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

This page is the practice ground for the parent topic. Before we solve anything, let us list every kind of question this chapter can throw at you, so no scenario ever surprises you in an exam.

Definition What the shorthand "R" means (read this first)

Everywhere below you will see $R$ and $R - X$ . This is chemist's shorthand:

$R$ = an alkyl group — a piece of an alkane with one H removed so it has a free "hand" to bond, e.g. $R = C H_{3}$ (methyl), $R = C_{2} H_{5}$ (ethyl), $R = C_{3} H_{7}$ (propyl). Picture a Lego brick with one stud exposed.
$R - X$ = an alkyl halide: that alkyl group joined to a halogen $X$ (where $X = C l, B r, I$ ). Example: $R - B r$ with $R = C_{2} H_{5}$ is $C_{2} H_{5} B r$ .
$R^{∙}$ = an alkyl free radical — the alkyl group carrying one lonely unpaired electron (the $∙$ dot), hungry to pair up. See free radicals.
$R C O O^{-}$ = a carboxylate ion — an alkyl group $R$ attached to $-COO^{-}$ (the deprotonated acid end).

So " $R - R$ " just means "two alkyl bricks welded stud-to-stud into one longer alkane."

The scenario matrix

Think of each row below as a "cell" — a distinct situation with its own trap. Every worked example that follows is tagged with the cell it lands in.

#	Cell (scenario class)	The trap / degenerate case it tests
A	Wurtz — symmetrical target	Cleanest case: split into two equal halves
B	Wurtz — mixed halides	Cross-coupling → count ALL products
C	Wurtz — impossible target	Odd-carbon / methane — no two equal halves
C+	Wurtz — 2°/3° halide side reactions	Elimination & rearrangement spoil the yield
D	Kolbe — carbon counting	The $2 (n - 1)$ rule, and the degenerate $C_{1}$ (formic) case
E	Kolbe — mixed salts	Two different carboxylates → mixture at anode
F	Hydrogenation — degree of unsaturation	How many $H_{2}$ ? alkene vs alkyne
G	Physical property — branching vs chain length	Ranking boiling points, isomers
H	Halogenation — product count (poly-substitution)	Excess reagent, how many products
I	Halogenation — selectivity (Cl₂ vs Br₂)	3° vs 1° H, statistical vs stability
J	Real-world / exam twist	Combining two methods; working backwards

Cell A — Wurtz, clean symmetrical target

Cell B — Wurtz, mixed halides (count the mixture)

Worked example React

$C H_{3} B r$ and $C H_{3} C H_{2} B r$ together under Wurtz. List all alkane products. Forecast: Two different radicals, $C H_{3}^{∙}$ and $C_{2} H_{5}^{∙}$ . Each can pair with itself or the other. Guess how many distinct alkanes appear.

Step 1. List the radicals formed: $C H_{3}^{∙}$ (call it M, 1C) and $C_{2} H_{5}^{∙}$ (call it E, 2C). Why this step? Coupling happens between radicals; naming them makes the pairing bookkeeping clean.

Step 2. Enumerate every pairing:

M + M $\to$ $C H_{3} C H_{3}$ (ethane, 2C)
E + E $\to$ $C H_{3} C H_{2} C H_{2} C H_{3}$ (n-butane, 4C)
M + E $\to$ $C H_{3} C H_{2} C H_{3}$ (propane, 3C)

Why this step? Radicals don't "know" which partner they came from — every combination is allowed.

Step 3. Count distinct products: 3 alkanes.

The figure below draws exactly this bookkeeping: the two starting radicals (left, coloured M and E), and the three arrows fanning out to the three alkanes. Notice the middle plum arrow — that is the cross-coupling term (M + E → propane) that only appears because two different radicals are present. Follow each coloured arrow from a radical to its product to see why self-coupling gives even carbons and cross-coupling gives the odd one.

Verify: Carbon accounting — the cross product (propane) sums $1 + 2 = 3$ . ✓ This is exactly the organometallic cross-coupling problem the parent warned about: mixed Wurtz = messy mixture.

Cell C — Wurtz, impossible target (degenerate case)

Worked example Can Wurtz make

propane ( $C_{3} H_{8}$ ) as a sole product? Can it make methane? Forecast: Propane is odd-carbon (3). Methane is 1 carbon. Does "split into two equal halves" work?

Step 1. For propane, try $3 \div 2 = 1.5$ carbons per half — not a whole number. Why this step? A symmetrical Wurtz needs two identical whole $R$ groups. Half-carbons don't exist.

Step 2. The only way to a $C_{3}$ chain by Wurtz is the mixed route (Cell B) — which also makes ethane and butane. So propane is never a clean Wurtz product. Why this step? Odd-carbon alkanes can only come as the cross term of a mixture.

Step 3. Methane needs $R - R$ with $R$ = "half a carbon" — impossible; $C H_{4}$ has no C–C bond to form.

Verify: Wurtz always makes even-numbered symmetrical alkanes cleanly ( $C_{2}, C_{4}, C_{6}, \dots$ ). Propane (odd) and methane (1C, no C–C) fail. ✓

Cell C+ — Wurtz with 2°/3° halides (side reactions ruin it)

Worked example Why does trying Wurtz on

tert-butyl bromide $(C H_{3})_{3} C B r$ give almost no clean $R - R$ alkane? Forecast: Wurtz looks like it should weld two tert-butyl groups into 2,2,3,3-tetramethylbutane. Guess what goes wrong before reading.

Step 1. Recall the mechanism: sodium turns $R - X$ into a reactive organosodium / carbanion-like species $R^{-}$ (see organometallics). Why this step? A 3° carbanion is very basic and sits on a very crowded carbon.

Step 2. Because that species is a strong base, instead of attacking a second $R - X$ (substitution/coupling) it rips off a neighbouring H — this is elimination, and it spits out an alkene (here isobutylene, $(C H_{3})_{2} C = C H_{2}$ ) rather than an alkane. Why this step? Bulky 2°/3° halides are crowded, so coupling is slow and the faster elimination pathway wins.

Step 3. A second spoiler: 2°/3° radicals/carbanions can rearrange (a group hops to a more stable position) before coupling, so even the alkane you do get is a scrambled skeleton, not the one you drew. Why this step? More-substituted intermediates seek greater stability, so the carbon skeleton shifts.

Verify: Practical rule — Wurtz works cleanly only for primary (1°) alkyl halides. With 2°/3° halides expect alkene by-products (elimination) + rearranged and mixed alkanes → low, messy yield. ✓

Cell D — Kolbe, carbon counting (and the formic-acid degenerate case)

Intuition WHERE the

$2 (n - 1)$ rule comes from (derive it, don't memorise) Kolbe electrolysis runs at the anode, the electrode where things get oxidised (lose electrons). Here is the whole chain, step by step:

The carboxylate $R C O O^{-}$ drifts to the positive anode and hands over its extra electron to the electrode: $R C O O^{-} \to R C O O^{∙} + e^{-}$ . That is where the $e^{-}$ in the half-equation comes from — it flows into the anode wire.
The carboxyl radical is unstable and snaps off $C O_{2}$ — the very stable $O = C = O$ molecule is the driving force: $R C O O^{∙} \to R^{∙} + C O_{2}$ . This is the key: the carbon of the $-COO$ group leaves as $C O_{2}$ , so each side loses exactly one carbon.
Two surviving $R^{∙}$ radicals couple: $R^{∙} + R^{∙} \to R - R$ .

So from a $C_{n}$ acid, each $R$ has $n - 1$ carbons (one lost as $C O_{2}$ ), and two of them join → $2 (n - 1)$ carbons total. That is the rule, and now you know why.

Worked example Which alkane forms when the sodium salt of

butanoic acid ( $C_{3} H_{7} C O O H$ , 4 carbons) is electrolysed? Then test formic acid ( $H C O O H$ , 1 carbon). Forecast: Use the derived rule: a $C_{n}$ acid gives a $2 (n - 1)$ -carbon alkane. Plug $n = 4$ , then $n = 1$ .

Step 1. Butanoic acid: $n = 4$ carbons. The $R$ group (after losing the $-COO$ carbon as $C O_{2}$ ) is $C_{3} H_{7}$ . Why this step? By the mechanism above, each carboxyl radical loses one carbon as $C O_{2}$ ; the surviving $R^{∙}$ has $n - 1 = 3$ carbons.

Step 2. Balance the anode half-reaction, then couple. Two carboxylates each surrender one electron to the anode (total $2 e^{-}$ into the wire), each drop a $C O_{2}$ , and the two $C_{3} H_{7}^{∙}$ radicals join: $2 C_{3} H_{7} C O O^{-} \to C_{6} H_{14} C_{3} H_{7} - C_{3} H_{7} + 2 C O_{2} + 2 e^{-}$ Why this step? Charge balance: left side has charge $- 2$ ; right side is neutral $+ 2 e^{-}$ , so $- 2 = - 2$ . The electrons are not "lost into space" — they travel through the external circuit to the cathode, where water is reduced to $H_{2}$ .

Step 3. Formic acid: $n = 1$ , so $R$ = " $H$ " (no carbon left after $C O_{2}$ ). Coupling gives $H - H = H_{2}$ , not an alkane. Why this step? $H C O O^{∙} \to H^{∙} + C O_{2}$ ; two H atoms make hydrogen gas, the degenerate case.

Verify: $2 (n - 1)$ with $n = 4 \Rightarrow 6$ ✓ (hexane). With $n = 1 \Rightarrow 0$ carbons ✓ — no alkane, only $H_{2} + C O_{2}$ . Charge on each side of Step 2 is $- 2$ ✓. See anode oxidation.

Cell E — Kolbe, mixed salts (mixture at the anode)

Cell F — Hydrogenation, degree of unsaturation

Cell G — Physical properties, branching vs chain length

Worked example Rank the boiling points:

n-pentane, neopentane (2,2-dimethylpropane), n-hexane. Forecast: Two forces at play — chain length (more carbons → higher b.p.) and branching (more spherical → lower b.p.). Guess the order.

Step 1. Compare by carbon count first: n-hexane ( $C_{6}$ ) > both $C_{5}$ molecules. Why this step? More carbons = larger surface = stronger London dispersion forces → higher boiling point.

Step 2. Between the two $C_{5}$ isomers, compare shape: n-pentane is a long "rod" (large surface contact); neopentane is a compact "ball." Why this step? A spherical molecule touches its neighbours over a smaller area → weaker dispersion → lower b.p.

Step 3. Order (highest → lowest): $n-hexane (69^{\circ} C) > n-pentane (36^{\circ} C) > neopentane (9.5^{\circ} C)$

Read the bar chart below as the answer made visible. The two teal/orange rods (n-hexane, n-pentane) stand tall because their stretched shape gives lots of surface contact; the plum ball (neopentane) is short because its spherical shape barely touches neighbours. The left arrow (branching lowers b.p.) compares the two $C_{5}$ isomers; the right arrow (more carbons) compares across chain length. Both effects live in one picture.

Verify: $69 > 36 > 9.5$ ✓. Chain length wins across different C-counts; branching decides between isomers. ✓

Cell H — Halogenation, product count (poly-substitution)

Worked example Chlorinate

methane with excess $C l_{2}$ (not excess alkane). How many chlorinated products are possible? Forecast: Each product still has H's that can be replaced again. Count how far substitution can go.

Step 1. Replace H one at a time: $C H_{4} \to C H_{3} C l \to C H_{2} C l_{2} \to C H C l_{3} \to C C l_{4}$ Why this step? Every chlorinated methane except $C C l_{4}$ still has a C–H bond for the chain to attack.

Step 2. Count the distinct chloromethanes formed: $C H_{3} C l$ , $C H_{2} C l_{2}$ , $C H C l_{3}$ , $C C l_{4}$ → 4 products. Why this step? Excess $C l_{2}$ drives substitution all the way; mono-substitution is NOT clean here.

Step 3. To favour $C H_{3} C l$ alone, you'd use excess methane instead (few $C l^{∙}$ per $C H_{4}$ ).

Verify: Methane has 4 H's → up to 4 replacements → 4 chlorinated products. ✓ (Matches the parent's "halogenation gives a mixture" warning.)

Cell I — Halogenation, selectivity (Cl₂ vs Br₂)

Worked example

Isobutane (2-methylpropane, $(C H_{3})_{3} C H$ ) is halogenated. It has 9 primary (1°) H's and 1 tertiary (3°) H. Which halogen gives mostly the tert-butyl product, and why? Forecast: One halogen "grabs anything" (statistical); the other "waits for the best radical" (selective). Which is which?

Step 1. Identify the H types: 9 equivalent 1° H (on the three $C H_{3}$ ) and 1 lone 3° H (on the central C). Why this step? The product depends on which H is abstracted — that fixes where the halogen goes.

Step 2. Radical stability (parent rule): $3^{\circ} > 2^{\circ} > 1^{\circ}$ . Abstracting the 3° H gives the most stable radical, stabilised by hyperconjugation + inductive donation. Why this step? A more stable radical means a lower-energy pathway.

Step 3. Bromine is less reactive → it is choosy and prefers the low-energy 3° route, giving mostly tert-butyl bromide. Chlorine is so reactive it grabs H almost at random — dominated by the 9 primary H's, so it gives mostly the 1° product.

Verify: Only 1 of the 10 H's is 3°, yet Br₂ still funnels to it because selectivity beats numbers. Reactivity order $F_{2} > C l_{2} > B r_{2} > I_{2}$ ; selectivity is the reverse. ✓

Cell J — Real-world / exam twist (combine methods, work backwards)

Worked example You are given only

ethene and lab reagents. Make n-butane in the fewest sensible steps. Forecast: Butane is $C_{4}$ . Ethene is $C_{2}$ . You must both add carbons and saturate. Plan backwards.

Step 1. Butane $C_{4}$ = two $C_{2}$ halves → Wurtz from ethyl bromide $C_{2} H_{5} B r$ (Cell A logic). Why this step? Symmetrical $C_{4}$ is the ideal Wurtz target — and ethyl bromide is a primary halide, so Wurtz is clean (recall Cell C+: 2°/3° halides misbehave).

Step 2. Get $C_{2} H_{5} B r$ from ethene: add HBr (a route into addition reactions): $C H_{2} = C H_{2} + H B r \to C H_{3} C H_{2} B r$ Why no Markovnikov worry here? Markovnikov's rule only decides which carbon gets the H when the two ends of the double bond are different. Ethene is symmetric — both carbons are identical $C H_{2}$ — so H and Br can add either way and give the same product, $C H_{3} C H_{2} B r$ . Regioselectivity simply doesn't apply.

Step 3. Couple with sodium: $2 C H_{3} C H_{2} B r + 2 N a ether C H_{3} C H_{2} C H_{2} C H_{3} + 2 N a B r$

Verify: Carbon count $2 + 2 = 4$ ✓ = n-butane. Two-step synthesis: addition then Wurtz. ✓

Recall Quick self-test on the matrix

Which cells give a mixture rather than one clean product? ::: B (mixed Wurtz), E (mixed Kolbe), H (poly-halogenation). What does the shorthand $R$ stand for? ::: An alkyl group — an alkane fragment with one bond free (e.g. $C H_{3}$ , $C_{2} H_{5}$ ). Why does Wurtz fail for 2°/3° halides? ::: The basic organosodium eliminates (gives alkene) or rearranges — low, messy yield; use only 1° halides. Where do the electrons in the Kolbe half-reaction go? ::: Into the anode/external circuit; the carboxylate is oxidised (loses $e^{-}$ ). A $C_{n}$ carboxylate gives a Kolbe alkane of how many carbons, and why? ::: $2 (n - 1)$ — each side drops its $-COO$ carbon as $C O_{2}$ , then two $R^{∙}$ couple. How many mol $H_{2}$ to saturate an alkyne? ::: 2 mol per triple bond. Between n-pentane and neopentane, which boils higher and why? ::: n-pentane — less branched, more surface contact, stronger dispersion. Which halogen gives the 3° product from isobutane? ::: Br₂ (more selective). Why is Markovnikov irrelevant for HBr + ethene? ::: Ethene is symmetric — both carbons identical — so either addition gives the same $C_{2} H_{5} B r$ .