Basic concepts of organic chemistry — OCR A-Level Chemistry

Organic formulae are written in several complementary ways, and each answers a different question. A molecular formula gives the total number of each atom, so ethanol is C₂H₆O. An empirical formula gives the simplest whole-number ratio of atoms; for ethanol the ratio 2:6:1 cannot be divided further, so the empirical formula is also C₂H₆O, whereas ethanoic acid, C₂H₄O₂, simplifies to CH₂O. A general formula applies to a homologous series, such as CₙH₂ₙ₊₂ for alkanes. A structural formula shows how atoms are joined, and a displayed formula draws every bond. Systematic IUPAC names are built from a stem for the carbon chain, a suffix for the principal functional group and prefixes for substituents, numbered to give the lowest locants. Learn to move between name, molecular, empirical, structural and displayed forms, and to check that each representation is consistent with the others.

IUPAC nomenclature gives every organic compound a single systematic name. Apply the rules in order. First find the longest carbon chain containing the principal functional group; this gives the stem (meth-, eth-, prop-, but- and so on). Identify and name any functional group, using its suffix where it takes priority (for example -ol, -oic acid, -al). Number the chain so the principal group, or the first point of difference, gets the lowest possible locant. Name and locate side chains as prefixes (methyl-, ethyl-, chloro-) in alphabetical order, not numerical order. Use di-, tri- and so on for repeated identical groups. Combine the parts into one word with hyphens and commas between numbers. Practise working from a displayed or skeletal formula to a name, and back again.

This criterion asks you to interpret and use the vocabulary that underpins all organic chemistry, so you can read a structure or a name and say precisely what each term means. The terms in this sub-section include homologous series, functional group, general formula, structural formula, displayed formula, skeletal formula and empirical formula, together with the idea of a homologous series whose members differ by CH₂. Interpretation means matching a definition to an example; use means applying the right term to a given structure or formula. For instance, given CH₃CH₂CH₂OH you should recognise the OH as the functional group, the compound as a member of the alcohols, and its displayed formula as one showing every bond. Practise converting one representation into another and naming the term each representation illustrates.

A general formula is the simplest algebraic expression that represents every member of a homologous series, using the letter n for the number of carbon atoms. For alkanes it is CₙH₂ₙ₊₂, so with n = 1 you get CH₄, with n = 2 you get C₂H₆ and with n = 3 you get C₃H₈. To use it, substitute the value of n and evaluate the hydrogen subscript: for n = 4, 2n + 2 = 10, giving C₄H₁₀. The formula must hold for every member of the series, which is why it is written algebraically rather than as a single molecular formula. Practise substituting several values of n and checking that each result matches a real alkane, and be ready to recognise the general formula of other series such as alkenes, CₙH₂ₙ.

A structural formula gives the minimal detail that shows how the atoms in a molecule are arranged. It lists each carbon with the atoms or groups attached to it, in order along the chain, but it does not draw every bond as a line. For butane, C₄H₁₀, the full form is CH₃CH₂CH₂CH₃, and the condensed form CH₃(CH₂)₂CH₃ is equally acceptable because the subscript 2 shows two CH₂ units repeated. To write one, identify the longest carbon chain, place the correct hydrogen count on each carbon so each carbon has four bonds, and enclose repeated identical groups in brackets with a subscript. Structural formulae differ from displayed formulae, which show every bond, and from molecular formulae such as C₄H₁₀, which show only the numbers of atoms.

A displayed formula shows every atom and every bond in a molecule as separate lines, so the relative positioning of atoms is explicit. For ethanol, the molecular formula is C₂H₆O. The structure is CH₃CH₂OH: one carbon is bonded to three hydrogen atoms and to the second carbon; the second carbon is bonded to two hydrogen atoms, to the first carbon and to the oxygen atom; the oxygen is bonded to that carbon and to one hydrogen. In a displayed formula each of these bonds is drawn as a line, and every hydrogen is shown. This distinguishes a displayed formula from a structural formula such as CH₃CH₂OH, which groups atoms, and from a skeletal formula, which omits hydrogens on carbon. When reading an MCQ, check that each carbon has exactly four bonds and that the OH group sits on the correct carbon.

A skeletal formula is a simplified organic formula. You remove the hydrogen atoms attached to carbon in the alkyl chains, so only the carbon skeleton and the associated functional groups are drawn. Each vertex and each end of a line is a carbon atom, and the hydrogen atoms on those carbons are understood rather than written. For butan-2-ol, CH₃CH(OH)CH₂CH₃, the skeleton is a four-carbon zigzag chain; the OH group sits on the second carbon from one end, so the line drawing shows a three-segment chain with OH attached to the middle vertex. Count vertices to confirm four carbons, then place the functional group on the correct carbon.

This reading introduces functional groups as the reactive parts of organic molecules that determine characteristic reactions. Work through the specification's named groups: alkene, alcohol, haloalkane, aldehyde, ketone, carboxylic acid, ester, amine, nitrile and acyl chloride, plus the benzene ring. For each, learn the displayed and skeletal representation, the suffix or prefix used in naming, and one reaction it undergoes. Build a table linking group, structure, name ending and a typical reaction, then test yourself by identifying groups in unfamiliar molecules. Use the specification wording as your scope; do not add groups outside it.

This row is a lead-in to the terminology that follows in section 4.1.1, so treat it as course guidance rather than a stand-alone assessed statement. Read the sub-items that follow (such as homologous series) as the actual terms you must interpret and use. For each term, learn both halves of the job: interpret it, meaning you can explain what it means and recognise a correct example, and use it, meaning you can apply it accurately when describing or naming a compound. Build a glossary in your own words, add one clear example and one near-miss for each term, and check your wording against the specification definition rather than a textbook paraphrase.

A homologous series is a family of organic compounds that share the same functional group, so they react in similar ways, and in which each successive member differs from the previous one by a CH₂ unit. That CH₂ step means relative molecular mass rises by 14 each time (C = 12, H₂ = 2, so 12 + 2 = 14) and the general formula stays the same. For example, methanol CH₃OH, ethanol C₂H₅OH and propan-1-ol C₃H₇OH form part of the alcohol series: each adds one carbon and two hydrogens. Members also show a gradual trend in physical properties such as boiling point as chain length increases. Check that a proposed pair really shares the functional group and differs by exactly CH₂ before calling them successive members.

A functional group is a group of atoms within a molecule that is responsible for the characteristic reactions of a compound. The definition has two parts that must both hold: the group is a defined cluster of atoms, and it governs how the compound typically reacts. So the hydroxyl group in ethanol, written as OH, makes ethanol behave as an alcohol, reacting with sodium to give hydrogen and with acidified dichromate(VI) to give ethanoic acid. The hydrocarbon chain attached to the group mainly tunes physical properties such as boiling point; the functional group drives the chemistry. Recognise the group from a displayed or skeletal formula, name it, and predict the reaction type it allows.

An alkyl group is a saturated hydrocarbon group with the general formula CₙH₂ₙ₊₁, obtained by removing one hydrogen atom from an alkane. The subscript relationship matters: for n carbon atoms there are 2n + 1 hydrogen atoms. So a methyl group has n = 1 and so CH₃, an ethyl group has n = 2 and so C₂H₅, and a propyl group has n = 3 and so C₃H₇. Check any candidate formula by substituting n and confirming the hydrogen count equals 2n + 1. Alkyl groups appear as branches on a carbon chain and as the carbon part of alcohols and haloalkanes. They are saturated, so they contain only single carbon–carbon bonds.

An aliphatic compound is one whose carbon skeleton is not aromatic: the carbon and hydrogen atoms are joined together in straight chains, branched chains or non-aromatic rings. So propane (a straight chain), 2-methylbutane (a branched chain) and cyclohexane (a non-aromatic ring) are all aliphatic, while benzene, with its delocalised ring, is not. When a multiple-choice option offers a structure, decide whether any aromatic ring is present; if none is, the compound is aliphatic. Check the whole structure, because a molecule can contain a ring and still be aliphatic provided that ring is not aromatic.

An alicyclic compound is a subset of the aliphatic compounds: it is aliphatic and its carbon atoms are arranged in a non-aromatic ring, which may or may not carry side chains. Cyclohexane is alicyclic with no side chain; methylcyclohexane is alicyclic with a side chain. Because alicyclic sits inside aliphatic, every alicyclic compound is also aliphatic, but not every aliphatic compound is alicyclic, since straight and branched chains are aliphatic without a ring. When choosing an option, confirm two things: a ring is present, and that ring is not aromatic.

An aromatic compound is one that contains a benzene ring, C₆H₆, drawn as a planar hexagon of six carbon atoms with one hydrogen on each carbon and delocalised π electrons above and below the ring. The ring is unsaturated because of its delocalised π system, yet it is not a simple alternating set of three C=C bonds: the six π electrons are shared over all six carbons. Recognise aromaticity by locating the ring in a displayed or skeletal formula, for example phenylamine C₆H₅NH₂ or methylbenzene C₆H₅CH₃, and distinguish it from a cycloalkane such as cyclohexane C₆H₁₂, which has no ring π system. In a multiple-choice question, test each option for the presence of a benzene ring rather than for the word 'ring' alone.

A saturated compound contains only single carbon–carbon bonds, so every carbon is bonded to four atoms by σ bonds, as in ethane CH₃CH₃ or cyclohexane C₆H₁₂. An unsaturated compound contains at least one multiple carbon–carbon bond: a C=C double bond as in ethene CH₂=CH₂, a C≡C triple bond as in ethyne HC≡CH, or an aromatic benzene ring whose delocalised π system also counts as unsaturation. To classify a structure, inspect every carbon–carbon link and ask whether any bond order exceeds one. A molecule can be unsaturated through a ring alone, for example benzene C₆H₆, even though no isolated double bond is drawn. In a multiple-choice question, test each option against all three sources of unsaturation rather than only looking for C=C.

A homologous series is a family of organic compounds with the same functional group, the same general formula and similar chemical properties, where successive members differ by a CH₂ unit. To predict a member's formula, take the general formula for that series and substitute the value of n for the member required. For alkanes the general formula is CₙH₂ₙ₊₂, so the member with n = 4 is C₄H₁₀. For alkenes CₙH₂ₙ gives C₄H₈ when n = 4. For alcohols CₙH₂ₙ₊₁OH gives C₄H₉OH when n = 4. Always identify the series first, then apply the correct general formula and substitute n carefully, checking the arithmetic in the subscript.

This topic heading introduces isomerism, the existence of compounds with the same molecular formula but different arrangements of atoms. Structural isomers differ in the order in which atoms are joined: chain isomers vary the carbon skeleton, position isomers place a functional group at a different carbon, and functional group isomers contain different functional groups. Stereoisomerism arises where atoms are joined in the same order but arranged differently in space, including E/Z isomerism about a C=C double bond and optical isomerism about a chiral carbon. Read the specification to see which types are required, and practise drawing and naming isomers of a given molecular formula.

Structural isomers are compounds that share the same molecular formula but have different structural formulae, meaning the atoms are joined in a different order or arrangement. For example, C₄H₁₀ has two structural isomers: butane, a straight chain, and 2-methylpropane, a branched chain. To determine possible structural formulae from a molecular formula, first work out the degree of unsaturation from the carbon and hydrogen count, then build every distinct carbon skeleton, and finally place functional groups or multiple bonds in each non-equivalent position. Check each candidate by counting atoms and confirming the valency of carbon (four bonds), hydrogen (one), oxygen (two) and nitrogen (three). Two drawings that can be rotated or reflected into each other are the same isomer, not two.

This section introduces reaction mechanisms as step-by-step accounts of how organic reactions occur, using curly arrows to show the movement of electron pairs. Read it to learn the conventions: a curly arrow starts at a bond or a lone pair and points to where the electron pair moves, forming a new bond or breaking an old one. Mechanisms distinguish bond fission by homolytic splitting, giving radicals, from heterolytic splitting, giving ions. Recognise the meaning of electrophile, nucleophile and radical, and how these labels connect to the arrow conventions. Use the section to build the habit of checking that every arrow obeys the octet and charge rules, and that the overall equation matches the sum of the steps.

This heading introduces how a covalent bond can break, and the specification splits the topic into two named types: homolytic fission and heterolytic fission. Read the two sub-statements together rather than in isolation. In homolytic fission the bonded pair is shared out, so each bonding atom receives one electron from the pair and two radicals form. In heterolytic fission the pair is not shared: one atom takes both electrons and the fragments become ions. Learn to draw each process with single-headed (half) curly arrows for homolytic fission and full curly arrows for heterolytic fission, and to label the products correctly as radicals or as ions. Check each product's electron count so you can justify why a radical has an unpaired electron and why an ion carries a charge.

Homolytic fission is the breaking of a covalent bond in which the shared pair is split evenly. Each of the two bonding atoms takes one electron from the bonded pair, so neither fragment gains or loses an electron overall. Because each atom keeps one of the two electrons, each fragment is left with an unpaired electron and is therefore a radical. For example, breaking a Cl–Cl bond homolytically gives two chlorine radicals, each written with a dot to show the unpaired electron. This contrasts with heterolytic fission, where one atom takes both electrons and ions form. In multiple-choice questions, look for the option that states one electron per atom and two radicals, and reject options that describe unequal sharing or ion formation.

Heterolytic fission is the breaking of a covalent bond in which both electrons of the shared pair move to one of the two bonded atoms. The bond splits unequally, so the atom that keeps the electron pair gains a lone pair and becomes negatively charged, while the other atom loses its share and becomes positively charged. For example, when the C–Br bond in bromoethane breaks heterolytically, the more electronegative bromine takes both electrons and forms Br⁻, and the carbon becomes a carbocation, C₂H₅⁺. Contrast this with homolytic fission, where each atom takes one electron and two radicals form. Recognise heterolytic fission by the paired-electron transfer and by the resulting ions of opposite charge.

A radical is a species that contains an unpaired electron, which makes it highly reactive. In mechanisms, a radical is shown with a dot beside its formula to represent that single unpaired electron, for example Cl• or •CH₃. The dot is not a charge and does not mean the species is an ion; it records one electron that is not paired. Radicals form when a bond breaks homolytically, so each atom takes one electron of the shared pair. Recognise a radical by the presence of the dot and by the unpaired electron it stands for, and distinguish it from an ion, which carries a full charge.

A curly arrow is a bookkeeping symbol for the movement of one electron pair during a reaction. The tail marks where that pair comes from: a lone pair on an atom, or the electron density of a covalent bond. The arrowhead points at the atom that receives the pair, so the pair ends up either as a new lone pair or as a new covalent bond. Heterolytic fission is shown by an arrow starting on a bond and pointing to one of the two atoms, which takes both electrons and becomes a negative ion while the other becomes positive. Bond formation is shown by an arrow from a lone pair on one atom to the atom it attacks. Draw the arrow once, from source to receiving atom, and never from a partial charge.

A reaction mechanism is a step-by-step diagram of how bonds break and form, and it must show electron movement explicitly. Each curly arrow represents one electron pair and runs from its source, such as a lone pair or a bond, to its destination, such as an atom or the gap where a new bond forms. Mechanisms also show relevant dipoles, because a polar bond such as C–Cl or C=O carries partial charges that direct where an attacking species sends its pair. Mark the dipole with δ⁺ and δ⁻ on the bonded atoms, choosing the more electronegative atom for δ⁻. Then draw arrows consistent with those charges: an electron-rich species attacks the δ⁺ centre. Check that every arrow starts and ends on a real electron source and destination, and that charges in the products follow from the arrows drawn.