Alkanes — preparation (Wurtz, Kolbe electrolysis, hydrogenation), properties, free-radical halogenation (Cl₂ - Br₂): Question bank

Everything here leans on four ideas you should already own: free radicals (an atom with one unpaired electron, hungry to pair it) from Free radicals and reaction intermediates, carbanions (a carbon carrying a lone pair and a negative charge) from Carbanions and organometallic reagents, van der Waals forces (weak stickiness between neighbouring molecules) from Van der Waals forces, and electrode redox from Electrolysis and redox at electrodes.

True or false — justify

Wurtz reaction can be used to prepare methane from a methyl halide.

False. Wurtz joins two alkyl groups (R–R). Methane has only one carbon, so there is no second half to weld to it — the smallest alkane Wurtz can build is ethane.

Kolbe electrolysis of sodium propanoate (

C H_{3} C H_{2} C O O^{-}

) gives butane.

True. Each propanoate loses

C O_{2}

to give the ethyl radical

C H_{3} C H_{2}^{∙}

; two ethyl radicals couple to

C H_{3} C H_{2}

–

C H_{2} C H_{3}

, which is n-butane.

In the Wurtz reaction sodium is oxidised.

True. Sodium donates an electron to the C–X bond (

N a \to N a^{+} + e^{-}

), so its oxidation number rises from

0

+ 1

— it is the reducing agent, so it is itself oxidised.

Chlorination of methane stops cleanly at

C H_{3} C l

False.

C H_{3} C l

still has three C–H bonds a

C l^{∙}

can attack, so it goes on to

C H_{2} C l_{2}, C H C l_{3}, C C l_{4}

. Only a large excess of methane biases toward mono-substitution.

Branched alkanes always boil higher than their straight-chain isomers.

False — the opposite. Branching makes the molecule more spherical, shrinking surface contact and weakening van der Waals forces, so n-pentane (36 °C) boils above neopentane (9.5 °C).

Fluorine is the most selective halogen in free-radical substitution.

False. Fluorine is the most reactive but the least selective — it grabs almost any H. Selectivity runs opposite to reactivity, so bromine is the selective one.

Iodine reacts readily with alkanes under UV light.

False. The I-abstraction step is endothermic and the overall reaction is reversible (HI reduces the alkyl iodide back), so

I_{2}

effectively does not substitute.

Catalytic hydrogenation adds the two H atoms to opposite faces of the double bond.

False. Both H atoms transfer from the same metal surface, so it is syn addition — both new C–H bonds form on the same face. See Alkenes — addition reactions.

Spot the error

"Wurtz mixing

C H_{3} B r

with

C_{2} H_{5} B r

is the best way to make propane."

Wrong: each radical can couple with itself or the other, producing ethane, propane and butane as a mixture. Cross-Wurtz gives poor, hard-to-separate yields — use it only for symmetrical alkanes.

"In Kolbe electrolysis the alkane forms at the cathode where reduction happens."

Wrong electrode. Carboxylate is an anion, so it migrates to and is oxidised at the anode; hydrogen gas (from water reduction) is what appears at the cathode.

"Alkanes are unreactive because their C–C and C–H bonds are polar and repel reagents."

Wrong reason. Those bonds are strong and non-polar with no π-electrons or lone pairs, so ionic reagents find no site to attack — that is why only radicals react.

"UV light is needed to give the electrons energy to jump orbitals in the alkane."

Wrong mechanism. UV energy matches the weak Cl–Cl (or Br–Br) bond energy and splits it homolytically into two radicals — it acts on the halogen, not the alkane.

"The regenerated

C l^{∙}

in propagation means chlorine is the catalyst."

Not a catalyst — a chain carrier.

C l^{∙}

is consumed and remade every cycle, but it is made from

C l_{2}

that is genuinely used up as reactant; it just isn't destroyed permanently the way a true reactant is in one step.

"Termination steps keep the reaction going."

Backwards: termination ends chains. Two radicals combine into a stable molecule with no unpaired electron left, so no new radical is passed on — the domino line stops.

"Tertiary radicals are more stable because they are more crowded."

Wrong cause. Their stability comes from more neighbouring alkyl groups donating electron density (hyperconjugation + inductive effect from Hyperconjugation), which soothes the electron-deficient radical centre — crowding is incidental.

Why questions

Why does bromination give mostly the product from the most substituted carbon while chlorination is less choosy?

Bromine's H-abstraction is only slightly exothermic, so its transition state resembles the radical and "feels" radical stability strongly — it waits for the 3° H. Chlorine's step is very exothermic, its transition state is reached early (before much radical character), so it barely distinguishes the H's.

Why is

C O_{2}

loss the driving force of Kolbe electrolysis?

C O_{2}

is an exceptionally stable molecule with strong C=O bonds. Its formation dumps a lot of energy and creates a gas that escapes, both of which pull the decarboxylation forward irreversibly.

Why must the ether in a Wurtz reaction be dry?

The intermediate organosodium behaves as a carbanion (a strong base). Any water would instantly protonate it, giving R–H instead of R–R and wasting the sodium.

Why does boiling point rise steadily as the alkane chain lengthens?

A longer chain has more surface along which neighbouring molecules touch, so more instantaneous London (van der Waals) contacts add up, and more thermal energy is needed to tear the molecules apart.

Why does a hydrogenation catalyst lower the activation energy instead of the reaction energy?

The metal surface adsorbs and splits

H_{2}

into reactive H atoms and holds the alkene nearby, offering an easier stepwise path over a lower barrier — but the start and end energies (hence overall energy released) are unchanged.

Why are alkanes insoluble in water but soluble in petrol or ether?

Alkanes are non-polar and cannot hydrogen-bond, so they can't break into water's H-bond network ("like dissolves like"); non-polar solvents interact through matching van der Waals forces instead.

Edge cases

What single alkane forms if Wurtz uses a mixture of

C H_{3} C H_{2} B r

and

C H_{3} C H_{2} B r

(both identical)?

Only n-butane. With one identical halide there is no cross-coupling possibility, so a single clean symmetrical product results — this is why Wurtz is restricted to same-R cases.

Can Kolbe electrolysis of sodium methanoate (

H C O O^{-}

, formate) make an alkane?

No usable alkane. The "R" group here is just H, so coupling two H-radicals gives

H_{2}

, not a carbon chain — Kolbe needs an alkyl R on the carboxylate.

What happens in free-radical chlorination if you use a huge excess of

C l_{2}

instead of excess methane?

Polysubstitution dominates: every C–H keeps meeting fresh

C l_{2}

, so the product shifts toward

C H C l_{3}

and

C C l_{4}

rather than

C H_{3} C l

Neopentane

C (C H_{3})_{4}

is chlorinated — how many different mono-chloro products are possible?

Exactly one. All twelve H atoms are equivalent (all on identical methyl groups), so substituting any of them gives the same product,

(C H_{3})_{3} C C H_{2} C l

At the very start of a halogenation, before any propagation, what is the only radical present?

Only the halogen radical from initiation (e.g.

C l^{∙}

). The alkyl radical cannot exist until a halogen radical abstracts an H in the first propagation step.

If two methyl radicals meet during methane chlorination, what forms and what is this step called?

They combine to ethane (

C H_{3}^{∙} + C H_{3}^{∙} \to C_{2} H_{6}

) — a termination step, since the unpaired electrons pair up and no chain carrier survives.

Recall One-line self-test before you close the page

Wurtz welds equal halves (dry ether), Kolbe kicks out $C O_{2}$ at the anode, hydrogenation fills double bonds syn with a metal, and halogenation is a light-started radical chain that never stops cleanly. ::: If you can say why each clause is true without peeking, you have beaten the traps.