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

SN1's slow step makes a carbocation, so SN1 rate tracks carbocation stability. Alkyl groups donate electron density (hyperconjugation + inductive) to soothe the $+$ charge, so more alkyl = more stable = faster ionisation: $({\text{CH}}_3{)}_3\text{CBr}(3^{\circ })>({\text{CH}}_3{)}_2\text{CHBr}(2^{\circ })>{\text{CH}}_3{\text{CH}}_2\text{Br}(1^{\circ })>{\text{CH}}_3\text{Br}(\text{methyl}).$ This is the exact reverse of the SN2 order (where methyl is fastest and $3^{\circ }$ is inert). The two mechanisms want opposite substrates.

• Slow SN2: although the reacting carbon is $1^{\circ }$, the neighbouring carbon carries three methyls (a $t$-butyl block). Backside approach passes right by that wall → severe steric hindrance to the incoming nucleophile → SN2 crawls.
• Slow SN1: ionisation would give a primary carbocation on that $C{H}_2$ carbon — very unstable → does not form easily.
• What the cation would do: if the $1^{\circ }$ cation does form, it does not wait to be captured. One of the three methyl groups on the adjacent quaternary carbon migrates with its bonding electrons onto the empty $C{H}_2^{+}$ carbon — a 1,2-methyl shift. WHY the shift happens: the migration turns an unstable primary cation into a stable tertiary cation $({\text{CH}}_3{)}_2{\text{C}}^{+}-{\text{CH}}_2{\text{CH}}_3$, a large drop in energy, so it occurs before any nucleophile arrives. Arrow-pushing: the $\text{C}-{\text{CH}}_3$ bonding pair swings over to fill the empty orbital, leaving the former quaternary carbon as the new cationic centre. Answer: primary → poor SN1 (unstable $1^{\circ }$ cation); neopentyl bulk → poor SN2 (blocked backside). Both direct channels choke, and any cation that forms rearranges by a 1,2-methyl shift to a $3^{\circ }$ cation before capture. Full theory: Carbocations — stability, hyperconjugation, rearrangements.

$3^{\circ }$ + protic solvent → SN1, rate $=k[R-X]$ only. The nucleophile (${\text{Cl}}^{-}$ or water) enters after the rate-determining ionisation, so its concentration is absent from the rate law. Answer: the rate of bromide disappearance is unchanged whether $[{\text{Cl}}^{-}]$ is $0$, $0.5$ or $2.0\text{M}$ (only the product ratio $R-Cl:R-OH$ shifts). Adding nucleophile does not speed a step it is not in.

Both are strong bases + heat → E2. The question is only which β-hydrogen the base removes. In the figure the cyan atoms are the carbon chain C1–C2–C3–C4 with Br (amber) on C2, and the two candidate β-hydrogens are marked with their arrows.

• Antiperiplanar requirement first. E2 needs the breaking C–H and the breaking C–X bonds to be antiperiplanar — that is, pointing in opposite directions on the same axis, at a dihedral angle of $180^{\circ }$. Their p-character orbitals then line up to overlap into the new $\pi$ bond with least electron repulsion. In an open-chain molecule like this, free rotation about the C1–C2 and C2–C3 bonds lets either β-carbon (C1 or C3) place one of its hydrogens antiperiplanar to Br. So geometry alone rules out neither — both β-H sets are reachable. That means the winner is decided by which H the base can physically grab, not by conformation.
• The two β-H sets (see figure). Amber arrow → the C1 (${\text{CH}}_3$) hydrogens give terminal but-1-ene (less substituted, Hofmann). Cyan arrow → the C3 (${\text{CH}}_2$) hydrogens give internal but-2-ene (more substituted, Zaitsev).
• Small base ($OE{t}^{-}$): slim enough to line up antiperiplanar to Br at either β-carbon → the transition state resembling the more stable, more substituted alkene wins → Zaitsev → but-2-ene.
• Bulky base ($OtB{u}^{-}$): too fat to reach the crowded internal C3 hydrogen even when it is antiperiplanar; it can only capture the exposed terminal C1 hydrogens → Hofmann → but-1-ene.
• cis/trans (E/Z) of the but-2-ene. But-2-ene can form as (E) (trans) or (Z) (cis). In the antiperiplanar E2 transition state leading to but-2-ene, the two methyl groups can end up either on the same side (cis) or opposite sides (trans). The transition state that places the bulky methyls on opposite sides is lower in energy (less steric strain), so (E)/trans-but-2-ene is the major geometric isomer, with a smaller amount of (Z)/cis. But-1-ene, being terminal, has no E/Z isomerism (its terminal $=C{H}_2$ carbon carries two identical H's). Answer: $OE{t}^{-}$ → major (E)-but-2-ene (Zaitsev, trans-selective). $OtB{u}^{-}$ → major but-1-ene (Hofmann; no E/Z possible).

Goal: put $OH$ on the terminal (less substituted) carbon → we need anti-Markovnikov placement.

• Step 1 — but-1-ene + $\text{HBr}$ with peroxides (ROOR): free-radical addition gives anti-Markovnikov Br → 1-bromobutane (Br on the terminal C). See Markovnikov & anti-Markovnikov addition.
  • WHY peroxides flip the regiochemistry (radical mechanism): With no peroxide, HBr adds ionically — the proton attaches first, making the more stable ($2^{\circ }$) carbocation, so $B{r}^{-}$ lands on the internal carbon → Markovnikov (2-bromobutane). With peroxides present, the mechanism changes to a radical chain: the weak $O-O$ bond of ROOR breaks on warming to give radicals, which pull an H off HBr to generate a bromine radical ($\text{Br}\cdot$). Now it is $\text{Br}\cdot$ (not $H^{+}$) that adds to the alkene first, and it adds to the terminal carbon because that produces the more stable, more substituted ($2^{\circ }$) carbon radical on the internal carbon. That internal radical then abstracts H from another HBr, placing the H on the internal carbon and Br on the terminal carbon → anti-Markovnikov. The switch happens because the first species to add changed from $H^{+}$ (ionic) to $\text{Br}\cdot$ (radical), and both pathways still route through the most stable intermediate — but that "most stable intermediate" now points the Br to the opposite end.
  • (Plain $HBr$ with no peroxide would give the wrong, Markovnikov 2-bromobutane.)
• Step 2 — 1-bromobutane + $\text{NaOH}$ (aq, dilute) via SN2: $1^{\circ }$ substrate + strong nucleophile → clean SN2 substitution → butan-1-ol. Product route also appears in Alcohols — preparation and reactions. Answer: but-1-ene $\stackrel{\text{HBr, ROOR}}{\to }$ 1-bromobutane $\stackrel{\text{NaOH, SN2}}{\to }$ butan-1-ol.

Substrate is $3^{\circ }$; solvent is protic; reagent is weak/neutral (water/ethanol). This is textbook SN1 + E1 (they share the same carbocation).

• (i) SN1 and E1, both branching from the same flat cation. The SN1 branch has two possible nucleophiles present: water and ethanol.
  • Water as nucleophile → its O attacks the cation, then the resulting oxonium ($R-\stackrel{+}{O}{\text{H}}_2$) loses a proton to solvent → the alcohol 3-methylhexan-3-ol.
  • Ethanol as nucleophile (etherification) → the O of ${\text{CH}}_3{\text{CH}}_2\text{OH}$ attacks the cation, giving a protonated ether $R-\stackrel{+}{O}(\text{H}){\text{CH}}_2{\text{CH}}_3$; loss of that proton to solvent gives a neutral ethyl ether. Mechanistically identical to the water case — a lone pair on oxygen captures the cation, then deprotonates. This is why a mixed protic solvent gives both an alcohol and an ether.
• (ii) The cation is planar (sp${}^2$); solvent attacks both faces about equally → racemisation — the optically active $(R)$ reactant gives a nearly 50:50 $(R)+(S)$ mixture (often a small excess of inversion because the departing $B{r}^{-}$ momentarily shields one face).
• (iii) E1 follows Zaitsev (no bulky base here) → the more substituted alkene dominates: loss of a β-H from the internal $C{H}_2$ (toward the propyl chain) gives the more substituted internal alkene (3-methylhex-2-ene) as major over the less substituted terminal alkene. cis/trans: since this internal double bond carries different groups on each end, it can form as (E) or (Z); the E1 pathway, like E2, favours the (E)/trans isomer because the transition state that places the two larger chains on opposite sides is less strained — so (E)/trans is the major geometric isomer.
• (iv) Rate $=k[R-X]$ — water is a nucleophile/solvent but enters after the slow ionisation, so extra water changes product ratios/solvent polarity slightly but does not appear in the rate law; the primary rate control is ionisation, not $[{\text{H}}_2\text{O}]$.

The oxygen landed on the central carbon, not the carbon that lost Br → the skeleton reorganised. That is the fingerprint of a carbocation rearrangement (an SN1-type ionisation must have happened despite the primary label, because hot polar conditions allow slow ionisation). Follow the figure left → right.

• Step 1 — ionise (left of figure): $B{r}^{-}$ leaves the $C{H}_2$ carbon, giving the shaky primary carbocation drawn in amber: $({\text{CH}}_3{)}_2\text{CH}-\stackrel{+}{C}{\text{H}}_2$.
• Step 2 — 1,2-hydride shift (the amber curved arrow): the H on the adjacent $CH$ (the α-neighbour) slides across with its bonding electrons into the empty orbital on the $C{H}_2^{+}$ carbon. WHY it moves: this converts the unstable primary cation into the stable tertiary cation $({\text{CH}}_3{)}_3{\text{C}}^{+}$ shown on the right — a 1,2-hydride shift driven purely by the large drop in energy from $1^{\circ }\to 3^{\circ }$. Arrow-pushing: the $\text{C}-\text{H}$ bonding pair swings into the empty p-orbital, and the carbon it left behind becomes the new cationic centre.
• Step 3 — capture and deprotonate (right of figure): a lone pair on the oxygen of water (or $O{H}^{-}$) attacks the new $3^{\circ }$ cation $({\text{CH}}_3{)}_3{\text{C}}^{+}$, forming a protonated alcohol $({\text{CH}}_3{)}_3\text{C}-\stackrel{+}{O}{\text{H}}_2$; loss of that proton to solvent gives the neutral product $({\text{CH}}_3{)}_3\text{COH}$, 2-methylpropan-2-ol. Answer — full pathway: ionise ($B{r}^{-}$ leaves the primary carbon) → 1,2-hydride shift rearranges the $1^{\circ }$ cation to the $3^{\circ }$ cation $({\text{CH}}_3{)}_3{\text{C}}^{+}$ → nucleophile (water/$O{H}^{-}$) attacks and deprotonates → 2-methylpropan-2-ol. The $OH$ appears at the more-substituted carbon because the cation moved there first, before capture. Full theory: Carbocations — stability, hyperconjugation, rearrangements.