Hydrocarbons

Shake an alkene with bromine water and the orange colour vanishes on contact. Benzene has three carbon-carbon double bonds drawn into it. Shake benzene with bromine water and nothing happens at all.

Why does a molecule with three double bonds refuse a reaction that a molecule with one performs instantly?

The answer is a number, and it can be measured with a calorimeter.

Hydrogenating cyclohexene, which has one double bond, releases 120 kJ mol${}^{-1}$. If benzene were simply cyclohexatriene, three isolated double bonds in a ring, it should release three times that.

$3\times 120=360\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}\text{predicted}$

Benzene releases 208.

The 152 kJ mol${}^{-1}$ shortfall is resonance energy: benzene is already 152 kJ mol${}^{-1}$ more stable than any drawing of it suggests, because its six pi electrons are delocalised over the whole ring rather than penned into three bonds.

That number decides the entire chapter's chemistry.

An alkene adds bromine because it has an ordinary pi bond and gains from swapping it for two sigma bonds. Benzene would have to spend 152 kJ mol${}^{-1}$ to add anything, because addition breaks the delocalised ring permanently. So benzene refuses to add, accepts the electrophile, and then throws out a proton to get its ring back: electrophilic substitution.

Four families, dozens of named reactions, and one question underneath all of it: what is the electron density doing, and what will the molecule protect?

1. Classification and Isomerism

Alkynes cannot show geometrical isomerism because the two carbons of a triple bond are $sp$ and linear, so each carries only one other group and there is no second arrangement to find.

Butane and 2-methylpropane are the first pair of chain isomers. Butene shows all three types at once: but-1-ene and but-2-ene are position isomers, 2-methylpropene is a chain isomer of both, and but-2-ene itself exists as cis and trans.

Illustration 1

A hydrocarbon has the molecular formula ${\mathrm{C}}_7{\mathrm{H}}_8$. How many degrees of unsaturation does it carry, and what does that allow it to be?

The degree of unsaturation counts how far a formula falls short of the saturated acyclic maximum, in units of one ring or one π bond:

$\text{DoU}=\frac{2C+2-H}{2}=\frac{2(7)+2-8}{2}=\frac{8}{2}=4$

Four is the signature of a benzene ring — three π bonds plus the ring itself — so ${\mathrm{C}}_7{\mathrm{H}}_8$ is toluene, and in practice a question giving this formula means toluene.

Run the count on two more formulae to see what it can and cannot do. ${\mathrm{C}}_5{\mathrm{H}}_{12}$ gives $(10+2-12)/2=0$, so it is saturated and acyclic: some isomer of pentane, with no further possibilities. ${\mathrm{C}}_5{\mathrm{H}}_{10}$ gives 1, and that single degree may be spent either on a double bond or on a ring — cyclopentane and the pentenes share the formula.

That ambiguity is the limit of the method. Degrees of unsaturation tell you how many rings and π bonds there are between them, never how the total is divided.

2. Alkanes

Preparation

Reactions

Alkanes are unreactive, which is why they were once called paraffins, meaning little affinity. Free radical halogenation is the characteristic reaction, in three stages: initiation (the halogen breaks homolytically under UV or heat), propagation (a halogen radical abstracts hydrogen to give an alkyl radical, which attacks another halogen molecule), and termination (any two radicals combine).

Reactivity runs ${\mathrm{F}}_2>\mathrm{C}{\mathrm{l}}_2>\mathrm{B}{\mathrm{r}}_2>{\mathrm{I}}_2$, with fluorination explosively violent and iodination not proceeding without an oxidising agent to remove the HI.

Illustration 2

2-Methylpropane has nine primary hydrogens and one tertiary. Chlorination gives 64 per cent primary product and 36 per cent tertiary. Bromination gives over 99 per cent tertiary. Both reactions form the same two possible radicals. Why the difference?

Divide out the statistics first, since there are nine primary hydrogens competing with one tertiary.

Chlorine can tell the two positions apart by a factor of 5. Bromine tells them apart by a factor of 1600, and that is what "selective" means quantitatively.

The cause is the energetics of the hydrogen abstraction step. Abstraction by a chlorine radical is exothermic, so the transition state is reached early, while the C-H bond is barely stretched and the radical has hardly begun to form. Whatever stability the tertiary radical will eventually have is not yet available to lower the barrier.

Abstraction by a bromine radical is endothermic, so the transition state comes late, with the C-H bond nearly broken and the radical almost fully formed. The tertiary radical's stability is fully felt in the barrier, so the tertiary route is enormously preferred.

Trap. "A more reactive reagent is a less selective one" is not a slogan. It is a statement about where the transition state sits, and it recurs throughout organic chemistry whenever a reaction has to choose between similar sites.

Combustion is complete in excess oxygen and gives carbon monoxide or soot when oxygen is limited. Pyrolysis or cracking breaks long chains into shorter alkanes and alkenes at high temperature.

Conformations of ethane

Rotation about a C-C sigma bond is nearly free but not entirely, because the hydrogens interact.

The staggered conformation of ethane, with the hydrogens as far apart as possible, is about 12.5 kJ mol${}^{-1}$ more stable than the eclipsed. The barrier is small enough that rotation is rapid at room temperature, so conformers cannot be separated.

Illustration 3

Butane's rotation profile has four distinct stationary points rather than ethane's two. Rank them and explain the ordering.

Rotating about the central C2-C3 bond swings two methyl groups past each other rather than two hydrogens, so the substituents differ in size and the profile becomes richer.

Read the pattern rather than the numbers. Staggered beats eclipsed by roughly 16 kJ mol${}^{-1}$ every time, which is the same torsional cost ethane pays. On top of that sits a steric penalty of about 3.8 kJ mol${}^{-1}$ whenever the two methyl groups are forced close, which is why gauche is above anti and fully eclipsed is above the other eclipsed forms.

Two separate effects, added. Conformational energy tracks how close the bulkiest groups are forced to come, and that principle governs every ring system met later.

3. Alkenes

Preparation

Electrophilic addition and Markovnikov's rule

The pi electrons attack the electrophile, generating a carbocation, which is then captured by the nucleophile.

Markovnikov's rule states that the hydrogen adds to the carbon already bearing more hydrogens. The rule is a description; the reason is that this route passes through the more stable carbocation.

Illustration 4

Propene with HBr gives 2-bromopropane, exactly as Markovnikov describes. Now try $\mathrm{C}{\mathrm{F}}_3-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ with HBr. The product is $\mathrm{C}{\mathrm{F}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$, which the rule as stated forbids.

Apply the rule literally. The terminal $\mathrm{C}{\mathrm{H}}_2$ has more hydrogens, so H should add there, leaving the positive charge on the carbon bearing $\mathrm{C}{\mathrm{F}}_3$.

That would be a secondary cation, which is normally preferred. But it sits directly next to a $\mathrm{C}{\mathrm{F}}_3$ group, three fluorines exerting one of the strongest $-I$ effects available. A group that drains electron density is the last thing a positive centre wants beside it.

The alternative is a primary cation, at the far carbon, one bond further from $\mathrm{C}{\mathrm{F}}_3$.

$\text{secondary but adjacent to }\mathrm{C}{\mathrm{F}}_3\text{loses to}\text{primary but further away}$

So H adds to the carbon with fewer hydrogens, the cation forms at the terminal carbon, and bromide lands there. The product is anti-Markovnikov by the rule and perfectly Markovnikov by the reason.

Trap. Whenever the rule and carbocation stability disagree, carbocation stability wins, because it is the actual mechanism and the rule is only a summary of it. Learn the reason and the rule comes free; learn the rule and this question is unanswerable.

The peroxide effect

In the presence of organic peroxides, HBr adds the other way round, giving the anti-Markovnikov product. This is the Kharasch effect.

Read the figure carefully, because the usual summary hides the point. The stability rule did not change. In both routes the intermediate forms at the secondary carbon, because a secondary cation and a secondary radical are both preferred over primary. What changed is which atom arrived first, so the bromine ends up at opposite ends.

The effect is observed only with HBr. With HCl the H-Cl bond is too strong for the propagation step to be favourable, and with HI the I-I bond is too weak to sustain the chain. This selectivity is a favourite exam point.

Other additions and tests

Halogens add across the double bond, and bromine water losing its orange colour is the standard test for unsaturation. Baeyer's reagent, cold dilute alkaline permanganate, is decolourised while forming a vicinal diol.

Ozonolysis cleaves the double bond entirely.

Illustration 5

Ozonolysis of 2-methylbut-2-ene is run twice, once with Zn and water and once with hydrogen peroxide. Give both sets of products, and explain why the workup matters.

The alkene is $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$. Cutting the double bond gives two fragments, one carrying two methyls and one carrying one.

Reductive workup, Zn and ${\mathrm{H}}_2\mathrm{O}$: propanone and ethanal. The zinc destroys the hydrogen peroxide that ozonolysis generates, so any aldehyde formed survives.

Oxidative workup, ${\mathrm{H}}_2{\mathrm{O}}_2$: propanone and ethanoic acid. There is nothing to protect the aldehyde, so it is oxidised further.

$(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{O}\text{unchanged},\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

Note which fragment changed. A ketone has no hydrogen on the carbonyl carbon and cannot be oxidised further, so propanone appears in both answers. Only the aldehyde fragment is at risk.

Ozonolysis is far more useful in reverse. Given the carbonyl products, joining their carbonyl carbons with a double bond reconstructs the original alkene, which is how the position of a double bond is determined. A question that specifies the workup is telling you whether an aldehyde fragment survived.

Polymerisation of ethene under high pressure gives polythene, addition polymerisation on an industrial scale.

4. Alkynes

Acidity, and what follows from it

The terminal hydrogen of an alkyne is on an $sp$ carbon, electronegative enough to make it weakly acidic. Internal alkynes have no such hydrogen and are not acidic.

Two tests follow. Ammoniacal silver nitrate gives a white precipitate of silver acetylide, and ammoniacal copper(I) chloride a red precipitate, with terminal alkynes only.

Illustration 6

Convert ethyne into pent-2-yne.

Ethyne's terminal acidity is not merely a test. It is a synthetic handle, because the anion it forms is a genuine carbon nucleophile.

$\mathrm{HC}\equiv \mathrm{CH}\stackrel{\mathrm{NaN}{\mathrm{H}}_2}{\to }\mathrm{HC}\equiv {\mathrm{C}}^{-}\mathrm{N}{\mathrm{a}}^{+}\stackrel{\mathrm{C}{\mathrm{H}}_3\mathrm{I}}{\to }\mathrm{HC}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_3$

That is propyne, and it still has a terminal hydrogen, so repeat with a different halide.

$\mathrm{HC}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{NaN}{\mathrm{H}}_2}{\to }{}^{-}\mathrm{C}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_3\stackrel{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{I}}{\to }{\mathrm{C}}_2{\mathrm{H}}_5-\mathrm{C}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_3$

Pent-2-yne, built one carbon fragment at a time from a two-carbon start.

Note what stops the sequence. Once both ends carry alkyl groups there is no acidic hydrogen left, so the chain cannot be extended further at that triple bond. Terminal acidity is what makes the reaction possible and its loss is what ends it, which is why the same property serves as a test and as a synthesis.

Preparation and addition

Ethyne is made industrially from calcium carbide and water, and in the laboratory by double dehydrohalogenation of a vicinal dihalide with alcoholic KOH followed by sodamide.

Addition follows Markovnikov's rule as for alkenes, in two stages, first to the alkene and then to the alkane.

Hydration is the special case worth learning properly. Water with dilute ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ and $\mathrm{HgS}{\mathrm{O}}_4$ gives an enol, which immediately tautomerises to the carbonyl compound. Ethyne therefore gives ethanal, and every other terminal alkyne gives a methyl ketone. The enol is never isolated, and forgetting the tautomerisation is the standard error.

Controlling the geometry of reduction

Illustration 7

From but-2-yne, make cis-but-2-ene and trans-but-2-ene separately. One starting material, two geometries, chosen by reagent.

Lindlar's catalyst, palladium on calcium carbonate poisoned with quinoline, gives the cis alkene. Both hydrogens are delivered from the metal surface, so they necessarily arrive on the same face, and the two methyl groups are left together. The poisoning stops the reduction at the alkene rather than continuing to the alkane.

Sodium in liquid ammonia gives the trans alkene. The mechanism is entirely different: sodium donates an electron to make a radical anion, which adopts the arrangement placing its two bulky groups as far apart as possible before protonation locks the geometry in.

The pattern is worth generalising. A reaction delivering two atoms at once fixes their relationship; a reaction delivering them one at a time lets the intermediate choose. That distinction reappears throughout organic chemistry, and naming the reagent for a required geometry is a standard question.

Illustration 8

But-1-yne and but-2-yne are isomers. Give a single test that separates them, and explain what the test is actually detecting.

Add ammoniacal silver nitrate. But-1-yne gives a white precipitate of the silver acetylide; but-2-yne gives nothing at all. Ammoniacal cuprous chloride does the same job with a red precipitate.

What the test detects is acidity, not the triple bond. Only but-1-yne has a hydrogen attached directly to an sp carbon:

$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}\text{against}\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_3$

An sp carbon is 50 per cent s in character, so its orbital lies closer to the nucleus and holds the resulting carbanion's lone pair tightly. That stabilisation is worth about 25 orders of magnitude in acidity against an alkane, which is enough for a metal ion to displace the hydrogen.

Both compounds contain a triple bond and only one responds, so the reagent is reporting on the C–H bond rather than on the C≡C bond. An internal alkyne is invisible to it, and so is any alkyne whose terminal hydrogen has already been substituted.

5. Aromatic Hydrocarbons

What makes a compound aromatic

Huckel's rule sets four conditions: cyclic, planar, fully conjugated, and $(4n+2)$ pi electrons with $n$ zero or a positive integer.

Benzene has six pi electrons, so $n=1$, and is aromatic. Cyclobutadiene has four, which fits $4n$, and is antiaromatic and highly unstable. Cyclooctatetraene has eight but is not planar, adopting a tub shape, so it is simply non-aromatic.

Illustration 9

Cyclopentadiene has $\mathrm{p}{K}_a$ 16, roughly the acidity of water and about $10^{34}$ times more acidic than propane. It is a plain hydrocarbon with no electronegative atom anywhere in it. Explain.

Look at the anion, as always with acidity.

Removing a proton from the $\mathrm{C}{\mathrm{H}}_2$ leaves a lone pair on that carbon. The ring already carries two double bonds, so it now has four pi electrons plus the new lone pair, making six, and the carbon becomes $s{p}^2$ so the whole ring is planar and fully conjugated.

Six pi electrons, cyclic, planar, conjugated: the cyclopentadienyl anion is aromatic.

The acid is an ordinary hydrocarbon; the base it produces is aromatic and unusually comfortable, so the proton leaves far more readily than any hydrocarbon proton has a right to.

The rule extends to ions in both directions.

The cycloheptatrienyl cation is the striking one: a carbocation stable enough to be bottled, because losing that hydride leaves six pi electrons in a planar seven-membered ring. Huckel's rule counts pi electrons, and it does not care whether the species is neutral.

The structure of benzene

All six C-C bonds are identical at 139 pm, between a single bond at 154 and a double at 134. The ring is planar with 120 degree angles and every carbon $s{p}^2$. The delocalisation is worth about 152 kJ mol${}^{-1}$, which is the number this chapter opened on.

Electrophilic substitution

Three steps: generate the electrophile, let the pi cloud attack it to form a resonance-stabilised arenium ion, and lose a proton to restore aromaticity.

Illustration 10

Propylbenzene cannot be made cleanly by Friedel-Crafts alkylation with 1-chloropropane. Give the actual product, then give a route that works.

The alkylation generates a primary propyl cation, which does not survive. A hydride shifts from the neighbouring carbon and it becomes the more stable secondary isopropyl cation before it ever reaches the ring.

$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2^{+}⟶\mathrm{C}{\mathrm{H}}_3\stackrel{+}{\mathrm{C}}\mathrm{HC}{\mathrm{H}}_3$

So the product is isopropylbenzene, not propylbenzene, and it is contaminated with polyalkylated material besides, because the alkyl group installed is activating and makes the product react faster than the benzene it came from.

Acylation fixes both defects at once.

${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COCl},\mathrm{AlC}{\mathrm{l}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Zn}-\mathrm{Hg},\mathrm{HCl}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$

The acylium ion $\mathrm{RC}{\mathrm{O}}^{+}$ is resonance-stabilised across carbon and oxygen, so it has no reason to rearrange. And the ketone it installs is deactivating, so the product is less reactive than the starting material and stops at one substitution.

Clemmensen reduction with zinc amalgam and HCl, or Wolff-Kishner with hydrazine and base, then converts the carbonyl to $\mathrm{C}{\mathrm{H}}_2$.

Trap. Friedel-Crafts fails altogether on two classes of ring. Nitrobenzene is too deactivated for the reaction to proceed at all. Aniline fails for a different reason: its nitrogen lone pair binds $\mathrm{AlC}{\mathrm{l}}_3$, creating a positively charged nitrogen that deactivates the ring severely. Being able to name the two different causes is worth more than knowing both fail.

Directive influence

Halogens are the exception and the reason is in GOC: a strong $-I$ effect withdraws density overall and slows the reaction, while lone pairs still direct by resonance to ortho and para. Two effects, two questions, two answers.

The reason in every case is the same: whichever positions give an arenium ion with the extra charge next to a stabilising group are the positions attacked.

A note on safety. Benzene and many polynuclear aromatic hydrocarbons such as benzpyrene are carcinogenic, which is why benzene has been largely replaced by toluene as a laboratory solvent.

Illustration 11

4-Nitrotoluene is brominated. The ring carries a methyl group, which is activating and ortho-para directing, and a nitro group, which is deactivating and meta directing. Predict where the bromine goes.

Number the ring with the methyl at C1, putting the nitro at C4.

The methyl sends incoming groups ortho and para to itself, meaning C2, C6 and C4. Para is already taken by the nitro group, leaving C2 and C6.

The nitro group sends incoming groups meta to itself, meaning C2 and C6.

The two directors agree, and bromination goes to C2, ortho to the methyl and meta to the nitro.

When the two disagree, the rule is that the activating group decides. It is not a convention but a consequence: an activating substituent raises the reactivity of its preferred positions by orders of magnitude, so nearly all the product forms there regardless of what the deactivating group would have preferred. The deactivating group still slows the whole reaction down; it simply does not get to choose where.

Summary

Benzene releases 208 kJ mol${}^{-1}$ on hydrogenation where three isolated double bonds would give 360, and the 152 kJ mol${}^{-1}$ shortfall is the resonance energy that makes it substitute rather than add.

Alkanes react only by radicals. Bromination is about 1600 times more selective per hydrogen than chlorination, because abstraction by bromine is endothermic and reaches a late transition state where radical stability is fully felt.

Conformational energy adds a torsional cost of about 16 kJ mol${}^{-1}$ for eclipsing to a steric cost of about 3.8 kJ mol${}^{-1}$ whenever bulky groups are forced close, which ranks butane's anti, gauche and two eclipsed forms.

Markovnikov's rule is a summary of carbocation stability, not a law. When they disagree, as with $\mathrm{C}{\mathrm{F}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, carbocation stability wins and the product looks anti-Markovnikov.

The peroxide effect inverts the product without changing the stability rule: a bromine radical adds first instead of a proton, and the intermediate still forms at the secondary carbon. It works only for HBr.

Ozonolysis with a reductive workup preserves aldehydes and with an oxidative workup carries them on to acids, while ketone fragments are unaffected either way. Run in reverse it locates a double bond.

Terminal alkyne acidity gives both the silver and copper tests and a route to longer chains through the acetylide anion, which stops working once both ends are substituted. Lindlar gives cis alkenes and sodium in liquid ammonia gives trans, because delivering two atoms at once fixes their relationship while delivering them one at a time lets the intermediate choose.

Huckel's rule counts pi electrons and ignores charge, so the cyclopentadienyl anion and the cycloheptatrienyl cation are both aromatic, which is why cyclopentadiene has $\mathrm{p}{K}_a$ 16.

Friedel-Crafts alkylation rearranges and polyalkylates; acylation followed by Clemmensen or Wolff-Kishner reduction avoids both. It fails on nitrobenzene for being too deactivated and on aniline because the amine binds the catalyst.

Activating groups direct ortho and para, deactivating groups direct meta, and halogens deactivate while still directing ortho and para, because rate and position are answered by different effects.