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Introduction to AS Level Organic Chemistry

Organic Chemistry Weightage: AS · Papers 1 and 2 (assumed in Paper 4) Topic 13
“Welcome to organic chemistry! Everything in the next nine organic topics rests on this one. Learn to read and draw the formulae, name the compounds, see the shape of each carbon atom, spot isomers, and describe how bonds break. Once these tools are automatic, the reactions that follow become patterns instead of facts to memorise.”
— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in Introduction to Organic Chemistry
  • Formulae and names: converting between empirical, molecular, structural, displayed and skeletal formulae, and naming compounds with up to six carbon atoms in the longest chain.
  • Isomers: drawing every structural isomer of a formula without repeating one, naming the type, and identifying cis/trans isomers and chiral centres.
  • Shape: the hybridisation, shape and bond angle at a carbon atom, and the number of σ and π bonds in a molecule.
  • Terminology: homolytic and heterolytic fission, free radical, nucleophile, electrophile, and the reaction types (addition, substitution, elimination, hydrolysis, condensation, oxidation, reduction).

Where this topic appears. Recent AS papers have asked for: pairs of structural isomers and counts of isomers of a given formula (including primary halogenoalkanes of C5H11Br), the number of chiral centres in a drug molecule, whether a molecule shows cis/trans isomerism, which of several molecules are chiral, the type of stereoisomerism in a table, the hybridisation of an atom, the nature of an attacking group (nucleophile) and a leaving group, and the repeat unit or monomer of an addition polymer. Almost every organic question in Papers 1, 2, 4 and 5 uses at least one skill from this topic.

1. Representing Organic Molecules

An organic compound is a compound of carbon (with H, O, N, halogens and so on), apart from the oxides of carbon, carbonates and hydrogencarbonates, which are treated as inorganic. A hydrocarbon contains only carbon and hydrogen. Carbon atoms bond to each other in chains and rings, which is why there are so many organic compounds.

empirical formulaDiagram showing: empirical formula; CH2​; simplest whole-number ratio of atoms; molecular formula; C3​H6​; actual number of each atom; structural formula; CH3​CH=CH2​; how the atoms are grouped on each C; displayed formula; C; C; C; H empirical formula CH2​ simplest whole-number ratio of atoms molecular formula C3​H6​ actual number of each atom structural formula CH3​CH=CH2​ how the atoms are grouped on each C displayed formula C C C H H H H H H every atom and every bond drawn skeletal formula each corner or end is a C
The five ways of showing propene, from least to most detail. Only the skeletal formula leaves out the C–H bonds; every other atom (such as O in an −OH group) must still be drawn.
Formula What it shows Propene
Empirical the simplest whole-number ratio of atoms CH2
Molecular the actual number of atoms of each element C3H6
Structural how the atoms are grouped on each carbon CH3CH=CH2
Displayed every atom and every bond all C–C, C=C and C–H bonds drawn
Skeletal carbon skeleton and functional groups only a zigzag with one double bond

From empirical to molecular formula. Divide each mass (or percentage) by Ar to get moles, divide by the smallest, then scale to whole numbers. Divide the relative molecular mass by the mass of the empirical formula to find the multiple.

Worked example. A compound is 54.5% C, 9.1% H and 36.4% O by mass, with Mr = 88.

⚠️ Examiner Trap: Hidden atoms in skeletal formulae. Each corner and each end of a line is a carbon atom with enough hydrogen atoms to make four bonds, but the H in an −OH group, and every other non-carbon atom, must be drawn. When you count the molecular formula from a skeletal formula, count the hidden hydrogen atoms too.

2. Homologous Series and Naming

A homologous series is a family of compounds with the same functional group and general formula, and each member differs from the next by CH2. The functional group is the atom or group of atoms that gives the compound its characteristic reactions.

alkeneDiagram showing: alkene; group: C=C; ending: -ene; ethene; CH2​=CH2​; halogenoalkane; group: C−X; ending: halo- (chloro-); chloromethane; CH3​Cl; alcohol; group: −OH; ending: -ol; ethanol alkene group: C=C ending: -ene ethene CH2​=CH2​ halogenoalkane group: C−X ending: halo- (chloro-) chloromethane CH3​Cl alcohol group: −OH ending: -ol ethanol CH3​CH2​OH aldehyde group: −CHO (end of chain) ending: -al ethanal CH3​CHO ketone group: C=O (inside chain) ending: -one propanone CH3​COCH3​ carboxylic acid group: −COOH ending: -oic acid ethanoic acid CH3​COOH ester group: −COO− ending: -oate methyl ethanoate CH3​COOCH3​ amine (primary) group: −NH2​ ending: -amine methylamine CH3​NH2​ nitrile group: −C≡N ending: -nitrile ethanenitrile CH3​CN
Functional groups required at AS Level. The functional group decides the characteristic reactions. Aldehydes always end the chain (no number needed), whereas the C=O of a ketone is inside the chain (it needs a number from pentanone onwards).
Homologous series General formula Example
alkane CnH2n+2 ethane, C2H6
alkene CnH2n ethene, C2H4
halogenoalkane CnH2n+1X chloromethane, CH3Cl
alcohol CnH2n+1OH methanol, CH3OH
aldehyde CnH2n+1CHO ethanal, CH3CHO
ketone CnH2n+1COCmH2m+1 propanone, CH3COCH3
carboxylic acid CnH2n+1COOH ethanoic acid, CH3COOH
ester CnH2n+1COOCmH2m+1 propyl ethanoate, CH3COOC3H7
amine (primary) CnH2n+1NH2 methylamine, CH3NH2
nitrile CnH2n+1CN ethanenitrile, CH3CN

These general formulae assume one functional group and an open chain. Cycloalkanes, CnH2n, do not follow the alkane formula.

Stems for the carbon chain: 1 meth-, 2 eth-, 3 prop-, 4 but-, 5 pent-, 6 hex- (then hept-, oct-, non-, dec-).

Rules for naming

  1. Find the longest carbon chain that contains the functional group. This gives the stem.
  2. Number the carbon atoms from the end that gives the lowest numbers, with the functional group taking the lowest number.
  3. Name each side chain as an alkyl group (methyl, ethyl, propyl). Use di-, tri-, tetra- for repeats, and put the groups in alphabetical order (ignore di- and tri- when ordering).
  4. Put commas between numbers and hyphens between numbers and words, for example 3-ethyl-2-methylpentane and 2,2,3-trimethylpentane.
  5. A number is not needed when the group can only be at the end of the chain (aldehydes and carboxylic acids, such as pentanal and butanoic acid) or when there is only one possible position (such as methanol or chloroethane).

Aryl compounds contain a benzene ring (for example propylbenzene). With one substituent on the ring no number is needed, because all six ring carbon atoms are equivalent.

⚠️ Examiner Trap: Name the longest chain, not the one drawn in a line. A structure drawn with a bend may still contain a longer straight chain. Count the longest continuous chain, and number from the end that gives the lowest numbers: CH3CH(CH3)CH2CH2CH3 is 2-methylpentane, not 4-methylpentane.

3. Bonding and Shapes (σ, π, Hybridisation)

A σ bond is formed by head-on overlap of orbitals and lies on the line between the two nuclei. Every single bond is a σ bond. A π bond is formed by sideways overlap of two p orbitals, with the electron density above and below the plane of the σ bonds. A double bond is one σ plus one π, and a triple bond is one σ plus two π.

carbon atom is sp3​Diagram showing: carbon atom is sp3​; carbon atom is sp2​; carbon atom is sp; C; H; H; H; H; C; C; H; H; H; H carbon atom is sp3​ carbon atom is sp2​ carbon atom is sp C H H H H C C H H H H p orbitals overlap C C H H two π bonds at right angles tetrahedral bond angle 109.5° bonds at each C: 4 σ, 0 π alkanes, e.g. ethane trigonal planar bond angle 120° bonds at each C: 3 σ, 1 π C=C, e.g. ethene linear bond angle 180° bonds at each C: 2 σ, 2 π C≡C, e.g. ethyne
Hybridisation decides the shape. A carbon atom with four σ bonds is sp3 (109.5°); with three σ bonds and one π it is sp2 (120°, planar); with two σ bonds and two π it is sp (180°, linear). Count the σ bonds to find the angle.

Worked example: propene, CH3CH=CH2. There are 8 σ bonds (2 C–C and 6 C–H) and 1 π bond. The CH3 carbon is sp3; the two carbon atoms of the double bond are sp2.

Rotation. There is free rotation about a σ bond, so a chain can twist into many shapes (all the same molecule). The π bond prevents rotation about C=C, which is the origin of cis/trans isomerism. The electron-rich π bond is also why alkenes are attacked by electrophiles.

4. Structural Isomerism

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

ISOMERSDiagram showing: ISOMERS; same molecular formula; STRUCTURAL ISOMERS; different structural formulae; STEREOISOMERS; same bonds, different 3D layout; chain; different; skeleton; butane; methylpropane; position; same group,; new place ISOMERS same molecular formula STRUCTURAL ISOMERS different structural formulae STEREOISOMERS same bonds, different 3D layout chain different skeleton butane methylpropane position same group, new place propan-1-ol propan-2-ol functional different group propanal propanone cis / trans restricted rotation about C=C but-2-ene optical chiral centre, mirror images CHBrClF Examples: C4​H10​ | C3​H8​O | C3​H6​O | C4​H8​ | CHBrClF
Isomers share a molecular formula. Structural isomers differ in how the atoms are joined (chain, position, functional group); stereoisomers have the same atoms joined in the same order but arranged differently in space.
Type What differs Example
Chain the carbon skeleton (branching) butane and methylpropane, C4H10
Position where the functional group (or a substituent) is attached propan-1-ol and propan-2-ol, C3H8O
Functional group the functional group itself propanal and propanone, C3H6O; propan-1-ol and methoxyethane, C3H8O

Isomers you must be able to list

Formula Structural isomers
C4H10 butane, methylpropane (2)
C5H12 pentane, 2-methylbutane, 2,2-dimethylpropane (3)
C3H7Cl 1-chloropropane, 2-chloropropane (2)
C3H6Br2 1,1-, 1,2-, 2,2- and 1,3-dibromopropane (4)
C3H6O propanal (aldehyde), propanone (ketone) (2, not counting cyclic or unsaturated forms)
C4H8 (alkenes) but-1-ene, but-2-ene, 2-methylpropene (3, plus cis/trans of but-2-ene); cyclobutane and methylcyclopropane are cyclic isomers

A reliable method. Start with the longest straight chain. Then shorten it by one carbon and place that carbon as a branch on every different position in turn. Repeat with a shorter chain. Move the functional group along each skeleton. Stop placing a group when the position gives a structure you have already drawn (check by finding the longest chain and the lowest numbers).

⚠️ Examiner Trap: The same molecule drawn twice. Because there is free rotation about C–C single bonds, bending or flipping a chain does not make a new isomer. The three drawings of 1,2-dibromopropane with the bromine atoms at different angles are one compound. Name each structure: two structures with the same name are the same compound.

5. Stereoisomerism

Stereoisomers have the same structural formula (the same atoms bonded in the same order) but a different arrangement of the atoms in space.

cis / trans: but-2-eneDiagram showing: cis / trans: but-2-ene; optical: CHBrClF (one chiral centre); C; C; CH3​; H; CH3​; H; cis-but-2-ene; C; C; CH3​; H; CH3​ cis / trans: but-2-ene optical: CHBrClF (one chiral centre) C C CH3​ H CH3​ H cis-but-2-ene C C CH3​ H CH3​ H trans-but-2-ene CH3​ groups on the same side (cis) or on opposite sides (trans) no rotation about C=C (π bond) C Br H F Cl C Br H Cl F mirror plane four different groups on one C non-superimposable mirror images (enantiomers); each rotates plane-polarised light oppositely
Cis/trans needs restricted rotation and two different groups on each carbon of the C=C. Optical isomerism needs a carbon bonded to four different atoms or groups (a chiral centre). The wedge shows a bond towards you and the dashes a bond away from you.

Cis/trans (geometrical) isomerism needs:

In the cis isomer the two matching groups are on the same side; in the trans isomer they are on opposite sides. The isomers are different compounds, with different physical properties (boiling point, melting point) and sometimes different reactivity. They cannot interconvert at room temperature because the π bond would have to break. Use cis and trans to describe where the matching (identical) groups on each carbon sit. Examples: but-2-ene, pent-2-ene, 1,2-dibromoethene. Not possible: propene, but-1-ene, 2-methylpropene (one carbon of the C=C carries two identical groups).

Optical isomerism arises when a molecule has a chiral centre: a carbon atom bonded to four different atoms or groups. The molecule and its mirror image cannot be superimposed (like left and right hands). The two forms are enantiomers. They have identical physical properties, except that each rotates the plane of plane-polarised light by the same angle in opposite directions.

⚠️ Examiner Trap: Chiral means four different groups, not four groups. Check each carbon atom in turn and write down its four groups. Two groups that look different but are the same chain read from the other direction (as in a symmetrical ring) make the carbon achiral. Include the hidden H in skeletal formulae, and put an asterisk on each centre you find.

6. Bond Fission, Carbocations and Reaction Types

Organic reactions involve breaking and making covalent bonds. A bond can break in two ways.

homolytic fissionDiagram showing: homolytic fission; heterolytic fission; Cl; Cl; →; Cl•​; +; Cl•​; each atom takes ONE electron; forms two free radicals (unpaired electron); needs UV light or heat; half-headed arrow = ONE electron; CH3​; Br homolytic fission heterolytic fission Cl Cl → Cl•​ + Cl•​ each atom takes ONE electron forms two free radicals (unpaired electron) needs UV light or heat half-headed arrow = ONE electron CH3​ Br → CH3​+​ + Br−​ more electronegative atom takes BOTH electrons forms carbocation (+) and anion (−) full-headed arrow = electron PAIR initiation, propagation, termination: free-radical substitution carbocation: electrophilic addition, nucleophilic substitution
Homolytic fission gives two free radicals (one electron each). Heterolytic fission gives ions: the more electronegative atom keeps both electrons, leaving a carbocation if carbon is the less electronegative atom.

Nucleophile: an electron-rich species that donates a pair of electrons to form a covalent bond (for example OH−, CN−, NH3, H2O). It is attracted to a δ+ or positive centre. Electrophile: an electron-deficient species that accepts a pair of electrons (for example H+, a carbocation, or the δ+ hydrogen of HBr).

primary (1°)Diagram showing: primary (1°); C+​; R; H; H; CH3​CH2​+​; 1 alkyl group; least stable; secondary (2°); C+​; R; R; H; (CH3​)2​CH+​ primary (1°) C+​ R H H CH3​CH2​+​ 1 alkyl group least stable secondary (2°) C+​ R R H (CH3​)2​CH+​ 2 alkyl groups tertiary (3°) C+​ R R R (CH3​)3​C+​ 3 alkyl groups most stable stability increases
Alkyl groups have a positive inductive effect: they push electron density towards the electron-deficient C+, spreading the charge. More alkyl groups mean a more stable carbocation, so tertiary > secondary > primary.

Alkyl groups donate electron density along the bond (the positive inductive effect), which reduces the charge on C+. The more alkyl groups, the more stable the carbocation: tertiary > secondary > primary. This explains the products of addition to alkenes (Topic 14) and the substitution mechanisms in Topic 15.

Types of reaction

Type What happens Example
Addition two molecules join to give one C2H4 + Br2 → C2H4Br2
Substitution one atom or group is replaced by another CH4 + Cl2 → CH3Cl + HCl
Elimination a small molecule is removed, forming a C=C C2H5OH → C2H4 + H2O
Hydrolysis breakdown by water (faster with acid or alkali) C2H5Br + NaOH → C2H5OH + NaBr
Condensation addition followed by loss of a small molecule such as H2O CH3COOH + C2H5OH → CH3COOC2H5 + H2O
Oxidation gain of O or loss of H (use [O]) C2H5OH + [O] → CH3CHO + H2O
Reduction gain of H or loss of O (use [H]) CH3COCH3 + 2[H] → CH3CH(OH)CH3

[O] and [H] stand for an oxygen or hydrogen atom supplied by the reagent. Equations written with them must still balance: complete oxidation of ethanol to ethanoic acid is C2H5OH + 2[O] → CH3COOH + H2O.

⚠️ Examiner Trap: Curly arrows start at electrons, not atoms. Draw a full-headed arrow starting at the middle of a bond or on a lone pair and ending at the atom that receives the pair. A single-headed (fish-hook) arrow is used only for radicals. If you write “nucleophile” or “electrophile”, say the pair of electrons is donated or accepted.

7. Quick Sheet and Checklist

Topic Key point
Empirical formula simplest ratio; moles ÷ smallest; scale to whole numbers
Skeletal formula each corner or end is C; draw H in −OH and all non-C atoms
Naming longest chain; lowest numbers; alkyl groups in alphabetical order
Alkane, alkene general formulae CnH2n+2; CnH2n
σ and π single = σ; double = σ + π; triple = σ + 2π
Hybridisation sp3 109.5°; sp2 120° planar; sp 180° linear
Structural isomers chain, position, functional group
Cis/trans restricted rotation and two different groups on each C=C carbon
Optical isomers chiral centre (four different groups); non-superimposable mirror images
Fission homolytic: radicals, UV; heterolytic: ions, polar bond
Nucleophile / electrophile donates / accepts a pair of electrons
Carbocation stability tertiary > secondary > primary

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