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

Definition The words and symbols we will use (each earned in the parent note)

A free radical is an atom or group with one unpaired electron — written with a raised dot, like $C l^{∙}$ . The $∙$ symbol just means "one lonely, unpaired electron here"; it is "hungry" for one more electron. See Free radicals and reaction intermediates.
A carbanion is a carbon carrying a negative charge (a lone pair it wants to share) — it behaves like a nucleophile. See Carbanions and organometallic reagents.
Homolysis = a bond splits evenly, one electron to each fragment ⇒ two radicals.
Symmetrical alkane = the product $R - R$ has the same group $R$ on both sides of the new C–C bond.
$R$ = any alkyl group (a carbon chain, e.g. $C H_{3}$ –, $C H_{3} C H_{2}$ –). $X$ = a halogen atom (F, Cl, Br, or I). So $R - X$ (an alkyl halide) just means "a carbon chain with one halogen stuck on it."
Reaction arrows: a plain arrow $\to$ means "turns into"; an arrow with words above it, like $dry ether$ , names the condition needed for that change.
Condition symbols above arrows: $Δ$ (Greek capital delta) = "heat the mixture"; $h ν$ = "shine light on it" ( $h ν$ is the energy of one particle of light, a photon — the light supplies the energy to break a bond). $N i$ , $P t$ , $P d$ named above an arrow = a metal catalyst.

Level 1 — Recognition

L1.1

Name the reagents and conditions that turn an alkyl halide into a higher symmetrical alkane, and write the general equation.

Recall Solution

This is the Wurtz reaction.

Reagent: sodium metal ( $N a$ ).
Solvent/condition: dry ether (moisture would destroy the reactive organosodium). $2 R - X + 2 N a dry ether R - R + 2 N a X$ (Recall from the key: $R$ = any alkyl chain, $X$ = a halogen atom, so $R - X$ is an alkyl halide.) WHY dry? The intermediate $R - N a$ is a strong base; a trace of water would just protonate it, giving $R - H$ instead of coupling.

L1.2

At which electrode does the alkane appear in Kolbe electrolysis, and what type of process (oxidation or reduction) occurs there? What competes at the other electrode?

Recall Solution

The alkane forms at the anode, where oxidation happens. See Electrolysis and redox at electrodes. Carboxylate loses an electron there: $R C O O^{-} \to R C O O^{∙} + e^{-}$ Losing an electron = being oxidised = happening at the anode. Simple rule: an-Ox (Anode = Oxidation). The other electrode (edge case that lowers yield): at the cathode (reduction), the $N a^{+}$ ions are not reduced — water is easier to reduce, so hydrogen gas is evolved there: $2 H_{2} O + 2 e^{-} \to H_{2} ↑ + 2 O H^{-}$ So the cell also builds up alkali ( $O H^{-}$ ) around the cathode. This, plus side reactions at the anode (some carboxylate simply gets over-oxidised to $C O_{2}$ + alkene/alcohol), is why real Kolbe yields are modest and the method works best with concentrated carboxylate solutions and a smooth Pt anode at high current density (which favours the radical-coupling path over the ionic side paths).

Level 2 — Application

L2.1

Write the full Wurtz equation to prepare n-hexane ( $C_{6} H_{14}$ ). What alkyl halide do you start from?

Recall Solution

n-Hexane has 6 carbons. Wurtz joins two equal halves, so each half is $C_{3}$ ⇒ start from n-propyl halide ( $C H_{3} C H_{2} C H_{2} X$ ). $2 C H_{3} C H_{2} C H_{2} B r + 2 N a dry ether C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} C H_{3} + 2 N a B r$ Check: $3 + 3 = 6$ carbons, one new C–C bond. ✔ Note on scope: n-propyl bromide is a primary halide — exactly the kind Wurtz handles reliably (see the L4 note on why primary halides are the safe choice).

L2.2

Write the Kolbe electrolysis of sodium propanoate ( $C H_{3} C H_{2} C O O N a$ ). Name the alkane and give its carbon count. Justify why the balanced net equation has two $C O_{2}$ and two electrons.

Recall Solution

First, the carbon-counting rule. Let $n$ be the number of carbons in the acid (counting its own $C O O H$ carbon). Kolbe removes one carbon as $C O_{2}$ from each side, then joins the two leftover $R$ groups, each with $(n - 1)$ carbons. So the alkane has $one side (n - 1) + other side (n - 1) = 2 (n - 1) carbons.$ Propanoic acid $C H_{3} C H_{2} C O O H$ has 3 carbons, so here $n = 3$ . Step by step, tracking every electron and carbon:

Dissociation: $C H_{3} C H_{2} C O O N a \to C H_{3} C H_{2} C O O^{-} + N a^{+}$
Anode oxidation (each ion gives up one electron): $C H_{3} C H_{2} C O O^{-} \to C H_{3} C H_{2} C O O^{∙} + e^{-}$
Decarboxylation (each radical burps out one $C O_{2}$ ): $C H_{3} C H_{2} C O O^{∙} \to C H_{3} C H_{2}^{∙} + C O_{2}$
Coupling (two radicals join into one C–C bond): $C H_{3} C H_{2}^{∙} + C H_{3} C H_{2}^{∙} \to C H_{3} C H_{2} - C H_{2} C H_{3}$

WHY two of everything: it takes two carboxylate ions to make two ethyl radicals, and only then can they couple into one alkane. Two ions ⇒ two electrons released (step 2 ×2) and two $C O_{2}$ molecules lost (step 3 ×2). Adding steps 1–4 (doubled) gives the balanced net equation: $2 C H_{3} C H_{2} C O O^{-} \to C_{4} H_{10} + 2 C O_{2} + 2 e^{-}$ Mass check: LHS carbons $= 2 \times 3 = 6$ ; RHS $= 4 (butane) + 2 (as C O_{2}) = 6$ . ✔ Charge check: LHS $= 2^{-}$ ; RHS $= 2 e^{-}$ carried away, neutral molecules otherwise. ✔ Product: n-butane, with $2 (n - 1) = 2 (3 - 1) = 4$ carbons. ✔

L2.3

Ethene reacts with $H_{2}$ over nickel. Write the reaction and state why the catalyst is essential.

Recall Solution

$C H_{2} = C H_{2} + H_{2} N i, Δ C H_{3} C H_{3}$ (Here $N i$ above the arrow = the metal catalyst; $Δ$ = heat, per the symbol key.) The catalyst surface ==adsorbs and splits $H_{2}$ == into reactive H atoms and holds the alkene close, lowering the activation energy. Both H atoms add to the same face (syn addition). This is the reverse idea of Alkenes — addition reactions.

Level 3 — Analysis

L3.1

Monochlorination of propane ( $C H_{3} C H_{2} C H_{3}$ ) gives two isomers. Name them and, using the number of each kind of hydrogen, estimate the statistical (no-selectivity) percentage of each.

Recall Solution

Propane has two kinds of H:

1° hydrogens on the two end $C H_{3}$ groups: $3 + 3 = 6$ of them.
2° hydrogens on the middle $C H_{2}$ : $2$ of them.

Total H replaceable = $6 + 2 = 8$ . If every H were equally reactive (pure statistics): $1-chloropropane = \frac{6}{8} \times 100% = 75%, 2-chloropropane = \frac{2}{8} \times 100% = 25%.$ So the two products are 1-chloropropane and 2-chloropropane. (Real $C l_{2}$ at $25^{\circ} C$ shifts this toward more 2-chloropropane because chlorine is slightly selective for the more stable 2° radical — but the count is the starting point.)

L3.2

The relative rate at which a $C l^{∙}$ removes a 2° H versus a 1° H in propane is about $3.8 : 1$ . Recompute the product ratio for monochlorination using these reactivities.

Recall Solution

"Amount of product" = (number of that H) × (per-H reactivity).

2-chloropropane (from 2° H): $2 \times 3.8 = 7.6$ "reactivity units".
1-chloropropane (from 1° H): $6 \times 1.0 = 6.0$ units.

Total $= 7.6 + 6.0 = 13.6$ . $2-chloropropane = \frac{7.6}{13.6} \times 100% \approx 55.9%, 1-chloropropane = \frac{6.0}{13.6} \times 100% \approx 44.1%.$ Notice how selectivity flipped the majority: with pure counting the 1° product won ( $75%$ ); once we weight by reactivity the 2° product edges ahead ( $\approx 56%$ ). The whole point is that counting H's and weighting by reactivity can name different majority products — always do the reactivity-weighted calculation.

L3.3

Why does bromination of propane give overwhelmingly 2-bromopropane (>90%), while chlorination is only mildly selective?

Recall Solution

Bromine is less reactive, so its transition state comes later and looks more like the product radical — it "feels" the stability difference strongly. It therefore strongly prefers the H that gives the most stable radical. Radical stability (from the parent note): $3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{∙}$ , driven by Hyperconjugation and inductive donation.

Chlorine per-H selectivity 2°:1° $\approx 3.8 : 1$ → mild preference.
Bromine per-H selectivity 2°:1° $\approx 82 : 1$ → dominant preference. With $82 : 1$ : $2 \times 82 = 164$ vs $6 \times 1 = 6$ , total $170$ ⇒ 2-bromopropane $= 164/170 \times 100% \approx 96.5%$ .

Level 4 — Synthesis

L4.1

Design a preparation of 2,3-dimethylbutane $(C H_{3})_{2} C H - C H (C H_{3})_{2}$ using the Wurtz reaction. Identify the halide, and comment honestly on how reliable this route is.

Recall Solution

Find the middle C–C bond and check the two halves are identical. $(C H_{3})_{2} C H ∣ C H (C H_{3})_{2}$ Each half is an isopropyl group, $(C H_{3})_{2} C H -$ . They are identical ⇒ the coupling logic of Wurtz applies. On paper you would write: $2 (C H_{3})_{2} C H B r + 2 N a dry ether (C H_{3})_{2} C H - C H (C H_{3})_{2} + 2 N a B r$ Carbon check: each isopropyl is $C_{3}$ , total $C_{6}$ , formula $C_{6} H_{14}$ . ✔ Honest caveat (scope of Wurtz): 2-bromopropane is a secondary halide, and Wurtz is only truly reliable for primary alkyl halides. Secondary (and tertiary) halides tend to suffer elimination (the strong base/organosodium pulls off a $β$ -H to give an alkene, here propene) and other side reactions, so the real yield of the coupled alkane is poor. So the clean textbook answer is "isopropyl bromide," but the honest lab answer is "Wurtz is a bad choice here — prefer primary halides; this secondary case will give lots of propene."

L4.2

You are given only sodium acetate ( $C H_{3} C O O N a$ ) and an electrolysis cell. Show how to obtain ethane, and explain why you cannot get methane the same way.

Recall Solution

Kolbe on acetate: $2 C H_{3} C O O^{-} \to C H_{3} - C H_{3} + 2 C O_{2} + 2 e^{-}$ Each acetate loses $C O_{2}$ to give $C H_{3}^{∙}$ ; two methyls couple → ethane. Why not methane? Kolbe couples two $R^{∙}$ fragments into $R - R$ . Methane ( $C H_{4}$ ) is a single carbon with no C–C bond to form — there is no way to build it by joining two radicals. (Same structural reason Wurtz can't make methane: it needs "two halves".)

L4.3

Starting from 1-butene ( $C H_{2} = C H C H_{2} C H_{3}$ ), design a route that delivers only the terminal monochloride, 1-chlorobutane ( $C H_{3} C H_{2} C H_{2} C H_{2} C l$ ). State each step and its condition, and give a concrete method that avoids the 2-chloro isomer.

Recall Solution

Step 1 — saturate the double bond (hydrogenation): $C H_{2} = C H C H_{2} C H_{3} + H_{2} N i, Δ C H_{3} C H_{2} C H_{2} C H_{3} (n-butane)$ Step 2 — free-radical chlorination (recall $h ν$ = shine light): $C H_{3} C H_{2} C H_{2} C H_{3} + C l_{2} h ν mixture of 1- and 2-chlorobutane + H C l$ Because $C l_{2}$ is not selective, this step alone gives a mixture, not the pure terminal chloride. To obtain only 1-chlorobutane you must add a concrete separation and recycle strategy:

Run at low conversion with excess n-butane so almost every product is a monochloride (double substitution suppressed).
Separate the two monochloride isomers by fractional distillation — they have different boiling points (1-chlorobutane $78^{\circ} C$ , 2-chlorobutane $68^{\circ} C$ , a $10^{\circ} C$ gap that fractionation resolves).
Collect the 1-chlorobutane fraction; recycle the unreacted n-butane back into Step 2.

WHY this is the honest answer: unlike an ionic reaction, radical chlorination cannot be steered to a single position by choice of reagent — the only route to a single terminal product is "make the mixture, then physically isolate the one you want and recycle the rest." So the deliverable = hydrogenate, monochlorinate under dilute conditions, then fractionally distil off pure 1-chlorobutane.

Level 5 — Mastery

L5.1

Monochlorination of 2-methylbutane $(C H_{3})_{2} C H - C H_{2} - C H_{3}$ is run with $C l_{2} / h ν$ . Using per-H reactivities $1^{\circ} : 2^{\circ} : 3^{\circ} = 1 : 3.8 : 5.0$ , find the percentage of each monochloride.

Recall Solution

First, number the carbons of 2-methylbutane so every "kind of H" has a unique home: $C 1 C H_{3} - C 2 ∣ C H - C 3 C H_{2} - C 4 C H_{3}, with a C 5 C H_{3} branch on C 2.$ There are four chemically distinct positions where an H can be replaced, giving four products. Note carefully: the two methyls C1 and C5 are equivalent (both are $C H_{3}$ groups bonded to the same central $C H$ ), so replacing an H on either gives the same product, 1-chloro-2-methylbutane. That is why their H's are pooled ( $3 + 3 = 6$ ). The terminal $C H_{3}$ (C4) is a different environment and gives a different product.

Product (IUPAC name)	comes from	H-type	H count	reactivity	units = count×react
1-chloro-2-methylbutane	C1 or C5 methyls	1°	6	1.0	6.0
2-chloro-2-methylbutane	C2 (the lone $C H$ )	3°	1	5.0	5.0
3-chloro-2-methylbutane	C3 ( $C H_{2}$ )	2°	2	3.8	7.6
1-chloro-3-methylbutane	C4 (terminal $C H_{3}$ )	1°	3	1.0	3.0

Total units $= 6.0 + 5.0 + 7.6 + 3.0 = 21.6$ . Now divide each product's units by the total and multiply by $100$ : $1-chloro-2-methylbutane = \frac{6.0}{21.6} \times 100 \approx 27.8%$ $2-chloro-2-methylbutane (3°) = \frac{5.0}{21.6} \times 100 \approx 23.1%$ $3-chloro-2-methylbutane (2°) = \frac{7.6}{21.6} \times 100 \approx 35.2%$ $1-chloro-3-methylbutane (terminal 1°) = \frac{3.0}{21.6} \times 100 \approx 13.9%$ Sanity: $27.8 + 23.1 + 35.2 + 13.9 = 100%$ ✔. The 2° product leads even though the 3° H is individually most reactive — because there are two 2° H's and only one 3° H.

L5.2

n-Pentane boils at $36^{\circ} C$ ; neopentane (2,2-dimethylpropane) boils at $9.5^{\circ} C$ . Both are $C_{5} H_{12}$ . Explain the $26.5^{\circ} C$ gap in terms of intermolecular forces, and predict which has the larger surface contact.

Recall Solution

Both molecules have the same mass and same formula, so the difference is shape, not size.

n-Pentane is a long, extended chain → large surface area → many contact points → strong van der Waals (London) forces → higher boiling point. See Van der Waals forces.
Neopentane is compact and near-spherical → minimal surface contact → weaker dispersion forces → lower boiling point. Difference $= 36 - 9.5 = 26.5^{\circ} C$ . n-Pentane has the larger surface contact. Branching = more spherical = weaker London forces = lower b.p.

L5.3

A student electrolyses a mixture of sodium acetate and sodium propanoate (Kolbe). List all alkanes formed and explain, by analogy with mixed Wurtz, why this gives a mixture.

Recall Solution

Both salts generate radicals at the anode:

Acetate → $C H_{3}^{∙}$ (call it $A$ ).
Propanoate → $C H_{3} C H_{2}^{∙}$ (call it $B$ ).

Any two radicals can couple: $A + A$ , $A + B$ , $B + B$ .

$A + A$ : $C H_{3} - C H_{3}$ → ethane ( $C_{2}$ )
$A + B$ : $C H_{3} - C H_{2} C H_{3}$ → propane ( $C_{3}$ )
$B + B$ : $C H_{3} C H_{2} - C H_{2} C H_{3}$ → n-butane ( $C_{4}$ )

Three alkanes result — exactly the same "cross-coupling" problem as mixed Wurtz: whenever two different radicals are present, every possible pairing happens, so no single clean product forms. This is a general rule — mixed radical couplings, whether by Wurtz or Kolbe, always give a mixture, and that is why both methods are only practical for symmetrical targets built from one kind of radical.

Recall Self-check summary

Which two named reactions fail on mixed/unsymmetrical substrates? ::: Wurtz and Kolbe — both couple radicals, so mixed inputs cross-couple into a mixture. Product amount in radical halogenation equals what product? ::: (Number of that H) × (per-H reactivity). Same formula, lower boiling point — what feature? ::: More branching → more spherical → weaker London forces. Kolbe carbon count from an acid with n carbons? ::: $2 (n - 1)$ carbons. What does $h ν$ above an arrow mean? ::: Shine light on the reaction (a photon supplies the bond-breaking energy). What does $Δ$ above an arrow mean? ::: Heat the mixture.