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

Before we start, one vocabulary refresh so no symbol is unearned:

$R - X$ = a carbon (call it the α-carbon) bonded to a halogen $X$ .
A β-hydrogen = an H on the carbon next door to the α-carbon; it is the one elimination steals.
Nu = nucleophile (electron-pair donor aiming at the α-carbon); base = proton-grabber aiming at a β-H. Same electrons, different target.
Rate law $rate = k [\dots]$ tells you which species must be present in the slowest step.

True or false — justify

Everything you touch flat is a carbocation :::- False. Only SN1 and E1 pass through a flat $s p^{2}$ carbocation; SN2 and E2 are concerted and never make a free cation. See Carbocations — stability, hyperconjugation, rearrangements.

SN2 rate depends on nucleophile concentration :::- True. The single transition state contains both $R - X$ and Nu, so $rate = k [R - X] [Nu]$ — doubling Nu doubles the rate.

SN1 rate depends on nucleophile concentration :::- False. The slow step (ionisation) contains only $R - X$ ; the nucleophile joins after the rate-determining step, so it cannot appear in $rate = k [R - X]$ .

A tertiary halide is the champion substrate for SN2 :::- False. Tertiary carbon is stuffed with bulky groups that block the backside approach; tertiary is essentially SN2-inert. It is the champion for SN1/E1 instead.

Every good leaving group is a strong base :::- False, exactly backwards. A good leaving group is a weak base — it is happy holding the electrons alone, so $I^{-} > B r^{-} > C l^{-} ≫ F^{-}$ . See Nucleophilicity vs Basicity.

Nucleophilicity and basicity always rise together :::- False. Basicity is affinity for $H^{+}$ ; nucleophilicity is affinity for carbon. Big, polarisable ions like $I^{-}$ are excellent nucleophiles but weak bases — Nucleophilicity vs Basicity.

The Zaitsev (more-substituted) alkene is always the major elimination product :::- False. With a bulky base ( $t$ -BuO $^{-}$ ) the base cannot reach the crowded β-H, so the less-substituted Hofmann alkene dominates.

SN1 of a single stereocentre gives pure inverted product :::- False. The flat carbocation is attacked from both faces, giving racemisation (often a slight inversion excess); only SN2 gives clean inversion. See Stereochemistry — R/S, optical activity, racemisation.

Polar protic solvents speed up SN2 :::- False. Protic solvents cage the nucleophile in a hydrogen-bond shell, blunting it; SN2 prefers polar aprotic solvents that leave Nu "naked". Protic solvents help SN1/E1 by stabilising the ions.

Molecularity and reaction order are the same thing :::- Not in general. Molecularity counts molecules in the elementary rate-determining step (SN1 is unimolecular, SN2 bimolecular); order is the experimental exponent sum. They coincide here only because each mechanism has one clean rate-determining step. See Reaction kinetics — molecularity vs order.

Free-radical HBr addition to an alkene follows Markovnikov :::- False. Peroxide-initiated (radical) HBr adds anti-Markovnikov; only the ionic HX addition follows Markovnikov & anti-Markovnikov addition.

Spot the error

"Adding more NaCN makes the SN1 reaction of $t$ -BuBr go faster." :::- The error: SN1's slow step is ionisation, which $CN^{-}$ is not part of; and $t$ -BuBr with a strong nucleophile that is also a base would actually favour E2/SN2 pathways, not accelerated SN1.

" $CH_{3} CH_{2} Br$ with NaCN gives a racemic product." :::- Ethyl bromide's α-carbon is not a stereocentre (two H's on it), so there is nothing to racemise; and the mechanism is SN2 anyway, which inverts rather than racemises.

"E2 works from any geometry as long as a β-H exists." :::- E2 needs the breaking C–H and C–X orbitals anti-periplanar ( $18 0^{\circ}$ dihedral) so they overlap into the new π bond; a β-H locked non-parallel (e.g. equatorial in some cyclohexanes) simply cannot eliminate by E2.

" $R - O H + H C l$ is preferred over $SOCl_{2}$ because it's simpler." :::- $H C l$ alone gives water and a reversible equilibrium; $SOCl_{2}$ converts $O H$ into a great leaving group and expels $S O_{2}$ and $H C l$ as gases, making it irreversible and clean. See Alcohols — preparation and reactions.

"The carbocation formed from a primary halide just reacts as-is." :::- Primary carbocations are far too unstable to form freely; if a primary substrate seems to give an SN1-like product, a hydride/methyl shift to a more stable cation has occurred first — Carbocations — stability, hyperconjugation, rearrangements.

"Finkelstein needs no driving force, it just swaps halogens." :::- It is driven by the precipitation of NaCl in acetone (NaI is soluble, NaCl is not); remove that precipitation and the equilibrium has no reason to move forward.

Why questions

Why does SN2 invert configuration? :::- The nucleophile attacks from the side opposite the leaving group, so the other three bonds flip through the planar transition state like an umbrella in wind — a clean $R \to S$ (or $S \to R$ ) inversion, called Walden inversion.

Why does a bulky base give the Hofmann product? :::- A big base cannot squeeze in to grab the crowded β-H that would give the substituted alkene, so it removes the most accessible (least hindered) β-H, yielding the less-substituted Hofmann alkene.

Why do polar aprotic solvents favour SN2? :::- They solvate the cation well but leave the anionic nucleophile poorly shielded and thus highly reactive ("naked"), so it strikes the α-carbon faster.

Why is $3^{\circ}$ best for SN1 but worst for SN2? :::- The two reactions want opposite things: SN1 wants a stable cation (three electron-donating alkyl groups help), while SN2 wants an open backside (three alkyl groups block it).

Why does heat push elimination over substitution? :::- Elimination increases the number of molecules (one substrate → alkene + HX-ish fragments), raising entropy; the $- T Δ S$ term makes elimination more favourable as temperature rises.

Why does the C–X bond make carbon electrophilic? :::- $X$ is more electronegative, pulling bonding electrons toward itself, so carbon carries a partial positive charge $δ^{+}$ that attracts electron-rich nucleophiles.

Edge cases

What happens to SN1 stereochemistry if the leaving group briefly lingers near one face? :::- It partially shields that face so the nucleophile prefers the opposite face, giving racemisation with a slight inversion excess rather than a perfect 50:50 mix.

Methyl halide ( $CH_{3} X$ ) — which mechanisms are even possible? :::- Only SN2 (and a trivial "no elimination" since there is no β-H). It cannot ionise to a methyl cation (far too unstable), so SN1/E1/E2 are all off the table.

A substrate with a good leaving group but no β-hydrogen and a strong base — what wins? :::- Elimination is impossible without a β-H, so the reaction must go substitution even though a strong base was present; geometry decides, not just reagent strength.

$2^{\circ}$ halide with a reagent that is both a decent nucleophile and a strong base — outcome? :::- The secondary substrate sits on the SN2/E2 border; the balance tips by conditions — heat and bulk favour E2, mild conditions and a good nucleophile favour SN2, so it is genuinely a competition.

Neopentyl halide $(CH_{3})_{3} C-CH_{2} X$ — primary carbon, yet SN2 is painfully slow. Why? :::- Its α-carbon is primary, but the adjacent quaternary carbon's three methyls physically block the backside approach, so steric hindrance stalls SN2 even without a tertiary α-carbon.

Zero base, weak neutral nucleophile, $1^{\circ}$ substrate, protic solvent, room temperature — what reaction? :::- Very little happens quickly: SN1/E1 are blocked (unstable $1^{\circ}$ cation) and there is no strong nucleophile/base for SN2/E2, so the primary halide is essentially unreactive under these mild conditions.

$t$ -butyl bromide with $t$ -BuO $^{-}$ (bulky strong base) — SN1, SN2, E1 or E2? :::- Tertiary blocks SN2; the strong bulky base drives concerted E2 (Hofmann-leaning), outcompeting the slower SN1/E1 ionisation route.