Organic Reaction Types and Mechanisms

AS · 15 min

There are hundreds of organic reactions on the AS syllabus, but they fall into a handful of types, and a reaction's type tells you most of what you need to know about it. This note gives you the vocabulary every organic question uses: how bonds break (homolytic and heterolytic fission), the species that attack molecules (free radicals, nucleophiles and electrophiles), the names of reaction types (addition, substitution, elimination, hydrolysis, condensation, oxidation, reduction), the curly-arrow language of mechanisms, and the inductive effect of alkyl groups. The terms are examined word for word in definitions and are used in almost every mechanism and synthesis question.

How covalent bonds break

A covalent bond is a shared pair of electrons. When it breaks, those two electrons have to go somewhere, and there are only two possibilities.

Definition

Homolytic fission is the breaking of a covalent bond in which each atom takes one electron from the bonding pair, forming two free radicals.

Heterolytic fission is the breaking of a covalent bond in which one atom takes both electrons from the bonding pair, forming a positive ion (cation) and a negative ion (anion).

A free radical is a species with one or more unpaired electrons.

Top: homolytic fission of chlorine, shown with half-headed (fish-hook) arrows, each representing one electron. Bottom: heterolytic fission of a polar C–Br bond; the full curly arrow represents the movement of both bonding electrons to bromine.

Homolytic fission happens most easily when the two atoms are identical or have similar electronegativities, so neither has a stronger claim on the electrons, and when energy is supplied by ultraviolet light or high temperature. The classic example is chlorine in sunlight:

\ce{Cl2 ->[UV] 2Cl^.}

The dot represents the unpaired electron. A radical is very reactive because it "wants" to pair that electron by forming a new bond.

Heterolytic fission happens with polar bonds. In C–Br, bromine is more electronegative than carbon, so the bonding pair is already pulled towards bromine ( $\ce{C^{\delta+}-Br^{\delta-}}$ ); when the bond breaks, bromine takes both electrons and becomes $\ce{Br-}$ . Heterolytic fission happens in solution, where the ions formed can be stabilised by the solvent.

Free-radical reactions

Free-radical reactions are chain reactions: one radical produces a product and a new radical, which carries the chain on. They have three stages.

Definition

Initiation: a step in which free radicals are formed, usually by homolytic fission using UV light. Radicals are produced and none are used up.
Propagation: steps in which a radical reacts with a molecule to form a product and a new radical, so the number of radicals stays the same and the chain continues.
Termination: a step in which two radicals combine to form a molecule, removing radicals and ending a chain.

You will use these terms for the free-radical substitution of alkanes in Alkanes.

Nucleophiles and electrophiles

Most AS mechanisms are ionic: they involve heterolytic fission, and an electron-rich species attacks an electron-poor site.

Definition

A nucleophile is a species that donates a pair of electrons to form a new covalent bond (an electron-pair donor). The word means "nucleus-loving": nucleophiles are attracted to positive or $\delta+$ centres.

An electrophile is a species that accepts a pair of electrons to form a new covalent bond (an electron-pair acceptor). The word means "electron-loving": electrophiles are attracted to electron-rich regions such as a C=C bond.

Every nucleophile has a lone pair of electrons, and many carry a negative charge. Every electrophile is positive, $\delta+$ , or can be made $\delta+$ by polarisation.

nucleophiles (lone pair donors)	electrophiles (electron pair acceptors)
$\ce{OH-}$ hydroxide ion	$\ce{H+}$ (and the $\delta+$ H in $\ce{H-Br}$ )
$\ce{CN-}$ cyanide ion	$\ce{Br2}$ when polarised by a C=C bond ( $\ce{Br^{\delta+}-Br^{\delta-}}$ )
$\ce{NH3}$ ammonia	carbocations such as $\ce{CH3CH2+}$
$\ce{H2O}$ water	the $\delta+$ carbon of a C=O group (attacked by nucleophiles)
$\ce{H-}$ hydride, from $\ce{NaBH4}$ or $\ce{LiAlH4}$

Notice the pattern. Nucleophiles attack $\delta+$ carbon atoms (in C–X bonds and C=O groups). Electrophiles attack C=C double bonds, which are regions of high electron density because of the $\pi$ electrons.

Types of reaction

Definition

Addition: two molecules react together to form a single product. Typical of unsaturated compounds (C=C, C=O).
Substitution: an atom or group in a molecule is replaced by a different atom or group.
Elimination: a small molecule (such as $\ce{H2O}$ or HBr) is removed from a larger molecule, leaving an unsaturated product (usually a C=C).
Hydrolysis: a reaction in which a bond is broken by reaction with water; it is often catalysed by acid or alkali.
Condensation: two molecules join together with the loss of a small molecule, such as water.
Oxidation (in organic chemistry): the addition of oxygen to, or the removal of hydrogen from, a molecule.
Reduction (in organic chemistry): the addition of hydrogen to, or the removal of oxygen from, a molecule.

A full description of a reaction combines the type of attacking species with the type of reaction:

Key result

mechanism name	attacking species	typical substrate	example
free-radical substitution	free radical, e.g. $\ce{Cl^.}$	alkane	$\ce{CH4 + Cl2 ->[UV] CH3Cl + HCl}$
electrophilic addition	electrophile, e.g. $\ce{Br2}$ , HBr	alkene	$\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}$
nucleophilic substitution	nucleophile, e.g. $\ce{OH-}$ , $\ce{CN-}$ , $\ce{NH3}$	halogenoalkane	$\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}$
nucleophilic addition	nucleophile, e.g. $\ce{CN-}$ , $\ce{H-}$	aldehyde or ketone	$\ce{CH3CHO + HCN -> CH3CH(OH)CN}$
elimination	base, e.g. $\ce{OH-}$ in ethanol	halogenoalkane, alcohol	$\ce{CH3CH2Br + OH- -> CH2=CH2 + H2O + Br-}$

Some reactions fit more than one label. The reaction of an ester with water, $\ce{CH3COOCH3 + H2O <=> CH3COOH + CH3OH}$ , is a hydrolysis. Its reverse, making an ester from an acid and an alcohol, is a condensation (water is lost as the two molecules join). The reaction of bromoethane with aqueous hydroxide is both a nucleophilic substitution and a hydrolysis.

Writing oxidation and reduction with [O] and [H]

Organic oxidising and reducing agents (acidified potassium dichromate(VI), sodium tetrahydridoborate) give complicated full equations. At AS you write simplified equations in which $\ce{[O]}$ stands for an oxygen atom supplied by the oxidising agent and $\ce{[H]}$ stands for a hydrogen atom supplied by the reducing agent. The equation must still balance in atoms.

\ce{CH3CH2OH + [O] -> CH3CHO + H2O}

\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}

\ce{CH3COCH3 + 2[H] -> CH3CH(OH)CH3}

\ce{CH3COOH + 4[H] -> CH3CH2OH + H2O}

Method

Balancing an [O] or [H] equation

Write the organic reactant and product.
Count the hydrogen atoms lost (oxidation) or the oxygen atoms lost (reduction). Hydrogen removed in oxidation leaves as $\ce{H2O}$ ; oxygen removed in reduction leaves as $\ce{H2O}$ .
Add enough $\ce{[O]}$ or $\ce{[H]}$ to balance O or H, including the water.
Check: every element balances on both sides.

For example, propan-1-ol to propanoic acid: $\ce{C3H8O -> C3H6O2}$ . Two H must leave (as one $\ce{H2O}$ ) and one O is gained by the organic molecule, so two oxygen atoms are needed in total: $\ce{CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O}$ .

Curly arrows: the language of mechanisms

A reaction mechanism shows, step by step, how bonds break and form. The movement of electrons is shown by curly arrows.

Key result

A full curly arrow shows the movement of a pair of electrons.
A half-headed (fish-hook) arrow shows the movement of a single electron, used in homolytic fission.
An arrow must start at the electrons that move: a lone pair (drawn as two dots) or a bond (start on the middle of the bond line).
An arrow must end where the electrons go: on an atom (forming a lone pair or ion), or between two atoms (forming a new bond).
Show partial charges ( $\delta+$ , $\delta-$ ) on polar bonds, full charges on ions, and the lone pair on any nucleophile.

When you describe a mechanism in words, as this course does in numbered steps, say exactly where each arrow starts and ends. "A curly arrow from the lone pair on the oxygen of $\ce{OH-}$ to the $\delta+$ carbon" earns the mark; "the $\ce{OH-}$ attacks" does not.

The inductive effect of alkyl groups

Alkyl groups such as $\ce{CH3}$ and $\ce{C2H5}$ are electron-donating: they push electron density through the $\sigma$ bonds towards a neighbouring atom. This is called a positive inductive effect.

The effect matters most for carbocations, ions in which a carbon atom carries a positive charge and has only three bonds. A carbocation is classified like an alcohol, by the number of carbons bonded to the positive carbon:

carbocation	example	alkyl groups on $\ce{C+}$	stability
primary	$\ce{CH3CH2+}$	1	least stable
secondary	$\ce{(CH3)2CH+}$	2	more stable
tertiary	$\ce{(CH3)3C+}$	3	most stable

Key result

Carbocation stability: tertiary > secondary > primary. Each alkyl group donates electron density towards the positive carbon, which spreads out (reduces) the positive charge and stabilises the ion. More alkyl groups, more stabilisation.

This one idea explains two major results at AS: Markovnikov addition to alkenes (the more stable carbocation forms; see Alkenes) and why tertiary halogenoalkanes react by the SN1 mechanism (see Nucleophilic substitution and elimination).

Worked examples

Routine: naming the reaction type

Classify each reaction as addition, substitution, elimination, hydrolysis, condensation, oxidation or reduction, and where possible name the mechanism.

(a) $\ce{C2H6 + Br2 -> C2H5Br + HBr}$ (in UV light) (b) $\ce{CH3CH=CH2 + HBr -> CH3CHBrCH3}$ (c) $\ce{CH3CH2CH2OH -> CH3CH=CH2 + H2O}$ (with heated $\ce{Al2O3}$ ) (d) $\ce{CH3COOH + CH3CH2OH -> CH3COOCH2CH3 + H2O}$

Solution

(a) Substitution (H replaced by Br) by free-radical substitution.

(b) Addition across the C=C: electrophilic addition.

(d) Condensation (two molecules join, water lost); also called esterification.

Routine: nucleophile or electrophile

State whether each species acts as a nucleophile, an electrophile or a free radical, giving a reason: (a) $\ce{NH3}$ (b) $\ce{H+}$ (c) $\ce{CN-}$ (d) $\ce{CH3^.}$ (e) the bromine molecule approaching an alkene.

Solution

(a) Nucleophile: the nitrogen has a lone pair it can donate.

(b) Electrophile: it has no electrons and accepts a pair.

(d) Free radical: one unpaired electron.

(e) Electrophile: the electron-rich C=C repels the electrons in Br–Br, polarising it to $\ce{Br^{\delta+}-Br^{\delta-}}$ ; the $\delta+$ bromine accepts a pair of electrons from the C=C.

Standard: writing [O] and [H] equations

Write balanced equations, using $\ce{[O]}$ or $\ce{[H]}$ , for:

(a) the oxidation of butan-2-ol to butanone; (b) the oxidation of propanal to propanoic acid; (c) the reduction of ethanal to ethanol; (d) the oxidation of butane-1,4-diol, $\ce{HOCH2CH2CH2CH2OH}$ , to butanedioic acid, $\ce{HOOCCH2CH2COOH}$ .

Solution

(a) $\ce{CH3CH(OH)CH2CH3 + [O] -> CH3COCH2CH3 + H2O}$ . Two H removed (as water); one [O] supplies the O in the water.

(b) $\ce{CH3CH2CHO + [O] -> CH3CH2COOH}$ . One O gained, no H lost.

(d) Each $\ce{-CH2OH}$ becomes $\ce{-COOH}$ , needing $2\ce{[O]}$ each:

\ce{HOCH2CH2CH2CH2OH + 4[O] -> HOOCCH2CH2COOH + 2H2O}

Check: left C4 H10 O6 (2 + 4); right C4 H(6 + 4) = 10, O(4 + 2) = 6. Balanced.

Standard: explaining fission

Hydrogen bromide can break up in two ways. (a) Write equations for the homolytic and heterolytic fission of HBr. (b) Explain which is more likely when HBr reacts with an alkene in a non-polar solvent at room temperature, in the dark.

Solution

(a) Homolytic: $\ce{H-Br -> H^. + Br^.}$ . Heterolytic: $\ce{H-Br -> H+ + Br-}$ (bromine, being more electronegative, takes both electrons).

(b) Heterolytic. The H–Br bond is polar ( $\ce{H^{\delta+}-Br^{\delta-}}$ ), and there is no UV light to supply the energy for homolytic fission. The electron-rich C=C attacks the $\delta+$ hydrogen, and the H–Br bond breaks heterolytically, with both electrons going to bromine to form $\ce{Br-}$ . This is the first step of electrophilic addition.

Exam-hard: carbocation stability and product

When HBr adds to 2-methylpropene, $\ce{CH2=C(CH3)2}$ , two different carbocations could form in the first step.

(a) Give the structures of both carbocations and classify each. (b) Explain, in terms of the inductive effect, which is more stable. (c) Hence name the major product.

Solution

(a) If $\ce{H+}$ adds to carbon 1 (the $\ce{CH2}$ end), the positive charge is on carbon 2: $\ce{(CH3)3C+}$ , a tertiary carbocation. If $\ce{H+}$ adds to carbon 2, the positive charge is on carbon 1: $\ce{(CH3)2CHCH2+}$ , a primary carbocation.

(b) Alkyl groups are electron-donating (positive inductive effect). The tertiary carbocation has three alkyl groups pushing electron density towards the positive carbon, which spreads the positive charge and stabilises it. The primary carbocation has only one alkyl group, so its charge is much less stabilised. The tertiary carbocation is more stable and forms faster.

Watch out

Nucleophile vs electrophile reversed. A nucleophile donates a lone pair; an electrophile accepts one. A useful check: nucleophiles have lone pairs; electrophiles are positive or $\delta+$ .
Curly arrows starting on an atom or a charge. An arrow from $\ce{OH-}$ must start on a lone pair on the oxygen, not on the minus sign. An arrow breaking a bond must start on the bond.
Using "free radical" for ions. A carbocation has no unpaired electron: it is an ion, not a radical.
[O] equations that don't balance. Count H and O atoms on both sides every time; oxidising an alcohol to a carboxylic acid needs $2\ce{[O]}$ and gives one $\ce{H2O}$ .
Calling every reaction with water "hydrolysis". Adding steam to an alkene is an addition (hydration), not a hydrolysis: no bond in the organic molecule is split.

Exam tip

Definitions of homolytic fission, heterolytic fission, free radical, nucleophile and electrophile are frequent 1-mark questions. Learn the key phrase in each: "each atom takes one electron", "one atom takes both electrons", "unpaired electron", "electron-pair donor", "electron-pair acceptor".
"Name the type of reaction" usually wants a two-word answer such as "nucleophilic substitution" or "electrophilic addition". "Substitution" alone is often not enough.
In mechanism questions, marks go for: correct dipoles, lone pairs on nucleophiles, curly arrows starting and ending in the right places, the intermediate with its charge, and the products.
In [O] and [H] equations, structural formulae are expected for the organic compounds.

Summary

Homolytic fission: each atom takes one electron, giving two free radicals (UV or high temperature; non-polar bonds).
Heterolytic fission: one atom takes both electrons, giving a cation and an anion (polar bonds).
Free radical: a species with an unpaired electron. Radical chain reactions have initiation, propagation and termination steps.
Nucleophile: electron-pair donor (has a lone pair), attacks $\delta+$ carbon. Electrophile: electron-pair acceptor, attacks C=C.
Reaction types: addition, substitution, elimination, hydrolysis, condensation, oxidation (+O or −H), reduction (+H or −O).
Use $\ce{[O]}$ and $\ce{[H]}$ for organic oxidation and reduction, and balance every atom.
Full curly arrow = two electrons; half arrow = one electron. Arrows start at a lone pair or bond and end at an atom or between atoms.
Alkyl groups are electron-donating; carbocation stability is tertiary > secondary > primary.

Practice

Question

Define (a) heterolytic fission, (b) free radical, (c) electrophile.
Classify each species as a nucleophile, electrophile or free radical: $\ce{Br^.}$ , $\ce{H2O}$ , $\ce{NO2+}$ , $\ce{OH-}$ , $\ce{(CH3)2CH+}$ .
Name the type of reaction (and, where relevant, the mechanism): (a) $\ce{CH3CH2CH2Br + KCN -> CH3CH2CH2CN + KBr}$ ; (b) $\ce{CH2=CH2 + H2O -> CH3CH2OH}$ ; (c) $\ce{CH3COOCH2CH3 + H2O -> CH3COOH + CH3CH2OH}$ ; (d) $\ce{CH3CHBrCH3 + NaOH -> CH3CH=CH2 + NaBr + H2O}$ (in ethanol).
Write equations using $\ce{[O]}$ or $\ce{[H]}$ for: (a) propan-1-ol to propanal; (b) propanone to propan-2-ol; (c) ethanal to ethanoic acid.
Explain why the carbon atom in a C–Cl bond is attacked by nucleophiles but the carbon atoms in a C=C bond are attacked by electrophiles.
What does a full curly arrow represent? Where must it start and where may it end?
Write equations for the initiation step for the reaction of bromine with methane, and for one possible termination step. State the conditions.
Place these carbocations in order of increasing stability and explain your order: $\ce{CH3CH2CH2+}$ , $\ce{(CH3)3C+}$ , $\ce{(CH3)2CH+}$ .
Ethyl ethanoate can be made from ethanol alone in two steps: some ethanol is oxidised to ethanoic acid, which then reacts with more ethanol. Write an equation for each step, using $\ce{[O]}$ for the first, and classify each step.
In the first step of the reaction between ethene and bromine, the bromine molecule becomes polarised, the C=C bond attacks one bromine atom, and the Br–Br bond breaks. (a) Explain why the Br–Br bond becomes polarised. (b) Describe the two curly arrows in this step, saying exactly where each starts and ends. (c) Is the Br–Br bond broken homolytically or heterolytically? Give the two species formed. (d) Explain why bromine is called an electrophile here even though $\ce{Br2}$ is a non-polar molecule.

Answers

(a) Breaking of a covalent bond in which one atom takes both electrons of the bonding pair, forming a cation and an anion. (b) A species with one or more unpaired electrons. (c) An electron-pair acceptor: a species that accepts a pair of electrons to form a new covalent bond.
$\ce{Br^.}$ : free radical. $\ce{H2O}$ : nucleophile (lone pairs on O). $\ce{NO2+}$ : electrophile (positive, accepts an electron pair). $\ce{OH-}$ : nucleophile. $\ce{(CH3)2CH+}$ : electrophile (a carbocation; it accepts a pair from $\ce{Br-}$ in electrophilic addition).
(a) Nucleophilic substitution. (b) Electrophilic addition (hydration). (c) Hydrolysis. (d) Elimination.
(a) $\ce{CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O}$ (b) $\ce{CH3COCH3 + 2[H] -> CH3CH(OH)CH3}$ (c) $\ce{CH3CHO + [O] -> CH3COOH}$
Chlorine is more electronegative than carbon, so the C–Cl bond is polar, $\ce{C^{\delta+}-Cl^{\delta-}}$ ; the $\delta+$ carbon attracts nucleophiles, which donate a lone pair to it. A C=C bond contains a $\pi$ bond, an exposed region of high electron density above and below the plane of the molecule, so it attracts electrophiles (electron-pair acceptors) and donates its $\pi$ electrons to them.
The movement of a pair of electrons. It must start at a lone pair or at a bond (the electrons that move). It ends on an atom (to form a lone pair or ion) or between two atoms (to form a new bond).
Initiation: $\ce{Br2 -> 2Br^.}$ , in ultraviolet light (homolytic fission). Termination, any one of: $\ce{Br^. + Br^. -> Br2}$ ; $\ce{CH3^. + Br^. -> CH3Br}$ ; $\ce{CH3^. + CH3^. -> C2H6}$ .
$\ce{CH3CH2CH2+}$ (primary) < $\ce{(CH3)2CH+}$ (secondary) < $\ce{(CH3)3C+}$ (tertiary). Alkyl groups donate electron density to the positive carbon (positive inductive effect), spreading out the charge; the more alkyl groups attached to the positive carbon, the greater the stabilisation.
Step 1: $\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}$ : oxidation. Step 2: $\ce{CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O}$ : condensation (esterification), catalysed by concentrated sulfuric acid.
(a) The high electron density of the $\pi$ bond in C=C repels the electrons in the Br–Br bond, pushing them towards the far bromine atom, so the near bromine becomes $\delta+$ and the far one $\delta-$ (an induced dipole). (b) Arrow 1 starts on the C=C double bond (the $\pi$ electrons) and ends on the $\delta+$ bromine atom, forming a new C–Br bond. Arrow 2 starts on the Br–Br bond and ends on the $\delta-$ bromine atom. (c) Heterolytically: a carbocation ( $\ce{CH2BrCH2+}$ ) and a bromide ion ( $\ce{Br-}$ ). (d) Although $\ce{Br2}$ has no permanent dipole, the C=C induces a dipole in it; the $\delta+$ bromine then accepts a pair of electrons (from the $\pi$ bond) to form a new covalent bond, which is the definition of an electrophile.

How covalent bonds break

Free-radical reactions

Nucleophiles and electrophiles

Types of reaction

Writing oxidation and reduction with [O] and [H]

Curly arrows: the language of mechanisms

The inductive effect of alkyl groups

Worked examples

Practice

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