4.2.4 · D5Hydrocarbons
Question bank — Alkenes — preparation (dehydration, dehydrohalogenation, Zaitsev's rule), addition reactions
True or false — justify
A "==C=C double bond==" here means a σ bond plus a π bond — an electron cloud sitting above and below the two carbons. Keep that picture in mind for every answer.
True or false: Dehydration and dehydrohalogenation are both β-eliminations.
True — in both, a leaving group ( as water, or ) leaves the α-carbon while an H leaves the neighbouring β-carbon, so a new C=C forms between them.
True or false: A 1° alcohol dehydrates faster than a 3° alcohol.
False — the rate follows carbocation stability , and the E1 step 2 makes that cation, so 3° is fastest and 1° needs the highest temperature.
True or false: Alcoholic KOH and aqueous KOH do the same thing to an alkyl halide.
False — in alcohol acts as a base (grabs β-H → elimination → alkene); in water it acts as a nucleophile (attacks carbon → substitution → alcohol). Same reagent, solvent decides the job.
True or false: Zaitsev's rule and Markovnikov's rule are two names for the same preference.
False — Zaitsev governs elimination (which alkene forms: the more substituted one). Markovnikov governs addition (where H and X go). Opposite directions of reaction.
True or false: More hydrogens on a carbocation carbon make it more stable.
False — it is the opposite. More alkyl groups (which means fewer H's) stabilise the positive charge through hyperconjugation and +I donation; see Carbocations — stability and rearrangement.
True or false: The peroxide (Kharasch) effect reverses the addition of HCl and HI as well as HBr.
False — it works only for HBr. The H–Cl bond is too strong and H–I too weak to sustain the free-radical chain, so those still add by Markovnikov.
True or false: Bromine water decolourising is a positive test for a C=C.
True — the π electrons attack , consuming the reddish-brown colour, so decolourisation signals unsaturation (an alkene or alkyne). See Alkynes — preparation and addition.
True or false: Catalytic hydrogenation adds the two H atoms to opposite faces of the double bond.
False — it is syn addition: both H atoms are delivered from the metal surface to the same face of the π bond.
True or false: Acid-catalysed hydration of propene gives a 1° alcohol.
False — water adds by Markovnikov, so lands on the more substituted carbon, giving the 2° alcohol propan-2-ol.
Spot the error
Each statement below is almost right. Find the flaw and correct it.
"OH⁻ leaves easily during dehydration, so no catalyst is needed."
Wrong — is a strong base and a terrible leaving group. The acid first protonates it to , which leaves as neutral water; that is why the catalyst is essential.
"2-bromobutane with alc. KOH gives but-1-ene as the major product."
Wrong — but-2-ene () is disubstituted and more stable, so by Zaitsev it is the major product; but-1-ene is minor.
"Propene + HBr gives 1-bromopropane as the major (ionic) product."
Wrong — without peroxide the mechanism is ionic, so H⁺ adds to the terminal CH₂ to give the more stable 2° cation, and Br⁻ lands on C2 → 2-bromopropane (Markovnikov).
"The peroxide effect changes the product because Br⁻ becomes a better nucleophile."
Wrong — the peroxide switches the mechanism from ionic to free-radical. Br• adds first to give the more stable carbon radical, reversing the orientation; nucleophilicity is irrelevant here.
"In electrophilic addition the alkene attacks Br⁻ first."
Wrong — the π electrons are the electron source, so the alkene attacks the electrophile (Br⁺ / the δ⁺ end of the reagent) first; the anion adds afterward.
"Zaitsev says remove the H that gives the least crowded double bond."
Wrong — Zaitsev favours the more substituted (more crowded with alkyl groups) alkene, because more hyperconjugation and +I lower its energy.
Why questions
Why does adding more alkyl groups to the double-bond carbons make an alkene more stable?
Each α C–H bond can overlap with the π system (hyperconjugation) and alkyl groups donate electron density (+I), spreading and lowering the energy of the π electrons.
Why must dehydration go through a carbocation while base-driven dehydrohalogenation often need not?
Acidic dehydration is E1: a poor leaving group is made good, then leaves to form a cation. Alcoholic KOH provides a strong base that can pull the β-H as the C–X bond breaks (E2), so no free cation is required. See Haloalkanes — SN1, SN2, E1, E2.
Why does the Markovnikov product dominate in ionic HX addition?
The reaction funnels through whichever carbocation is more stable (); H⁺ therefore adds to give that cation, placing X on the more substituted carbon.
Why does only HBr, and not HCl or HI, show the peroxide effect?
The radical chain needs a bond that can break homolytically and propagate; H–Cl is too strong to initiate cleanly and H–I too weak to hold the chain, leaving HBr as the only one that sustains it.
Why can a 1° alcohol sometimes give an alkene that looks rearranged?
The initial 1° cation is unstable and can undergo a hydride/alkyl shift to a more stable 2° or 3° cation before losing a β-H — see Carbocations — stability and rearrangement.
Why is hydrogenation used industrially to make vegetable ghee?
It converts unsaturated (double-bond-rich) liquid oils into saturated (double-bond-free) semi-solid fats by adding H₂ across each C=C; the saturated product is alkane-like and solid.
Edge cases
For a symmetrical alkene like but-2-ene, does Markovnikov's rule still choose a product?
No choice is needed — both carbons of the C=C carry the same substitution, so H and X give the same product whichever way they add; the rule is silent because there is no "more substituted" carbon.
For ethene (CH₂=CH₂) reacting with HBr, is the product Markovnikov or anti-Markovnikov?
Neither label applies — ethene is symmetric, so both carbons are equivalent and only one product (bromoethane) is possible.
If a dehydration could give only a terminal (monosubstituted) alkene, does Zaitsev fail?
No — Zaitsev only picks among available alkenes. If the substrate has just one β-H position, that lone possible alkene is the product regardless of substitution.
For 2-methylpropan-2-ol (a 3° alcohol), why is dehydration especially easy?
Loss of protonated water gives a stable 3° carbocation directly, so the slow step is fast; only 2-methylprop-1-ene can form, so there is no product competition either.
What happens to bromine water with a fully saturated alkane like ethane?
It stays reddish-brown — no π electrons exist to attack , so no decolourisation, confirming the absence of unsaturation.
In the free-radical HBr addition, what happens if no peroxide (radical initiator) is present?
The chain never starts, so the reaction defaults to the ordinary ionic Markovnikov pathway; the peroxide effect requires the initiator.
Recall One-line self-check
Elimination builds the C=C (Zaitsev: more substituted wins); addition destroys it (Markovnikov: H to the H-rich carbon). Solvent switches KOH between base and nucleophile; peroxide switches HBr between ionic and radical. If you can say why each switch happens, you own this chapter.