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

This page is the practice arena for the parent topic. We will not re-teach the mechanisms — instead we will hunt down every kind of case the topic can produce and solve one representative of each. If you can do all cells below, nothing on an exam can surprise you.

Before we start, one reminder of the vocabulary we are using (each was built in the parent note): $R - X$ is a carbon bonded to a halogen; a nucleophile ( $Nu$ ) attacks the $δ^{+}$ carbon; a base grabs a β-hydrogen (an H on the carbon next to the C–X carbon); a carbocation is a positively charged, flat, three-bond carbon. We link out to Carbocations — stability, hyperconjugation, rearrangements, Nucleophilicity vs Basicity, Stereochemistry — R/S, optical activity, racemisation and Reaction kinetics — molecularity vs order where those ideas deepen.

The scenario matrix

Every problem in this chapter is a point in a small grid. The rows are the input classes (what kind of substrate/reagent you were handed); the columns are the behaviours that can result. A complete study covers every cell.

Case class (input)	What can go wrong / vary	Winning pathway(s)	Covered by
$1^{\circ}$ substrate, strong Nu	steric access open	pure SN2	Ex 1
$3^{\circ}$ substrate, weak Nu, protic	cation forms, branches	SN1 + E1 mixture	Ex 2
$3^{\circ}$ substrate, strong base	no SN2 access, base drives it	pure E2	Ex 2b
$2^{\circ}$ substrate, bulky strong base	Zaitsev vs Hofmann fork	E2 (Hofmann)	Ex 3
$2^{\circ}$ substrate, small strong base	Zaitsev fork	E2 (Zaitsev)	Ex 4
$2^{\circ}$ substrate, weak Nu, protic	SN1/E1 competition	SN1 + E1	Ex 4b
Degenerate: methyl / no β-H	elimination impossible	SN only	Ex 5
Chiral centre (sign of rotation)	inversion vs racemisation	stereochemistry	Ex 6
Solvent flip (aprotic ↔ protic)	same substrate, opposite winner	SN2 ↔ SN1	Ex 7
Limiting/kinetics (double $[Nu]$ )	rate response tells order	rate-law logic	Ex 8
Real-world word problem	choose reagent for a target	synthesis design	Ex 9
Exam twist (rearrangement)	cation shifts before capture	SN1 + hydride shift	Ex 10

Case 1 — pure SN2 (open $1^{\circ}$ carbon)

Worked example

$CH_{3} CH_{2} CH_{2} Br + NaN_{3} (DMSO)$ Forecast: guess the mechanism and the rate law before reading on.

Step 1 — classify the substrate. 1-bromopropane: the C–Br carbon carries only one other carbon → it is $1^{\circ}$ . Why this step? $1^{\circ}$ carbons make terrible carbocations, so SN1/E1 are ruled out immediately.

Step 2 — classify the reagent. Azide $N_{3}^{-}$ is a strong, charged nucleophile but a weak base. Why this step? Strong Nu + weak base = substitution over elimination; charged Nu in a $1^{\circ}$ substrate = SN2.

Step 3 — classify the solvent. DMSO is polar aprotic — it leaves $N_{3}^{-}$ "naked" and reactive. Why this step? Aprotic solvents accelerate SN2; see Nucleophilicity vs Basicity.

Step 4 — write product and rate law. Backside attack replaces Br: product $CH_{3} CH_{2} CH_{2} N_{3}$ , and $rate = k [RBr] [N_{3}^{-}]$

Verify: doubling $[N_{3}^{-}]$ should double the rate (order 1 in Nu). Order overall $= 1 + 1 = 2$ . ✓ (checked numerically in VERIFY).

Case 2 — SN1 + E1 mixture ( $3^{\circ}$ , protic, weak Nu)

Worked example

$(CH_{3})_{3} CCl$ in warm aqueous ethanol Forecast: one product or several?

Step 1 — substrate. tert-butyl chloride is $3^{\circ}$ . Why? $3^{\circ}$ gives the most stable carbocation → ionisation is easy.

Step 2 — reagent + solvent. Water/ethanol are weak, neutral nucleophiles and polar protic solvents. Why? Protic solvents solvate and stabilise $R^{+}$ and $C l^{-}$ , so the slow ionisation step is favoured — the hallmark of SN1/E1.

Step 3 — the cation branches. Once $(CH_{3})_{3} C^{+}$ forms it has two fates: water captures the carbon (SN1 → alcohol) or loses a β-H (E1 → alkene). Why? Both pathways share the same intermediate, so both appear.

Step 4 — name products. SN1: $(CH_{3})_{3} COH$ . E1: $(CH_{3})_{2} C=CH_{2}$ (the only alkene possible → automatically "Zaitsev").

Verify: rate law is $rate = k [RCl]$ (first order, independent of $[H_{2} O]$ ). Count atoms: $C_{4} H_{9} Cl \to C_{4} H_{10} O$ balances with $+ H_{2} O, - HCl$ . ✓

Case 2b — pure E2 on a $3^{\circ}$ substrate (strong base, no SN2 possible)

Worked example

$(CH_{3})_{3} CBr + NaOEt$ (hot ethanol) Forecast: the substrate is $3^{\circ}$ and the reagent is a strong base — SN2 is impossible, but SN1 and E2 both look plausible. Which dominates?

Step 1 — substrate + reagent. tert-butyl bromide is $3^{\circ}$ ; ethoxide $OEt^{-}$ is a strong, negatively charged base. Why? $3^{\circ}$ blocks backside attack ⇒ SN2 is dead. A strong charged base does not wait for ionisation — it actively rips off a β-H.

Step 2 — E2 outruns SN1. A strong base makes the bimolecular elimination fast, so E2 wins over the slower unimolecular SN1/E1 branch. Why? SN1/E1 depend on the slow spontaneous ionisation; a strong base provides a faster, base-driven route (its concentration enters the rate law and speeds things up).

Step 3 — product + rate law. Removing a β-H from a methyl gives 2-methylpropene $(CH_{3})_{2} C=CH_{2}$ (the only alkene possible): $rate = k [RBr] [OEt^{-}]$

Verify: doubling $[OEt^{-}]$ doubles the E2 rate (first order in base ⇒ overall order 2). Atom balance: $C_{4} H_{9} Br \to C_{4} H_{8}$ with $- HBr$ : carbons $4 = 4$ , hydrogens $9 = 8 + 1$ . ✓

Case 3 — E2 Hofmann (bulky base fork)

Worked example 2-bromobutane + potassium

tert-butoxide (hot) Forecast: which alkene is major?

Step 1 — substrate + conditions. $2^{\circ}$ substrate, strong bulky base ( $t$ -BuO $^{-}$ ), heat. Why? Strong base + heat = elimination; bulk = Hofmann steering.

Step 2 — locate the β-hydrogens. 2-bromobutane $CH_{3} -CHBr-CH_{2} -CH_{3}$ has β-H's on C1 (the $CH_{3}$ , 3 H's, exposed) and on C3 (the $CH_{2}$ , more hindered). Why? E2 removes a β-H; which one decides which alkene forms.

Step 3 — bulky base picks the accessible H. $t$ -BuO $^{-}$ cannot reach the crowded C3 H's easily, so it grabs a C1 H. Why? Transition-state accessibility beats product stability with bulky bases.

Step 4 — product. Removing a C1 H gives 1-butene ( $CH_{2} =CH-CH_{2} -CH_{3}$ ), the less-substituted = Hofmann major; 2-butene is minor.

Verify: count degrees of substitution — 1-butene is mono-substituted, 2-butene di-substituted; the Hofmann rule predicts the less substituted one wins here. ✓ (See figure below.)

Read the figure (s01). The navy circle is C2 (where Br sits, violet arrow shows Br leaving). Follow the orange arrow to C1: those are the exposed β-H's a bulky base can reach, giving 1-butene (Hofmann). Follow the magenta arrow to C3: those β-H's are hindered, reachable only by a small base, giving 2-butene (Zaitsev). The whole Case 3 vs Case 4 fork is literally "which arrow does the base follow" — that is what to observe.

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

Case 4 — E2 Zaitsev (small base, same substrate)

Worked example 2-bromobutane + sodium ethoxide

$NaOEt$ (hot) Forecast: same substrate as Case 3 — will the answer change?

Step 1 — reagent difference. $OEt^{-}$ is a strong base but small. Why? A small base can reach the more hindered β-H, so accessibility no longer forces Hofmann.

Step 2 — thermodynamics decides. Now the more stable, more-substituted alkene forms preferentially. Why? With no steric block, the reaction follows the lower-energy transition state, which resembles the more stable alkene (Carbocations — stability, hyperconjugation, rearrangements logic of hyperconjugation applies to alkenes too).

Step 3 — product. Major = 2-butene ( $CH_{3} -CH=CH-CH_{3}$ , di-substituted, Zaitsev); minor = 1-butene.

Verify: contrast with Case 3 — identical substrate, opposite major product, driven only by base size. That is the whole Zaitsev↔Hofmann fork. ✓

Case 4b — $2^{\circ}$ substrate, weak Nu, protic solvent (SN1/E1 competition)

Worked example 2-bromobutane in warm 80% aqueous ethanol (no added base)

Forecast: same substrate again, but now no strong reagent — just a weak nucleophilic solvent. What happens at a $2^{\circ}$ carbon with nothing to push it?

Step 1 — classify. $2^{\circ}$ substrate, weak neutral nucleophile (water/ethanol), polar protic solvent, mild heat. Why? No strong Nu and no strong base means neither SN2 nor E2 can get going; the only route left is spontaneous ionisation → SN1/E1.

Step 2 — form the $2^{\circ}$ cation. $CH_{3} -CH^{+} -CH_{2} -CH_{3}$ forms slowly (the rate-determining step). It is less stable than a $3^{\circ}$ cation, so this reaction is sluggish — a signature of the $2^{\circ}$ SN1/E1 corner. Why? $2^{\circ}$ cations are borderline; the protic solvent's stabilisation is what makes the pathway viable at all.

Step 3 — the cation branches (both products appear), written as balanced equations.

SN1 with water (hydrolysis) → 2-butanol: $C_{4} H_{9} Br + H_{2} O \to CH_{3} CH(OH)CH_{2} CH_{3} + HBr$
SN1 with ethanol (solvolysis) → the mixed ether 2-ethoxybutane: $C_{4} H_{9} Br + C_{2} H_{5} OH \to CH_{3} CH(OC_{2} H_{5}) CH_{2} CH_{3} + HBr$ (the ethanol oxygen attacks the cation, so the ether joins a butyl to an ethyl group).
E1 → alkenes, Zaitsev major: $C_{4} H_{9} Br \to CH_{3} CH=CHCH_{3} (2-butene, major) + HBr$ Why? All three share the same carbocation, so a $2^{\circ}$ substrate in a protic nucleophilic solvent always gives this SN1(alcohol)/SN1(ether)/E1 mixture — the competition this cell warns about.

Verify: rate law $= k [RBr]$ (first order, Nu-independent) confirms SN1/E1, not SN2/E2. Check each balance: 2-butanol $C_{4} H_{9} Br + H_{2} O \to C_{4} H_{10} O + HBr$ (C $4 = 4$ , H $9 + 2 = 10 + 1$ ); 2-ethoxybutane $C_{4} H_{9} Br + C_{2} H_{6} O \to C_{6} H_{14} O + HBr$ (C $4 + 2 = 6$ , H $9 + 6 = 14 + 1$ ); E1 $C_{4} H_{9} Br \to C_{4} H_{8} + HBr$ (H $9 = 8 + 1$ ). ✓

Case 5 — degenerate: no β-hydrogen (elimination impossible)

Worked example

$CH_{3} Br + OH^{-}$ and separately $(CH_{3})_{3} C-CH_{2} Br$ (neopentyl) Forecast: can either give an alkene?

Step 1 — count β-hydrogens. Methyl bromide has no carbon next to the C–Br carbon, so zero β-H. Neopentyl's β-carbon is the quaternary $C (CH_{3})_{3}$ — also no β-H. Why? E1/E2 both require a β-H to form C=C. No β-H ⇒ elimination is structurally impossible.

Step 2 — methyl case. $OH^{-}$ attacks the open methyl carbon → SN2 only, giving $CH_{3} OH$ . Why? Methyl is the least hindered possible carbon; strong Nu ⇒ textbook SN2.

Step 3 — neopentyl case (the trap). Backside attack is blocked by the bulky $t$ -butyl group, so SN2 is very slow; and SN1 is bad because a $1^{\circ}$ cation is unstable — so neopentyl halides are notoriously unreactive in all four pathways. Why? This degenerate case shows both mechanisms can fail simultaneously.

Verify: methyl → $CH_{3} OH$ (SN2, inversion meaningless as methyl is not a stereocentre). Neopentyl gives essentially no fast reaction. ✓

Case 6 — stereochemistry: inversion vs racemisation (sign of rotation)

Worked example (R)-2-bromobutane, two conditions

Forecast: which condition scrambles the chirality?

Step 1 — condition A: $CN^{-}$ in acetone (SN2). Why? Strong Nu, $2^{\circ}$ , aprotic → SN2. Backside attack ⇒ Walden inversion: (R) → (S) product, single clean stereochemistry.

Step 2 — condition B: warm aqueous ethanol (SN1). Why? Weak Nu, protic → SN1. The carbocation is flat (sp²); water attacks both faces → racemisation (≈50:50, slight inversion excess).

Step 3 — quantify. Idealised: SN2 gives 0% retention (100% inverted); pure SN1 gives 50% each enantiomer ⇒ net optical rotation ≈ 0. Why? Equal enantiomers cancel their rotations (Stereochemistry — R/S, optical activity, racemisation).

Verify: enantiomeric excess $ee = ∣% R - % S ∣$ . SN2: $ee = 100%$ . Pure SN1: $ee = 0%$ . ✓ (checked in VERIFY).

Read the figure (s02). The left panel is SN2: the magenta arrow shows Nu attacking from behind the carbon while the violet arrow shows X leaving from the front; the orange arrows are the three spectator bonds flipping "inside-out like an umbrella" — this is why one stereocentre inverts, $R \to S$ . The right panel is SN1: the orange carbon is the flat $sp^{2}$ cation, and Nu can strike from top (magenta) or bottom (violet) with equal chance — attack from both faces is exactly why you get a racemic 50:50 mix. Observe that inversion needs one geometry; racemisation needs a flat carbon.

Case 7 — solvent flip changes the winner

Worked example Same 2-bromopropane +

$Br^{-*}$ (labelled), (a) DMSO vs (b) water/methanol Forecast: does merely swapping solvent change the mechanism?

Step 1 — substrate. $2^{\circ}$ — the "swing" substrate that can go either SN1 or SN2. Why? $2^{\circ}$ is the only class where solvent alone can tip the balance.

Step 2 — aprotic (DMSO). A polar aprotic solvent cannot hydrogen-bond to an anion, so $Br^{-}$ is left "naked", small, and aggressive — it stays a strong nucleophile. It therefore reaches the $2^{\circ}$ carbon and displaces the leaving group in one step → SN2, with rate law $rate = k [RBr] [Br^{-}]$ Why this prediction? SN2 is a single bimolecular step whose transition state contains both the substrate and the nucleophile, so both concentrations appear in the rate law (overall order 2). The reason DMSO chooses SN2 is that it boosts nucleophile strength without stabilising a cation — no incentive to ionise.

Step 3 — protic (water/methanol). A polar protic solvent hydrogen-bonds around the anion (a "cage"), making $Br^{-}$ a weak, sluggish nucleophile; but the same solvent excellently stabilises ions, so the substrate spontaneously ionises to a carbocation → SN1, with rate law $rate = k [RBr]$ Why this prediction? SN1's slow step is unimolecular ionisation (nucleophile absent), so only $[RBr]$ appears (overall order 1). The protic solvent both weakens the Nu (killing SN2) and stabilises $R^{+}$ and $Br^{-}$ (feeding SN1).

Verify: the observed order changes from 2 to 1 purely from solvent — a classic exam data-table clue. If doubling $[Br^{-}]$ doubles rate ⇒ SN2; if it does nothing ⇒ SN1. ✓ (checked in VERIFY).

Case 8 — kinetics / limiting behaviour (double the nucleophile)

Worked example For

$RX + Nu^{-}$ , doubling $[Nu^{-}]$ doubles the rate in run A but does nothing in run B. Identify each. Forecast: which is SN1, which is SN2?

Step 1 — write both candidate laws. SN2: $r = k [RX] [Nu]$ . SN1: $r = k [RX]$ . Why this step? The rate law is the fingerprint of the rate-determining step (Reaction kinetics — molecularity vs order).

Step 2 — apply the doubling test. Run A: rate $\times 2$ when $[Nu] \times 2$ ⇒ first order in Nu ⇒ SN2. Run B: rate unchanged ⇒ zero order in Nu ⇒ SN1. Why this step? Only a species in the slow step appears in the law; Nu is absent from SN1's slow ionisation, so its concentration cannot affect the rate.

Step 3 — limiting sanity. As $[Nu] \to 0$ : SN2 rate $\to 0$ (it needs the collision); SN1 rate stays $k [RX]$ (Nu-independent), though with no added Nu the eventual product just comes from solvent. Why this step? Checking the extreme value confirms which term the nucleophile lives in.

Verify: SN2 factor $= [Nu]_{new} / [Nu]_{old} = 2$ ; SN1 factor $= 1$ . ✓ (checked in VERIFY).

Case 9 — real-world synthesis design (word problem)

Worked example You need to convert 1-butanol into pentanenitrile (

$CH_{3} CH_{2} CH_{2} CH_{2} CN$ ). Design a 2-step route and choose conditions to avoid elimination. Forecast: what reagent turns –OH into a good leaving group cleanly?

Step 1 — make the halide. $CH_{3} (CH_{2})_{3} OH + SOCl_{2} \to CH_{3} (CH_{2})_{3} Cl + SO_{2} + HCl$ . Why this step? $SOCl_{2}$ converts –OH into a great leaving group and its by-products ( $SO_{2}$ , $HCl$ ) leave as gases → irreversible, clean (Alcohols — preparation and reactions).

Step 2 — substitute with cyanide. $CH_{3} (CH_{2})_{3} Cl + NaCN$ in DMSO → SN2 → the nitrile. Why this step? $1^{\circ}$ substrate + strong Nu + aprotic solvent + weak-base Nu = substitution over elimination, avoiding the alkene.

Step 3 — why not use a strong base/heat? Those would push E2 and waste substrate as 1-butene. Why this step? Keeping the reagent a good nucleophile but poor base is the design lever.

Verify — clean atom bookkeeping. The 4-carbon butyl chain is preserved from butanol → chlorobutane → product; the $CN^{-}$ then adds one new carbon, so the product $CH_{3} (CH_{2})_{3} CN$ has 5 carbons total and is named pentanenitrile (the nitrile carbon counts as C1 in IUPAC nitrile naming). Balanced substitution: $C_{4} H_{9} Cl + CN^{-} \to C_{5} H_{9} N + Cl^{-}$ — carbons $4 + 1 = 5$ , hydrogens $9 = 9$ . ✓ (checked in VERIFY).

Case 10 — exam twist: carbocation rearrangement

Worked example 2-bromo-3-methylbutane

$(CH_{3})_{2} CH-CHBr-CH_{3}$ in warm methanol. Predict the major substitution product. Forecast: does the methoxy group end up where the Br was?

Step 1 — conditions ⇒ SN1. $2^{\circ}$ substrate, protic solvent, weak neutral Nu (methanol) → ionisation to a $2^{\circ}$ carbocation. Why this step? SN1 goes through a free carbocation — the stage where rearrangement can happen.

Step 2 — spot the better cation. The $2^{\circ}$ cation sits next to a carbon bearing an H that, if it migrates (1,2-hydride shift), yields a $3^{\circ}$ cation — more stable. Why this step? Carbocations rearrange toward greater stability (Carbocations — stability, hyperconjugation, rearrangements).

Step 3 — capture both cations. Methanol traps the new $3^{\circ}$ centre → major ether $(CH_{3})_{2} C(OCH_{3}) -CH_{2} CH_{3}$ (2-methoxy-2-methylbutane; methoxy has moved to the more substituted carbon). A fraction of the original $2^{\circ}$ cation is captured before it rearranges → minor ether $(CH_{3})_{2} CH-CH(OCH_{3}) -CH_{3}$ (unrearranged, methoxy where Br was). Why this step? The nucleophile traps whatever cation is present at the moment of attack; the more stable rearranged cation lives longer and dominates, but capture is fast enough to catch some unrearranged cation too.

Verify. The tell-tale sign of rearrangement is product connectivity ≠ starting connectivity, seen in the major product. Rate law is still $rate = k [RBr]$ (first order) because rearrangement happens after the slow ionisation step, so it cannot appear in the rate law. Atom balance for the major ether: $C_{5} H_{11} Br + CH_{3} OH \to C_{6} H_{14} O + HBr$ — carbons $5 + 1 = 6$ , hydrogens $11 + 4 = 14 + 1$ . ✓ (checked in VERIFY).

Recall Self-test the whole matrix

Which single factor flips Case 3 into Case 4? ::: The base size (bulky $t$ -BuO $^{-}$ → Hofmann; small $OEt^{-}$ → Zaitsev). In Case 2b, why does E2 beat SN1 on a $3^{\circ}$ substrate? ::: A strong charged base drives fast bimolecular elimination, outrunning the slow spontaneous ionisation of SN1; SN2 is blocked by bulk. In Case 4b, why is a $2^{\circ}$ substrate in protic solvent slow and messy? ::: The $2^{\circ}$ cation is only borderline stable, so ionisation is sluggish, and the shared cation branches into SN1 (alcohol) + SN1 (ether) + E1. In Case 6, why does SN1 give zero net rotation? ::: The flat carbocation is attacked from both faces equally → racemic mixture cancels. In Case 7, what does swapping DMSO for water do? ::: Turns SN2 (order 2) into SN1 (order 1) by weakening the nucleophile and stabilising the cation. In Case 8, run B's rate ignores $[Nu]$ — what mechanism? ::: SN1 (Nu absent from the slow ionisation step). In Case 10, does rearrangement change the rate law? ::: No — it happens after the rate-determining ionisation, so rate $= k [RX]$ still.