdict.
First, the jargon (read this before the traps)
A picture of all four knobs at once
The figure below is the whole decision map: read your four knobs across it and the arrow lands you on a mechanism. Refer back to it while you work the traps.
A picture of backside attack vs the solvent cage
These two pictures carry the two most-tested visual ideas, so you never have to leave the page to "see" them.
True or false — justify
Tertiary halides are the fastest substrates for SN2 because they are the most electron-rich.
False — SN2 rate depends on the nucleophile physically reaching the carbon; three alkyl groups wall it off, so 3° is effectively unreactive in SN2 (see the blocked 180° line in the figure above). More in SN2 Reaction Mechanism.
A better leaving group speeds up SN1 but has no effect on SN2.
False — the C–X bond breaks in the rate-determining step of all four mechanisms, so a good LG accelerates SN1, SN2, E1 and E2 alike.
Polar solvents always speed up SN2 because SN2 involves charged species.
False — polar protic solvent H-bond-cages the nucleophile (right panel above) and slows SN2; you want polar aprotic, same polarity but no cage. See Hydrogen Bonding and Solvation.
I⁻ is a better nucleophile than F⁻ in water, therefore I⁻ is also the stronger base.
False — nucleophilicity (kinetic, attacks carbon) and basicity (grabs H⁺) are different; in water I⁻ is the better nucleophile but F⁻ is the stronger base.
Heating a reaction always increases substitution yield because molecules move faster.
False — heat favours elimination, because elimination has the larger positive (one substrate splits into two fragments — more disorder) and that entropy term grows with temperature. See E1 and E2 Elimination.
A strong base is automatically a strong nucleophile.
False — t-BuO⁻ (tert-butoxide) is a strong base but a weak nucleophile because its bulk blocks approach to carbon while it can still reach a small H⁺.
Fluoride, being the smallest and most electronegative halide, is the best leaving group.
False — F⁻ is a strong base (HF is a weak acid, small ), so it holds electrons badly on departure and is the worst halide leaving group.
In aprotic solvent the halide nucleophilicity order reverses toward the basicity order (F⁻ > Cl⁻ > Br⁻ > I⁻).
True — with no H-bond cage stripping the anion, the small charge-dense F⁻ is now the most reactive, tracking its higher basicity.
Spot the error
"CH₃CH₂F reacts faster than CH₃CH₂I with NaCN in DMSO because F is more electronegative and pulls electrons."
Error — reactivity here is set by the leaving group, and I⁻ ≫ F⁻ as a LG, so the iodide is far faster; the fluoride is essentially inert.
"An alcohol reacts with HBr by SN1 because OH⁻ is a good leaving group."
Error — OH⁻ is a bad leaving group (strong base); the alcohol must first be protonated to –OH₂⁺, which leaves as neutral water. See Acid Strength and pKa.
"Acetone is polar protic, so it stabilises carbocations and pushes toward SN1."
Error — acetone has no O–H/N–H, so it is polar aprotic; it solvates cations but leaves anions bare, pushing toward SN2, not SN1.
"A 1° halide with NaOEt in ethanol must give SN1 because the solvent is protic."
Error — 1° substrates cannot form a stable cation, so SN1 is off the table; strong EtO⁻ drives SN2 (with some E2), the protic solvent notwithstanding.
"Rearrangement can occur in an SN2 reaction if a more stable cation is available."
Error — SN2 has no carbocation intermediate (concerted, one step); only the cation pathways (SN1/E1) rearrange. See Carbocation Stability and Rearrangement.
"NaI in acetone converts R–Cl to R–I by SN1 — the Finkelstein reaction."
Error — Finkelstein is SN2: strong nucleophile I⁻ in aprotic acetone attacks a 1°/2° carbon in one step. See Finkelstein and Williamson Syntheses.
"Since TsO⁻ (tosylate) is a huge ion it must be a poor leaving group."
Error — size is not the criterion; TsO⁻ is a very weak base (charge delocalised over three oxygens), so it is an excellent leaving group.
Why questions
Why does increasing alkyl substitution flip a substrate from SN2-only to SN1-favoured?
More groups add steric bulk that blocks backside attack (kills SN2) while stabilising the C⁺ by hyperconjugation/induction (helps SN1) — two opposite effects that cross over.
Why does polar aprotic solvent make a nucleophile more reactive?
It solvates the counter-cation but cannot H-bond the anion, leaving the nucleophile "naked" and hungry so it attacks carbon far faster.
Why is a weak base a good leaving group?
A weak base can comfortably hold the negative charge after departure, lowering the transition-state energy of bond breaking; low basicity ⇔ its conjugate acid HX has a high .
Why does a bulky strong base like t-BuO⁻ give E2 rather than SN2?
It is too fat to reach the crowded carbon for substitution, but a peripheral β-hydrogen is still accessible, so it plucks that proton and eliminates.
Why do neutral weak nucleophiles (H₂O, ROH) push a 3° halide toward SN1 not SN2?
They are too weak to force a bond in a rate-determining step, so the substrate ionises on its own first; the 3° cation is stable enough to make this feasible.
Why does a good leaving group help even the concerted SN2 step?
Even though SN2 is one step, the C–X bond is partly broken in its transition state, so a stable (weak-base) LG lowers that transition-state energy.
Edge cases
Methyl halide (CH₃X) with a strong base — SN2 or E2?
SN2 only — there is no β-carbon, so no β-hydrogen exists to eliminate; elimination is geometrically impossible.
A 3° halide with a strong bulky base and heat — what dominates?
E2 — SN2 is blocked by bulk, and the strong base plus heat favour bimolecular elimination over the SN1/E1 pathway.
A 1° halide can never form a carbocation, so is SN1 strictly zero for it?
Essentially yes for simple 1° cases; the 1° cation is far too unstable, so 1° substrates run SN2 (or E2), never a clean SN1.
But a 1° benzylic or allylic halide — does the "1° never does SN1" rule still hold?
No — the cation here is stabilised by resonance into the ring/double bond, so these 1° substrates can do SN1; they are the exception that blurs the SN1/SN2 line and often do both readily. See Carbocation Stability and Rearrangement.
A substrate with a poor leaving group but an unusually acidic β-hydrogen (e.g. β to a carbonyl) — which elimination route opens up?
E1cB — the base removes the acidic β-H first to give a stabilised carbanion, which then expels the poor LG in a second step; this is the case E1/E2 cannot handle. See E1 and E2 Elimination.
How does E1cB differ from E1 in the order of bond changes?
In E1 the LG leaves first (making a cation); in E1cB the β-proton leaves first (making a carbanion) — opposite ends go first, so E1cB needs an acidic β-H and tolerates a poor LG.
A benzylic 2° halide with a moderate nucleophile — why is the mechanism ambiguous?
Resonance makes the cation both easy to form (favours SN1) while the carbon is also fairly open (allows SN2), so conditions (solvent, Nu strength) tip the balance rather than the substrate alone.
An ambident nucleophile like CN⁻ (attacks via C or N) — what decides which end bonds?
Hard/soft matching: the soft carbon end of CN⁻ prefers the soft SN2 carbon (giving R–CN), while the harder nitrogen end can dominate under more SN1-like/ionic conditions (giving isonitrile R–NC).
Nitrite ion NO₂⁻ (ambident: O or N) with a 1° halide in aprotic solvent — likely product?
SN2 conditions favour attack through the softer nitrogen, giving the nitro compound R–NO₂ rather than the O-bonded nitrite ester.
Same substrate, same nucleophile, only the solvent switched from DMSO to methanol — what changes?
The SN2 rate drops sharply, because methanol (protic) H-bond-cages the nucleophile that DMSO (aprotic) had left bare (right panel of the solvent figure).
A 2-haloethyl sulfide or amine reacts far faster and with retention — why does it break the SN1/SN2 rules?
Neighbouring-group participation (anchimeric assistance): the sulfur/nitrogen lone pair loops in to displace the LG intramolecularly, forming a cyclic intermediate; the external nucleophile then opens it, giving two inversions = net retention and a big rate boost.
Why does anchimeric assistance often outrun a normal SN2 even with a weak external nucleophile?
The internal nucleophile is already tethered next to the carbon (effective concentration is enormous), so its intramolecular attack is far faster than any bimolecular collision.
If the leaving group is water (from a protonated alcohol) vs iodide, both are "excellent" — is the mechanism identical?
Not necessarily; the LG only sets how easily C–X breaks, while substrate class and nucleophile still decide SN1 vs SN2 vs E1 vs E2.
Adding a trace of a very good nucleophile to a reaction that was drifting toward SN1 — what happens?
The strong nucleophile can now intercept in a rate-determining bimolecular step, pulling the pathway toward SN2 (or E2 if it is basic and bulky).
Recall One-line self-test
Cover every answer above and reason each aloud in under ten seconds. Passing bar ::: verdict and the correct one-line reason, not the verdict alone.