Hydrocarbons

Propene with hydrogen bromide gives $2$-bromopropane. Add a trace of peroxide and you get $1$-bromopropane instead. Yet peroxide has no effect whatever on the addition of hydrogen chloride or hydrogen iodide. Why does the trick work only for bromine?

Because a radical chain needs both of its propagation steps to be favourable, and only bromine manages that.

The chain has two steps: the halogen radical adds to the alkene, and the resulting carbon radical abstracts hydrogen from the hydrogen halide.

For chlorine, the addition step is fine but abstracting hydrogen from the very strong $\text{H}-\text{Cl}$ bond is endothermic. For iodine, hydrogen abstraction is easy but the addition of the bulky, weakly bonding iodine radical is endothermic. Only bromine sits at the point where both steps release energy, so only bromine sustains the chain.

That is the level at which Advanced examines this chapter. Not which reagent gives which product, but why — and the "why" almost always fixes the regiochemistry and the stereochemistry together.

1. Free radical substitution: reactivity against selectivity

Halogenation of an alkane proceeds by a radical chain, and the halogen chosen decides how selective it is:

The pattern is general: the more reactive a reagent, the less selective it is. A very exothermic step has an early transition state that barely resembles the product radical, so the stability differences between radicals hardly show. Bromination's hydrogen abstraction is nearly thermoneutral, its transition state resembles the radical closely, and the stability differences are expressed in full.

Illustration 1

$2$-Methylbutane is halogenated. Predict the major product with chlorine and with bromine.

The molecule has one tertiary, two secondary and nine primary hydrogens, on three primary carbons.

Chlorine: weighting by count and by relative rate gives a spread of products with no single one dominant; the primary positions contribute heavily simply because there are nine of them.

Bromine: the tertiary position, despite being a single hydrogen, is favoured by a factor of $1600$, so $2$-bromo-$2$-methylbutane is overwhelmingly the major product.

Selectivity beats statistics for bromine and loses to it for chlorine, which is why bromination is used synthetically and chlorination is not.

2. Cyclohexane conformations

The chair is essentially strain free, with all bonds staggered. Each carbon carries one axial bond, parallel to the ring axis, and one equatorial bond, roughly in the ring plane. A ring flip converts every axial position into an equatorial one and vice versa.

An axial substituent suffers $1,3$-diaxial repulsion from the two axial hydrogens on the same face, so bulky groups prefer the equatorial position. The preference is worth about $7.3$ kJ mol${}^{-1}$ for methyl and around $21$ for tertiary butyl — large enough that a tertiary butyl group effectively locks the ring in one conformation.

Illustration 2

Which is more stable: cis- or trans-$1,4$-dimethylcyclohexane? Answer the same for the $1,2$ isomers.

For $1,4$: the trans isomer can place both methyls equatorial, while cis must have one axial whichever way it flips. Trans is more stable.

For $1,2$: the cis isomer can achieve one equatorial and one axial, while trans can place both equatorial. Trans is more stable here too.

The rule alternates with the substitution pattern. For $1,2$ and $1,4$ the trans isomer wins; for $1,3$ it is the cis isomer that can put both groups equatorial.

Illustration 3

A disubstituted cyclohexane carries a tertiary butyl group and a methyl group in a $1,4$ relationship. Explain why the ring conformation is fixed regardless of which isomer is present.

A tertiary butyl group costs about $21$ kJ mol${}^{-1}$ when axial, far more than a methyl group's $7.3$.

The ring therefore always adopts the conformation with the tertiary butyl group equatorial, whatever that forces on the methyl.

In the trans isomer both end up equatorial; in the cis isomer the methyl is forced axial and stays there.

A tertiary butyl group is described as a conformational anchor. It is used deliberately in mechanistic studies to hold a ring in one arrangement so that axial and equatorial reactivities can be compared.

3. Electrophilic addition: getting the regiochemistry right

Markovnikov's rule is a consequence, not a postulate: the electrophile adds so as to generate the more stable carbocation.

The last two are the synthetically important pair. Oxymercuration gives the Markovnikov alcohol without rearrangement, because the intermediate is a bridged mercurinium ion rather than a free cation. Hydroboration gives the anti-Markovnikov alcohol because boron, the electropositive atom, adds to the less hindered carbon.

Illustration 4

Give the product of treating $3,3$-dimethylbut-1-ene with (a) aqueous acid and (b) borane followed by alkaline hydrogen peroxide.

(a) Protonation gives a secondary cation, which rearranges by a methyl shift to the tertiary cation. Water then attacks, giving $2,3$-dimethylbutan-2-ol — a rearranged skeleton.

(b) Boron adds to the terminal carbon and hydrogen to the internal one, both from the same face, with no cationic intermediate. Oxidation replaces boron by hydroxyl, giving $3,3$-dimethylbutan-1-ol with the skeleton intact.

The two routes give alcohols on opposite carbons and different skeletons. Choosing between them is the standard synthesis question in this chapter.

4. Electrophilic addition: the stereochemistry

The pattern is that a cyclic intermediate forces one face on the incoming group, and whether the result is syn or anti depends on whether that group was delivered by the ring or attacked it from outside.

Illustration 5

Give the stereochemical outcome of treating cyclohexene with (a) bromine and (b) osmium tetroxide.

(a) Bromine forms a bridged bromonium ion on one face; bromide then attacks the opposite face, so the two bromines end up trans. The product is trans-$1,2$-dibromocyclohexane, as a racemic mixture.

(b) Osmium tetroxide forms a cyclic ester delivering both oxygens from the same face, so hydrolysis gives cis-cyclohexane-$1,2$-diol, a meso compound.

Same alkene, opposite stereochemistry, from the geometry of the intermediate alone. Neither result would follow from a free carbocation, which is why these reactions are used as mechanistic evidence.

5. Alkynes: acidity and selective reduction

A terminal alkyne has $p{K}_a$ about $25$, far more acidic than an alkene ($44$) or an alkane ($50$), because its carbanion sits in an $sp$ orbital with fifty per cent $s$ character and holds the pair close to the nucleus. Sodium amide deprotonates it and the acetylide is a useful nucleophile.

Reduction is stereoselective and the choice of reagent decides the geometry:

${\text{H}}_2/\text{Lindlar}\to \text{cis alkene},{\text{Na in liquid NH}}_3\to \text{trans alkene}$

Lindlar's poisoned palladium delivers both hydrogens from the catalyst surface, hence syn addition and a cis alkene. The dissolving-metal reduction proceeds through a vinyl radical anion, which adopts the less crowded trans geometry before picking up its second proton.

Hydration also splits two ways. Mercuric ion catalysis gives the Markovnikov enol, which tautomerises to a ketone; hydroboration gives the anti-Markovnikov enol, which tautomerises to an aldehyde.

Illustration 6

Give the products of hydrating but-1-yne (a) with dilute acid and mercuric sulphate and (b) by hydroboration followed by oxidation.

(a) Markovnikov addition puts the hydroxyl on carbon two, giving an enol that tautomerises to butan-2-one.

(b) Boron adds to the terminal carbon, so the hydroxyl ends up there, giving an enol that tautomerises to butanal.

Only a terminal alkyne gives an aldehyde by this route, and only through hydroboration. Ethyne itself is the sole alkyne whose acid-catalysed hydration gives an aldehyde, since both carbons are terminal.

Illustration 7

Show how but-$1$-yne can be converted into hex-$3$-yne, and state why the same approach fails for a secondary halide.

Deprotonate the terminal alkyne with sodium amide to give the acetylide, which is a strong nucleophile.

Treat it with bromoethane. The acetylide displaces bromide in an $\text{SN}2$ reaction, giving hex-$3$-yne.

With a secondary halide the acetylide acts as a base rather than a nucleophile, abstracting a proton and giving an alkene by elimination.

Acetylide alkylation works only with primary halides. That restriction is the standard limitation of the method and is examined regularly.

6. Aromatic substitution and the Friedel-Crafts limits

Electrophilic aromatic substitution proceeds through an arenium ion, and the substituent already present decides both the rate and the position:

The halogens are the exception that carries most of the marks: deactivating by induction, ortho-para directing by resonance.

Friedel-Crafts reactions have three limitations that are examined constantly. Alkylation rearranges, since it goes through a carbocation, so a primary halide gives a branched product. Alkylation also polysubstitutes, because the alkyl product is more activated than the starting material. And neither version works on a strongly deactivated ring such as nitrobenzene, or on aniline, whose nitrogen complexes irreversibly with the aluminium chloride.

Acylation avoids the first two problems: the acylium ion is resonance stabilised and does not rearrange, and the ketone product is deactivated so substitution stops cleanly at one. Reducing the ketone afterwards is therefore the standard route to a straight-chain alkylbenzene.

Illustration 8

How would you prepare propylbenzene from benzene, and why does the obvious route fail?

Friedel-Crafts alkylation with $1$-chloropropane gives mainly isopropylbenzene, because the primary cation rearranges by a hydride shift to the secondary one before attacking the ring.

The correct route is acylation with propanoyl chloride, giving phenyl propyl ketone with no rearrangement, followed by Clemmensen or Wolff-Kishner reduction of the carbonyl to a methylene group.

Two steps beat one here. The acylium ion cannot rearrange because its positive charge is delocalised onto oxygen, which is exactly the stabilisation an alkyl cation lacks.

Illustration 9

Explain why toluene undergoes nitration about $25$ times faster than benzene while nitrobenzene reacts about $10^5$ times slower.

The rate is governed by the stability of the arenium ion intermediate.

A methyl group donates electron density by induction and hyperconjugation, stabilising the positive intermediate and lowering the activation energy.

A nitro group withdraws density strongly by both induction and resonance, destabilising the intermediate severely.

The nitro group's effect is a hundred thousandfold, which is why a second nitration of nitrobenzene needs fuming acid and high temperature, and stops at the meta position.

Illustration 10

A benzene ring carries both a methyl group and a nitro group in a meta relationship. Predict where a third substituent will enter.

The methyl group is activating and directs ortho and para to itself.

The nitro group is deactivating and directs meta to itself.

For a meta-disubstituted ring these two instructions coincide at the position between them and at the positions ortho and para to the methyl.

The strongly activating methyl group controls the outcome, and the position between the two groups is sterically crowded, so substitution occurs mainly para to the methyl.

When two substituents disagree, the more strongly activating one wins. Steric hindrance then decides among the positions it favours.

7. Oxidative cleavage and structure determination

Ozonolysis breaks a carbon-carbon double bond and tells you exactly where it was:

$\text{reductive work-up}\left(\text{Zn}/{\text{H}}_2\text{O}\right)\to \text{aldehydes and ketones}$ $\text{oxidative work-up}\left({\text{H}}_2{\text{O}}_2\right)\to \text{acids and ketones}$

The difference matters: a $\text{CH}$ fragment gives an aldehyde under reductive conditions and a carboxylic acid under oxidative ones, while a fully substituted carbon gives a ketone either way. Terminal ${\text{CH}}_2$ gives methanal or, oxidatively, carbon dioxide.

Hot concentrated permanganate cleaves alkenes similarly but always oxidatively.

Illustration 11

An alkene ${\text{C}}_6{\text{H}}_{12}$ gives propanone and propanal on reductive ozonolysis. Identify it.

Propanone contributes a ${\left({\text{CH}}_3\right)}_2\text{C}$ fragment; propanal contributes a ${\text{CH}}_3{\text{CH}}_2\text{CH}$ fragment.

Rejoining them across a double bond gives ${\left({\text{CH}}_3\right)}_2\text{C}={\text{CHCH}}_2{\text{CH}}_3$, that is $2$-methylpent-$2$-ene.

Checking the formula: ${\text{C}}_6{\text{H}}_{12}$, correct.

Ozonolysis is the classic locating tool. Each carbonyl compound names the fragment that was attached to one end of the double bond, so rejoining them reconstructs the alkene uniquely.

Illustration 12

The same alkene is subjected to oxidative ozonolysis. What changes?

The ${\left({\text{CH}}_3\right)}_2\text{C}$ fragment still gives propanone, since that carbon carries no hydrogen and cannot be oxidised further.

The ${\text{CH}}_3{\text{CH}}_2\text{CH}$ fragment now gives propanoic acid rather than propanal.

Only fragments bearing a hydrogen change. Comparing the two work-ups therefore reveals which end of the double bond was mono-substituted, which is extra structural information for free.

Illustration 13

Explain why Friedel-Crafts alkylation of benzene with excess methyl chloride gives substantial amounts of polymethylbenzenes.

Each methyl group added is electron donating, so the product ring is more activated than benzene itself.

The monosubstituted product therefore competes successfully with the remaining benzene for the electrophile.

Substitution continues, giving di, tri and higher methylated products even when the halide is not in large excess.

Acylation does not suffer this, because the ketone formed is deactivated and stops competing, which is why acylation followed by reduction is the controlled route.

Summary

• The peroxide effect works only for hydrogen bromide, because only there are both radical propagation steps exothermic.
• Reactivity and selectivity are inversely related: chlorine is $5:3.8:1$ across $3^{\circ }:2^{\circ }:1^{\circ }$, bromine $1600:82:1$.
• Cyclohexane: bulky groups prefer equatorial, avoiding $1,3$-diaxial repulsion; tertiary butyl locks the ring.
• For $1,2$ and $1,4$ dimethylcyclohexane the trans isomer is more stable; for $1,3$ it is the cis.
• Markovnikov addition means the more stable carbocation, and any such route may rearrange.
• Oxymercuration gives the Markovnikov alcohol without rearrangement; hydroboration gives the anti-Markovnikov one.
• ${\text{Br}}_2$ adds anti through a bromonium ion; ${\text{H}}_2$, ${\text{BH}}_3$ and ${\text{OsO}}_4$ all add syn.
• Peracid then hydrolysis gives the anti diol; cold permanganate or osmium tetroxide gives the syn diol.
• A tertiary butyl group anchors a cyclohexane ring in one conformation; acetylide alkylation works only with primary halides.
• When two ring substituents disagree on direction, the more strongly activating one decides, and sterics choose among its positions.
• Terminal alkynes have $p{K}_a$ $25$ because the carbanion sits in an $sp$ orbital with fifty per cent $s$ character.
• Lindlar gives the cis alkene, sodium in liquid ammonia the trans.
• Alkyne hydration: mercuric ion gives a ketone, hydroboration gives an aldehyde.
• Halogens are deactivating but ortho-para directing — the standard exception.
• Friedel-Crafts alkylation rearranges and polysubstitutes; acylation does neither, so acylate then reduce.
• Ozonolysis locates a double bond; reductive work-up gives aldehydes, oxidative gives acids from the same fragments.