Lesson 7 of 34

Look for accessible electron-rich and electron-poor sites

Bond polarity, delocalisation and access explain why similar-looking compounds react differently.

A-Level 9476 (2026-2027)

I can explain alkane unreactivity.Syllabus
Syllabus 9476, 11.3(c). Explain alkane unreactivity.
I can explain alkene reactivity towards electrophiles.Syllabus
Syllabus 9476, 11.3(d). Explain alkene reactivity towards electrophiles.
I can compare benzene and alkene reactivity through delocalisation.Syllabus
Syllabus 9476, 11.3(e). Compare benzene and alkene reactivity through delocalisation.
I can interpret halogenoalkane reactivity, especially hydrolysis.Syllabus
Syllabus 9476, 11.3(f). Interpret halogenoalkane reactivity, especially hydrolysis.
I can explain chlorobenzene resistance to nucleophilic substitution.Syllabus
Syllabus 9476, 11.3(g). Explain chlorobenzene resistance to nucleophilic substitution.
I can explain carbonyl reactivity towards nucleophiles.Syllabus
Syllabus 9476, 11.3(h). Explain carbonyl reactivity towards nucleophiles.
I can apply reaction and reactivity concepts to mechanisms.Syllabus
Syllabus 9476, 11.3(i). Apply reaction and reactivity concepts to mechanisms.
I can track electron flow in polar mechanisms.Syllabus
Syllabus 9476, 11.3(j). Track electron flow in polar mechanisms.

Look for accessible electron-rich and electron-poor sites

Bond polarity, delocalisation and access explain why similar-looking compounds react differently.

Make a guess. You are not marked.

1-chlorobutane and 1-iodobutane are each warmed with aqueous silver nitrate in ethanol. Which gives a precipitate faster?

1-chlorobutane, because chloride is a smaller leaving group
1-chlorobutane, because the C-Cl bond is more polar
1-iodobutane, because C-I is the weakest bond

Show the answer

1-iodobutane, because C-I is the weakest bond

The C-I bond is the weakest carbon-halogen bond, so it breaks most easily and releases iodide fastest.

Structure → reactive site → consequence
Family	Electronic structure	Characteristic response
Alkane	Strong, nearly non-polar C-C and C-H sigma bonds; no readily available pi cloud or strongly polar centre.	Generally unreactive towards polar reagents under mild conditions; radical initiation or combustion conditions can open other pathways.
Alkene	Accessible, localised pi electron density above and below the carbon framework.	Donates electron density to electrophiles in addition reactions.
Benzene	Six pi electrons delocalised over the entire aromatic ring.	More resistant to addition than an alkene; substitution can restore the stabilised aromatic system after temporary disruption.
Halogenoalkane	Polar C-X bond with partially positive carbon.	A nucleophile can replace X; bond strength and steric/carbocation factors affect the pathway.
Carbonyl compound	C=O is polar, with partially positive carbon and partially negative oxygen.	Nucleophile attacks the carbon while pi electrons move onto oxygen.

For similar halogenoalkanes, hydrolysis gets easier from RCl to RBr to RI. This is because the C-Cl bond is the strongest and the C-I bond is the weakest. Bond polarity alone does not explain the trend. Before you compare rates, make sure the carbon chain, solvent and mechanism are the same.

In chlorobenzene, a lone pair on chlorine overlaps with the π electrons of the ring. This gives the C-Cl bond some double-bond character, so it is harder to break than in a halogenoalkane. The flat ring also blocks a nucleophile from attacking the back of that carbon. Losing chloride first would make a very unstable phenyl cation. So chlorobenzene does not undergo nucleophilic substitution under the conditions that hydrolyse halogenoalkanes.

Benzene has electrons that can attack an electrophile. But breaking up its delocalised ring costs a lot of stabilisation energy. So benzene needs a stronger electrophile, often made with a catalyst, than an alkene does. Afterwards, the intermediate loses H⁺ and the delocalised ring is restored. That is why benzene undergoes substitution, not addition.

Exam-style question

Ethene decolourises bromine rapidly at room temperature in the dark, but ethane does not react with bromine under these conditions. Explain this difference. [3]

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Check your understanding

Cyclohexene decolourises bromine at room temperature. Benzene reacts only when AlBr₃ is added, and forms C₆H₅Br, not C₆H₆Br₂. Which idea explains both facts?

Benzene's six pi electrons are delocalised, so it cannot polarise Br₂ alone, and substitution keeps the stable ring.
AlBr₃ removes hydrogen atoms from benzene first, so that bromine can then bond to the free carbons.
Benzene has fewer pi electrons than cyclohexene, so it attracts bromine molecules more weakly.

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Benzene's six pi electrons are delocalised, so it cannot polarise Br₂ alone, and substitution keeps the stable ring.

Spread-out pi density makes benzene a weaker electron donor than a localised C=C, so a Lewis acid is needed to make Br⁺. Losing H⁺ from the intermediate restores the delocalised ring, which is more stable than any addition product.

Check your understanding

1-chlorobutane and chlorobenzene are each warmed with NaOH(aq), acidified with dilute HNO₃, and AgNO₃(aq) is added. What is seen?

A white precipitate with 1-chlorobutane only, because chlorobenzene's C-Cl bond has partial double-bond character.
A white precipitate with chlorobenzene only, because its ring pulls electrons away from the chlorine.
A white precipitate with both, because each compound contains a chlorine atom bonded to a carbon.

Show the answer

A white precipitate with 1-chlorobutane only, because chlorobenzene's C-Cl bond has partial double-bond character.

OH^- substitutes Cl in 1-chlorobutane and releases Cl^-, which gives AgCl. In chlorobenzene a Cl lone pair is delocalised into the ring, strengthening C-Cl, and the electron-rich ring repels the nucleophile, so no Cl^- is released.