Why a Carbocation Is Flat but a Carbanion Is a Pyramid
Distinguish homolytic and heterolytic bond fission and the free radicals, carbocations and carbanions they form, then learn how organic compounds are purified, tested for their elements and analysed quantitatively.
What happens to a covalent bond when an organic reaction begins?
Every organic reaction starts with a bond breaking, and the way it breaks decides which short-lived species form next — a radical, a positive ion or a negative ion. Before any of that chemistry can be studied, the compound itself must be purified and its elements identified.
This lesson covers bond fission and reactive intermediates, purification and elemental analysis, and the calculations that give the percentage of each element.
This lesson covers bond fission and reactive intermediates, purification and elemental analysis, and the calculations that give the percentage of each element.
What is the difference between homolytic and heterolytic fission, and what intermediates form?
In homolytic fission a covalent bond breaks evenly, each atom keeping one electron and forming free radicals, while in heterolytic fission one atom takes both electrons, forming a carbocation or a carbanion.
Homolytic fission:
- Favoured by heat, light or peroxides, and by non-polar bonds:
- Free radicals such as have an unpaired electron, are neutral and very reactive
- Stability: tertiary > secondary > primary > methyl
Heterolytic fission:
- Favoured by polar bonds and polar solvents:
- Carbocation — carbon with only six electrons and a positive charge, hybridised and planar; an electrophile
- Carbanion — carbon with eight electrons, including a lone pair, and a negative charge, hybridised and pyramidal; a nucleophile
Stability orders:
- Carbocations: tertiary > secondary > primary > methyl, because alkyl groups donate electrons by the +I effect and hyperconjugation
- Carbanions: methyl > primary > secondary > tertiary, because alkyl groups push more electrons onto an already negative carbon
Electrophiles and nucleophiles. Electrophiles such as , and carbocations seek electrons; nucleophiles such as , , and carbanions donate an electron pair.
An everyday example. Cooking oil left in a hot kitchen turns rancid partly through free radical reactions, as oxygen attacks the oil molecules in a chain of radical steps.
The substance. Carbanions follow the reverse stability order of carbocations — the same electron-donating alkyl groups that calm a positive charge make a negative charge worse.
Homolytic fission:
- Favoured by heat, light or peroxides, and by non-polar bonds:
- Free radicals such as have an unpaired electron, are neutral and very reactive
- Stability: tertiary > secondary > primary > methyl
Heterolytic fission:
- Favoured by polar bonds and polar solvents:
- Carbocation — carbon with only six electrons and a positive charge, hybridised and planar; an electrophile
- Carbanion — carbon with eight electrons, including a lone pair, and a negative charge, hybridised and pyramidal; a nucleophile
Stability orders:
- Carbocations: tertiary > secondary > primary > methyl, because alkyl groups donate electrons by the +I effect and hyperconjugation
- Carbanions: methyl > primary > secondary > tertiary, because alkyl groups push more electrons onto an already negative carbon
Electrophiles and nucleophiles. Electrophiles such as , and carbocations seek electrons; nucleophiles such as , , and carbanions donate an electron pair.
An everyday example. Cooking oil left in a hot kitchen turns rancid partly through free radical reactions, as oxygen attacks the oil molecules in a chain of radical steps.
The substance. Carbanions follow the reverse stability order of carbocations — the same electron-donating alkyl groups that calm a positive charge make a negative charge worse.
How are organic compounds purified and tested for nitrogen, sulphur and halogens?
Organic compounds are purified by crystallisation, sublimation, distillation or chromatography, chosen according to their physical properties, and their elements are detected by converting them into simple ions, as in Lassaigne's test.
Purification methods:
- Crystallisation — dissolving in a hot solvent and cooling, as for impure benzoic acid in hot water
- Simple distillation — for liquids with widely different boiling points, such as chloroform and aniline
- Fractional distillation — for liquids with close boiling points, using a fractionating column
- Steam distillation — for steam-volatile liquids that do not mix with water, such as aniline
- Chromatography — separating components by their different adsorption or partition between a stationary and a mobile phase; in thin-layer chromatography each spot has a retention factor,
Detecting elements:
- Carbon and hydrogen — heating with copper(II) oxide gives carbon dioxide, which turns lime water milky, and water, which turns anhydrous copper sulphate blue
- Lassaigne's test — fusing with sodium converts nitrogen, sulphur and halogens into NaCN, and NaX
- Nitrogen — the extract boiled with iron(II) sulphate and acidified gives Prussian blue
- Sulphur — sodium nitroprusside gives a violet colour, or lead acetate a black precipitate
- Halogens — after boiling with nitric acid, silver nitrate gives white AgCl, pale yellow AgBr or yellow AgI
An everyday example. The refinery complex at Jamnagar separates crude oil into petrol, kerosene and diesel by fractional distillation, the same principle used in a school laboratory.
The substance. Steam distillation lets a liquid distil below its own boiling point — the liquid and water together reach atmospheric pressure at a lower temperature than either would alone.
Purification methods:
- Crystallisation — dissolving in a hot solvent and cooling, as for impure benzoic acid in hot water
- Simple distillation — for liquids with widely different boiling points, such as chloroform and aniline
- Fractional distillation — for liquids with close boiling points, using a fractionating column
- Steam distillation — for steam-volatile liquids that do not mix with water, such as aniline
- Chromatography — separating components by their different adsorption or partition between a stationary and a mobile phase; in thin-layer chromatography each spot has a retention factor,
Detecting elements:
- Carbon and hydrogen — heating with copper(II) oxide gives carbon dioxide, which turns lime water milky, and water, which turns anhydrous copper sulphate blue
- Lassaigne's test — fusing with sodium converts nitrogen, sulphur and halogens into NaCN, and NaX
- Nitrogen — the extract boiled with iron(II) sulphate and acidified gives Prussian blue
- Sulphur — sodium nitroprusside gives a violet colour, or lead acetate a black precipitate
- Halogens — after boiling with nitric acid, silver nitrate gives white AgCl, pale yellow AgBr or yellow AgI
An everyday example. The refinery complex at Jamnagar separates crude oil into petrol, kerosene and diesel by fractional distillation, the same principle used in a school laboratory.
The substance. Steam distillation lets a liquid distil below its own boiling point — the liquid and water together reach atmospheric pressure at a lower temperature than either would alone.
Formula
How do you calculate the percentage of carbon, nitrogen and halogen in an organic compound?
Quantitative analysis turns each element into a product that can be weighed or titrated: carbon and hydrogen are weighed as carbon dioxide and water, nitrogen is found by the Kjeldahl method, and halogens by the Carius method.
Here m is the mass of compound in grams, N the normality of the acid and V the volume of acid in mL neutralised by the ammonia; every result is a percentage by mass.
Worked example 1 — carbon and hydrogen. 0.20 g of a compound gives 0.44 g of carbon dioxide and 0.18 g of water:
So the compound is 60 per cent carbon and 10 per cent hydrogen.
Worked example 2 — Kjeldahl. The ammonia from 0.30 g of urea neutralises 20 mL of 0.50 N sulphuric acid:
That matches urea, , in which nitrogen makes up 28 of 60 mass units.
Worked example 3 — Carius. 0.25 g of a compound gives 0.574 g of silver chloride:
An everyday example. Food-testing laboratories estimate the protein in milk powder and pulses by measuring their nitrogen with the Kjeldahl method.
The substance. The Kjeldahl method fails for nitro and azo compounds and for nitrogen in a ring, as in pyridine — that nitrogen is not converted into ammonium sulphate, so the Dumas method is used instead.
Here m is the mass of compound in grams, N the normality of the acid and V the volume of acid in mL neutralised by the ammonia; every result is a percentage by mass.
Worked example 1 — carbon and hydrogen. 0.20 g of a compound gives 0.44 g of carbon dioxide and 0.18 g of water:
So the compound is 60 per cent carbon and 10 per cent hydrogen.
Worked example 2 — Kjeldahl. The ammonia from 0.30 g of urea neutralises 20 mL of 0.50 N sulphuric acid:
That matches urea, , in which nitrogen makes up 28 of 60 mass units.
Worked example 3 — Carius. 0.25 g of a compound gives 0.574 g of silver chloride:
An everyday example. Food-testing laboratories estimate the protein in milk powder and pulses by measuring their nitrogen with the Kjeldahl method.
The substance. The Kjeldahl method fails for nitro and azo compounds and for nitrogen in a ring, as in pyridine — that nitrogen is not converted into ammonium sulphate, so the Dumas method is used instead.
Exam tip
What earns full marks on reaction intermediates and organic analysis?
For each reactive intermediate, state its hybridisation, its shape and the number of electrons around carbon — a three-part answer that covers every mark.
- Free radical: 7 electrons, neutral; carbocation: 6 electrons, , planar; carbanion: 8 electrons, , pyramidal
- Carbocation stability rises with alkyl groups; carbanion stability falls
- Lassaigne's test: Prussian blue for N, violet for S, silver halide precipitates for halogens
- Kjeldahl:
The trap. Adding silver nitrate straight to the sodium fusion extract. Boil it with nitric acid first — otherwise cyanide and sulphide ions also form precipitates and mimic a halogen.
- Free radical: 7 electrons, neutral; carbocation: 6 electrons, , planar; carbanion: 8 electrons, , pyramidal
- Carbocation stability rises with alkyl groups; carbanion stability falls
- Lassaigne's test: Prussian blue for N, violet for S, silver halide precipitates for halogens
- Kjeldahl:
The trap. Adding silver nitrate straight to the sodium fusion extract. Boil it with nitric acid first — otherwise cyanide and sulphide ions also form precipitates and mimic a halogen.
Did you know
Why does black sketch-pen ink split into colours on wet filter paper?
Put a dot of black water-based ink near the bottom of a strip of filter paper and dip the edge in water. As water climbs the paper, the black dot stretches into separate coloured bands.
Black ink is often a mixture of several dyes. Each dye divides differently between the water moving up the paper and the water held in its fibres, so each travels a different distance.
That is paper chromatography, done with nothing more than a sketch pen and a glass of water.
Black ink is often a mixture of several dyes. Each dye divides differently between the water moving up the paper and the water held in its fibres, so each travels a different distance.
That is paper chromatography, done with nothing more than a sketch pen and a glass of water.
Exam relevance
How do JEE Main and NEET test reactive intermediates, purification and organic analysis?
Organic Chemistry: Some Basic Principles and Techniques is a recurring chapter in both JEE Main and NEET, and reactive intermediates reappear in every reaction mechanism that follows.
What gets asked. Stability orders of carbocations, carbanions and free radicals, hybridisation and shape of intermediates, matching purification methods to mixtures, colours in Lassaigne's test, and percentage composition from the Kjeldahl and Carius methods.
Question types. Arrange-in-order, match-the-column and single-correct questions, with numerical-value questions on percentage composition in JEE Main.
Why it matters later. Carbocations drive the mechanisms in Hydrocarbons and Haloalkanes and Haloarenes, and free radicals explain the halogenation of alkanes in the next chapter.
The trap that costs marks. Applying the Kjeldahl method to a nitro compound or pyridine — that nitrogen is not converted into ammonia.
What gets asked. Stability orders of carbocations, carbanions and free radicals, hybridisation and shape of intermediates, matching purification methods to mixtures, colours in Lassaigne's test, and percentage composition from the Kjeldahl and Carius methods.
Question types. Arrange-in-order, match-the-column and single-correct questions, with numerical-value questions on percentage composition in JEE Main.
Why it matters later. Carbocations drive the mechanisms in Hydrocarbons and Haloalkanes and Haloarenes, and free radicals explain the halogenation of alkanes in the next chapter.
The trap that costs marks. Applying the Kjeldahl method to a nitro compound or pyridine — that nitrogen is not converted into ammonia.
Key takeaways
What must you be able to do from this lesson?
- Bond fission: homolytic fission gives free radicals; heterolytic fission gives carbocations and carbanions, with opposite stability orders
- Purification and detection: crystallisation, distillation and chromatography, and Lassaigne's test for nitrogen, sulphur and halogens
- Quantitative analysis: carbon and hydrogen from combustion, nitrogen by Kjeldahl, halogens by Carius
In a Carius determination, 0.30 g of a compound gave 0.47 g of silver bromide — what percentage of the compound is bromine?
- Purification and detection: crystallisation, distillation and chromatography, and Lassaigne's test for nitrogen, sulphur and halogens
- Quantitative analysis: carbon and hydrogen from combustion, nitrogen by Kjeldahl, halogens by Carius
In a Carius determination, 0.30 g of a compound gave 0.47 g of silver bromide — what percentage of the compound is bromine?