Introduction to organic chemistry

In a nutshell

Organic chemistry is the chemistry of carbon compounds. Before you can learn a single organic reaction, you need the shared language every later topic uses.

This subtopic gives you that toolkit: how to write a formula six different ways, how to name a compound with IUPAC rules, how to show a mechanism, and how to spot and draw isomers. Naming and drawing marks turn up on almost every organic question.

Assumed knowledge: Bonding, Amount of substance.

Core content

Nomenclature: six ways to represent a compound

A hydrocarbon is a compound containing carbon and hydrogen only. The word "only" is what earns the mark.

AQA expects you to recognise and write six kinds of formula. Here they are for butane and butan-1-ol:

How to draw each one so it scores:

• Displayed formula: show every bond. Each CH₃ must be drawn out as a carbon with three separate C-H bonds, and an alcohol group must be drawn as C-O-H, with the O-H bond shown.
• Structural formula: write a branch or group in brackets straight after the carbon it is attached to, for example CH₃CH(CH₃)CH₃ or CH₃CH(OH)CH₃. Repeated units can be grouped, as in CH₃(CH₂)₂CH₃.
• Skeletal formula: every line end and every corner is a carbon atom. Hydrogens on carbon are not drawn (each carbon has enough H to make four bonds). Any other atom, and the H of an OH, is drawn.

Nomenclature: homologous series and functional groups

A functional group is an atom or group of atoms in a molecule that is responsible for its characteristic chemical reactions.

A homologous series is a series of compounds containing the same functional group. Members of a homologous series:

• have the same functional group
• have the same general formula
• differ from the next member by CH₂
• have similar chemical properties (they react in the same way, because they share the functional group)
• show a gradual trend in physical properties, for example boiling point rises as the chain gets longer.

Physical properties change gradually, they are not the same. Chemical properties are similar.

The functional groups you will meet at AS:

General formulae worth knowing:

• alkanes: C_nH_2n+2
• alkenes and cycloalkanes: C_nH_2n (so an alkene and a cycloalkane with the same number of carbons are isomers)
• alcohols: C_nH_2n+1OH
• halogenoalkanes: C_nH_2n+1X

Nomenclature: IUPAC naming rules

AQA tests naming for chains and rings of up to six carbons. The stem tells you the number of carbons in the main chain:

Name any compound with the same five steps:

1. Find the longest carbon chain that contains the functional group. This gives the stem. (A ring gives cyclo- + stem.)
2. Choose the suffix from the main functional group. If there are two, the higher-priority group takes the suffix: carboxylic acid > aldehyde > ketone > alcohol. Halogens and alkyl branches are always prefixes.
3. Number the chain from the end that gives the suffix group the lowest number. Then give the C=C the lowest number it can have, then the prefixes. In aldehydes, carboxylic acids and nitriles the functional-group carbon is always carbon 1, so no number is written for it.
4. Add prefixes in alphabetical order (bromo before chloro before methyl). Use di-, tri-, tetra- for repeats, but ignore them when alphabetising, so it is 2,2-dichloro-1-fluoro..., not 1-fluoro-2,2-dichloro...
5. Punctuate: commas between numbers, hyphens between numbers and letters.

Two spelling rules for the stem:

• If the suffix starts with a vowel, drop the final "e" of -ane: propanol, propanal, propanone, propanoic acid.
• If the suffix starts with a consonant, keep the "e": propanenitrile, ethane-1,2-diol.

Each example below shows one rule:

Numbers are left out only when there is no other possible position, for example methylpropane and propanone.

Reaction mechanisms: free-radical steps

A free radical is a species with an unpaired electron. The unpaired electron is shown as a dot, for example Cl• and •CH₃.

Curly arrows are not needed for radical mechanisms. You write balanced equations for three kinds of step instead:

Using the reaction of methane with chlorine (taught in full in Alkanes):

• Initiation: Cl₂ → 2Cl•
• Propagation: Cl• + CH₄ → HCl + •CH₃
• Propagation: •CH₃ + Cl₂ → CH₃Cl + Cl•
• Termination: •CH₃ + Cl• → CH₃Cl, or 2•CH₃ → C₂H₆

Check every step: atoms balance, no charges appear, and the two propagation steps added together give the overall equation (CH₄ + Cl₂ → CH₃Cl + HCl) with no radicals left in it.

Reaction mechanisms: curly arrows

All the other AS mechanisms move pairs of electrons, shown with curly arrows.

• A curly arrow shows the movement of one pair of electrons.
• Bond formation: the arrow starts from a lone pair or from an existing covalent bond, and points to where the new bond forms.
• Bond breaking: the arrow starts from the bond that breaks, and points to the atom that takes both electrons.

To "outline a mechanism" you must draw it: the structures of the species involved, every relevant lone pair, the δ+ and δ− charges, any full charges, and the curly arrows. A description in words does not earn the marks.

Still don't get it? · what a curly arrow means

Imagine two people both holding the same rope. A curly arrow is a note saying "this rope gets picked up from here and handed over to there." In chemistry the rope is a pair of electrons.

Step by step:

1. Electrons in a bond or a lone pair always come in pairs. One curly arrow moves one pair.
2. The tail of the arrow sits where the pair is now: on a lone pair, or in the middle of a bond.
3. The head of the arrow points to where the pair goes: to an atom (it now owns the pair), or into the gap between two atoms (a new bond forms there).
4. If an arrow leaves a bond, that bond is gone. If an arrow lands between two atoms, a new bond is there.
5. Then fix the charges: an atom that takes the whole pair from a broken bond becomes one unit more negative (Br becomes Br⁻); an atom that uses its lone pair to make a new bond becomes one unit more positive (OH⁻ becomes a neutral O in the product).

In the exam wording: a bond forms with a curly arrow from a lone pair or from another covalent bond; a bond breaks with a curly arrow starting from the bond. If the tail of your arrow starts on an atom, a charge or a hydrogen instead of a lone pair or a bond, the mark is lost.

Isomerism: structural isomers

Structural isomers are compounds with the same molecular formula but different structural formulae.

There are three types:

The four structural isomers of C₄H₉Br show chain and position isomerism together:

Isomerism: stereoisomers and E-Z isomerism

Stereoisomers are compounds with the same structural formula but with their atoms arranged differently in space.

E-Z isomerism is a type of stereoisomerism found in some alkenes. It needs both of these:

1. Restricted rotation about the C=C double bond (the planar double bond cannot rotate).
2. Two different atoms or groups attached to each carbon of the C=C.

But-2-ene has both, so it has an E isomer and a Z isomer. But-1-ene has two H atoms on one carbon of the C=C, so it has no E-Z isomers.

Because the double bond cannot rotate, E-but-2-ene cannot turn into Z-but-2-ene. They are different compounds with the same structural formula, so they are stereoisomers, not structural isomers.

Isomerism: the CIP priority rules

To label an isomer E or Z, rank the two groups on each carbon of the C=C using the Cahn-Ingold-Prelog (CIP) priority rules:

1. Look at the atom directly attached to the C=C carbon. The atom with the higher atomic number has the higher priority (Br 35 > Cl 17 > O 8 > N 7 > C 6 > H 1).
2. If those atoms are the same, move one bond further out. List the atoms attached to each of them in order of decreasing atomic number and compare the lists atom by atom. The first point of difference decides.

Then:

• Z (from zusammen, together): the two higher-priority groups are on the same side of the C=C.
• E (from entgegen, opposite): the two higher-priority groups are on opposite sides.

Examples of rule 2:

Still don't get it? · E-Z isomers and CIP priorities

Picture two flags stuck into a plank of wood with two nails. With one nail you could spin the plank round and swap which side each flag is on. With two nails it is stuck: whichever side the flags start on, they stay there.

Step by step:

1. A C-C single bond is like the one-nail plank: the two ends can rotate, so there is only one compound.
2. A C=C double bond is like the two-nail plank: the two ends cannot rotate.
3. If each carbon of the C=C carries two different groups, the groups can be arranged two ways that cannot be swapped. That gives two compounds.
4. If one carbon carries two identical groups (two H, say), both arrangements look the same, so there is only one compound.
5. To name the two compounds, pick the "heavier" group on each carbon: compare the atomic numbers of the atoms joined directly to the C=C. If they tie, compare the next atoms out.
6. Heavier groups on the same side is Z ("Zame Zide"); on opposite sides is E.

In exam wording: E-Z isomerism arises because there is restricted rotation about the C=C and each carbon of the C=C has two different groups attached. Both points are needed. Priority is decided by higher atomic number of the attached atom, then of the atoms one bond further away.

Worked examples

1. Naming a compound from its structure

Name CH₃CH₂CH(CH₃)CH(OH)CH₃.

1. Longest chain containing the -OH: five carbons, so the stem is pent-.
2. Suffix: the only functional group is -OH, so -ol. The suffix starts with a vowel, so the stem is pentan.
3. Number from the end nearer the -OH. From the right-hand end the -OH is on carbon 2; from the left it would be on carbon 4. Use the right: the -OH is on C2 and the CH₃ branch is on C3.
4. Prefix: one methyl on carbon 3, so 3-methyl.
5. Assemble: 3-methylpentan-2-ol.

Where marks are lost: numbering from the branch end ("3-methylpentan-4-ol"). The suffix group always wins the lowest number.

2. Drawing a structure from its name

Draw the structural formula of 3-chloro-2-methylbutanoic acid.

1. Stem and suffix: butanoic acid, a four-carbon chain ending in -COOH.
2. Carbon 1 is the COOH carbon. Number away from it: C1 is COOH, then C2, C3, C4.
3. Place the prefixes: a methyl on C2, a chlorine on C3.
4. Fill in hydrogens so every carbon has four bonds.

Answer: CH₃CHClCH(CH₃)COOH, molecular formula C₅H₉ClO₂.

Where marks are lost: counting carbon 1 as the first carbon after the COOH (or CHO) group. That turns 3-methylbutanal into 2-methylbutanal.

3. Empirical and molecular formula of a hydrocarbon

A hydrocarbon contains 85.7% carbon by mass. Its relative molecular mass, M_r, is 56.0. Find its empirical and molecular formulae. (A_r: C = 12.0, H = 1.0)

A hydrocarbon contains only C and H, so the rest is hydrogen:

$\%\,\text{H} = 100 - 85.7 = 14.3\,\%$

Divide each percentage by A_r (this gives moles in 100 g):

$\text{C}: \frac{85.7}{12.0} = 7.14 \qquad \text{H}: \frac{14.3}{1.0} = 14.3$

Divide by the smaller value:

$\text{C}: \frac{7.14}{7.14} = 1.00 \qquad \text{H}: \frac{14.3}{7.14} = 2.00$

Empirical formula = CH₂, with empirical formula mass:

$12.0 + (2 \times 1.0) = 14.0$

Number of empirical units in one molecule:

$\frac{M_r}{\text{empirical formula mass}} = \frac{56.0}{14.0} = 4$

Molecular formula = 4 × CH₂ = C₄H₈.

C₄H₈ fits C_nH_2n, so it could be an alkene (such as but-1-ene) or a cycloalkane (such as cyclobutane): the data alone cannot tell you which.

Where marks are lost: giving the molecular formula when the empirical formula is asked for (C₄H₈ instead of CH₂), or rounding a ratio such as 1 : 1.25 to 1 : 1 instead of multiplying up to 4 : 5.

4. Finding every structural isomer systematically

How many aldehydes have the molecular formula C₅H₁₀O?

Work from the longest chain down, remembering the CHO carbon is always carbon 1:

1. Five-carbon chain: CH₃CH₂CH₂CH₂CHO, pentanal.
2. Four-carbon chain plus one methyl. The methyl can go on C2 or C3 (on C4 it would just extend the chain):
   • CH₃CH₂CH(CH₃)CHO, 2-methylbutanal
   • CH₃CH(CH₃)CH₂CHO, 3-methylbutanal
3. Three-carbon chain plus two methyls. Both must go on C2: (CH₃)₃CCHO, 2,2-dimethylpropanal.

Answer: 4 aldehydes.

Where marks are lost: stopping too early, or counting the same molecule twice drawn a different way round (an "ethyl on carbon 2 of propanal" is really 2-methylbutanal). Name each structure you draw: two identical names means a duplicate.

5. Deciding E or Z with the CIP rules

One isomer of 1-chloro-2-methylbut-1-ene, ClCH=C(CH₃)CH₂CH₃, has the Cl atom and the ethyl group on the same side of the C=C. Is it E or Z?

1. Carbon 1 carries Cl and H. Cl (atomic number 17) beats H (1), so Cl is the priority group.
2. Carbon 2 carries CH₃ and CH₂CH₃. Both attach through C, a tie. One bond further out: CH₃ has (H, H, H) and CH₂CH₃ has (C, H, H). C beats H, so CH₂CH₃ is the priority group.
3. Compare sides: the two priority groups, Cl and CH₂CH₃, are on the same side.

Answer: Z-1-chloro-2-methylbut-1-ene.

Where marks are lost: deciding by the size of the whole group, or by "more atoms", instead of comparing atomic numbers one bond at a time.

6. Writing free-radical steps for a new reaction

Ethane reacts with chlorine in UV light to form chloroethane. Write equations for the initiation, the two propagation steps and one termination step.

• Initiation (Cl-Cl bond breaks): Cl₂ → 2Cl•
• First propagation (Cl• removes an H): Cl• + CH₃CH₃ → HCl + •CH₂CH₃
• Second propagation (the ethyl radical takes a Cl and regenerates Cl•): •CH₂CH₃ + Cl₂ → CH₃CH₂Cl + Cl•
• Termination (two radicals join): •CH₂CH₃ + Cl• → CH₃CH₂Cl

Check: adding the two propagation steps and cancelling Cl• gives CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl, which balances and has no radicals.

Where marks are lost: missing the dot on a radical, writing a termination step where a propagation step is asked for, or making H• and CH₃CH₂Cl in the first propagation step instead of HCl and the ethyl radical.

Common exam mistakes

• Homologous series: students write "same molecular formula" or "same empirical formula"; members have the same general formula. Writing "same chemical and physical properties" is treated as a contradiction: chemical properties are similar, physical properties show a trend.
• Hydrocarbon: "contains carbon and hydrogen" without the word only does not score.
• Molecular vs structural formula: asked for the molecular formula of hexane, students write CH₃CH₂CH₂CH₂CH₂CH₃; it is C₆H₁₄. Writing both formulae is penalised, because the examiner will not choose for you.
• Empirical formula: students give the molecular formula (C₄H₁₀) when the empirical formula (C₂H₅) is asked for.
• Displayed formulae: leaving a group as "CH₃" or "OH", or drawing an alcohol as C-H-O. A displayed formula must show every bond, including every C-H and the O-H.
• IUPAC names: wrong locant numbers, prefixes out of alphabetical order (it is 3-chloro-2,2-dimethyl..., chloro before methyl), a missing "di", or non-IUPAC names such as "2-methylpropan-3-ol" (correct: 2-methylpropan-1-ol). Missing hyphens and commas are usually ignored; wrong numbers, order and spelling are not.
• Aldehyde numbering: drawing 2-methylbutanal when asked for 3-methylbutanal, because the CHO carbon was not counted as carbon 1.
• Structural isomer definition: students leave out "same molecular formula", or write "different arrangement of atoms in space". That phrase describes stereoisomers and is rejected here: structural isomers have different structural formulae.
• Stereoisomer definition: fewer than a third of students score both marks. The first mark needs same structural formula (not "same structure", and not "same molecular formula"); the second needs atoms or groups arranged differently in space.
• Explaining E-Z isomerism: students give only "no rotation about the C=C". The second mark needs two different groups on each carbon of the C=C.
• E vs Z is not structural: many students say E-but-2-ene is a structural isomer of Z-but-2-ene. They have the same structural formula, so they are stereoisomers.
• Type of isomerism: "functional" alone does not score; write functional group isomerism (both words). Chain and position isomerism are also often swapped.
• Counting isomers: most wrong answers are one too few or one too many. Work systematically from the longest chain, and remember cycloalkanes are isomers of alkenes.
• Radical equations: dots missing from radicals, charges on radicals, radicals left in an overall equation, or a termination step written as the second propagation step.
• Curly arrows: arrows that start on an atom or a charge instead of a lone pair or a bond; a negative charge put on NH₃; a full + charge put on a carbon that is only δ+; a bond drawn as a line and two dots together. Describing the mechanism in words instead of drawing it rarely scores.

Key definitions

• Hydrocarbon: a compound containing carbon and hydrogen only.
• Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula: the actual number of atoms of each element in a molecule.
• General formula: an algebraic formula that represents every member of a homologous series (e.g. C_nH_2n+2 for alkanes).
• Structural formula: shows the arrangement of atoms in a molecule carbon by carbon, without showing every bond (e.g. CH₃CH(OH)CH₃).
• Displayed formula: shows every atom and every covalent bond in a molecule.
• Skeletal formula: shows the carbon skeleton as lines, with hydrogen atoms on carbon left out and functional groups drawn on.
• Functional group: an atom or group of atoms responsible for the characteristic chemical reactions of a compound.
• Homologous series: a series of compounds with the same functional group and the same general formula, in which each member differs from the next by CH₂, with similar chemical properties and a trend in physical properties.
• Free radical: a species with an unpaired electron (shown by a dot).
• Structural isomers: compounds with the same molecular formula but different structural formulae.
• Chain isomers: structural isomers with different carbon skeletons.
• Position isomers: structural isomers with the same functional group in a different position on the same carbon skeleton.
• Functional group isomers: structural isomers with different functional groups.
• Stereoisomers: compounds with the same structural formula but with atoms (or groups) arranged differently in space.
• E-Z isomerism: a form of stereoisomerism caused by restricted rotation about the C=C double bond, when each carbon of the C=C has two different groups attached.

Specification

• I can write the empirical, molecular, general, structural, displayed and skeletal formulae of an organic compound.
• I can draw structural, displayed and skeletal formulae for a given organic compound.
• I can state the characteristics of a homologous series: same functional group, same general formula, differ by CH₂, similar chemical properties, trend in physical properties.
• I can apply IUPAC rules to name organic compounds with chains and rings of up to six carbon atoms.
• I can apply IUPAC rules to draw the structure of a compound from its name (chains and rings of up to six carbons).
• I can represent a radical with a dot, and write balanced equations for the initiation, propagation and termination steps of a free-radical mechanism (without curly arrows).
• I can state that bond formation is shown by a curly arrow from a lone pair or from another covalent bond, and bond breaking by a curly arrow starting from the bond.
• I can outline a mechanism by drawing the species involved and curly arrows for the movement of electron pairs.
• I can define structural isomer, and draw chain, position and functional group isomers.
• I can define stereoisomer.
• I can explain that E-Z isomerism arises from restricted rotation about the planar C=C, and draw the structural formulas of E and Z isomers.
• I can apply the CIP priority rules to label an isomer E or Z.

Related notes

• Bonding
• Amount of substance
• Alkanes
• Halogenoalkanes
• Alkenes
• Alcohols
• Optical isomerism (A-level only, the second kind of stereoisomerism)