Shapes of carbon compounds, structural formulae and functional groups
Link to Std 9–10 : You learnt that carbon forms millions of compounds because it makes four covalent bonds; now we see the three-dimensional shapes and the short ways to draw them.
A few Lego blocks can build hundreds of models. Carbon atoms join in chains, branches and rings in the same way, and a small group of atoms decides how the molecule behaves.
A functional group is an atom or group of atoms that decides the chemical properties of an organic compound (for example –OH, –CHO, –COOH). A homologous series is a family of compounds with the same functional group, in which each member differs from the next by CH₂.
In the figure : The sp³ carbon is tetrahedral (109.5°), the sp² carbon is planar (120°) and the sp carbon is linear (180°). Below, butane written in the bond-line style.
Rule 1: Shape follows hybridisation — Four single bonds give sp³ (tetrahedral); one double bond gives sp² (planar); one triple bond gives sp (linear).
- Easy : Methane (CH₄) is tetrahedral with bond angle 109.5°.
- Medium : Ethene (C₂H₄) is planar with H–C–H angles near 120°.
- Exam-type : Ethyne (C₂H₂) is linear, with bond angle 180°.
Rule 2: Ways of writing structures — Complete (all bonds drawn), condensed (CH₃CH₂CH₂CH₃) and bond-line (zig-zag lines, with a carbon at each corner and end and hydrogens understood).
- Easy : Butane as condensed formula is CH₃CH₂CH₂CH₃; the bond-line form is a zig-zag with 4 carbons.
- Medium : In a bond-line drawing of pentane the zig-zag has 5 corners and ends, so 5 carbons.
- Exam-type : Each carbon must show four bonds, so hydrogens are counted as 4 minus the lines touching the carbon.
Rule 3: Functional groups and families — –OH (alcohol), –CHO (aldehyde), >C=O (ketone), –COOH (carboxylic acid), –NH₂ (amine), –X (halide), C=C (alkene), C≡C (alkyne).
- Easy : Ethanol (CH₃CH₂OH) has the –OH group, so it is an alcohol.
- Medium : Propanal (CH₃CH₂CHO) is an aldehyde, while propanone (CH₃COCH₃) is a ketone.
- Exam-type : Ethanoic acid (CH₃COOH) has the –COOH group, a carboxylic acid.
Rule 4: Homologous series — Successive members differ by CH₂ (14 mass units) and show a gradual change in physical properties with similar chemical properties.
- Easy : CH₄, C₂H₆, C₃H₈ and C₄H₁₀ are members of the alkane series.
- Medium : The molecular mass changes by 14 from one member to the next: 16, 30, 44, 58.
- Exam-type : General formulae: alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alkynes CₙH₂ₙ₋₂.
Carbon always makes four bonds; the shape depends on how many of them are double or triple.
Forgetting the hydrogens in a bond-line drawing. Every carbon must reach four bonds.
Memory clue: "4-3-2 sets 109°, 120°, 180°" — four groups around carbon, three groups, or two groups.
Spot the difference
| Point | Aldehyde | Ketone |
|---|---|---|
| Group | –CHO at the end | >C=O inside the chain |
| Example | Propanal | Propanone |
| Formula | CH₃CH₂CHO | CH₃COCH₃ |
Practice questions
- 1What is a functional group?
- 2What is the shape of the carbon atom in methane?
- 3Name the functional group in CH₃COOH.
- 4What is the molecular mass difference between consecutive members of a homologous series?
- 5Write the condensed formula of butane.
- 6State the general formula of alkenes.
- 7How many hydrogen atoms are on a carbon in a bond-line drawing that touches 2 lines?
- 8Write the molecular formula of the bond-line drawing of a zig-zag with 5 carbons and only single bonds.
Show answers (try them yourself first)
- 1An atom or group of atoms that decides the chemical properties of a compound.
- 2Tetrahedral.
- 3The carboxylic acid group, –COOH.
- 414 units (the mass of CH₂).
- 5CH₃CH₂CH₂CH₃.
- 6CₙH₂ₙ.
- 74 − 2 = 2 hydrogen atoms.
- 8C₅H₁₂ (the alkane formula 2n + 2 with n = 5).
IUPAC nomenclature
Link to Std 9–10 : You named a few alkanes and alcohols; here we use a complete set of steps that works for any chain.
A postal address works because it follows a fixed order: house, street, city. An IUPAC name follows a fixed order too, so everyone can build the structure from the name.
An IUPAC name has three parts: a prefix (branches and substituents), a root (the number of carbons in the longest chain) and a suffix (the principal functional group), as in 2-methylpropan-1-ol.
In the figure : 3-Ethyl-2-methylpentane: the longest chain has 5 carbons, a methyl group sits on carbon 2 and an ethyl group on carbon 3.
Rule 1: Root words — Meth (1), eth (2), prop (3), but (4), pent (5), hex (6), hept (7), oct (8), non (9), dec (10).
- Easy : CH₃CH₂CH₃ has 3 carbons, so it is propane.
- Medium : A chain of 6 carbons with a double bond is hexene.
- Exam-type : A chain of 8 carbons with only single bonds is octane (C₈H₁₈).
Rule 2: Longest chain and lowest numbers — Choose the longest chain containing the principal functional group; number from the end that gives the lowest locants to the functional group, then to the multiple bond, then to the branches.
- Easy : CH₃CH(CH₃)CH₂CH₃ has a 4-carbon chain with a methyl on carbon 2: 2-methylbutane.
- Medium : C(CH₃)₄ has a 3-carbon chain with two methyls on carbon 2: 2,2-dimethylpropane.
- Exam-type : CH₃CH(CH₃)CH(C₂H₅)CH₂CH₃ has a 5-carbon chain, a methyl on 2 and an ethyl on 3: 3-ethyl-2-methylpentane.
Rule 3: Functional groups as suffix — –OH becomes "-ol", –CHO "-al", >C=O "-one", –COOH "-oic acid", C=C "-ene", C≡C "-yne". Halogens are prefixes (chloro, bromo).
- Easy : CH₃CH(OH)CH₃ is propan-2-ol.
- Medium : CH₃CH₂CH₂COOH is butanoic acid (the carboxyl carbon is carbon 1).
- Exam-type : CH₃CH₂COCH₂CH₂CH₃ has a ketone on carbon 3 of a 6-carbon chain: hexan-3-one.
Rule 4: Multiple bonds and halogens — Number the double or triple bond from the end that gives it the lowest number; name the halogen as a prefix and order prefixes alphabetically.
- Easy : CH₃CH=CHCH₃ is but-2-ene.
- Medium : CH₃CH₂C≡CH is but-1-yne.
- Exam-type : CH₃CHClCH₃ is 2-chloropropane; CH₃CH₂Br is bromoethane.
Longest chain, lowest numbers, alphabetical prefixes.
Numbering from the wrong end. Check that the functional group gets the lowest number.
Memory clue: "Prefix – Root – Suffix" reads branch, chain, group.
Spot the difference
| Point | Prefix | Suffix |
|---|---|---|
| Shows | Branches and halogens | Principal functional group |
| Example | 2-methyl, chloro | -ol, -oic acid |
| Order | Alphabetical | At the end of the name |
Practice questions
- 1Name CH₃CH₂CH₂CH₂CH₃.
- 2Give the IUPAC name of CH₃CH₂OH.
- 3What is the root word for 4 carbon atoms?
- 4Name CH₃CH(CH₃)CH₂CH₃.
- 5Name CH₃CH=CHCH₃.
- 6Name CH₃CH₂CH₂COOH.
- 7Name CH₃CH₂COCH₂CH₂CH₃.
- 8Name CH₃CH(CH₃)CH(C₂H₅)CH₂CH₃.
Show answers (try them yourself first)
- 1Pentane.
- 2Ethanol.
- 3But.
- 42-Methylbutane.
- 5But-2-ene.
- 6Butanoic acid.
- 7Hexan-3-one.
- 83-Ethyl-2-methylpentane.
Isomerism
Link to Std 9–10 : You know that the same elements can combine to form different compounds; isomers are the clearest case: one formula, many structures.
The same six Lego blocks can be joined into a straight line or into a branched shape. The blocks are the same but the models are different.
Isomers are compounds with the same molecular formula but different structures or arrangements in space. In structural isomerism the atoms are connected differently; in stereoisomerism the connections are the same but the arrangement in space differs.
In the figure : Structural isomerism has four kinds (chain, position, functional, metamerism); stereoisomerism has geometrical and optical kinds.
Rule 1: Chain and position isomerism — Chain isomers differ in the carbon skeleton. Position isomers differ in the place of the functional group or multiple bond.
- Easy : C₄H₁₀ has two chain isomers: butane and 2-methylpropane.
- Medium : C₅H₁₂ has three chain isomers: pentane, 2-methylbutane and 2,2-dimethylpropane.
- Exam-type : C₄H₉Cl has four isomers in all: 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane and 2-chloro-2-methylpropane.
Rule 2: Functional and metamerism — Functional isomers have different functional groups. Metamers have the same group but different alkyl groups on each side of it.
- Easy : C₂H₆O has ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃).
- Medium : C₃H₈O has propan-1-ol, propan-2-ol and methoxyethane.
- Exam-type : Diethyl ether (C₂H₅OC₂H₅) and methyl propyl ether (CH₃OC₃H₇) are metamers of formula C₄H₁₀O.
Rule 3: Geometrical isomerism — A double bond cannot rotate, so groups on its two carbons may lie on the same side (cis) or opposite sides (trans). Each carbon of the double bond needs two different groups.
- Easy : But-2-ene exists as cis and trans forms.
- Medium : But-1-ene has no cis–trans forms because one carbon of the double bond carries two H atoms.
- Exam-type : Trans isomers are usually more stable and have a lower dipole moment than cis because of less crowding and cancelled dipoles.
Rule 4: Optical isomerism (introduction) — A carbon with four different groups (a chiral carbon) gives two non-superimposable mirror-image forms.
- Easy : In CH₃CH(OH)COOH the middle carbon has four different groups, so it is chiral.
- Medium : The two mirror-image forms rotate plane-polarised light in opposite directions.
- Exam-type : CH₃CH₂OH has no chiral carbon, since each carbon carries at least two identical groups.
Isomers have the same molecular formula but different properties.
Counting a rotated drawing of the same molecule as a new isomer.
Memory clue: "Chain, Position, Function, Meta" are the four structural types.
Spot the difference
| Point | Structural isomers | Stereoisomers |
|---|---|---|
| Connection of atoms | Different | Same |
| Example | Butane, 2-methylpropane | cis- and trans-but-2-ene |
| Differ in | Structure | Arrangement in space |
Practice questions
- 1Define isomers.
- 2How many isomers does C₄H₁₀ have?
- 3Give one pair of functional isomers.
- 4Draw and name the three isomers of C₅H₁₂.
- 5Why does but-1-ene not show cis–trans isomerism?
- 6Name the isomers of C₃H₈O.
- 7How many isomers does C₄H₉Cl have? Name them.
- 8Which carbon in CH₃CH(OH)COOH is chiral and why?
Show answers (try them yourself first)
- 1Compounds with the same molecular formula but different structures.
- 2Two: butane and 2-methylpropane.
- 3Ethanol and methoxymethane (C₂H₆O).
- 4Pentane, 2-methylbutane, 2,2-dimethylpropane.
- 5One carbon of the double bond has two identical hydrogens.
- 6Propan-1-ol, propan-2-ol and methoxyethane.
- 7Four: 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane, 2-chloro-2-methylpropane.
- 8The middle carbon, because it is bonded to four different groups: –H, –OH, –CH₃ and –COOH.
Electronic effects: inductive, resonance and hyperconjugation
Link to Std 9–10 : You know that atoms differ in electronegativity; this pull shifts electrons through a molecule and decides how strong an acid or how stable an intermediate is.
In a tug of war, a strong person at one end drags the whole rope towards them. An electronegative atom pulls the electrons along the chain in a similar way.
The inductive effect is the permanent shift of sigma electrons along a chain due to an electronegative atom or group. Resonance is the delocalisation of pi electrons or lone pairs. Hyperconjugation is the delocalisation of electrons of a C–H sigma bond into a nearby empty or pi orbital.
In the figure : The stability of carbocations grows from CH₃⁺ to primary, secondary and tertiary, because more alkyl groups give more electron release and more hyperconjugation.
Rule 1: Inductive effect — Groups such as –Cl, –NO₂ and –COOH pull electrons (–I effect); alkyl groups push electrons (+I effect). The effect weakens quickly with distance.
- Easy : In Cl–CH₂–, chlorine pulls electrons towards itself (–I).
- Medium : ClCH₂COOH is a stronger acid than CH₃COOH because Cl pulls electrons and stabilises the carboxylate ion.
- Exam-type : Acid strength: CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH, because more Cl atoms give a larger –I effect.
Rule 2: Resonance effect — A group may release electrons into a pi system (+R, for example –OH, –NH₂) or withdraw them (–R, for example –NO₂, –CHO).
- Easy : In phenol, the lone pair of oxygen is delocalised into the ring (+R effect).
- Medium : Phenoxide ion is stabilised by resonance, so phenol is more acidic than ethanol.
- Exam-type : In nitrobenzene the –NO₂ group withdraws electron density from the ring (–R), lowering its reactivity.
Rule 3: Hyperconjugation — The more C–H bonds next to a positive centre (α-hydrogens), the more stable the carbocation.
- Easy : CH₃⁺ has 0 α-hydrogens.
- Medium : The ethyl cation has 3, isopropyl has 6 and tert-butyl has 9 α-hydrogens.
- Exam-type : Stability: tert-butyl > isopropyl > ethyl > methyl cation, in line with the number of α-hydrogens.
Rule 4: Electromeric effect and summary — The electromeric effect is a temporary shift of pi electrons to one atom under the attack of a reagent. It appears only during the reaction.
- Easy : In C=O the pi electrons shift to oxygen when a nucleophile attacks.
- Medium : The inductive and resonance effects are permanent; the electromeric effect is temporary.
- Exam-type : In C=C attacked by an electrophile (E⁺), the pi pair moves to the carbon that will bond to E⁺.
Electron-withdrawing groups increase acidity; electron-releasing groups decrease it.
Forgetting that the inductive effect fades with distance. It is strongest on the nearest atom.
Memory clue: "Alkyl pushes, halogen pulls".
Spot the difference
| Point | Inductive effect | Resonance effect |
|---|---|---|
| Electrons | Sigma | Pi or lone pair |
| Distance | Fades quickly | Acts across the conjugated system |
| Nature | Permanent | Permanent |
Practice questions
- 1Define the inductive effect.
- 2Name one group with –I effect and one with +I effect.
- 3How many α-hydrogens are in a tert-butyl cation?
- 4Why is ClCH₂COOH a stronger acid than CH₃COOH?
- 5Arrange CH₃⁺, CH₃CH₂⁺ and (CH₃)₂CH⁺ in increasing stability.
- 6Why is phenol more acidic than ethanol?
- 7Arrange CCl₃COOH, CH₃COOH and CH₂ClCOOH in decreasing acid strength.
- 8Distinguish between the inductive and the electromeric effect.
Show answers (try them yourself first)
- 1The permanent shift of sigma electrons along a chain due to an electronegative atom or group.
- 2–Cl (–I) and –CH₃ (+I).
- 3Nine.
- 4Cl pulls electrons away from the carboxylate ion and stabilises it.
- 5CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺.
- 6The phenoxide ion is stabilised by resonance with the ring.
- 7CCl₃COOH > CH₂ClCOOH > CH₃COOH.
- 8The inductive effect is a permanent shift of sigma electrons; the electromeric effect is a temporary shift of pi electrons during a reaction.
Bond fission, reactive intermediates and types of reactions
Link to Std 9–10 : You know that a reaction breaks old bonds and forms new ones; now we study how the bond breaks and what short-lived particles form.
A chocolate bar can be broken into two equal pieces, or one person may keep the bigger piece. Bonds also break either evenly or unevenly.
In homolytic fission each atom takes one electron of the shared pair and forms a free radical. In heterolytic fission one atom takes both electrons and ions form (a carbocation and a carbanion). An electrophile seeks electrons; a nucleophile donates an electron pair.
In the figure : Homolytic fission gives two free radicals; heterolytic fission gives a positive and a negative ion.
Rule 1: Types of bond fission — Homolytic fission occurs in non-polar bonds, in heat or light. Heterolytic fission occurs in polar bonds, in polar solvents.
- Easy : Cl–Cl in sunlight breaks into 2 Cl• (homolytic).
- Medium : (CH₃)₃C–Br in a polar solvent gives (CH₃)₃C⁺ and Br⁻ (heterolytic).
- Exam-type : In heterolytic fission the more electronegative atom takes the electron pair.
Rule 2: Electrophiles and nucleophiles — Electrophiles are electron-poor (H⁺, NO₂⁺, carbocations, AlCl₃). Nucleophiles are electron-rich (OH⁻, CN⁻, NH₃, H₂O).
- Easy : H⁺ is an electrophile; OH⁻ is a nucleophile.
- Medium : NH₃ is a nucleophile because the nitrogen has a lone pair to donate.
- Exam-type : In the reaction CH₃Br + OH⁻ → CH₃OH + Br⁻, OH⁻ is the nucleophile attacking the electron-poor carbon.
Rule 3: Stability of intermediates — Carbocation: 3° > 2° > 1° > methyl. Free radical: the same order. Carbanion: methyl > 1° > 2° > 3°, because alkyl groups push electrons onto a carbon that already has too many.
- Easy : The tertiary carbocation is the most stable.
- Medium : The tertiary carbanion is the least stable.
- Exam-type : A carbocation has 6 electrons and is sp² hybridised and planar; a carbanion is pyramidal and a free radical is near planar.
Rule 4: Types of organic reactions — Substitution (an atom or group replaced), addition (to a multiple bond), elimination (a small molecule lost to form a double bond) and rearrangement.
- Easy : CH₄ + Cl₂ → CH₃Cl + HCl is a substitution.
- Medium : CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br is an addition.
- Exam-type : CH₃CH₂OH → CH₂=CH₂ + H₂O (with conc. H₂SO₄) is an elimination.
Electrophile means "electron-loving" (it needs electrons); nucleophile means "nucleus-loving" (it has electrons).
Calling a carbocation stable because it is positive. It is stabilised only by the groups around it.
Memory clue: "Nucleophile Gives, Electrophile Takes" an electron pair.
Spot the difference
| Point | Homolytic | Heterolytic |
|---|---|---|
| Electrons | One to each atom | Both to one atom |
| Product | Free radicals | Ions |
| Conditions | Light, heat | Polar solvent |
Practice questions
- 1Define homolytic fission.
- 2Is NH₃ an electrophile or a nucleophile?
- 3Which carbocation is the most stable: primary, secondary or tertiary?
- 4Give the products of heterolytic fission of (CH₃)₃C–Br.
- 5Classify CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.
- 6Classify CH₃CH₂OH → CH₂=CH₂ + H₂O.
- 7Arrange methyl, primary and tertiary carbanions in decreasing stability.
- 8Why is a carbocation planar?
Show answers (try them yourself first)
- 1Breaking of a bond so that each atom keeps one electron of the shared pair.
- 2Nucleophile.
- 3Tertiary.
- 4(CH₃)₃C⁺ and Br⁻.
- 5Addition.
- 6Elimination.
- 7Methyl > primary > tertiary.
- 8The carbon has 3 bonds and is sp² hybridised; the empty p-orbital lies perpendicular to the plane.
Purification and analysis of organic compounds
Link to Std 9–10 : You separated mixtures by filtration and distillation; the same ideas, with a few extra methods, are used to purify an organic compound before it is analysed.
To get pure sugar crystals from syrup, the syrup is cooled slowly. To see the colours in ink, a drop is placed on a piece of blotting paper. These are crystallisation and chromatography.
A pure organic compound is analysed to find the elements (qualitative analysis) and their percentages (quantitative analysis), from which the empirical and molecular formulae follow. Rf in chromatography is the distance moved by the spot ÷ the distance moved by the solvent.
In the figure : Five purification methods and what each separates.
Rule 1: Choosing the method — Crystallisation for solids with different solubility, sublimation for solids that vaporise (camphor, naphthalene), simple distillation for liquids with well-separated boiling points, fractional distillation when they are close, chromatography for small amounts of similar compounds.
- Easy : Camphor is separated from salt by sublimation.
- Medium : Ethanol and water (boiling at 78 °C and 100 °C) need fractional distillation.
- Exam-type : Rf = 3.0 cm ÷ 7.5 cm = 0.40 for a spot that moved 3.0 cm while the solvent front moved 7.5 cm.
Rule 2: Qualitative analysis — Lassaigne's test: fuse the compound with sodium to convert N, S and halogens to NaCN, Na₂S and NaX, then test the extract. Carbon and hydrogen are tested by heating with copper(II) oxide to get CO₂ (turns lime water milky) and H₂O (turns anhydrous copper sulphate blue).
- Easy : A Prussian blue colour in Lassaigne's test shows nitrogen.
- Medium : A black precipitate with lead acetate shows sulphur.
- Exam-type : A curdy white precipitate with silver nitrate (soluble in ammonia) shows chlorine.
Rule 3: Percentage of C, H, N and halogen — % C = (12 × mass of CO₂ × 100) ÷ (44 × mass of compound); % H = (2 × mass of H₂O × 100) ÷ (18 × mass of compound); Dumas for N; Carius for halogens.
- Easy : A 0.20 g sample gives 0.44 g CO₂: % C = 12 × 0.44 × 100 ÷ (44 × 0.20) = 60%.
- Medium : The same sample gives 0.12 g water: % H = 2 × 0.12 × 100 ÷ (18 × 0.20) = 6.67%.
- Exam-type : A 0.30 g compound gives 0.287 g AgCl in the Carius method: % Cl = 35.5 × 0.287 × 100 ÷ (143.5 × 0.30) = 23.7%.
Rule 4: Nitrogen and the formula — Dumas: % N = 28 × V × 100 ÷ (22 400 × w), with V the volume of N₂ in mL at STP. Kjeldahl: % N = 1.4 × N × V ÷ w. Divide each percentage by the atomic mass to get the simplest ratio.
- Easy : 0.30 g gives 40 mL of N₂ at STP: % N = 28 × 40 × 100 ÷ (22 400 × 0.30) = 16.7%.
- Medium : Kjeldahl with 0.50 g and 20 mL of 1.0 N acid used: % N = 1.4 × 1.0 × 20 ÷ 0.50 = 56%.
- Exam-type : A compound has C 40%, H 6.7% and O 53.3%. Ratio: C 40 ÷ 12 = 3.33, H 6.7 ÷ 1 = 6.7, O 53.3 ÷ 16 = 3.33, so the ratio is 1 : 2 : 1 and the empirical formula is CH₂O.
A pure compound has a sharp, fixed melting point or boiling point.
Using the mass of the compound instead of the mass of CO₂ or H₂O in the percentage formulas.
Memory clue: "12 over 44 for C, 2 over 18 for H" — these are the fractions of C in CO₂ and H in H₂O.
Spot the difference
| Point | Simple distillation | Fractional distillation |
|---|---|---|
| Boiling points | Far apart | Close together |
| Column | Not needed | Fractionating column used |
| Example | Water from salt water | Ethanol from water |
Practice questions
- 1Name the method to separate camphor from sand.
- 2What does a Prussian blue colour in Lassaigne's test show?
- 3Define Rf.
- 4Find Rf when a spot moves 3.0 cm and the solvent front 7.5 cm.
- 5A 0.20 g compound gives 0.44 g CO₂. Find % C.
- 6The same compound gives 0.12 g water. Find % H.
- 70.30 g of a compound gives 40 mL N₂ at STP. Find % N.
- 8A compound has C 40%, H 6.7%, O 53.3%. Find the empirical formula.
Show answers (try them yourself first)
- 1Sublimation.
- 2Nitrogen.
- 3Rf = distance moved by the spot ÷ distance moved by the solvent front.
- 43.0 ÷ 7.5 = 0.40.
- 512 × 0.44 × 100 ÷ (44 × 0.20) = 60%.
- 62 × 0.12 × 100 ÷ (18 × 0.20) = 6.67%.
- 728 × 40 × 100 ÷ (22 400 × 0.30) = 16.7%.
- 8C : H : O = 3.33 : 6.7 : 3.33 = 1 : 2 : 1, so CH₂O.