Alkenes — preparation (dehydration, dehydrohalogenation, Zaitsev's rule), addition reactions: Worked examples

Definition The vocabulary you need first (α, β, Δ, E1, E2)

Before any example, pin down five symbols the parent note used casually:

α-carbon (alpha): the carbon that carries the leaving group (the –OH or –Br). Think "α = the one letting go."
β-carbon (beta): any carbon directly next to the α-carbon. A β-H is a hydrogen sitting on that neighbour. Elimination always pulls the leaving group off α and an H off β, so the new double bond forms between α and β.
Δ (delta): the triangle drawn on a reaction arrow simply means "heat applied." $Δ$ = "warm it up."
E1 (Elimination, unimolecular): a two-step elimination — first the leaving group departs to make a carbocation (slow), then a base pulls the β-H. Acid-catalysed dehydration goes by E1.
E2 (Elimination, bimolecular): a one-step elimination — the base pulls the β-H at the same time as the leaving group departs, in a single concerted motion. Alcoholic-KOH dehydrohalogenation goes by E2. (More on both in Haloalkanes — SN1, SN2, E1, E2.)

The scenario matrix

Every question in this topic falls into one of these case classes. Think of it like the quadrants of a graph: we must visit each cell so you never meet a scenario we did not show.

Cell	Case class	The one thing being tested	Example
P1	Prep · dehydration, only ONE alkene possible	E1 mechanism, good vs bad leaving group	Ex 1
P2	Prep · dehydration/DHX with a choice of β-H	Zaitsev (more substituted wins)	Ex 2
P3	Prep · degenerate — symmetric substrate, one product	Recognising "no choice"	Ex 3
P4	Prep · same reagent (KOH) solvent switch	Aqueous vs alcoholic (subst. vs elim.)	Ex 4
A1	Add · symmetric alkene + symmetric reagent	No orientation to worry about (test for unsaturation)	Ex 5
A2	Add · unsymmetric alkene + HX (ionic)	Markovnikov via carbocation stability	Ex 6
A3	Add · HBr + peroxide	Anti-Markovnikov via radical	Ex 7
A4	Add · limiting/trap — peroxide with HCl or HI	The rule's boundary (fails)	Ex 8
W1	Word problem — real-world (hydrogenation of oil)	Reading chemistry from a story	Ex 9
X1	Exam twist — carbocation rearrangement	The "gotcha" the naive rule misses	Ex 10

Each example below is tagged with its cell. The map below groups those ten cells into the two grand themes — preparation (build the C=C by elimination) on the left, addition (destroy the C=C) on the right. Use it to locate any question you meet: read across to the cell, then jump to that example.

Figure — Alkenes — preparation (dehydration, dehydrohalogenation, Zaitsev's rule), addition reactions

Figure 1 — the scenario map. Blue box = preparation cells P1–P4; orange box = addition cells A1–A4, W1, X1. The grey double arrow between them is the reminder that addition is literally the reverse idea of preparation.

The examples

Cell P1 — dehydration, single product

Worked example Ex 1 · Dehydrate propan-1-ol

$CH_{3} -CH_{2} -CH_{2} -OH conc. H_{2} S O_{4} Δ ?$ (Recall $Δ$ on the arrow just means "heat.")

Forecast: which two atoms leave, and what single alkene appears?

Protonate the –OH. $- OH + H^{+} \to - O + H_{2}$ . The carbon carrying it is the α-carbon, C1. Why this step? $OH^{-}$ is a strong base = a terrible leaving group. Protonating it makes the leaving group neutral water, which drifts off happily.
Lose water → carbocation on the α-carbon (C1). $- O + H_{2} \to C^{+} + H_{2} O$ . Why this step? This is the slow (rate-determining) step of E1 (the two-step, carbocation route). C1 is now electron-hungry.
Remove a β-H from C2. C2 is the only carbon adjacent to α, so it is the only β-carbon. Its C–H electrons swing in to build the π bond, ejecting $H^{+}$ (catalyst back). Why this step? There is only one β-carbon (C2), so there is no Zaitsev choice — one alkene only.

Answer: $CH_{3} -CH = CH_{2}$ (propene).

Verify: Carbon count in = 3, out = 3 (propene C₃) ✓. Atoms removed = one H (from C2) + one OH (from C1) = the elements of H₂O ✓.

Cell P2 — Zaitsev choice

Worked example Ex 2 · Dehydrohalogenation of 2-bromopentane

$CH_{3} -CHBr-CH_{2} -CH_{2} -CH_{3} alc. KOH ?$ This goes by E2 (concerted, one step — no carbocation).

Forecast: Br sits on C2. Which neighbour's H do we pull — C1 or C3 — and why?

Identify the β-carbons. Br is on C2 (the α-carbon). Its neighbours are C1 and C3 — both carry H, so both are β-carbons. Two possible alkenes. Why this step? An elimination needs a β-H on a carbon next to the leaving group; here two such carbons exist → a genuine choice.
Draw both products and count alkyl groups on the C=C (using the mono-/di-substituted counting defined above; note every substituent separated by a dash).
- Pull the β-H from C1 → pent-1-ene, $CH_{2} = CH-CH_{2} -CH_{2} -CH_{3}$ : the double-bond carbons carry 1 alkyl group (monosubstituted).
- Pull the β-H from C3 → pent-2-ene, $CH_{3} -CH = CH-CH_{2} -CH_{3}$ : the double-bond carbons carry 2 alkyl groups (disubstituted). Why this step? Zaitsev ranks alkenes by substitution because more alkyl groups = more hyperconjugation + inductive donation = lower energy (see Hyperconjugation and Inductive effect).
Pick the more substituted alkene as major. Why this step? "The rich get richer" — the transition state leading to the more stable alkene is itself lower in energy.
Now pick its geometry (cis vs trans). Pent-2-ene has a group on each double-bond carbon, so it exists as two stereoisomers: cis (both alkyl groups on the same side of the C=C) and trans (on opposite sides). The trans isomer is the major one. There are two reasons, and the mechanistic one comes first: (i) mechanism — E2 requires the departing β-H and the leaving Br to be anti-periplanar (H and Br on exactly opposite sides, 180° apart, so their orbitals line up to form the π bond in one motion); the anti-periplanar arrangement that fits this chain places the two alkyl chains on opposite sides, i.e. it produces the trans alkene. (ii) stability — trans is also lower in energy because its two chains point away from each other (less crowding). Mechanism and stability agree here, both favouring trans. Why this step? Stereochemistry of an E2 is set by the geometry the mechanism demands (anti-periplanar), not by product stability alone — stability only breaks ties when the mechanism allows both.

Answer (major): trans-pent-2-ene (disubstituted, 2 alkyl groups, trans); minor = cis-pent-2-ene, then pent-1-ene (monosubstituted, 1 alkyl group).

Verify: Alkyl-group count: major = 2, minor (pent-1-ene) = 1, and 2 > 1 ✓. Carbon count preserved: 5 in, 5 out ✓.

Cell P3 — degenerate (symmetric) substrate

Worked example Ex 3 · Dehydrate the symmetric alcohol cyclohexanol

$cyclohexanol H_{3} P O_{4} Δ ?$ ( $Δ$ = heat.)

Forecast: does symmetry give us a Zaitsev choice or not?

Locate the leaving group and β-carbons. –OH is on one ring carbon (the α-carbon); the two adjacent ring carbons are the β-carbons, and they are equivalent by symmetry (a mirror plane through the α-carbon). Why this step? Before applying Zaitsev, always ask do the choices actually differ?
Both β-H removals give the same molecule. Whether you pull the left or right β-H, you get the identical ring double bond. (No cis/trans issue either: in a small ring the double bond is locked, so no separate stereoisomers form.) Why this step? This is the degenerate case — the "choice" collapses to one product, so Zaitsev is silent (nothing to compare).

Answer: cyclohexene (single product; Zaitsev not needed).

Verify: Number of distinct alkene products = 1 (both pathways coincide) ✓. Carbon count 6 → 6 ✓.

Cell P4 — same reagent, solvent decides

Worked example Ex 4 · 2-bromobutane with KOH — two solvents

$CH_{3} -CHBr-CH_{2} -CH_{3} + KOH \to ?$ Predict the product for (a) aqueous KOH and (b) alcoholic KOH.

Forecast: same reagent both times — will the products really differ?

(a) Aqueous KOH: OH⁻ behaves as a nucleophile. Water solvates OH⁻, damping its basicity but leaving it a good nucleophile → it attacks the carbon, displacing Br⁻ (substitution). Why this step? See Haloalkanes — SN1, SN2, E1, E2: nucleophile + C = substitution. Product (a): butan-2-ol, $CH_{3} -CH(OH)-CH_{2} -CH_{3}$ .
(b) Alcoholic KOH: OR⁻/OH⁻ behaves as a base. In alcohol the ion is a strong base → it grabs a β-H off a carbon next to the α-carbon (C2), driving a concerted E2 elimination. Why this step? Base + β-H = elimination; then Zaitsev picks the alkene.
Apply Zaitsev, then geometry. But-2-ene (disubstituted, 2 alkyl groups) beats but-1-ene (monosubstituted, 1 alkyl group). But-2-ene again has a group on each double-bond carbon → cis and trans exist. As in Ex 2, the E2 mechanism requires the β-H and Br to be anti-periplanar (opposite sides, 180° apart); the anti-periplanar arrangement here delivers the trans alkene, which is also the more stable (chains apart, least crowded). Major = trans-but-2-ene. Why this step? Same two-tier logic as Ex 2: substitution first, then the anti-periplanar geometry of E2 (backed by trans-stability) fixes trans over cis. Product (b), major: trans-but-2-ene.

Answer: (a) butan-2-ol (substitution, SN2); (b) trans-but-2-ene (E2 elimination, Zaitsev major).

Verify: (a) still C₄ ✓ and has one O (alcohol) with no C=C. (b) but-2-ene alkyl count = 2 > but-1-ene = 1 ✓.

Cell A1 — symmetric alkene + symmetric reagent

Worked example Ex 5 · Bromine water test on ethene

$CH_{2} = CH_{2} + Br_{2} (aq, reddish-brown) \to ?$

Forecast: which carbon gets which Br? (Trick: is there even a choice?)

Check symmetry. Both alkene carbons are identical ( $CH_{2}$ ), and both ends of $Br_{2}$ are identical. No orientation question — Markovnikov is irrelevant here. Why this step? Orientation rules only matter when both partners are unsymmetric.
π electrons attack Br₂ → cyclic bromonium ion → Br⁻ opens it (anti). Why this step? The mechanism is the same regardless of symmetry; anti addition puts one Br on each carbon.
Observable: the reddish-brown colour disappears as Br₂ is consumed. Why this step? This decolourisation is the classic test for unsaturation (a C=C or C≡C is present).

Answer: 1,2-dibromoethane, $BrCH_{2} -CH_{2} Br$ ; colour decolourises.

Verify: Two Br atoms added, one per carbon → each carbon gains exactly 1 Br ✓. Carbon count 2 → 2 ✓.

Cell A2 — unsymmetric alkene + HX (ionic Markovnikov)

Worked example Ex 6 · But-1-ene + HCl

$CH_{3} -CH_{2} -CH = CH_{2} + HCl \to ?$

Forecast: where does H go, where does Cl go?

H⁺ adds first (electrophile). It can land on C1 (terminal $CH_{2}$ ) or C2. Why this step? The π electrons grab the proton; which carbon it lands on decides which carbon becomes positive.
Compare the two carbocations.
- H⁺ to C1 → positive charge on C2 = a 2° (secondary) cation (two carbons attached).
- H⁺ to C2 → positive charge on C1 = a 1° (primary) cation (one carbon attached). Why this step? Reaction flows through the more stable cation; 2° > 1° by hyperconjugation/+I (Carbocations — stability and rearrangement).
Cl⁻ attacks the 2° cation (on C2). Why this step? This is exactly Markovnikov: H went to the carbon with more H's (C1), Cl to the more substituted carbon (C2).

Answer: 2-chlorobutane, $CH_{3} -CH_{2} -CHCl-CH_{3}$ .

Verify: Cation degree chosen = 2, rejected = 1, and 2 > 1 ✓. Cl lands on the carbon that had fewer H's (Markovnikov) ✓.

The figure below traces exactly this branching: the alkene at top splits into the green (favoured, 2°) path on the left and the red (rejected, 1°) path on the right, then Cl⁻ closes onto the green cation to give the major product.

Figure 2 — Ex 6 decision tree. Green boxes/arrows = the pathway through the more stable 2° cation that actually happens; red dashed arrow = the rejected 1° route. The bottom-left green box is the Markovnikov major product, 2-chlorobutane.

Cell A3 — HBr + peroxide (anti-Markovnikov radical)

Worked example Ex 7 · Propene + HBr

with peroxide $CH_{3} -CH = CH_{2} + HBr peroxide ?$

Forecast: with peroxide the rule flips — where does Br go now?

Mechanism switches to free-radical. Peroxide breaks to give radicals that make Br•; the electrophile adding first is now Br•, not H⁺. Why this step? Different first-adding species ⇒ different intermediate ⇒ possibly different orientation.
Br• adds to give the more stable carbon radical. Br• onto C1 (terminal) leaves the unpaired electron on C2 = 2° radical (stable). Br• onto C2 would leave a 1° radical (unstable). Why this step? Radical stability ordering is 3° > 2° > 1°, same logic as cations.
The 2° radical grabs H from HBr, regenerating Br• (chain continues). Br is now on C1 — the carbon with more H's. Why this step? This grabbing step both places the H on C2 (completing anti-Markovnikov: Br on the less substituted carbon, opposite to Ex 6) and regenerates Br•, which keeps the chain reaction going.

Answer: 1-bromopropane, $CH_{3} -CH_{2} -CH_{2} Br$ .

Verify: Radical degree chosen = 2, rejected = 1, so 2 > 1 ✓. Br carbon here (C1) has more H's — the reverse of Markovnikov Ex 6 ✓.

Cell A4 — the limiting/boundary case

Worked example Ex 8 · Propene + HCl

with peroxide (the trap) $CH_{3} -CH = CH_{2} + HCl peroxide ?$

Forecast: does peroxide flip HCl too? (Careful!)

Test whether a radical chain can survive with HCl. The peroxide effect needs both radical steps to be exothermic enough to sustain the chain. For H–Cl the bond is too strong — the step that would make Cl• (or abstract H) fails energetically. Why this step? The rule has a domain: it works only where the chain propagates. HCl (too strong) and HI (too weak) both break the chain.
So no radical pathway ⇒ fall back to the normal ionic (Markovnikov) route. Why this step? When the special mechanism is unavailable, the default takes over — exactly like Ex 6.

Answer: 2-chloropropane (Markovnikov) — peroxide has no effect with HCl.

Verify: Peroxide anti-Markovnikov works only for HBr; for HCl the product equals the normal ionic Markovnikov product (Cl on the 2° carbon) ✓.

Cell W1 — real-world word problem

Worked example Ex 9 · From oil to

ghee A vegetable oil molecule has 3 C=C double bonds (it is "polyunsaturated"). It is stirred with $H_{2}$ gas over a Ni catalyst until fully hardened into a fat. How many H₂ molecules add, and what physical change do you expect?

Forecast: guess how many H₂ per molecule before reading on.

Each C=C consumes exactly one H₂. Hydrogenation adds H across every double bond (syn addition on the metal surface). Why this step? One π bond ⇒ one pair of H atoms ⇒ one H₂ molecule.
Multiply by the number of double bonds: $3 C=C \times 1 H_{2} = 3 H_{2}$ . Why this step? The double bonds react independently; total H₂ = number of unsaturations.
Physical change: removing double bonds turns a bent, kinked (liquid) chain into a straight, tightly packing (solid) chain → the oil solidifies to fat/ghee. Why this step? Saturation raises melting point; this is the industrial basis of vanaspati ghee.

Answer: 3 molecules of H₂; product is a saturated solid fat.

Verify: H₂ needed = number of C=C = 3 ✓. After reaction, C=C count = 0 (fully saturated) ✓.

Cell X1 — exam twist: carbocation rearrangement

Worked example Ex 10 · 3,3-dimethylbut-1-ene + HBr (ionic)

$(CH_{3})_{3} C-CH = CH_{2} + HBr \to ?$ The naive Markovnikov answer is a trap. Find the real major product.

Forecast: apply plain Markovnikov first, then ask "can this cation improve itself?"

H⁺ adds to C1 (terminal), cation forms on C2. By Markovnikov this is a 2° cation, sitting right next to C3, the carbon bearing three methyl groups. Why this step? Standard first step — but we must not stop here.
A methyl group (with its bonding electrons) shifts from C3 to C2 — a 1,2-methyl shift. Why this step? Moving a methyl from the crowded C3 onto the positive C2 relocates the positive charge onto C3, which is now attached to three carbons → the cation upgrades from 2° to 3° (more stable). Carbocations rearrange whenever a more stable cation is just one shift away (Carbocations — stability and rearrangement).
Br⁻ attacks the new 3° carbon (the former C3). Why this step? The nucleophile bonds where the (now more stable) positive charge actually sits — not where Markovnikov first placed it.

Answer: 2-bromo-2,3-dimethylbutane — the product from the rearranged 3° cation, not the un-rearranged 2° bromide the naive rule predicts.

Verify: Cation before shift = 2°, after shift = 3°, and 3° > 2° ✓ (rearrangement is favourable). Carbon count 6 → 6 ✓.

Recall Self-test: name the cell before you solve

Which cell handles a symmetric substrate with no product choice? ::: P3 (degenerate) Which cell is the boundary where a rule fails? ::: A4 (peroxide + HCl/HI) Which rule governs addition orientation, and which governs elimination? ::: Markovnikov = addition; Zaitsev = elimination What is the α-carbon vs the β-carbon? ::: α carries the leaving group; β is the neighbour that loses the H Is dehydration E1 or E2, and is alcoholic-KOH dehydrohalogenation E1 or E2? ::: Dehydration = E1 (carbocation, two steps); alc-KOH = E2 (concerted, one step) In an E2, what geometry must the β-H and leaving group adopt? ::: anti-periplanar (opposite sides, 180° apart) — this is what actually sets trans vs cis In Ex 10, why doesn't plain Markovnikov give the answer? ::: The 2° cation rearranges (methyl shift) to a more stable 3° cation

Related: Alkanes — preparation and properties · Alkynes — preparation and addition · Aromatic hydrocarbons — electrophilic substitution