Friedel–Crafts alkylation fails with aniline because:

The lone pair on nitrogen complexes with AlCl₃ and consumes the catalyst

The ring is too deactivated by the amino group

The lone pair on nitrogen complexes with AlCl₃ and consumes the catalyst

Aniline is insoluble in the reaction medium

To prepare m-nitrochlorobenzene from benzene, the correct order is:

Either order gives the same product

Bromination of 2-methylpropane gives mainly the tertiary bromide because:

Bromine radicals are less reactive and therefore selective for the hydrogen giving the most stable radical

Bromine radicals are highly reactive and therefore unselective

Bromine radicals are less reactive and therefore selective for the hydrogen giving the most stable radical

There are more tertiary hydrogens than primary ones

Tertiary C–H bonds are the strongest

Hydrocarbons — Alkanes, Alkenes, Alkynes and Aromatics

The four families of hydrocarbons behave differently for one reason: what is available at the reactive site. An alkane has only strong sigma bonds and no electron-rich region, so it reacts only with radicals under harsh conditions. An alkene or alkyne has a loose pi cloud that attracts electrophiles. An arene has a pi cloud too, but a delocalised one it will not give up — so it substitutes rather than adds. Read every reaction in this chapter as a consequence of that one sentence and you will not need to memorise which family does which, because the reagent and the family together already determine the outcome.

Alkanes and Free-Radical Substitution

Alkanes are prepared by Wurtz coupling of alkyl halides with sodium (useful only for symmetrical alkanes, since a mixed pair gives three products), by Kolbe electrolysis of a carboxylate salt, by decarboxylation of a sodium salt with soda lime, and by hydrogenation of alkenes. Their only important reaction is halogenation, which proceeds by a radical chain: initiation by light or heat, then propagation, then termination.

🎯 Why chlorination gives a mixture but bromination is selective

The chlorine radical is highly reactive and therefore unselective — it abstracts almost any hydrogen it meets, so chlorinating propane gives both the 1- and 2-chloro products in comparable amounts. The bromine radical is much less reactive, so it waits for the easiest hydrogen, which is the one giving the most stable radical: tertiary before secondary before primary. Bromination of 2-methylpropane therefore gives overwhelmingly the tertiary bromide. The general principle is worth carrying beyond this chapter: a more reactive reagent is a less selective one, because it has too much energy to be fussy about which site it attacks.

Two structural points recur. Conformations of ethane interconvert freely by rotation about the C–C sigma bond, with the staggered form about 12 kJ/mol lower than the eclipsed — a barrier far too small to allow separation, which is why conformers are not isomers. And alkanes' boiling points rise with chain length but fall with branching, because a branched molecule is more spherical and has less surface available for van der Waals contact: n-pentane boils at 36 °C, neopentane at 9.5 °C.

Alkenes and Alkynes — Electrophilic Addition

The additions worth knowing, with their regiochemistry
Reagent	Product and rule
HBr	Markovnikov — H to the carbon with more hydrogens, so Br to the more substituted carbon
HBr with peroxide	Anti-Markovnikov (Kharasch) — radical mechanism reverses the outcome
H₂O with acid	Markovnikov alcohol, via a carbocation
B₂H₆ then H₂O₂/OH⁻	Anti-Markovnikov alcohol, and syn addition
Cold dilute KMnO₄	syn-diol; decolourises the reagent — Baeyer test for unsaturation
O₃ then Zn/H₂O	Cleaves the double bond to two carbonyls — locates the bond position

🎯 Markovnikov is not a rule to memorise — it is carbocation stability

When H⁺ adds to propene it can go to either carbon, giving either a primary or a secondary carbocation. The secondary one is more stable, so it forms overwhelmingly, and the bromide then attaches there — which is exactly what Markovnikov's rule describes. State it that way and two things follow for free. Peroxides reverse it because they switch the mechanism to radicals, and radical stability puts the bromine on the less substituted carbon first. And hydroboration also reverses it, but for a different reason: boron and hydrogen add together in one step with boron going to the less hindered carbon, so no carbocation is involved at all. Two anti-Markovnikov routes, two different explanations — which is why a question naming peroxide and one naming borane are not the same question.

Alkynes add electrophiles too but more slowly, because the pi electrons of a triple bond are held closer to the two carbons. Their distinctive chemistry is acidity: a terminal alkyne's hydrogen sits on an sp carbon with 50 % s character, so it is acidic enough to be removed by sodamide, giving an acetylide that reacts with alkyl halides to extend the chain. Terminal alkynes therefore give a precipitate with ammoniacal AgNO₃ or Cu₂Cl₂ while internal alkynes and alkenes do not — the standard way to distinguish them.

🧠 Ozonolysis works backwards to locate a double bond

Ozonolysis cleaves a double bond and puts a carbonyl oxygen on each of the two carbons that were joined. So if you are told the products, you can reconstruct the starting alkene by simply removing the two oxygens and joining the carbons that carried them. Products of acetone and acetaldehyde, for instance, must come from 2-methylbut-2-ene. This reverse reading is how almost every ozonolysis question in the exam is set, and it is far quicker than working forwards from candidate structures. One detail matters: reductive workup with Zn stops at the aldehyde, whereas an oxidative workup with H₂O₂ takes it on to the carboxylic acid.

Aromatics — Substitution, Not Addition

A ring is aromatic if it is planar, fully conjugated and holds (4n + 2) pi electrons — Hückel's rule. Benzene qualifies with six; cyclobutadiene with four is antiaromatic and unstable; cyclooctatetraene with eight escapes by puckering into a tub shape and behaving as a normal alkene. The resonance energy of about 150 kJ/mol is why benzene substitutes rather than adds: addition would permanently destroy the delocalisation, whereas substitution restores it.

The five electrophilic substitutions and their reagents
Reaction	Reagent	Electrophile
Nitration	conc. HNO₃ + conc. H₂SO₄	NO₂⁺
Sulphonation	conc. H₂SO₄	SO₃ — and the reaction is reversible
Halogenation	X₂ with AlCl₃ or FeCl₃	X⁺
Friedel–Crafts alkylation	RCl with AlCl₃	R⁺ — prone to rearrangement and polysubstitution
Friedel–Crafts acylation	RCOCl with AlCl₃	RCO⁺ — no rearrangement, cleaner

⚠️ Friedel–Crafts fails on nitrobenzene and on aniline, for opposite reasons

On nitrobenzene the ring is too electron-poor: the strongly withdrawing nitro group deactivates it so severely that the weak electrophile cannot attack, and this is why nitrobenzene is used as a solvent for Friedel–Crafts reactions on other substrates. On aniline the problem is the reverse — the lone pair on nitrogen is a Lewis base and complexes irreversibly with AlCl₃, so the catalyst is consumed and the amino group is converted into a deactivating anilinium ion. Two failures, opposite causes, and the standard workaround for aniline is to acetylate the nitrogen first, run the substitution, then hydrolyse the amide back. A question asking "why does this substrate fail" is asking which of the two situations applies.

Directing effects follow from where the substituent puts electron density. Ortho/para directors are electron-donating: −NH₂, −OH, −OR, −R, and also the halogens, which direct ortho/para while still deactivating. Meta directors are electron-withdrawing: −NO₂, −CN, −COOH, −CHO, −SO₃H. The practical consequence is that the order of steps in a synthesis matters: to make m-nitrochlorobenzene you must nitrate first and then chlorinate, since chlorine would send the nitro group ortho or para instead.

Key takeaways

Wurtz coupling is useful only for symmetrical alkanes, since a mixed pair gives three products.
Chlorine radicals are unselective, bromine radicals selective — a more reactive reagent is a less selective one.
Boiling point rises with chain length but falls with branching, because branched molecules touch less.
Markovnikov addition is just carbocation stability; peroxides and borane both reverse it, for different reasons.
Ozonolysis products read backwards give the original alkene; Zn workup stops at aldehyde, H₂O₂ goes to acid.
Terminal alkynes are acidic and precipitate with ammoniacal AgNO₃ — internal alkynes do not.
Aromaticity needs planarity, full conjugation and (4n + 2) pi electrons; that is why benzene substitutes, not adds.
Friedel–Crafts fails on nitrobenzene (too deactivated) and on aniline (the amine kills the catalyst).

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Addition of HBr to propene in the presence of peroxide gives mainly:
1. 2-bromopropane
2. 1-bromopropane
3. A 1 : 1 mixture of both
4. 1,2-dibromopropane
Show answer
Answer: B — 1-bromopropane
Peroxide switches the mechanism from ionic to free-radical, and the bromine radical adds to give the more stable secondary carbon radical, which places bromine on the terminal carbon — the anti-Markovnikov product. Without peroxide the reaction goes through the more stable secondary carbocation and gives 2-bromopropane instead, which is why the peroxide in the question is not decoration.
Ozonolysis of an alkene gives acetone and acetaldehyde. The alkene is:
1. But-2-ene
2. 2-methylbut-2-ene
3. 2-methylpropene
4. Pent-2-ene
Show answer
Answer: B — 2-methylbut-2-ene
Read the cleavage backwards: remove the carbonyl oxygen from each product and join the two carbons that carried them. Acetone contributes (CH₃)₂C and acetaldehyde contributes CHCH₃, which joined give (CH₃)₂C=CHCH₃, that is 2-methylbut-2-ene. But-2-ene would give two acetaldehydes and 2-methylpropene would give acetone plus formaldehyde, so neither fits.
Which compound gives a white precipitate with ammoniacal AgNO₃?
1. But-2-yne
2. But-1-yne
3. But-1-ene
4. Butane
Show answer
Answer: B — But-1-yne
Only a terminal alkyne has a hydrogen on an sp carbon, whose 50 % s character makes it acidic enough to be replaced by silver, giving the insoluble acetylide. But-2-yne has its triple bond internal with no such hydrogen, and the alkene and alkane have no acidic hydrogen at all. This test therefore distinguishes terminal from internal alkynes, which no addition reaction can do.
Friedel–Crafts alkylation fails with aniline because:
1. The ring is too deactivated by the amino group
2. The lone pair on nitrogen complexes with AlCl₃ and consumes the catalyst
3. Aniline is insoluble in the reaction medium
4. Aniline is not aromatic
Show answer
Answer: B — The lone pair on nitrogen complexes with AlCl₃ and consumes the catalyst
The amino group is strongly activating, so deactivation cannot be the problem — that is nitrobenzene's failure mode, not aniline's. Instead the nitrogen lone pair acts as a Lewis base towards AlCl₃, tying up the catalyst and converting the amine into a positively charged anilinium group that deactivates the ring. The workaround is to acetylate the nitrogen first, substitute, then hydrolyse the amide back.
Which of the following is aromatic by Hückel's rule?
1. Cyclobutadiene
2. Benzene
3. Cyclooctatetraene
4. Cyclohexane
Show answer
Answer: B — Benzene
Benzene is planar, fully conjugated and has six pi electrons, which fits (4n + 2) with n = 1. Cyclobutadiene has four pi electrons and is antiaromatic and unstable; cyclooctatetraene has eight, which also fails the rule, and it escapes antiaromaticity by puckering into a tub and behaving as an ordinary alkene. Cyclohexane has no pi system at all, so the rule does not apply to it.
To prepare m-nitrochlorobenzene from benzene, the correct order is:
1. Chlorinate first, then nitrate
2. Nitrate first, then chlorinate
3. Either order gives the same product
4. Sulphonate first, then nitrate
Show answer
Answer: B — Nitrate first, then chlorinate
The nitro group is a meta director, so once it is on the ring the incoming chlorine goes meta as required. Chlorinating first would fail, because chlorine is an ortho/para director and would send the nitro group to those positions instead. This is the standard demonstration that in aromatic synthesis the sequence of steps is part of the answer, not an incidental detail.
Bromination of 2-methylpropane gives mainly the tertiary bromide because:
1. Bromine radicals are highly reactive and therefore unselective
2. Bromine radicals are less reactive and therefore selective for the hydrogen giving the most stable radical
3. There are more tertiary hydrogens than primary ones
4. Tertiary C–H bonds are the strongest
Show answer
Answer: B — Bromine radicals are less reactive and therefore selective for the hydrogen giving the most stable radical
The bromine radical has a modest appetite, so it can afford to be choosy and abstracts preferentially the hydrogen whose removal gives the most stable radical — here the tertiary one. Note the arithmetic works against this: 2-methylpropane has nine primary hydrogens and only one tertiary, so selectivity has to be strong to overcome the statistics. Chlorine radicals, being far more reactive, are unselective and give a mixture.
Hydroboration–oxidation of propene gives:
1. Propan-2-ol
2. Propan-1-ol
3. Propanal
4. Acetone
Show answer
Answer: B — Propan-1-ol
Boron and hydrogen add across the double bond in a single concerted step, with the bulkier boron going to the less hindered terminal carbon; oxidation then replaces boron with OH, giving the anti-Markovnikov primary alcohol. Acid-catalysed hydration would instead go through a secondary carbocation and give propan-2-ol, so the two routes are complementary rather than equivalent.