Characteristic organic reactions

1. Overview

Organic chemistry has its own vocabulary: families of compounds, the two ways a covalent bond can break, the species that attack molecules, and the types of reaction and mechanism. This section also shows how curly arrows track electron pairs. Every later organic topic uses these terms, so learn them precisely.

Key Definitions

• Homologous series: a family of organic compounds with the same functional group and the same general formula, in which each member differs from the next by CH₂; members have similar chemical properties and a gradual trend in physical properties such as boiling point.
• Saturated and unsaturated: a saturated compound has only single bonds between its carbon atoms; an unsaturated compound has at least one carbon-carbon multiple bond, such as C=C.
• Homolytic fission: the breaking of a covalent bond in which each of the two atoms takes one electron from the shared pair, so two free radicals form.
• Heterolytic fission: the breaking of a covalent bond in which one atom takes both electrons of the shared pair; from a neutral molecule this gives a positive ion and a negative ion.
• Free radical: a species with one or more unpaired electrons, shown by a dot, for example Cl• or CH₃•.
• Nucleophile: an electron-pair donor; a species with a lone pair that it uses to form a new covalent bond, usually with a carbon atom that carries a partial positive charge.
• Electrophile: an electron-pair acceptor; a species that accepts a pair of electrons to form a new covalent bond, usually from an electron-rich region such as a C=C bond.
• Hydrolysis: a reaction in which a bond in a molecule is broken by reaction with water, often in the presence of a dilute acid or alkali; for example, an ester and water give a carboxylic acid and an alcohol.
• Condensation reaction: a reaction in which two molecules join to form a larger molecule and a small molecule, such as water, is lost.
• Curly arrow: an arrow with a full head that shows the movement of a pair of electrons in a mechanism; its tail starts at a lone pair or a covalent bond and its head points to where the pair ends up.

Content

Homologous series

The alcohols, CₙH₂ₙ₊₁OH, are a homologous series. Methanol, CH₃OH, ethanol, C₂H₅OH, and propan-1-ol, C₃H₇OH, each have one more CH₂ than the one before, so each $M_{\text{r}}$ is 14.0 higher. All three have the OH group, so they react in similar ways. Their boiling points rise along the series, because longer molecules have more electrons and so stronger instantaneous dipole-induced dipole forces.

A general formula gives any member: with $n = 4$, an alcohol is C₄H₉OH. It does not prove the series, though: cyclohexane, C₆H₁₂, fits the alkene formula CₙH₂ₙ, but its carbon atoms form a ring of single bonds, so it is not an alkene.

Saturated and unsaturated

A saturated compound has only single bonds between its carbon atoms. An unsaturated compound has at least one carbon-carbon multiple bond, usually C=C. Only carbon-carbon bonds count:

• Hexane, CH₃(CH₂)₄CH₃, is saturated; hex-1-ene, CH₂=CH(CH₂)₃CH₃, is unsaturated.
• Propanoic acid, CH₃CH₂COOH, has a C=O bond but is saturated; propenoic acid, CH₂=CHCOOH, is unsaturated.

One bond of a C=C can open so that new atoms bond to both carbons. So alkenes undergo addition reactions (and decolourise aqueous bromine, 14.2), while alkanes, which cannot gain atoms without losing others, react by substitution.

Breaking a bond: homolytic and heterolytic fission

A covalent bond is a shared pair of electrons. When it breaks, the pair can be shared out in two ways ("homo" means same, "hetero" means different).

Homolytic fission: each atom takes one electron, so two free radicals form.

H-Br → H• + Br•

Heterolytic fission: one atom takes both electrons. From a neutral molecule, this gives a positive ion and a negative ion.

H-Br → H⁺ + Br⁻

The more electronegative atom usually takes the pair. Bromine (2.6 in the Data Booklet) is more electronegative than hydrogen (2.1), so bromine becomes Br⁻.

Homolytic fission usually needs ultraviolet (UV) light or a high temperature, and it starts free-radical reactions (14.1). Heterolytic fission happens in reactions of ions and polar molecules, such as halogenoalkanes (15.1).

Free radicals and chain reactions

A free radical has one or more unpaired electrons, shown by a dot: Cl•, CH₃•. Radicals are very reactive, because the unpaired electron readily pairs with an electron from another species to form a bond.

Free-radical reactions are chain reactions with three kinds of step:

Step	What happens	Radicals in → out
Initiation	homolytic fission of a molecule gives two radicals, e.g. Cl₂ → 2Cl• in UV light	0 → 2
Propagation	a radical and a molecule give a new molecule and a new radical	1 → 1
Termination	two radicals join to form a molecule	2 → 0

Each propagation step makes the radical that the next one uses, so one initiation step can lead to many product molecules until a termination step ends the chain. The propagation steps add up to the overall equation. The full mechanism for alkanes is in 14.1.

Nucleophiles and electrophiles

A nucleophile ("nucleus-loving") is an electron-pair donor. It uses a lone pair to form a new covalent bond, usually with a carbon atom that carries a partial positive charge (δ+). It may be a negative ion, such as OH⁻ or CN⁻, or a neutral molecule with a lone pair, such as H₂O or NH₃.

The δ+ carbon is in a polar bond. In C-Cl, C-Br and C=O, the other atom is more electronegative than carbon (Data Booklet: C 2.5, Cl 3.0, O 3.5), so it pulls the bonding pair towards itself.

An electrophile ("electron-loving") is an electron-pair acceptor. It accepts a pair of electrons to form a new covalent bond, usually from an electron-rich region such as a C=C bond. Electrophiles include positive ions, such as H⁺, and molecules with a δ+ atom, such as the H of Hδ+-Brδ−. Even non-polar Br₂ can act as one: near a C=C bond, the electron-rich double bond repels the Br-Br bonding pair, so the nearer Br atom becomes δ+.

In the name of a mechanism, nucleophilic or electrophilic describes the attacking species. In the electrophilic addition of bromine to an alkene, Br₂ is the electrophile and the alkene is the molecule it attacks.

Five types of reaction

Each name describes what happens overall, not how it happens. R is an alkyl group.

• Addition: two molecules join to make one, across a multiple bond. Alkene + Br₂ → dibromoalkane.
• Substitution: one atom or group is replaced by another. RBr + OH⁻ → ROH + Br⁻.
• Elimination: a small molecule (e.g. HBr, H₂O) is lost from neighbouring carbons, leaving a C=C. RCH₂CH₂Br → RCH=CH₂ + HBr.
• Hydrolysis: a bond is broken by reaction with water. Ester + water → acid + alcohol.
• Condensation: two molecules join and a small molecule (often H₂O) is lost. Acid + alcohol → ester + water.

Three points catch people out:

• Two names can fit one reaction. Heating a halogenoalkane with NaOH(aq) is a substitution, also called a hydrolysis because the C-halogen bond is broken to give an alcohol.
• Hydration is not hydrolysis. In hydration, water adds across a C=C bond and nothing splits off, so it is an addition. Dehydration (alcohol → alkene + water) is an elimination.
• Esterification is one particular condensation. When asked for the type of reaction, write "condensation".

Worked example 1: naming the type of reaction

Question: State the type of each reaction, choosing from addition, substitution, elimination, hydrolysis and condensation. Where two names fit, give both.

• (a) hex-1-ene + H₂ (nickel catalyst, heat) → hexane
• (b) 1-bromohexane + NaOH(aq), heat → hexan-1-ol + NaBr
• (c) bromocyclohexane + NaOH in ethanol, heat → cyclohexene + NaBr + H₂O
• (d) hexanoic acid + methanol (concentrated H₂SO₄ catalyst) → methyl hexanoate + H₂O
• (e) methyl hexanoate + H₂O (dilute acid, heat) → hexanoic acid + methanol

Method: compare the organic molecules before and after, and look for a small molecule joining or leaving.

	Clue	Type
(a)	two molecules → one; the C=C becomes C-C	addition
(b)	Br replaced by OH; the C-Br bond is broken	substitution (also hydrolysis)
(c)	H and Br lost from neighbouring carbons; a C=C forms	elimination
(d)	two molecules join and H₂O is lost	condensation
(e)	one molecule + H₂O → two molecules	hydrolysis

Answer: (a) addition; (b) substitution, also called hydrolysis; (c) elimination; (d) condensation; (e) hydrolysis.

Oxidation and reduction in organic chemistry

In organic reactions it is usually easiest to track oxygen and hydrogen. Oxidation is gain of oxygen or loss of hydrogen; reduction is gain of hydrogen or loss of oxygen. This agrees with oxidation numbers (6.1): when ethanol, CH₃CH₂OH, is oxidised to ethanal, CH₃CHO, the oxidation number of the functional-group carbon rises from −1 to +1.

Organic equations use a shorthand for the reagent:

• [O] is one oxygen atom from the oxidising agent, e.g. acidified K₂Cr₂O₇ or acidified KMnO₄.
• [H] is one hydrogen atom from the reducing agent, e.g. NaBH₄ or LiAlH₄.

Write the organic reactant and product, then balance with [O] or [H]. Hydrogen removed in an oxidation, or oxygen removed in a reduction, leaves as H₂O; never write H₂ or O₂ as a product. When the reagent is hydrogen gas itself (hydrogenation with a nickel catalyst), write H₂ rather than 2[H].

Worked example 2: equations with [O] and [H]

Question: Use [O] or [H] to write balanced equations for (a) the oxidation of hexan-1-ol to hexanal, distilling the aldehyde off as it forms; (b) the oxidation of hexan-1-ol to hexanoic acid under reflux with excess acidified K₂Cr₂O₇; (c) the reduction of pentan-3-one by NaBH₄. (d) Is the reaction of but-2-ene with steam to form butan-2-ol an oxidation, a reduction or neither?

(a) Hexan-1-ol loses two hydrogen atoms to become hexanal. One [O] removes them as water:

CH₃(CH₂)₄CH₂OH + [O] → CH₃(CH₂)₄CHO + H₂O

Check: C₆H₁₄O₂ on each side.

(b) A second [O] turns the CHO group into COOH:

CH₃(CH₂)₄CH₂OH + 2[O] → CH₃(CH₂)₄COOH + H₂O

Check: C₆H₁₄O₃ on each side.

(c) The C=O group gains two hydrogen atoms and becomes CH(OH). No oxygen is removed, so no water forms:

CH₃CH₂COCH₂CH₃ + 2[H] → CH₃CH₂CH(OH)CH₂CH₃

Check: C₅H₁₂O on each side.

(d) CH₃CH=CHCH₃ + H₂O → CH₃CH(OH)CH₂CH₃

The molecule gains exactly one water molecule, so no [O] or [H] is needed. Adding the atoms of water is overall neither oxidation nor reduction; the same is true of hydrolysis.

Answer: (a), (b) and (c) are the three equations above, with 1[O], 2[O] and 2[H]. (d) Neither: it is an addition of water (hydration).

Naming a mechanism

A mechanism shows, step by step, which bonds break, which form and where the electrons go. Its name has two parts: what attacks, then what happens overall.

Mechanism	Attacked by	Found in
free-radical substitution	Cl• or Br•	alkanes, in UV light (14.1)
electrophilic addition	Br₂, HBr	alkenes (14.2)
nucleophilic substitution	OH⁻, CN⁻, NH₃	halogenoalkanes (15.1)
nucleophilic addition	CN⁻	aldehydes and ketones + HCN (17.1)

Each pairing follows from structure. The electron-rich C=C of an alkene attracts electrophiles. Nucleophiles attack a δ+ carbon: in C-X (X is a halogen), X leaves, so it is substitution; at C=O nothing leaves, so it is addition. Alkanes have strong C-H bonds with almost no polarity, so polar reagents leave them alone, but radicals react.

Curly arrows

In a mechanism with heterolytic steps, each curly arrow shows a pair of electrons moving. The rules:

1. One arrow moves one pair. Draw a full arrowhead (both sides of the head).
2. The tail starts where the electrons are: at a lone pair (drawn as two dots) or at the middle of a covalent bond, including a C=C bond. Never start at an atom's symbol (a hydrogen atom included) or at a charge sign.
3. The head points to where the pair goes: to the atom it will form a new bond with, or, when a bond breaks, to the atom that keeps the pair.
4. Show the evidence: lone pairs on the attacking atom, partial charges (δ+ and δ−) on polar bonds that react, and full charges on ions and intermediates.

Radical steps are written as equations with a dot on each radical (Worked example 3). A half-headed ("fish-hook") arrow moves one electron, so never use one for a pair.

Worked example 3: steps in a radical chain

Question: Cyclohexane, C₆H₁₂, reacts with chlorine in UV light to give chlorocyclohexane, C₆H₁₁Cl, and hydrogen chloride. Five steps are listed in a jumbled order.

Step	Equation
A	C₆H₁₁• + Cl₂ → C₆H₁₁Cl + Cl•
B	Cl₂ → 2Cl•
C	C₆H₁₁• + Cl• → C₆H₁₁Cl
D	C₆H₁₂ + Cl• → C₆H₁₁• + HCl
E	2C₆H₁₁• → C₁₂H₂₂

(a) Name the type of each step, and the type of fission in the initiation step. (b) Show that the propagation steps give the overall equation.

(a) Count the radicals on each side. B: 0 → 2, initiation; the Cl-Cl bond splits evenly, one electron to each atom, so this is homolytic fission. D and A: 1 → 1, propagation. C and E: 2 → 0, termination.

(b) Add D and A:

C₆H₁₂ + Cl• + C₆H₁₁• + Cl₂ → C₆H₁₁• + HCl + C₆H₁₁Cl + Cl•

Cl• and C₆H₁₁• appear on both sides and cancel, leaving C₆H₁₂ + Cl₂ → C₆H₁₁Cl + HCl. The Cl• used in D is made again in A, so the chain keeps going.

Answer: (a) B is initiation (homolytic fission); D and A are propagation; C and E are termination. (b) C₆H₁₂ + Cl₂ → C₆H₁₁Cl + HCl.

Worked example 4: naming mechanisms and placing the first arrows

Question: For each reaction, name the mechanism, identify the attacking species, and describe the curly arrows in the first step.

• (a) 1-bromopentane heated with KCN in ethanol → hexanenitrile, CH₃(CH₂)₄CN
• (b) but-2-ene + Br₂ → 2,3-dibromobutane
• (c) butanal + HCN, with KCN as a catalyst → 2-hydroxypentanenitrile, CH₃CH₂CH₂CH(OH)CN

(a) Nucleophilic substitution: CN⁻ is the nucleophile, and CN replaces Br. Arrow 1 runs from the lone pair on the carbon atom of CN⁻ to the δ+ carbon of C-Br. Arrow 2 runs from the C-Br bond to the δ− bromine, which leaves as Br⁻ (15.1).

(b) Electrophilic addition: Br₂ is the electrophile, and both Br atoms add across the C=C. Arrow 1 runs from the C=C bond to the nearer, δ+ bromine of the polarised Br₂. Arrow 2 runs from the Br-Br bond to the far, δ− bromine, which becomes Br⁻. A carbocation (a positive ion with the charge on a carbon atom) forms, and Br⁻ then attacks it (14.2).

(c) Nucleophilic addition: CN⁻ is the nucleophile; nothing leaves, and H and CN add across the C=O. Arrow 1 runs from the lone pair on the carbon of CN⁻ to the δ+ carbon of C=O. Arrow 2 runs from the C=O bond to the δ− oxygen, which becomes O⁻. The O⁻ then takes H⁺ from HCN, re-forming CN⁻ (17.1).

Answer: (a) nucleophilic substitution, by CN⁻; (b) electrophilic addition, by Br₂; (c) nucleophilic addition, by CN⁻.

Key Equations

R and R′ are alkyl groups; a dot (•) is an unpaired electron; one [O] is one oxygen atom from the oxidising agent and one [H] is one hydrogen atom from the reducing agent.

Homolytic fission (each atom takes one electron): X-Y → X• + Y•

Heterolytic fission (Y, usually the more electronegative atom, takes both electrons): X-Y → X⁺ + Y⁻

Oxidation with [O]:

• primary alcohol to aldehyde: RCH₂OH + [O] → RCHO + H₂O
• primary alcohol to carboxylic acid: RCH₂OH + 2[O] → RCOOH + H₂O
• secondary alcohol to ketone: RCH(OH)R′ + [O] → RCOR′ + H₂O

Reduction with [H]:

• aldehyde to primary alcohol: RCHO + 2[H] → RCH₂OH
• ketone to secondary alcohol: RCOR′ + 2[H] → RCH(OH)R′

Condensation (forward) and hydrolysis (reverse) of an ester: RCOOH + R′OH ⇌ RCOOR′ + H₂O

Common Mistakes to Avoid

• Mistake: starting a curly arrow at an atom's symbol or a charge sign, or drawing it the wrong way round. Fix: start at the electrons. For OH⁻, start at a lone pair drawn on the O, not at the "−"; with HBr and an alkene, arrow 1 runs from the C=C bond to the δ+ H, not from the H.
• Mistake: leaving out lone pairs and dipoles, or putting full charges on a polar molecule. Fix: show the lone pair on the carbon of CN⁻ (not the nitrogen) and δ+ and δ− on C-Br. HBr is Hδ+-Brδ−, not H⁺ and Br⁻.
• Mistake: mixing up homolytic and heterolytic fission, or reading Cl• as a chloride ion. Fix: check the products. Homolytic fission of the C-Cl bond in chlorocyclohexane gives C₆H₁₁• + Cl•, two neutral radicals (Cl• has seven outer electrons, one unpaired); heterolytic fission gives C₆H₁₁⁺ + Cl⁻, two ions (Cl⁻ has eight).
• Mistake: writing "hydration" or "displacement" for a hydrolysis or substitution, or "esterification" for the type of reaction. Fix: 1-bromohexane + NaOH(aq) is substitution (also hydrolysis); but-2-ene + steam is an addition (hydration); acid + alcohol → ester + water is condensation.
• Mistake: unbalanced [O] and [H] equations, with H₂ or O₂ as a product, or [H⁺] written for [H]. Fix: put the balancing number in front and add H₂O when oxidation removes hydrogen: CH₃(CH₂)₄CH₂OH + 2[O] → CH₃(CH₂)₄COOH + H₂O.

Exam Tips

• State the type of reaction with one term from the list: addition, substitution, elimination, hydrolysis, condensation, oxidation or reduction. Do not hedge: a wrong term next to a right one makes the answer contradictory.
• Name the mechanism in full, with both parts, e.g. "nucleophilic substitution" or "electrophilic addition". "Substitution" on its own names a type of reaction, not a mechanism.
• Draw or complete a mechanism with all four pieces of evidence: curly arrows, δ+ and δ− on polar bonds, lone pairs on nucleophiles, and the charge on each ion and intermediate.
• Construct an equation using [O] or [H] with structural formulas, the balancing number in front of [O] or [H], and H₂O where oxidation removes hydrogen.
• Define homolytic and heterolytic fission by where the two bonding electrons go and what forms (radicals or ions). To identify a step in a radical chain, count the radicals on each side.