Chapter 9: Hydrocarbons

Chain isomers differ in the way the carbon skeleton (the chain) is built. For $$\mathrm{C_6H_{14}}$$ we keep the molecular formula fixed and vary only the arrangement of the six carbon atoms.

1. Continuous (unbranched) chain of 6 carbons:
$$\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3}$$
IUPAC name: Hexane (n-hexane)

2. A 5-carbon chain with one methyl branch. The branch can sit on C-2 or on C-3 (a branch on C-1 or C-4/C-5 would simply lengthen the main chain or duplicate C-2).
Branch on C-2: $$\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH_3}$$
IUPAC name: 2-Methylpentane

Branch on C-3: $$\mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3}$$
IUPAC name: 3-Methylpentane

3. A 4-carbon chain with two methyl branches. Both methyls on C-2 (a dimethyl-2,3 arrangement on a butane chain is also possible).
Both on C-2: $$\mathrm{CH_3-C(CH_3)_2-CH_2-CH_3}$$
IUPAC name: 2,2-Dimethylbutane

One methyl on C-2 and one on C-3: $$\mathrm{CH_3-CH(CH_3)-CH(CH_3)-CH_3}$$
IUPAC name: 2,3-Dimethylbutane

Thus $$\mathrm{C_6H_{14}}$$ has five chain isomers.

An alkyl group $$\mathrm{C_5H_{11}-}$$ is derived from a pentane $$(\mathrm{C_5H_{12}})$$ skeleton by removing one hydrogen atom. The carbon skeleton can be a straight chain (pentane) or a branched chain (2-methylbutane, 2,2-dimethylpropane). Removing an H from each distinct carbon gives the various isomeric alkyl groups. Attaching $$\mathrm{-OH}$$ at that same carbon gives the corresponding alcohol.

From pentane $$\mathrm{CH_3-CH_2-CH_2-CH_2-CH_3}$$:

• $$\mathrm{-OH}$$ at C-1 $$\rightarrow$$ $$\mathrm{CH_3CH_2CH_2CH_2CH_2-OH}$$ : Pentan-1-ol
• $$\mathrm{-OH}$$ at C-2 $$\rightarrow$$ $$\mathrm{CH_3CH_2CH_2CH(OH)CH_3}$$ : Pentan-2-ol
• $$\mathrm{-OH}$$ at C-3 $$\rightarrow$$ $$\mathrm{CH_3CH_2CH(OH)CH_2CH_3}$$ : Pentan-3-ol

From 2-methylbutane $$\mathrm{(CH_3)_2CH-CH_2-CH_3}$$:

• $$\mathrm{-OH}$$ at a terminal methyl (C-1) $$\rightarrow$$ $$\mathrm{(CH_3)_2CH-CH_2-CH_2OH}$$ : 3-Methylbutan-1-ol
• $$\mathrm{-OH}$$ at the branch carbon (C-2) $$\rightarrow$$ $$\mathrm{(CH_3)_2C(OH)-CH_2-CH_3}$$ : 2-Methylbutan-2-ol
• $$\mathrm{-OH}$$ at C-3 $$\rightarrow$$ $$\mathrm{(CH_3)_2CH-CH(OH)-CH_3}$$ : 3-Methylbutan-2-ol
• $$\mathrm{-OH}$$ at the C-4 methyl $$\rightarrow$$ $$\mathrm{CH_3-CH(CH_2OH)-CH_2-CH_3}$$ : 2-Methylbutan-1-ol

From 2,2-dimethylpropane (neopentane) $$\mathrm{(CH_3)_4C}$$: all four methyl groups are equivalent, so only one alkyl group (neopentyl) is possible.

• $$\mathrm{-OH}$$ on a methyl $$\rightarrow$$ $$\mathrm{(CH_3)_3C-CH_2OH}$$ : 2,2-Dimethylpropan-1-ol

So eight distinct alcohols are obtained, derived from the three pentane skeletons.

Alkanes can be prepared by decarboxylation: the sodium salt of a carboxylic acid, on heating with soda lime ($$\mathrm{NaOH}$$ + $$\mathrm{CaO}$$), loses $$\mathrm{CO_2}$$ and gives an alkane with one carbon fewer than the acid.

Propane has 3 carbon atoms. So the carboxylic acid required must have $$3 + 1 = 4$$ carbon atoms — that is butanoic acid ($$\mathrm{CH_3CH_2CH_2COOH}$$). Its sodium salt is sodium butanoate.

Chemical equation:

$$\mathrm{CH_3CH_2CH_2COONa + NaOH \xrightarrow[\Delta]{CaO} CH_3CH_2CH_3 + Na_2CO_3}$$

Thus the sodium salt of butanoic acid (sodium butanoate) is needed.

Method. Every single bond is one $$\sigma$$ bond. A double bond is one $$\sigma$$ + one $$\pi$$; a triple bond is one $$\sigma$$ + two $$\pi$$. For an open-chain (acyclic) molecule with $$n$$ carbon atoms the number of C–C linkages is $$n-1$$ (each linkage contributes one $$\sigma$$). The number of C–H $$\sigma$$ bonds equals the number of H atoms.

(i) 2,8-Dimethyldeca-3,6-diene — molecular formula $$\mathrm{C_{12}H_{22}}$$.
C–C $$\sigma$$ bonds $$= 12-1 = 11$$; C–H $$\sigma$$ bonds $$= 22$$.
Total $$\sigma = 11 + 22 = 33$$; $$\pi = 2$$ (two C=C).

(ii) Octa-1,3,5,7-tetraene — molecular formula $$\mathrm{C_8H_{10}}$$.
C–C $$\sigma$$ bonds $$= 8-1 = 7$$; C–H $$\sigma$$ bonds $$= 10$$.
Total $$\sigma = 7 + 10 = 17$$; $$\pi = 4$$ (four C=C).

(iii) 2-Propylpent-1-ene — molecular formula $$\mathrm{C_8H_{16}}$$.
C–C $$\sigma$$ bonds $$= 8-1 = 7$$; C–H $$\sigma$$ bonds $$= 16$$.
Total $$\sigma = 7 + 16 = 23$$; $$\pi = 1$$ (one C=C).

(iv) 5-Ethyl-2,7-dimethylundec-4-ene — molecular formula $$\mathrm{C_{15}H_{30}}$$.
C–C $$\sigma$$ bonds $$= 15-1 = 14$$; C–H $$\sigma$$ bonds $$= 30$$.
Total $$\sigma = 14 + 30 = 44$$; $$\pi = 1$$ (one C=C).

For alkenes of formula $$\mathrm{C_5H_{10}}$$ we vary (a) the position of the C=C double bond and (b) the carbon skeleton. This gives the following five structural isomers:

1. $$\mathrm{CH_2=CH-CH_2-CH_2-CH_3}$$ — Pent-1-ene

2. $$\mathrm{CH_3-CH=CH-CH_2-CH_3}$$ — Pent-2-ene (a position isomer of pent-1-ene)

3. $$\mathrm{CH_2=C(CH_3)-CH_2-CH_3}$$ — 2-Methylbut-1-ene

4. $$\mathrm{CH_2=CH-CH(CH_3)-CH_3}$$ — 3-Methylbut-1-ene

5. $$\mathrm{CH_3-C(CH_3)=CH-CH_3}$$ — 2-Methylbut-2-ene

Isomers 1 and 2 are position isomers; isomers with a methyl branch (3, 4, 5) are chain isomers of the straight-chain pentenes. (Pent-2-ene additionally shows cis–trans isomerism, but that is geometrical, not structural, isomerism.)

The alkyne series is ethyne (1st), propyne (2nd), but-1-yne (3rd), pentyne (4th), hexyne (5th). So the 5th member has molecular formula $$\mathrm{C_6H_{10}}$$.

Straight-chain isomers (vary position of the triple bond on a 6-carbon chain):

1. $$\mathrm{HC\equiv C-CH_2-CH_2-CH_2-CH_3}$$ — Hex-1-yne
2. $$\mathrm{CH_3-C\equiv C-CH_2-CH_2-CH_3}$$ — Hex-2-yne
3. $$\mathrm{CH_3-CH_2-C\equiv C-CH_2-CH_3}$$ — Hex-3-yne

Branched isomers (5-carbon chain with one methyl, or 4-carbon chain with two methyls):

4. $$\mathrm{HC\equiv C-CH(CH_3)-CH_2-CH_3}$$ — 3-Methylpent-1-yne
5. $$\mathrm{HC\equiv C-CH_2-CH(CH_3)-CH_3}$$ — 4-Methylpent-1-yne
6. $$\mathrm{CH_3-C\equiv C-CH(CH_3)-CH_3}$$ — 4-Methylpent-2-yne
7. $$\mathrm{HC\equiv C-C(CH_3)_2-CH_3}$$ — 3,3-Dimethylbut-1-yne

So there are seven isomers.

Types of isomerism:

• Position isomerism — between isomers having the same carbon skeleton but a different position of the triple bond, e.g. hex-1-yne, hex-2-yne and hex-3-yne; or 4-methylpent-1-yne and 4-methylpent-2-yne.
• Chain isomerism — between isomers having different carbon skeletons, e.g. hex-1-yne (straight chain) with 3-methylpent-1-yne or with 3,3-dimethylbut-1-yne.

The conversion is carried out in three stages.

Step 1 — Ethanoic acid to methane (decarboxylation). The sodium salt of ethanoic acid, heated with soda lime ($$\mathrm{NaOH}$$ + $$\mathrm{CaO}$$), loses $$\mathrm{CO_2}$$ to give methane.

$$\mathrm{CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3}$$

Step 2 — Methane to ethyne. Methane, on being heated to about $$1773\ \mathrm{K}$$, gives ethyne (acetylene).

$$\mathrm{2CH_4 \xrightarrow{1773\ K} CH\equiv CH + 3H_2}$$

Step 3 — Ethyne to benzene (cyclic polymerisation). Three molecules of ethyne, on passing through a red-hot iron tube at about $$873\ \mathrm{K}$$, undergo cyclic polymerisation to benzene.

$$\mathrm{3\,CH\equiv CH \xrightarrow[873\ K]{Red\ hot\ Fe\ tube} C_6H_6}$$

In the Wurtz reaction an alkyl halide reacts with sodium; the two alkyl groups couple to form a new C–C bond.

If two different alkyl halides $$\mathrm{R-X}$$ and $$\mathrm{R'-X}$$ are used together, the alkyl groups couple in all possible combinations. Three different alkanes are formed:

$$\mathrm{R-R}$$, $$\mathrm{R'-R'}$$ and $$\mathrm{R-R'}$$

This produces a mixture of three alkanes which have similar physical properties and are therefore very difficult to separate. Hence, the Wurtz reaction with two different alkyl halides is not a useful synthetic method — it is best used with a single alkyl halide to make a symmetrical alkane.

In Kolbe's electrolytic method, an aqueous solution of the sodium (or potassium) salt of a carboxylic acid is electrolysed. At the anode the carboxylate ion is discharged and loses $$\mathrm{CO_2}$$, giving an alkyl free radical; two alkyl radicals then combine to give an alkane.

$$\mathrm{2\,RCOO^- \longrightarrow 2\,R^{\bullet} + 2\,CO_2}$$
$$\mathrm{R^{\bullet} + R^{\bullet} \longrightarrow R-R}$$

Because the alkane is formed by the union of two alkyl groups, the smallest possible alkane obtainable is ethane ($$\mathrm{CH_3-CH_3}$$), formed when two methyl radicals combine (from sodium acetate).

Methane ($$\mathrm{CH_4}$$) has only one carbon atom and so cannot arise from the coupling of two alkyl groups. There is no carboxylate that would yield it. Hence methane cannot be prepared by Kolbe's electrolytic method.

Toluene contains seven carbon atoms (a benzene ring of 6 carbons plus a methyl group). It can be obtained from an alkane by aromatization (also called reforming) — the alkane is cyclised and dehydrogenated over a catalyst at high temperature and pressure.

For this, the straight-chain alkane must also have seven carbon atoms, i.e. n-heptane. On heating n-heptane at about $$773\ \mathrm{K}$$ and $$10$$–$$20\ \mathrm{atm}$$ over a catalyst such as $$\mathrm{V_2O_5}$$, $$\mathrm{Mo_2O_3}$$ or $$\mathrm{Cr_2O_3}$$ supported on alumina, it cyclises and loses hydrogen to give toluene.

$$\mathrm{CH_3(CH_2)_5CH_3 \xrightarrow[773\ K,\ 10{-}20\ atm]{Cr_2O_3/Al_2O_3} C_6H_5CH_3 + 4H_2}$$

So n-heptane is suggested for the preparation of toluene.

Propene is $$\mathrm{CH_3-CH=CH_2}$$.

Condition for geometrical (cis–trans) isomerism: each of the two doubly-bonded carbon atoms must carry two different groups.

Examine the two alkene carbons of propene:

• One carbon ($$\mathrm{=CH-}$$) carries a $$\mathrm{CH_3}$$ group and an $$\mathrm{H}$$ — two different groups.
• The terminal carbon ($$\mathrm{=CH_2}$$) carries two $$\mathrm{H}$$ atoms — two identical groups.

Since one of the doubly-bonded carbons bears two identical atoms, interchanging them does not create a new spatial arrangement. Therefore propene does not show geometrical isomerism.

The reaction of benzene with an alkyl halide and anhydrous $$\mathrm{AlCl_3}$$ is a Friedel–Crafts alkylation. The Lewis acid first generates a carbocation from the alkyl halide.

1-Chloropropane gives a primary carbocation:

$$\mathrm{CH_3CH_2CH_2-Cl \xrightarrow{AlCl_3} CH_3CH_2\overset{+}{C}H_2\ (1^\circ\ carbocation)}$$

A primary carbocation is unstable. It rearranges by a 1,2-hydride shift: a hydrogen atom (with its bonding electrons) moves from the middle carbon to the positively charged carbon, converting the unstable primary cation into a much more stable secondary (isopropyl) carbocation:

$$\mathrm{CH_3CH_2\overset{+}{C}H_2 \longrightarrow CH_3-\overset{+}{C}H-CH_3\ (2^\circ\ carbocation)}$$

It is this more stable isopropyl cation that attacks the benzene ring. Hence the product is isopropylbenzene (cumene) and not n-propylbenzene.

Chlorination of methane proceeds by a free-radical chain mechanism, which involves three kinds of steps.

Initiation: a chlorine molecule absorbs energy (ultraviolet light or heat) and undergoes homolysis into two chlorine atoms (radicals).

$$\mathrm{Cl_2 \xrightarrow{h\nu} 2\,\overset{\bullet}{C}l}$$

Propagation: a chlorine radical abstracts a hydrogen atom from methane to give a methyl radical; the methyl radical then reacts with $$\mathrm{Cl_2}$$ to give chloromethane and regenerate a chlorine radical.

$$\mathrm{CH_4 + \overset{\bullet}{C}l \longrightarrow \overset{\bullet}{C}H_3 + HCl}$$
$$\mathrm{\overset{\bullet}{C}H_3 + Cl_2 \longrightarrow CH_3Cl + \overset{\bullet}{C}l}$$

Termination: the chain ends when two radicals combine with each other. One such termination step is the combination of two methyl radicals:

$$\mathrm{\overset{\bullet}{C}H_3 + \overset{\bullet}{C}H_3 \longrightarrow CH_3-CH_3}$$

This coupling of two methyl radicals produces a molecule of ethane. Hence ethane is obtained as a by-product during the chlorination of methane.

In ozonolysis the $$\mathrm{C=C}$$ double bond breaks and each alkene carbon is converted to a carbonyl carbon. To find the parent alkene we do the reverse: join the carbonyl carbons of the two products by a double bond.

Ethanal is $$\mathrm{CH_3-CHO}$$ — its carbonyl carbon carries a $$\mathrm{CH_3}$$ group and an $$\mathrm{H}$$. It came from the fragment $$\mathrm{CH_3-CH=}$$.

Pentan-3-one is $$\mathrm{(C_2H_5)_2C=O}$$ — its carbonyl carbon carries two ethyl groups. It came from the fragment $$\mathrm{=C(C_2H_5)_2}$$.

Joining the two fragments through a double bond gives alkene 'A':

$$\mathrm{CH_3-CH=C(C_2H_5)_2}$$

Naming 'A': the longest chain through the double bond has 5 carbons (pent-2-ene), with an ethyl group on C-3.

IUPAC name of 'A': 3-Ethylpent-2-ene.

Step 1 — find the molecular formula. 'A' has three C–C $$\sigma$$ bonds and one C–C $$\pi$$ bond. A double bond is one $$\sigma$$ + one $$\pi$$, so the $$\pi$$ bond is part of one of these three C–C linkages. There are therefore 3 carbon–carbon linkages in all, which means a chain of $$3+1 = 4$$ carbon atoms.

There are 8 C–H $$\sigma$$ bonds, i.e. 8 hydrogen atoms. So the molecular formula of 'A' is $$\mathrm{C_4H_8}$$ — consistent with an alkene $$\mathrm{C_nH_{2n}}$$.

Step 2 — use the ozonolysis data. 'A' on ozonolysis gives two moles of a single aldehyde of molar mass $$44\ \mathrm{u}$$. The aldehyde of molar mass $$44$$ is ethanal, $$\mathrm{CH_3CHO}$$, since $$12\times2 + 1\times4 + 16 = 44$$.

Obtaining two moles of the same aldehyde means the alkene is symmetrical about the double bond, with a $$\mathrm{CH_3-CH=}$$ unit on each side:

$$\mathrm{CH_3-CH=CH-CH_3}$$

This is $$\mathrm{C_4H_8}$$, which matches the bond count of Step 1.

IUPAC name of 'A': But-2-ene.

In ozonolysis the $$\mathrm{C=C}$$ bond breaks and each alkene carbon becomes a carbonyl carbon. To reconstruct the alkene, join the carbonyl carbons of the two products through a double bond.

Propanal is $$\mathrm{CH_3CH_2-CHO}$$ — its carbonyl carbon carries an ethyl group and an $$\mathrm{H}$$. It came from the fragment $$\mathrm{CH_3CH_2-CH=}$$.

Pentan-3-one is $$\mathrm{(C_2H_5)_2C=O}$$ — its carbonyl carbon carries two ethyl groups. It came from the fragment $$\mathrm{=C(C_2H_5)_2}$$.

Joining the two fragments through a double bond gives the alkene:

$$\mathrm{CH_3CH_2-CH=C(C_2H_5)_2}$$

This alkene is 3-ethylhex-3-ene.

Hex-2-ene is $$\mathrm{CH_3-CH=CH-CH_2-CH_2-CH_3}$$. Each carbon of the $$\mathrm{C=C}$$ bond carries two different groups (one carbon bears $$\mathrm{CH_3}$$ and $$\mathrm{H}$$; the other bears the n-propyl group $$\mathrm{-CH_2CH_2CH_3}$$ and $$\mathrm{H}$$), so geometrical isomerism is possible.

cis-hex-2-ene: draw the $$\mathrm{C=C}$$ horizontally with the $$\mathrm{CH_3}$$ group and the $$\mathrm{n}$$-propyl group on the same side of the double bond; the two $$\mathrm{H}$$ atoms then lie together on the opposite side.

trans-hex-2-ene: draw the $$\mathrm{CH_3}$$ group and the $$\mathrm{n}$$-propyl group on opposite sides of the double bond; the two $$\mathrm{H}$$ atoms are likewise on opposite sides.

Which has the higher boiling point? In the cis isomer the two alkyl groups lie on the same side, so their bond dipoles do not cancel and the molecule has a net dipole moment — it is polar. In the trans isomer the alkyl groups are on opposite sides and their dipoles largely cancel, so it is almost non-polar.

The polar cis isomer has stronger intermolecular (dipole–dipole) attractions, so more energy is needed to separate its molecules. Hence cis-hex-2-ene has the higher boiling point.

Although benzene is conventionally drawn with three double bonds, the six $$\pi$$ electrons are not localised as three separate double bonds.

Each of the six carbon atoms is $$sp^2$$ hybridised and carries one unhybridised $$p$$-orbital perpendicular to the plane of the ring. These six $$p$$-orbitals overlap sideways with one another all round the ring.

This overlap produces two continuous rings of delocalised $$\pi$$ electron charge — one lying above and one below the plane of the carbon hexagon. The $$\pi$$ electrons therefore belong to the molecule as a whole, not to individual pairs of carbon atoms.

This complete delocalisation lowers the energy of the molecule. The actual benzene molecule is more stable than any single Kekulé structure (with fixed double bonds) by an amount called the resonance energy (about $$150\ \mathrm{kJ\,mol^{-1}}$$).

Because of this large resonance (delocalisation) energy, benzene is extraordinarily stable. It resists addition reactions (which would destroy the delocalised system) and instead prefers substitution reactions (which preserve it).

A system is said to be aromatic only if it satisfies all of the following conditions:

1. It must be cyclic and planar. All the atoms of the ring lie in one plane, so that the $$p$$-orbitals can overlap effectively.
2. It must have complete delocalisation of $$\pi$$ electrons in the ring. Every atom of the ring is $$sp^2$$ (or $$sp$$) hybridised and carries a $$p$$-orbital, giving an uninterrupted (fully conjugated) closed loop of overlapping $$p$$-orbitals.
3. It must obey Hückel's rule. The cyclic $$\pi$$ system must contain $$(4n+2)$$ $$\pi$$ electrons, where $$n = 0, 1, 2, 3, \ldots$$ — that is, $$2, 6, 10, \ldots$$ $$\pi$$ electrons. Benzene, with $$6$$ $$\pi$$ electrons, corresponds to $$n = 1$$.

First write out the structure, labelling the backbone carbons C-1 to C-5:

$$\mathrm{\overset{C1}{C}H_3 - \overset{C2}{C}H_2 - \overset{C3}{C}(CH_3)_2 - \overset{C4}{C}H_2 - \overset{C5}{C}H(CH_3)_2}$$

Besides the five backbone carbons there are four branch methyl carbons — two attached to C-3 and two attached to C-5.

A carbon is classified by how many other carbon atoms are directly attached to it: primary $$(1^\circ)$$ — bonded to one C; secondary $$(2^\circ)$$ — bonded to two C; tertiary $$(3^\circ)$$ — bonded to three C; quaternary $$(4^\circ)$$ — bonded to four C.

Summary:

• Primary $$(1^\circ)$$ carbons: 5 in all — C-1 together with the four branch methyl groups — each bonded to 3 H atoms.
• Secondary $$(2^\circ)$$ carbons: 2 — C-2 and C-4 — each bonded to 2 H atoms.
• Tertiary $$(3^\circ)$$ carbon: 1 — C-5 — bonded to 1 H atom.
• (C-3 is a quaternary carbon, bonded to 0 H atoms.)

The boiling point of an alkane depends on the strength of the intermolecular van der Waals (London dispersion) forces, which in turn depend on how large an area of contact the molecules can present to one another.

As an alkane chain becomes more branched, the molecule becomes more compact and approaches a spherical shape. A sphere has the smallest surface area for a given volume, so branched molecules touch one another over a smaller area.

A smaller area of contact means weaker van der Waals forces, so less energy is needed to separate the molecules, and the boiling point is lower.

Therefore, for isomers of the same molecular formula, boiling point decreases as branching increases. For example, among the isomers of $$\mathrm{C_5H_{12}}$$:

n-pentane (b.p. $$309\ \mathrm{K}$$) $$>$$ 2-methylbutane (b.p. $$301\ \mathrm{K}$$) $$>$$ 2,2-dimethylpropane (b.p. $$282.5\ \mathrm{K}$$).

Without peroxide — Markovnikov addition (ionic mechanism). In the absence of peroxide, HBr adds to propene by an ionic (electrophilic) mechanism that follows Markovnikov's rule: the hydrogen adds to the doubly-bonded carbon that already carries more hydrogen atoms.

Mechanism:

Step 1 — HBr supplies the electrophile $$\mathrm{H^+}$$, which adds to the double bond. Addition could give either a secondary or a primary carbocation:

$$\mathrm{CH_3-CH=CH_2 + H^+ \longrightarrow CH_3-\overset{+}{C}H-CH_3}$$ (secondary — more stable)
or $$\mathrm{CH_3-CH_2-\overset{+}{C}H_2}$$ (primary — less stable)

The secondary carbocation is more stable (greater stabilisation by hyperconjugation and the inductive effect), so it is formed preferentially.

Step 2 — the bromide ion $$\mathrm{Br^-}$$ attacks this carbocation:

$$\mathrm{CH_3-\overset{+}{C}H-CH_3 + Br^- \longrightarrow CH_3-CHBr-CH_3}$$

Product: 2-bromopropane.

With benzoyl peroxide — anti-Markovnikov addition (free-radical mechanism, the peroxide or Kharasch effect). In the presence of peroxide, HBr adds by a free-radical chain mechanism, and the bromine adds to the terminal carbon (anti-Markovnikov).

Mechanism:

Initiation — the peroxide breaks homolytically, and the radicals formed react with HBr to give bromine atoms:

$$\mathrm{(C_6H_5COO)_2 \xrightarrow{\Delta} 2\,C_6H_5\overset{\bullet}{C}OO \longrightarrow 2\,\overset{\bullet}{C}_6H_5 + 2\,CO_2}$$
$$\mathrm{\overset{\bullet}{C}_6H_5 + HBr \longrightarrow C_6H_6 + \overset{\bullet}{B}r}$$

Propagation — the bromine radical adds to the double bond. It adds to the terminal carbon because this produces the more stable secondary carbon radical:

$$\mathrm{CH_3-CH=CH_2 + \overset{\bullet}{B}r \longrightarrow CH_3-\overset{\bullet}{C}H-CH_2Br}$$ (secondary radical — more stable)
$$\mathrm{CH_3-\overset{\bullet}{C}H-CH_2Br + HBr \longrightarrow CH_3-CH_2-CH_2Br + \overset{\bullet}{B}r}$$

The regenerated $$\mathrm{\overset{\bullet}{B}r}$$ continues the chain. Product: 1-bromopropane.

Conclusion. The change of mechanism reverses the orientation of addition: the ionic route goes through the more stable carbocation (Br on C-2), while the free-radical route goes through the more stable carbon radical (Br on C-1). The peroxide effect is observed only with HBr (not with HCl or HI).

The two Kekulé structures. Kekulé represented benzene as a six-membered ring with three alternating double bonds. Because the double bonds can be placed in two alternative ways, benzene is a resonance hybrid of two Kekulé structures (I and II) in which the double bonds occupy different positions.

For o-xylene (1,2-dimethylbenzene) the two Kekulé forms differ in whether the bond between the two methyl-bearing carbons (C-1 and C-2) is a double bond or a single bond.

Ozonolysis of structure I (double bonds at C1–C2, C3–C4, C5–C6). Cleaving each $$\mathrm{C=C}$$ converts the carbons to carbonyl groups and gives:

• from the C1–C6 fragment (one $$\mathrm{CH_3}$$, one $$\mathrm{H}$$): methylglyoxal, $$\mathrm{CH_3-CO-CHO}$$
• from the C2–C3 fragment (one $$\mathrm{CH_3}$$, one $$\mathrm{H}$$): methylglyoxal, $$\mathrm{CH_3-CO-CHO}$$
• from the C4–C5 fragment (two $$\mathrm{H}$$): glyoxal, $$\mathrm{OHC-CHO}$$

Ozonolysis of structure II (double bonds at C2–C3, C4–C5, C6–C1) gives:

• from the C1–C2 fragment (two $$\mathrm{CH_3}$$): dimethylglyoxal, $$\mathrm{CH_3-CO-CO-CH_3}$$
• from the C3–C4 fragment (two $$\mathrm{H}$$): glyoxal, $$\mathrm{OHC-CHO}$$
• from the C5–C6 fragment (two $$\mathrm{H}$$): glyoxal, $$\mathrm{OHC-CHO}$$

Observed result. In practice, ozonolysis of o-xylene gives a mixture of three products: glyoxal $$\mathrm{(OHC-CHO)}$$, methylglyoxal $$\mathrm{(CH_3COCHO)}$$ and dimethylglyoxal $$\mathrm{(CH_3COCOCH_3)}$$.

How this supports the Kekulé structure. A single fixed Kekulé structure could give only two of these three products. Obtaining all three shows that the bond between the methyl-bearing carbons is a double bond in some molecules and a single bond in others — i.e. benzene is a resonance hybrid of the two Kekulé structures, with the double bonds continually interchanging their positions. This is exactly what Kekulé proposed (two oscillating forms), so the result supports the Kekulé structure for benzene.

The acidic behaviour compared here is the tendency of a C–H bond to release its hydrogen as $$\mathrm{H^+}$$. This depends on the hybridisation of the carbon holding that hydrogen — specifically on the s-character of its hybrid orbital.

An orbital with greater s-character holds its electrons closer to the nucleus, making that carbon effectively more electronegative. This makes the C–H bond more polar, the proton easier to remove, and the resulting carbanion more stable.

Greater s-character $$\Rightarrow$$ more acidic. Hence the decreasing order of acidic behaviour is:

$$\text{Ethyne} \;>\; \text{Benzene} \;>\; \textit{n}\text{-Hexane}$$

In benzene each carbon is $$sp^2$$ hybridised, and the six unhybridised $$p$$-orbitals overlap to form a delocalised $$\pi$$ electron cloud lying above and below the plane of the ring.

This $$\pi$$ cloud makes the ring a region of high electron density.

Electrophilic substitution (occurs easily): an electrophile $$\mathrm{E^+}$$ is electron-deficient. It is strongly attracted to the electron-rich $$\pi$$ cloud of benzene and therefore readily attacks the ring. Substitution (rather than addition) then takes place because it restores the stable delocalised $$\pi$$ system. Hence benzene undergoes electrophilic substitution easily.

Nucleophilic substitution (occurs with difficulty): a nucleophile is electron-rich (it carries a lone pair or a negative charge). It is repelled by the electron-rich $$\pi$$ cloud of benzene, so it cannot easily approach and attack the ring. Hence benzene undergoes nucleophilic substitution only with great difficulty.

On hydrogenation a $$\mathrm{C=C}$$ double bond becomes a $$\mathrm{C-C}$$ single bond (two H atoms add across it). So an alkene gives 2-methylbutane on hydrogenation only if it has the carbon skeleton of 2-methylbutane with one double bond placed somewhere within it.

The 2-methylbutane skeleton is $$\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}$$ (C-1 to C-4 with a methyl branch on C-2). Place the double bond at each possible position:

1. Double bond between C-1 and C-2:
$$\mathrm{CH_2=C(CH_3)-CH_2-CH_3}$$ — 2-Methylbut-1-ene

2. Double bond between C-2 and C-3:
$$\mathrm{CH_3-C(CH_3)=CH-CH_3}$$ — 2-Methylbut-2-ene

3. Double bond between C-3 and C-4:
$$\mathrm{(CH_3)_2CH-CH=CH_2}$$ — 3-Methylbut-1-ene

(Placing a double bond from C-2 to the branch methyl gives the same compound as case 1, since the two methyl groups on C-2 are equivalent.)

Each of these three alkenes, on catalytic hydrogenation, gives 2-methylbutane.

Nitration is an electrophilic substitution carried out by the nitronium ion $$\mathrm{NO_2^+}$$. It occurs more easily the more electron-rich the ring is.

• Toluene bears a $$\mathrm{-CH_3}$$ group, which is electron-donating (by the inductive effect and hyperconjugation). It increases the electron density of the ring — the ring is activated.
• Benzene has no substituent — it is the reference compound, neither activated nor deactivated.
• m-Dinitrobenzene bears two $$\mathrm{-NO_2}$$ groups, which are strongly electron-withdrawing. They decrease the electron density of the ring — the ring is strongly deactivated.

The more electron-rich the ring, the faster it reacts with $$\mathrm{NO_2^+}$$. Hence the ease of nitration is:

$$\text{Toluene} > \text{Benzene} > \textit{m}\text{-dinitrobenzene}$$

Therefore toluene undergoes nitration most easily, because its electron-releasing methyl group makes the ring the most electron-rich (most activated) of the three.

Ethylation of benzene is a Friedel–Crafts alkylation: benzene reacts with an ethyl halide (e.g. ethyl chloride) in the presence of a Lewis acid catalyst. The Lewis acid is needed to generate the ethyl carbocation electrophile from the ethyl halide.

Apart from anhydrous aluminium chloride $$\mathrm{(AlCl_3)}$$, another Lewis acid that works well for this purpose is anhydrous ferric chloride, $$\mathrm{FeCl_3}$$.

(Other Lewis acids such as anhydrous $$\mathrm{AlBr_3}$$, $$\mathrm{FeBr_3}$$ or $$\mathrm{BF_3}$$ can also be used.)

In the Wurtz reaction an alkyl halide reacts with sodium; two alkyl groups join together through a new C–C bond:

$$\mathrm{2\,R-X + 2\,Na \longrightarrow R-R + 2\,NaX}$$

When a single alkyl halide $$\mathrm{R-X}$$ is used, the product $$\mathrm{R-R}$$ contains twice the number of carbon atoms of the group R. So a single alkyl halide can give only an alkane with an even number of carbon atoms (ethane, butane, hexane, and so on).

To obtain an alkane with an odd number of carbon atoms, one is forced to couple two different alkyl halides. But then the two kinds of alkyl groups combine in all possible ways and a mixture of three alkanes is produced. These three alkanes have similar physical properties and are very difficult to separate, so the yield of the desired alkane is poor. This is why the Wurtz reaction is not preferred for odd-carbon alkanes.

Example — preparation of propane (3 carbon atoms, odd). One must take methyl bromide and ethyl bromide together. The coupling then gives three products:

$$\mathrm{CH_3Br + CH_3Br \xrightarrow{2Na} CH_3-CH_3}$$  (ethane)
$$\mathrm{C_2H_5Br + C_2H_5Br \xrightarrow{2Na} C_2H_5-C_2H_5}$$  (butane)
$$\mathrm{CH_3Br + C_2H_5Br \xrightarrow{2Na} CH_3-C_2H_5}$$  (propane)

The desired propane is obtained only as part of a hard-to-separate mixture with ethane and butane, in low yield. Hence the Wurtz reaction is not preferred for preparing alkanes with an odd number of carbon atoms.