1. Overview
Every organic molecule is built on a skeleton of carbon atoms that forms a straight chain, a branched chain or a ring. This section describes those skeletons, then uses the hybridisation of each carbon or nitrogen atom (sp³, sp² or sp) to explain the shape and bond angles around it and how its σ and π bonds are arranged. It ends with planar molecules, such as ethene, and applies the basics from topics 3.4 (σ and π bonds, hybridisation) and 3.5 (VSEPR theory) to whole organic molecules.
Key Definitions
- Straight-chained (unbranched) molecule: an organic molecule with no ring in which no carbon atom is bonded to more than two other carbon atoms, so there are no side chains.
- Branched molecule: an organic molecule with no ring in which at least one carbon atom is bonded to three or four other carbon atoms, so a side chain, such as a methyl group, comes off the main chain.
- Cyclic molecule: a molecule in which some of the atoms are joined in a closed ring, as in cyclohexane, C₆H₁₂.
- sp³ hybridised carbon: a carbon atom whose four sp³ hybrid orbitals point to the corners of a tetrahedron, about 109.5° apart, and form four σ bonds.
- sp² hybridised carbon: a carbon atom whose three sp² hybrid orbitals form three σ bonds in one plane, about 120° apart, leaving one unhybridised p orbital at 90° to that plane, which forms the π bond in C=C and C=O.
- sp hybridised carbon: a carbon atom whose two sp hybrid orbitals form two σ bonds at 180°, leaving two unhybridised p orbitals, which form two π bonds, as in a nitrile group, -C≡N.
- Sigma (σ) bond: a covalent bond formed by the direct, head-on overlap of two orbitals along the line joining the two bonded nuclei.
- Pi (π) bond: a covalent bond formed by the sideways overlap of two parallel p orbitals on adjacent atoms, with its electron density above and below the σ bond.
- Planar: describes a molecule, or a group of atoms within one, in which all the atoms lie in the same flat plane, as in ethene.
Content
Straight chains, branches and rings
The carbon atoms of an organic molecule, joined to one another, form its carbon skeleton. To describe it, count how many other carbon atoms each carbon is bonded to.
| Type | Test | Example |
|---|---|---|
| straight-chained | no ring; no C bonded to more than 2 other C | hexane, CH₃(CH₂)₄CH₃ |
| branched | no ring; at least one C bonded to 3 or 4 other C | 2-methylpentane, (CH₃)₂CH(CH₂)₂CH₃ |
| cyclic | atoms joined in a closed ring | cyclohexane, C₆H₁₂ |
- "Straight" means unbranched, not straight in shape. Each carbon in hexane is sp³, so every C-C-C angle is about 109.5° and the chain is a zigzag, as a skeletal formula shows.
- Only carbon side chains are branches. In pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃, no carbon atom is bonded to more than two other carbon atoms, so the molecule is straight-chained. The same is true of Cl, NH₂ or =O on a middle carbon.
- Rings can carry side chains, as in methylcyclohexane (a CH₃ group on a six-carbon ring). A cycloalkane (one ring, single bonds only) is CₙH₂ₙ, like an alkene, so C₆H₁₂ may be cyclic or a hexene (topic 13.4). The benzene ring is studied at A Level (topic 29.3).
Worked example 1: straight-chained, branched or cyclic?
Question: Describe each molecule as straight-chained, branched or cyclic.
- A CH₃CH₂CH(CH₃)CH₂CH₂OH
- B (CH₃CH₂)₂CHCl
- C CH₃CH₂CH₂CH₂COOCH₃
- D (CH₃)₃CCH₂CH₃
- E methylcyclopentane: a ring of five carbon atoms with one CH₃ group attached
Step 1: look for a ring. Only E has one, so E is cyclic.
Step 2: write out any brackets. B is CH₃CH₂CHClCH₂CH₃. In D, the second carbon carries three CH₃ groups and one CH₂CH₃ group.
Step 3: find the carbon with the most carbon neighbours.
| Molecule | Most C neighbours on one C | Type |
|---|---|---|
| A | 3 (the CH carbon) | branched |
| B | 2 (the CHCl carbon) | straight-chained |
| C | 2 | straight-chained |
| D | 4 (the second carbon) | branched |
In C, the ester's O atom joins a five-carbon chain to a CH₃ group; neither part branches.
Answer: A branched (3-methylpentan-1-ol); B straight-chained (3-chloropentane); C straight-chained (methyl pentanoate); D branched (2,2-dimethylbutane); E cyclic. B only looks branched because of its brackets.
Shapes and bond angles around carbon
An organic molecule has no single shape name: describe the shape around each atom in turn. Around a carbon atom, each hybrid orbital (topic 3.4) holds the bonding pair of one σ bond, and these pairs repel and point as far apart as possible, as VSEPR theory predicts (topic 3.5). π bonds use the leftover p orbitals, so they do not change the shape.
| Bonds on the carbon | Hybridisation | Shape and angle |
|---|---|---|
| four single | sp³ | tetrahedral, 109.5° |
| one double, two single | sp² | trigonal planar, 120° |
| one triple and one single, or two double | sp | linear, 180° |
- sp³: any carbon with four single bonds: every carbon in an alkane, every CH₃ group, and a carbon carrying -OH, -Cl or -NH₂ with no double bond.
- sp²: a carbon with one double and two single bonds: both carbons of a C=C bond in an alkene, and the C=O carbon in aldehydes, ketones, carboxylic acids and esters.
- sp: the carbon of a nitrile group, -C≡N, both carbons of ethyne, H-C≡C-H, and any carbon with two double bonds, as in CO₂. The two sp orbitals point 180° apart, so the C-C≡N unit of a nitrile is a straight line.
Which angle to write. Real angles can differ from these by a few degrees when the groups around a carbon differ, but unless a question gives data, write 109.5°, 120° or 180°.
Nitrogen and oxygen atoms. Lone pairs count too. For a nitrogen atom, the quick test from topic 3.4 (σ bonds plus lone pairs) gives its hybridisation. For an oxygen atom, give the shape and angle from its bonding pairs and lone pairs (topic 3.5). Lone pairs repel more strongly than bonding pairs, so they close the angle.
| Atom | σ bonds + lone pairs | Shape around it |
|---|---|---|
| N of an amine, -NH₂ | 3 + 1 = 4: sp³ | pyramidal, about 107° |
| N of a nitrile, -C≡N | 1 + 1 = 2: sp | linear; lone pair points away from C |
| O of an alcohol, -OH | 2 + 2 | non-linear, about 104.5° |
Chains and rings in three dimensions.
- Single bonds in a chain rotate freely. σ overlap is symmetrical around the line joining the two nuclei, so one end of a C-C bond can turn without reducing the overlap. A long chain is flexible, but its C-C-C angles stay close to 109.5°.
- A six-membered ring of sp³ carbons is not flat, even though it is drawn as a flat hexagon. A flat six-membered ring would need an average angle of 120°, but sp³ carbons need about 109.5°, so the cyclohexane ring puckers.
σ and π bonds: how they are arranged
Every pair of bonded atoms shares exactly one σ bond, from head-on overlap of an orbital on each atom (for carbon, one of its hybrid orbitals; for hydrogen, its 1s orbital). A double bond adds one π bond and a triple bond two, from sideways overlap of unhybridised p orbitals (topic 3.4). The σ bonds set the shape; a π bond does not change it, but it locks the atoms around it in place.
| Hybridisation | σ bonds | π bonds |
|---|---|---|
| sp³ | four, tetrahedral | none |
| sp² | three, in one plane, 120° apart | one, above and below that plane |
| sp | two, in a line, 180° apart | two, at 90° to each other and to the line |
In a C=O group, the sp² carbon forms its σ bond to oxygen by head-on overlap of an sp² orbital with an orbital on oxygen, and the π bond by sideways overlap of its p orbital with a p orbital on oxygen. The carbon and its three bonded atoms lie in one plane, with the π electron density above and below it.
In a C≡N group, the σ bond lies along the C-N line, and the two π bonds surround it at 90° to each other.
Worked example 2: shape and bonding in pent-4-en-2-one
Question: Pent-4-en-2-one is CH₃COCH₂CH=CH₂. Number the carbon atoms C1 (the CH₃ group) to C5 (the =CH₂ group). (a) State the hybridisation of each carbon atom and the bond angle around it. (b) Deduce the numbers of σ and π bonds in one molecule. (c) Describe how the σ and π bonds are arranged around C4.
Step 1: hybridisation from the bonds on each carbon.
| Carbon | Bonds on it | Hybridisation, angle |
|---|---|---|
| C1 (CH₃) | four single | sp³, 109.5° |
| C2 (C=O) | one double, two single | sp², 120° |
| C3 (CH₂) | four single | sp³, 109.5° |
| C4 (=CH-) | one double, two single | sp², 120° |
| C5 (=CH₂) | one double, two single | sp², 120° |
Step 2: count the σ bonds, one for each pair of bonded atoms.
- C-H: 3 (on C1) + 2 (on C3) + 1 (on C4) + 2 (on C5) = 8
- carbon-carbon: C1-C2, C2-C3, C3-C4 and C4=C5 = 4
- carbon-oxygen: C2=O = 1
Total: 8 + 4 + 1 = 13 σ bonds. Check: C₅H₈O has 14 atoms and no ring, so it has σ bonds (see Key Equations).
Step 3: count the π bonds: one in C=O and one in C=C, so 2.
Step 4: the bonds around C4. C4 uses its three sp² orbitals for σ bonds to C3, C5 and H, in one plane at about 120°. Its unhybridised p orbital, at 90° to that plane, overlaps sideways with the p orbital on C5 to form the π bond, above and below the plane.
Answer: (a) C1 and C3 sp³ (109.5°); C2, C4 and C5 sp² (120°). (b) 13 σ and 2 π bonds. (c) As in Step 4: three σ bonds in one plane at 120°; one π bond above and below that plane.
Planar molecules and restricted rotation
Trigonal planar describes the three bonds around one atom, as at every sp² carbon. Planar describes a whole molecule, or a group of atoms, lying in one flat plane. A molecule with a trigonal planar carbon is not necessarily planar.
Why ethene is planar.
- Each carbon is sp², so its three σ bonds lie in one plane, and its p orbital stands at 90° to that plane.
- The π bond forms only when the two p orbitals are parallel, so that they overlap sideways.
- Both σ planes contain the C-C axis, and parallel p orbitals mean the planes are parallel too, so they are the same plane. So all six atoms of C₂H₄ lie in one plane, with the π electron density above and below it.
Restricted rotation. Turning one end of a C=C bond through 90° would leave the two p orbitals at right angles, with no sideways overlap, so the π bond would break. That needs roughly the energy of the π bond, about 260 kJ mol⁻¹ (C=C 610 minus C-C 350, topic 3.4), which collisions at room temperature almost never supply. So there is no rotation about a C=C bond. This restricted rotation is the origin of cis/trans isomerism (topic 13.4).
What a double bond fixes. The two C=C carbons and the four atoms bonded to them lie in one plane; the rest of the molecule may not. Likewise, a C=O carbon and its three bonded atoms lie in one plane, and an sp carbon holds its two bonded atoms in a straight line.
When is a whole molecule planar? A flat plane through an sp³ carbon can contain at most two of its four bonded atoms. So any molecule containing an sp³ carbon atom is non-planar, such as propene (its CH₃ group) and cyclohexane. A molecule with one C=C bond, in which every other atom is bonded to one of the two C=C carbons, is planar: ethene, chloroethene (CH₂=CHCl) and tetrafluoroethene (CF₂=CF₂). Ethyne and HCN have all their atoms on one line: call them linear.
Worked example 3: planar, linear or neither?
Question: (a) 3-Methylpent-2-ene is CH₃CH=C(CH₃)CH₂CH₃. How many of its six carbon atoms must lie in the same plane? Is the molecule planar? (b) Which atoms in butanenitrile, CH₃CH₂CH₂C≡N, lie in a straight line? (c) Explain why methanal, HCHO, is planar but ethanal, CH₃CHO, is not.
(a) Step 1: find the double bond. Number the main chain C1 to C5 from the left. C2 and C3 form the C=C bond, so both are sp².
Step 2: list the atoms the π bond fixes: C2, C3 and the four atoms bonded to them, which are C1 and H (on C2) and the branch carbon and C4 (on C3). That puts five carbon atoms in one plane.
Step 3: check the rest. C5 is bonded to C4, and the ethyl group can turn about the C3-C4 single bond, so C5 is not held in the plane. The sp³ carbons (C1, C4, C5 and the branch carbon) make the molecule non-planar.
(b) The nitrile carbon is sp, so its σ bonds to the next carbon and to nitrogen are 180° apart. The other three carbons are sp³, so the rest of the chain zigzags at about 109.5°.
(c) In methanal, the sp² carbon's three σ bonds, to two H atoms and to O, lie in one plane at about 120°, so all four atoms share that plane. In ethanal, the CH₃ carbon is sp³: a plane through it and the C=O carbon can contain at most one of its three H atoms.
Answer: (a) Five; the molecule is non-planar. (b) The nitrile carbon, the carbon bonded to it and the nitrogen (C-C≡N angle 180°). (c) Methanal's only carbon is sp², so it is planar; ethanal's sp³ CH₃ carbon puts at least two H atoms out of the plane.
Key Equations
Number of σ bonds ( = σ bonds in the molecule; = atoms in the molecule; = rings; counts, no unit). Each pair of bonded atoms shares one σ bond, so a molecule with no ring has one fewer σ bond than atoms, and each ring adds one:
For methylcyclopentane, C₆H₁₂: σ bonds (6 C-C and 12 C-H).
Number of π bonds ( = π bonds in the molecule; , = numbers of double and triple bonds; counts, no unit):
Hybridisation of a C or N atom ( = number of hybrid orbitals on the atom; = σ bonds the atom forms; = lone pairs on the atom; π bonds do not count; counts, no unit):
| σ bonds + lone pairs | Hybridisation | Angle, no lone pairs |
|---|---|---|
| 4 | sp³ | 109.5° (tetrahedral) |
| 3 | sp² | 120° (trigonal planar) |
| 2 | sp | 180° (linear) |
Common Mistakes to Avoid
- Mistake: calling a molecule branched because a functional group sits on a middle carbon. Fix: only a carbon side chain is a branch. CH₃CHClCH₂CH₂CH₃ (2-chloropentane) is straight-chained; CH₃CH(CH₃)CH₂CH₂Cl (1-chloro-3-methylbutane) is branched.
- Mistake: thinking a straight chain is straight in shape, or drawing C-C-C angles of 180°. Fix: chain carbons with only single bonds are sp³, so C-C-C is about 109.5° and the chain zigzags. Three atoms lie in a straight line around an sp carbon, as in C-C≡N.
- Mistake: calling a whole molecule planar because it contains a C=C bond. Fix: the double bond fixes only its two carbons and their four bonded atoms. In but-1-ene, CH₂=CHCH₂CH₃, the two ethyl-group carbons are sp³, so the molecule is non-planar.
- Mistake: giving the nitrogen of a nitrile as sp³. Fix: count σ bonds plus lone pairs: N in -C≡N has 1 + 1 = 2, so it is sp and C-C≡N is linear (180°). An amine nitrogen (3 + 1 = 4) is sp³.
- Mistake: counting the orbitals of hydrogen atoms when asked how many hybrid orbitals a molecule uses. Fix: hydrogen bonds through its 1s orbital and is never hybridised. In 3-methylpent-2-ene, the four sp³ carbons use 4 × 4 = 16 sp³ orbitals; the twelve H atoms add none.
- Mistake: writing "the double bond cannot move" or "it cannot rotate because it is strong". Fix: write "there is no rotation about the C=C bond, because rotation would turn the p orbitals out of line and break the π bond". A C-C single bond in a chain is strong too, yet it rotates freely.
Exam Tips
- State the hybridisation of each C or N atom from its σ bonds plus lone pairs. When a table asks how many carbons are sp, sp² and sp³, check that the numbers add up to the total number of carbons. The Data Booklet's carbon-13 NMR table also labels carbonyl carbons sp² and nitrile carbons sp.
- Describe the shape around a named atom with a shape name and angle, such as "trigonal planar, 120°". For a whole molecule, say whether it is planar and which atoms lie in the plane.
- Explain a shape in three linked steps: count the σ bonds (and lone pairs) on the atom to give its hybridisation; say that the electron pairs in the hybrid orbitals repel and get as far apart as possible; then give the angle.
- Describe σ and π bonds by naming the orbitals and the overlap: "head-on overlap of sp² orbitals gives the σ bond; sideways overlap of p orbitals gives the π bond, above and below the plane". For an alkene, say it is planar.
- Draw sp³ centres with wedges (topic 3.5), a C=C unit flat on the page with its four attached atoms at about 120°, and C-C≡N as a straight line.