Chapter 8 · Organic Chemistry – Some Basic Principles and Techniques

8.1 Carbon's Bonding, Structural Formulae and Classification

Carbon forms four bonds, links to itself in chains and rings, and makes single, double and triple bonds. This section shows how hybridisation fixes the shapes of organic molecules, how chemists write their structures — complete, condensed, bond-line and wedge–dash — and how millions of compounds are sorted into families by skeleton and functional group.

Until 1828 most chemists believed that the substances made by living things — sugar, urea, fats, dyes — could only be made by living things, because they needed a mysterious "vital force". In that year Friedrich Wöhler warmed ammonium cyanate, a salt made from minerals, and obtained urea, a compound until then isolated only from urine. Hermann Kolbe made acetic acid from its elements in 1845 and Marcellin Berthelot made methane in 1856. The vital force was dead; organic chemistry became the chemistry of carbon compounds, and today tens of millions of them are known — far more than the compounds of all the other elements put together.

Why carbon? A carbon atom forms four strong covalent bonds, it bonds to other carbon atoms to build chains and rings of almost any length, and it forms stable multiple bonds with carbon, oxygen and nitrogen. Before we can name organic compounds or follow their reactions, we need three tools: the link between hybridisation and shape (which you met in Section 4.3), a quick way to draw structures, and a system for sorting compounds into families.

Tetravalence of Carbon and the Shapes of Organic Molecules

📘 Definition 8.1 — Organic compound and catenation
Organic chemistry is the chemistry of hydrocarbons (compounds of carbon and hydrogen only) and of the compounds derived from them by replacing hydrogen atoms with other atoms or groups. A few simple carbon compounds — the oxides $\mathrm{CO}$ and $\mathrm{CO_{2}}$, carbonates, hydrogencarbonates, cyanides and carbides — are traditionally treated as inorganic.
Catenation is the ability of atoms of an element to bond to one another to form long chains and rings. Carbon shows it to a far greater extent than any other element, because the $\mathrm{C{-}C}$ bond is short and strong (bond enthalpy about $348\ \mathrm{kJ\,mol^{-1}}$).

Carbon ($1s^{2}2s^{2}2p^{2}$) is tetravalent: it shares its four valence electrons in four covalent bonds. In valence bond language (Section 4.3) the carbon atom mixes its $2s$ orbital with one, two or all three of its $2p$ orbitals. The hybrid orbitals form σ bonds; any $2p$ orbitals left over overlap sideways to form π bonds. So the type of hybridisation can be read straight off the bonds a carbon atom makes.

📐 Rule 8.1 — Hybridisation and shape of a carbon atom
Count the atoms bonded to the carbon atom (each σ bond uses one hybrid orbital):
  • 4 σ bonds (only single bonds): $sp^{3}$, tetrahedral, bond angles $109.5^{\circ}$ — e.g. every carbon of $\mathrm{CH_{4}}$, $\mathrm{CH_{3}CH_{3}}$.
  • 3 σ bonds + 1 π bond (one double bond): $sp^{2}$, trigonal planar, $120^{\circ}$ — e.g. $\mathrm{CH_{2}{=}CH_{2}}$, the carbonyl carbon of $\mathrm{HCHO}$.
  • 2 σ bonds + 2 π bonds (a triple bond, or two double bonds): $sp$, linear, $180^{\circ}$ — e.g. $\mathrm{HC{\equiv}CH}$, the central carbon of $\mathrm{CH_{2}{=}C{=}CH_{2}}$.
The $s$ character of the hybrid orbitals is $25\%$ ($sp^{3}$), $33\%$ ($sp^{2}$) and $50\%$ ($sp$). An $s$ orbital holds its electrons closer to the nucleus than a $p$ orbital, so the more $s$ character, the shorter and stronger the bonds formed and the more electronegative the carbon atom: $sp\gt sp^{2}\gt sp^{3}$.
HybridisationExampleShape round CBond angle$s$ character$\mathrm{C{-}C}$ bond length
$sp^{3}$ethane, $\mathrm{C{-}C}$tetrahedral$109.5^{\circ}$$25\%$$154\ \mathrm{pm}$
$sp^{2}$ethene, $\mathrm{C{=}C}$trigonal planar$120^{\circ}$$33\%$$134\ \mathrm{pm}$
$sp$ethyne, $\mathrm{C{\equiv}C}$linear$180^{\circ}$$50\%$$120\ \mathrm{pm}$

The π bond has two important consequences. Sideways overlap works only while the two $p$ orbitals stay parallel, so the groups on a double bond cannot rotate freely about it (the origin of geometrical isomerism in Section 8.2), and the atoms of $\mathrm{C{=}C}$ and the four atoms attached to them all lie in one plane. And because the π electrons sit above and below the bond axis, away from the nuclei, they are easily reached by electron-seeking reagents: a multiple bond is a reactive centre of an organic molecule.

✏️ Example 1 — Hybridisation along a chain
Give the hybridisation of each carbon atom (numbered from the left) in (a) $\mathrm{CH_{2}{=}CH{-}CH_{3}}$; (b) $\mathrm{CH_{2}{=}C{=}CH_{2}}$; (c) $\mathrm{CH_{3}{-}C{\equiv}N}$; (d) $\mathrm{CH_{3}COOH}$; (e) $\mathrm{HCONH_{2}}$; (f) benzene, $\mathrm{C_{6}H_{6}}$.
Method Count σ bonds (atoms attached) round each carbon: 4 → $sp^{3}$, 3 → $sp^{2}$, 2 → $sp$.
(a), (b) Propene: $\mathrm{C1}$ $sp^{2}$ (2 H + C2), $\mathrm{C2}$ $sp^{2}$ (H, C1, C3), $\mathrm{C3}$ $sp^{3}$. Propa-1,2-diene: $\mathrm{C1}$ $sp^{2}$, $\mathrm{C2}$ $sp$ (two double bonds, only two σ bonds), $\mathrm{C3}$ $sp^{2}$.
(c), (d) Ethanenitrile: $\mathrm{CH_{3}}$ carbon $sp^{3}$, nitrile carbon $sp$ (σ to C and to N, plus two π). Ethanoic acid: $\mathrm{CH_{3}}$ carbon $sp^{3}$; the $\mathrm{COOH}$ carbon has three σ bonds ($\mathrm{C}$, $=\mathrm{O}$, $\mathrm{{-}OH}$), so $sp^{2}$.
(e), (f) Methanamide: the single carbon is bonded to H, O and N (one double bond) → $sp^{2}$. Benzene: each carbon is bonded to two carbons and one hydrogen → all six $sp^{2}$, and the ring is planar.
Check In (b) the two $\mathrm{CH_{2}}$ planes are at right angles: the central $sp$ carbon uses one $p$ orbital for each π bond, and those two $p$ orbitals are perpendicular.
✏️ Example 2 — Counting σ and π bonds
Count the σ and π bonds in (a) but-1-en-3-yne, $\mathrm{CH_{2}{=}CH{-}C{\equiv}CH}$; (b) benzene; (c) propenenitrile, $\mathrm{CH_{2}{=}CH{-}C{\equiv}N}$; (d) tetracyanoethene, $\mathrm{(NC)_{2}C{=}C(CN)_{2}}$. (e) In (a), which carbon–carbon bond is the shortest?
Rule Every single bond is 1σ; a double bond is 1σ + 1π; a triple bond is 1σ + 2π. Count bonds to H too.
(a) $\mathrm{C{-}H}$: 4σ; $\mathrm{C{=}C}$: 1σ + 1π; $\mathrm{C{-}C}$: 1σ; $\mathrm{C{\equiv}C}$: 1σ + 2π. Total 7σ, 3π.
(b) 6 $\mathrm{C{-}H}$ + 6 $\mathrm{C{-}C}$ σ bonds in the ring = 12σ; three double bonds in a Kekulé structure = 3π.
(c) $\mathrm{C{-}H}$: 3σ; $\mathrm{C{=}C}$: 1σ + 1π; $\mathrm{C{-}C}$: 1σ; $\mathrm{C{\equiv}N}$: 1σ + 2π. Total 6σ, 3π.
(d) $\mathrm{C{=}C}$: 1σ + 1π; four $\mathrm{C{-}CN}$ single bonds: 4σ; four $\mathrm{C{\equiv}N}$: 4σ + 8π. Total 9σ, 9π.
(e) The triple bond $\mathrm{C3{\equiv}C4}$ (about $120\ \mathrm{pm}$): highest bond order and both carbons $sp$. The $\mathrm{C2{-}C3}$ single bond is shorter than the $154\ \mathrm{pm}$ of ethane because it joins $sp^{2}$ and $sp$ carbons.
Check Molecular formula of (d) is $\mathrm{C_{6}N_{4}}$: 10 atoms, a connected molecule with no rings needs $10-1=9$ σ bonds ✓.

Representing Organic Molecules: Structural Formulae

A molecular formula such as $\mathrm{C_{3}H_{8}O}$ says which atoms are present but not how they are joined — and three different compounds share that formula (Section 8.2). Organic chemists therefore write structural formulae, at several levels of detail.

📘 Definition 8.2 — Structural formulae
  1. Complete (expanded) structural formula — every atom and every bond is shown; a dash is a shared pair, $=$ a double and $\equiv$ a triple bond. Lone pairs may be added (a Lewis structure).
  2. Condensed structural formula — the bonds to hydrogen, and often the $\mathrm{C{-}C}$ bonds, are left out and the atoms on each carbon are grouped: $\mathrm{CH_{3}CH_{2}CH_{2}OH}$. Identical groups in a row are bracketed with a subscript: $\mathrm{CH_{3}(CH_{2})_{4}CH_{3}}$; identical groups on one atom too: $\mathrm{(CH_{3})_{3}CH}$.
  3. Bond-line (skeletal) formula — carbon chains are drawn as zig-zag lines. Every line end and every vertex is a carbon atom; hydrogen atoms on carbon are not shown (each carbon carries enough H to make four bonds); all other atoms, and the hydrogens attached to them, are written in.
  4. Three-dimensional (wedge–dash) formula — bonds in the plane of the paper are plain lines, a solid wedge points towards the viewer and a hashed wedge points away.
🧭 Problem-solving strategy — Reading a bond-line formula
  1. Put a carbon atom at every line end and every vertex (unless another atom is written there).
  2. For each carbon, add up its bonds (a double line counts 2, a triple line 3). Its number of hydrogens is $4-(\text{bonds drawn})$.
  3. Copy the hetero atoms ($\mathrm{O}$, $\mathrm{N}$, halogens) and the hydrogens written on them.
  4. Count atoms for the molecular formula; then check that every carbon has four bonds, every oxygen two and every nitrogen three.
✏️ Example 3 — From bond-line to condensed and molecular formulae
Write the condensed and the molecular formula of each compound.
(a) Five carbons in the chain, one branch carbon on C3. Ends: $\mathrm{CH_{3}}$; C2 and C4: two lines → $\mathrm{CH_{2}}$; C3: three lines → $\mathrm{CH}$. Condensed $\mathrm{CH_{3}CH_{2}CH(CH_{3})CH_{2}CH_{3}}$; molecular formula $\mathrm{C_{6}H_{14}}$.
(b) Four carbons; the last one has two lines (one to C, one to O) → $\mathrm{CH_{2}}$. $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}OH}$, i.e. $\mathrm{C_{4}H_{10}O}$.
(c) Six ring carbons. The two on the double bond have $2+1=3$ bonds → 1 H each; the other four have 2 bonds → 2 H each. $\mathrm{C_{6}H_{10}}$ (cyclohexene).
(d) C2 has 2 single bonds + 1 double bond = 4 → no H. $\mathrm{CH_{3}COCH_{2}CH_{3}}$, i.e. $\mathrm{C_{4}H_{8}O}$ (butan-2-one).
Check Saturated acyclic compounds fit $\mathrm{C}_{n}\mathrm{H}_{2n+2}$(O): (a) $\mathrm{C_{6}H_{14}}$ ✓, (b) $\mathrm{C_{4}H_{10}O}$ ✓. One ring or one double bond removes 2 H: (c) $\mathrm{C_{6}H_{10}}$ has both (ring + $\mathrm{C{=}C}$) → $14-4$ ✓; (d) one $\mathrm{C{=}O}$ → $10-2=8$ ✓.
✏️ Example 4 — Expanding condensed formulae
Find the molecular formula and molar mass ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{N}=14$, $\mathrm{O}=16\ \mathrm{g\,mol^{-1}}$) of (a) $\mathrm{CH_{3}(CH_{2})_{4}CH_{3}}$; (b) $\mathrm{(CH_{3})_{3}COH}$; (c) $\mathrm{CH_{3}CH{=}CHCHO}$; (d) $\mathrm{(CH_{3}CH_{2})_{2}NH}$; (e) $\mathrm{HOOC(CH_{2})_{4}COOH}$. For (b), describe its bond-line formula.
(a) $2\ \mathrm{CH_{3}}+4\ \mathrm{CH_{2}}$: $\mathrm{C_{6}H_{14}}$, $M=6(12)+14(1)=86\ \mathrm{g\,mol^{-1}}$.
(b) Three $\mathrm{CH_{3}}$ on one carbon that also carries $\mathrm{OH}$: $\mathrm{C_{4}H_{10}O}$, $M=48+10+16=74\ \mathrm{g\,mol^{-1}}$. Bond-line: a central vertex with four lines — three bare line ends (the methyls) and one ending in $\mathrm{OH}$.
(c) $\mathrm{CH_{3}}$, $\mathrm{CH}$, $\mathrm{CH}$, $\mathrm{CHO}$: $\mathrm{C_{4}H_{6}O}$, $M=48+6+16=70\ \mathrm{g\,mol^{-1}}$.
(d) Two ethyl groups on $\mathrm{NH}$: $\mathrm{C_{4}H_{11}N}$, $M=48+11+14=73\ \mathrm{g\,mol^{-1}}$.
(e) $2\ \mathrm{COOH}+4\ \mathrm{CH_{2}}$: $\mathrm{C_{6}H_{10}O_{4}}$, $M=72+10+64=146\ \mathrm{g\,mol^{-1}}$ (hexanedioic acid, used to make nylon).
Check (a) and (b) are saturated: $2(6)+2=14$ ✓ and $2(4)+2=10$ ✓.
🚀 Beyond the board syllabus (problem-solving tool) — degree of unsaturation

Not named in the CBSE, JEE Main, NEET or JEE Advanced syllabus, but it is a quick check used in structure and isomer problems in all three entrance exams.

For a compound $\mathrm{C}_{c}\mathrm{H}_{h}\mathrm{N}_{n}\mathrm{X}_{x}\mathrm{O}_{o}$ ($\mathrm{X}$ = halogen) the degree of unsaturation (double-bond equivalent) is $$\text{DU}=c-\tfrac{h}{2}-\tfrac{x}{2}+\tfrac{n}{2}+1$$ Oxygen and sulphur do not appear. Each ring or double bond counts 1, each triple bond 2. Examples: $\mathrm{C_{6}H_{6}}$: $6-3+1=4$ (one ring + three $\mathrm{C{=}C}$); $\mathrm{C_{6}H_{10}}$: $2$; $\mathrm{C_{2}H_{3}N}$ ($\mathrm{CH_{3}CN}$): $2-1.5+0.5+1=2$ (one $\mathrm{C{\equiv}N}$). JEE and NEET use it to rule out structures before naming them.
Interactive 1 — Structural Formula Translator

Choose a molecule and slide between the three ways of writing it. In the complete formula every hydrogen is drawn; in the condensed formula they are grouped on their carbons; in the bond-line formula they vanish and each carbon becomes a corner or a line end. The lab counts the atoms, the σ and π bonds and the hybridisation of every carbon (red $sp^{3}$, blue $sp^{2}$, green $sp$). Press ▶ Play to step through all the molecules.

complete
molecular formula—
molar mass—
σ / π bonds—
carbons sp³ · sp² · sp—

Classification of Organic Compounds

With so many compounds known, chemists sort them twice: once by the shape of the carbon skeleton, and once by the reactive group that the skeleton carries.

📘 Definition 8.3 — Classification by the carbon skeleton
  1. Acyclic (open-chain or aliphatic) compounds contain straight or branched chains of atoms, but no ring: $\mathrm{CH_{3}CH_{2}CH_{3}}$, $\mathrm{(CH_{3})_{2}CHCH_{2}OH}$.
  2. Cyclic (closed-chain) compounds contain one or more rings:
    • Alicyclic compounds have rings of carbon atoms that behave like aliphatic compounds: cyclopropane, cyclohexane, cyclohexene. A ring of carbon atoms only is called homocyclic.
    • Aromatic compounds are a special class of planar, unsaturated rings with exceptional stability (Chapter 9). Benzenoid aromatic compounds contain one or more benzene rings — benzene, methylbenzene (toluene), naphthalene; non-benzenoid aromatic compounds such as tropolone do not.
    • Heterocyclic compounds have at least one atom other than carbon (N, O or S) in the ring. They may be aromatic — pyridine, furan, thiophene — or alicyclic, such as tetrahydrofuran.
📘 Definition 8.4 — Functional group and homologous series
A functional group is an atom or group of atoms, joined in a specific way, that gives a family of compounds its characteristic chemical properties: $\mathrm{{-}OH}$ (hydroxyl), $\mathrm{{-}CHO}$ (aldehyde), $\mathrm{{-}COOH}$ (carboxyl), $\mathrm{{-}NH_{2}}$ (amino), and the carbon–carbon multiple bonds $\mathrm{C{=}C}$ and $\mathrm{C{\equiv}C}$.
A homologous series is a family of compounds with the same functional group in which each member differs from the next by a $\mathrm{{-}CH_{2}{-}}$ unit. The members are called homologues.
ClassFunctional groupGeneral formulaExample (IUPAC name, common name)
alkane— (only $\mathrm{C{-}C}$, $\mathrm{C{-}H}$)$\mathrm{C}_{n}\mathrm{H}_{2n+2}$$\mathrm{CH_{3}CH_{3}}$ ethane
alkene$\mathrm{C{=}C}$$\mathrm{C}_{n}\mathrm{H}_{2n}$$\mathrm{CH_{2}{=}CH_{2}}$ ethene (ethylene)
alkyne$\mathrm{C{\equiv}C}$$\mathrm{C}_{n}\mathrm{H}_{2n-2}$$\mathrm{HC{\equiv}CH}$ ethyne (acetylene)
haloalkane$\mathrm{{-}X}$ (F, Cl, Br, I)$\mathrm{C}_{n}\mathrm{H}_{2n+1}\mathrm{X}$$\mathrm{CH_{3}CH_{2}Cl}$ chloroethane
alcohol$\mathrm{{-}OH}$$\mathrm{C}_{n}\mathrm{H}_{2n+1}\mathrm{OH}$$\mathrm{CH_{3}CH_{2}OH}$ ethanol
ether$\mathrm{{-}O{-}}$$\mathrm{R{-}O{-}R'}$$\mathrm{CH_{3}OCH_{3}}$ methoxymethane (dimethyl ether)
aldehyde$\mathrm{{-}CHO}$$\mathrm{C}_{n}\mathrm{H}_{2n}\mathrm{O}$$\mathrm{CH_{3}CHO}$ ethanal (acetaldehyde)
ketone$\mathrm{\gt C{=}O}$$\mathrm{C}_{n}\mathrm{H}_{2n}\mathrm{O}$$\mathrm{CH_{3}COCH_{3}}$ propanone (acetone)
carboxylic acid$\mathrm{{-}COOH}$$\mathrm{C}_{n}\mathrm{H}_{2n}\mathrm{O_{2}}$$\mathrm{CH_{3}COOH}$ ethanoic acid (acetic acid)
ester$\mathrm{{-}COOR}$$\mathrm{C}_{n}\mathrm{H}_{2n}\mathrm{O_{2}}$$\mathrm{CH_{3}COOCH_{3}}$ methyl ethanoate
acid chloride$\mathrm{{-}COCl}$—$\mathrm{CH_{3}COCl}$ ethanoyl chloride
amide$\mathrm{{-}CONH_{2}}$—$\mathrm{CH_{3}CONH_{2}}$ ethanamide
amine$\mathrm{{-}NH_{2}}$$\mathrm{C}_{n}\mathrm{H}_{2n+3}\mathrm{N}$$\mathrm{CH_{3}NH_{2}}$ methanamine (methylamine)
nitrile$\mathrm{{-}C{\equiv}N}$—$\mathrm{CH_{3}CN}$ ethanenitrile (acetonitrile)
nitro compound$\mathrm{{-}NO_{2}}$—$\mathrm{CH_{3}NO_{2}}$ nitromethane
📐 Rule 8.2 — Characteristics of a homologous series
  1. All members can be represented by one general formula (alkanes $\mathrm{C}_{n}\mathrm{H}_{2n+2}$, alkanols $\mathrm{C}_{n}\mathrm{H}_{2n+1}\mathrm{OH}$, …).
  2. Successive members differ by $\mathrm{CH_{2}}$, i.e. by $14\ \mathrm{u}$ in molecular mass.
  3. All members contain the same functional group and have similar chemical properties.
  4. Physical properties change gradually along the series: boiling point and density rise with molar mass, solubility in water falls as the hydrocarbon part grows.
  5. Members can usually be prepared by the same general methods.
✏️ Example 5 — Classifying compounds
Classify each compound by its skeleton and name its functional group (if any): (a) cyclohexanol; (b) pyridine; (c) naphthalene; (d) $\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$; (e) tetrahydrofuran (a five-membered ring of four $\mathrm{CH_{2}}$ groups and one O); (f) $\mathrm{CH_{3}COOCH_{2}CH_{3}}$.
(a), (b) Cyclohexanol: alicyclic (a ring of six $sp^{3}$ carbons), alcohol ($\mathrm{{-}OH}$). Pyridine: heterocyclic aromatic (N in a benzene-like ring); no functional group in the sense of the table — the ring N itself is the reactive site.
(c), (d) Naphthalene: benzenoid aromatic (two fused benzene rings), a hydrocarbon. Propan-1-amine: acyclic, amine ($\mathrm{{-}NH_{2}}$).
(e) Heterocyclic but not aromatic — all four carbons are $sp^{3}$ and the ring has no double bonds — so it is an alicyclic heterocycle; functionally a cyclic ether.
(f) Acyclic; ester ($\mathrm{{-}COO{-}}$): ethyl ethanoate (ethyl acetate).
✏️ Example 6 — Working with a homologous series
(a) Write the formula and molar mass of the fifth member of the alkanol series, starting from methanol. (b) Which of $\mathrm{HCOOH}$, $\mathrm{CH_{3}CH_{2}COOH}$, $\mathrm{CH_{3}COOCH_{3}}$ and $\mathrm{C_{6}H_{5}COOH}$ are homologues of ethanoic acid? (c) By how much do the molar masses of the first and the sixth alkene differ? Name the sixth. (d) An alkyne has $M=68\ \mathrm{g\,mol^{-1}}$. Find its molecular formula. ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$.)
(a) $\mathrm{C}_{n}\mathrm{H}_{2n+1}\mathrm{OH}$ with $n=5$: $\mathrm{C_{5}H_{11}OH}$ ($\mathrm{C_{5}H_{12}O}$), $M=60+12+16=88\ \mathrm{g\,mol^{-1}}$.
(b) $\mathrm{HCOOH}$ and $\mathrm{CH_{3}CH_{2}COOH}$ (same $\mathrm{{-}COOH}$ group, differ from $\mathrm{CH_{3}COOH}$ by $\mathrm{CH_{2}}$). $\mathrm{CH_{3}COOCH_{3}}$ is an ester (an isomer of propanoic acid, not a homologue); $\mathrm{C_{6}H_{5}COOH}$ is aromatic and differs by $\mathrm{C_{5}H_{2}}$, not by a whole number of $\mathrm{CH_{2}}$ units.
(c) Five $\mathrm{CH_{2}}$ steps: $5\times14=70\ \mathrm{g\,mol^{-1}}$. First: ethene $\mathrm{C_{2}H_{4}}$ ($28$); sixth: $\mathrm{C_{7}H_{14}}$ ($98$), hept-1-ene (or any other heptene).
(d) $\mathrm{C}_{n}\mathrm{H}_{2n-2}$: $12n+2n-2=68$ → $14n=70$ → $n=5$: $\mathrm{C_{5}H_{8}}$.
Check $5(12)+8(1)=68$ ✓; $98-28=70$ ✓.
Interactive 2 — Homologous Series Explorer

Choose a series and the number of carbon atoms $n$. The lab draws the straight-chain member in bond-line form, gives its name, formula and molar mass, and plots the boiling points of the whole series against $n$ (grey: the alkanes, for comparison). Each step adds one $\mathrm{CH_{2}}$ — exactly $14\ \mathrm{g\,mol^{-1}}$ — and the boiling point climbs smoothly. Press ▶ Play to grow the chain.

3
name—
formula—
molar mass—
boiling point (1 atm)—

Exercises

I
Hybridisation, shape and bonds.
  • 1 In its ground state carbon has only two unpaired electrons. Explain how it forms four bonds of equal strength in methane.
  • 2 Give the hybridisation of each carbon (from the left) in but-2-enal, $\mathrm{CH_{3}{-}CH{=}CH{-}CHO}$.
  • 3 Give the hybridisation of each carbon (from the left) in $\mathrm{HC{\equiv}C{-}CH_{2}{-}CH{=}CH_{2}}$.
  • 4 Count the σ and π bonds in but-2-yne, $\mathrm{CH_{3}{-}C{\equiv}C{-}CH_{3}}$.
  • 5 Count the σ and π bonds in buta-1,2-diene, $\mathrm{CH_{2}{=}C{=}CH{-}CH_{3}}$.
  • 6 Count the σ and π bonds in cyclohexa-1,3-diene, $\mathrm{C_{6}H_{8}}$ (a six-membered ring with two double bonds).
  • 7 Arrange the carbon–carbon bonds of ethane, ethene, ethyne and benzene in order of increasing length.
  • 8 Why is a $\mathrm{C{-}H}$ bond to an $sp$ carbon shorter than a $\mathrm{C{-}H}$ bond to an $sp^{3}$ carbon?
1 One $2s$ electron is promoted to the empty $2p$ orbital ($2s^{1}2p^{3}$, four unpaired electrons); the four orbitals mix into four equivalent $sp^{3}$ hybrids pointing to the corners of a tetrahedron, each overlapping a hydrogen $1s$ orbital. The energy released by forming two extra bonds more than repays the promotion energy.
2 $\mathrm{C1}$ $sp^{3}$; $\mathrm{C2}$, $\mathrm{C3}$ $sp^{2}$ ($\mathrm{C{=}C}$); $\mathrm{C4}$ $sp^{2}$ ($\mathrm{C{=}O}$).
3 $sp$, $sp$, $sp^{3}$, $sp^{2}$, $sp^{2}$.
4 $\mathrm{C{-}H}$ 6, $\mathrm{C{-}C}$ 2, $\mathrm{C{\equiv}C}$ 1σ + 2π: 9σ, 2π. (Check: 10 atoms, no ring → 9 σ.)
5 $\mathrm{C{-}H}$ 6, two $\mathrm{C{=}C}$ (2σ + 2π), one $\mathrm{C{-}C}$: 9σ, 2π.
6 8 $\mathrm{C{-}H}$ + 6 ring $\mathrm{C{-}C}$ σ = 14σ; 2π. (Check: 14 atoms and one ring → $13+1=14$ σ bonds.)
7 ethyne ($120\ \mathrm{pm}$) $\lt$ ethene ($134\ \mathrm{pm}$) $\lt$ benzene ($139\ \mathrm{pm}$, between single and double) $\lt$ ethane ($154\ \mathrm{pm}$).
8 An $sp$ hybrid has $50\%$ $s$ character (against $25\%$ for $sp^{3}$); $s$ electrons stay closer to the nucleus, so the hybrid is shorter and the bond it forms is shorter and stronger.
II
Writing structures.
  • 1 How many $\mathrm{C{-}H}$ bonds are drawn in the complete structural formula of $\mathrm{CH_{3}CH(CH_{3})CH_{2}CH_{3}}$?
  • 2 A chain of six carbon atoms carries an $\mathrm{OH}$ group on the second carbon. Write its condensed and molecular formulae.
  • 3 Write the molecular formula of $\mathrm{(CH_{3})_{2}CHCH_{2}CH(CH_{3})_{2}}$.
  • 4 Find the molecular formula and molar mass of $\mathrm{CH_{3}COCH_{2}CH_{2}COOH}$ ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$).
  • 5 A bond-line formula is a zig-zag of four lines (five carbons) with a double line between the second and third carbons. Write its condensed and molecular formulae.
  • 6 A bond-line formula shows a hexagon of single lines with one corner joined to $\mathrm{OH}$. Write its molecular formula.
  • 7 A bond-line formula shows a benzene ring with one extra line from a corner (no label at its end). Name the compound and give its molecular formula.
  • 8 In a wedge–dash drawing of $\mathrm{CH_{2}BrCl}$, Br and Cl are on plain lines, one H on a wedge and one on a hashed wedge. Which atoms lie in the plane of the paper?
1 $\mathrm{C_{5}H_{12}}$: every H is on carbon → 12 $\mathrm{C{-}H}$ bonds.
2 $\mathrm{CH_{3}CH(OH)CH_{2}CH_{2}CH_{2}CH_{3}}$ (or $\mathrm{CH_{3}CH(OH)(CH_{2})_{3}CH_{3}}$); $\mathrm{C_{6}H_{14}O}$.
3 $2+1+1+1+2=7$ carbons; H: $6+1+2+1+6=16$ → $\mathrm{C_{7}H_{16}}$.
4 $\mathrm{C_{5}H_{8}O_{3}}$; $M=60+8+48=116\ \mathrm{g\,mol^{-1}}$.
5 $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$; $\mathrm{C_{5}H_{10}}$ (pent-2-ene).
6 Cyclohexanol: 5 $\mathrm{CH_{2}}$ + 1 $\mathrm{CH(OH)}$ → $\mathrm{C_{6}H_{12}O}$.
7 Methylbenzene (toluene), $\mathrm{C_{6}H_{5}CH_{3}}$ = $\mathrm{C_{7}H_{8}}$: the unlabelled line end is a $\mathrm{CH_{3}}$ group.
8 C, Br and Cl (plain lines are in the plane); one H is in front of the plane and one behind.
III
Classification, functional groups and homologous series.
  • 1 Classify by skeleton: cyclopentane, thiophene, methylbenzene, $\mathrm{CH_{3}CH_{2}OCH_{2}CH_{3}}$.
  • 2 Name the functional groups in $\mathrm{CH_{2}{=}CHCOOH}$.
  • 3 Name the functional groups in $\mathrm{HOCH_{2}CH_{2}NH_{2}}$.
  • 4 Which of $\mathrm{CH_{3}OH}$, $\mathrm{C_{2}H_{5}OH}$, $\mathrm{CH_{3}OCH_{3}}$, $\mathrm{C_{3}H_{7}OH}$ and $\mathrm{CH_{3}CHO}$ belong to one homologous series?
  • 5 Write the condensed formula and molar mass of the third member of the aldehyde series (methanal is the first).
  • 6 Give the general formula of the alkanoic acids and the molecular formula of the member with $M=102\ \mathrm{g\,mol^{-1}}$.
  • 7 Write the next higher homologue of $\mathrm{C_{4}H_{9}NH_{2}}$ and its molar mass ($\mathrm{N}=14$).
  • 8 Ethanol and propan-1-ol react in almost the same way, but boil at $78\ ^{\circ}\mathrm{C}$ and $97\ ^{\circ}\mathrm{C}$. Explain both facts.
1 Cyclopentane: alicyclic. Thiophene: heterocyclic aromatic (S in a five-membered aromatic ring). Methylbenzene: benzenoid aromatic. Ethoxyethane: acyclic (an ether).
2 Carbon–carbon double bond (alkene) and carboxyl, $\mathrm{{-}COOH}$ (propenoic acid).
3 Hydroxyl ($\mathrm{{-}OH}$, alcohol) and amino ($\mathrm{{-}NH_{2}}$, amine).
4 $\mathrm{CH_{3}OH}$, $\mathrm{C_{2}H_{5}OH}$, $\mathrm{C_{3}H_{7}OH}$ (alkanols, each differing by $\mathrm{CH_{2}}$). $\mathrm{CH_{3}OCH_{3}}$ is an ether, $\mathrm{CH_{3}CHO}$ an aldehyde.
5 $\mathrm{CH_{3}CH_{2}CHO}$ (propanal), $\mathrm{C_{3}H_{6}O}$: $M=36+6+16=58\ \mathrm{g\,mol^{-1}}$.
6 $\mathrm{C}_{n}\mathrm{H}_{2n}\mathrm{O_{2}}$ (i.e. $\mathrm{C}_{n-1}\mathrm{H}_{2n-1}\mathrm{COOH}$): $14n+32=102$ → $n=5$, $\mathrm{C_{5}H_{10}O_{2}}$ (pentanoic acid).
7 $\mathrm{C_{5}H_{11}NH_{2}}$ ($\mathrm{C_{5}H_{13}N}$): $M=60+13+14=87\ \mathrm{g\,mol^{-1}}$.
8 Same functional group ($\mathrm{{-}OH}$) → same chemistry. Propan-1-ol has one more $\mathrm{CH_{2}}$: a larger molecule with stronger intermolecular (dispersion) forces, so a higher boiling point — the gradation of Rule 8.2.
IV
Putting it together.
  • 1 Applied. Paracetamol is $\mathrm{HO{-}C_{6}H_{4}{-}NHCOCH_{3}}$. Name its functional groups and find its molar mass ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{N}=14$, $\mathrm{O}=16$).
  • 2 Applied. Aspirin is $\mathrm{CH_{3}COO{-}C_{6}H_{4}{-}COOH}$. Name its functional groups, and find its molecular formula and molar mass.
  • 3 Benzene ($\mathrm{C_{6}H_{6}}$) is flat but cyclohexane ($\mathrm{C_{6}H_{12}}$) is not. Explain using hybridisation.
  • 4 How many $sp^{3}$, $sp^{2}$ and $sp$ carbons are there in but-2-enenitrile, $\mathrm{CH_{3}{-}CH{=}CH{-}C{\equiv}N}$?
  • 5 Count the σ and π bonds in but-2-enenitrile.
  • 6 Why can the two ends of ethane rotate freely about the $\mathrm{C{-}C}$ bond, while the two ends of ethene cannot?
  • 7 When ethene is polymerised to polythene, $n\,\mathrm{CH_{2}{=}CH_{2}}\rightarrow\mathrm{({-}CH_{2}{-}CH_{2}{-})}_{n}$, how does the hybridisation of carbon change?
  • 8 In which of ethene, ethane, benzene and propene do all the atoms lie in one plane?
1 Hydroxyl on a benzene ring (a phenol) and an amide ($\mathrm{{-}NHCO{-}}$). $\mathrm{C_{8}H_{9}NO_{2}}$: $M=96+9+14+32=151\ \mathrm{g\,mol^{-1}}$.
2 Ester ($\mathrm{{-}OCOCH_{3}}$) and carboxylic acid ($\mathrm{{-}COOH}$) on a benzene ring. $\mathrm{C_{9}H_{8}O_{4}}$: $M=108+8+64=180\ \mathrm{g\,mol^{-1}}$.
3 Every benzene carbon is $sp^{2}$ (trigonal planar, $120^{\circ}$ — exactly the angle of a regular hexagon), and the π overlap holds the ring flat. Cyclohexane carbons are $sp^{3}$ ($109.5^{\circ}$); a flat hexagon would force $120^{\circ}$, so the ring puckers.
4 $sp^{3}$: 1 ($\mathrm{CH_{3}}$); $sp^{2}$: 2 ($\mathrm{CH{=}CH}$); $sp$: 1 ($\mathrm{C{\equiv}N}$).
5 $\mathrm{C{-}H}$ 5, $\mathrm{C{-}C}$ 2, $\mathrm{C{=}C}$ 1σ + 1π, $\mathrm{C{\equiv}N}$ 1σ + 2π: 9σ, 3π.
6 A σ bond is symmetrical about the axis: rotation does not change the overlap. A π bond needs the two $p$ orbitals parallel; rotating one end would destroy the overlap, costing the π bond energy.
7 $sp^{2}$ (trigonal, in ethene) → $sp^{3}$ (tetrahedral, in polythene): the π bond is used to form a new σ bond to the next unit.
8 Ethene and benzene. Ethane has tetrahedral carbons; in propene the $\mathrm{CH_{3}}$ carbon is $sp^{3}$, so at least two of its hydrogens lie out of the plane.
🎯 Exam corner BoardJEE MainNEET

E1. NEET The hybridisations of the three carbon atoms (in order) in $\mathrm{CH_{2}{=}C{=}CH_{2}}$ are (A) $sp^{2}, sp^{2}, sp^{2}$ (B) $sp^{2}, sp, sp^{2}$ (C) $sp, sp^{2}, sp$ (D) $sp^{3}, sp, sp^{3}$

Idea The end carbons have 3 σ bonds (2 H + C) → $sp^{2}$; the central carbon has only 2 σ bonds (and 2 π) → $sp$.
Answer (B).

E2. JEE Main Numerical answer. How many σ bonds are there in hexa-1,3-dien-5-yne, $\mathrm{CH_{2}{=}CH{-}CH{=}CH{-}C{\equiv}CH}$?

Count $\mathrm{C{-}H}$: $2+1+1+1+1=6$; carbon–carbon: five bonds between six carbons, each with one σ → 5. Total $6+5=11$.
Check $\mathrm{C_{6}H_{6}}$ has 12 atoms and no ring: $12-1=11$ σ bonds ✓ (and $1+1+2=4$ π bonds).
Answer 11.

E3. Board · 3 marks What is a homologous series? State two of its characteristics, and write the condensed formula of the next homologue of $\mathrm{CH_{3}CH_{2}CHO}$.

1 mark A family of compounds with the same functional group in which successive members differ by a $\mathrm{CH_{2}}$ unit.
1 mark Any two: one general formula; successive molar masses differ by $14\ \mathrm{g\,mol^{-1}}$; similar chemical properties; gradual change in physical properties; similar methods of preparation.
1 mark $\mathrm{CH_{3}CH_{2}CH_{2}CHO}$ (butanal).

E4. JEE Main In which compound are all the carbon atoms $sp^{2}$ hybridised? (A) $\mathrm{CH_{2}{=}CH{-}CH_{3}}$ (B) $\mathrm{CH_{2}{=}CH{-}CH{=}CH_{2}}$ (C) $\mathrm{HC{\equiv}C{-}CH{=}CH_{2}}$ (D) $\mathrm{CH_{2}{=}C{=}CH_{2}}$

Check each (A) has an $sp^{3}$ $\mathrm{CH_{3}}$. (B) every carbon is on exactly one double bond → all $sp^{2}$. (C) the triple-bond carbons are $sp$. (D) the central carbon is $sp$.
Answer (B) buta-1,3-diene.

E5. NEET Which pair consists of members of the same homologous series? (A) $\mathrm{C_{2}H_{5}OH}$ and $\mathrm{CH_{3}OCH_{3}}$ (B) $\mathrm{CH_{3}CHO}$ and $\mathrm{CH_{3}COCH_{3}}$ (C) $\mathrm{C_{2}H_{6}}$ and $\mathrm{C_{3}H_{8}}$ (D) $\mathrm{C_{2}H_{4}}$ and $\mathrm{C_{2}H_{2}}$

Idea Homologues share a functional group and differ by $\mathrm{CH_{2}}$. (A) alcohol vs ether (same formula — isomers). (B) aldehyde vs ketone. (D) alkene vs alkyne, differing by $\mathrm{H_{2}}$.
Answer (C) ethane and propane, both alkanes, differing by $\mathrm{CH_{2}}$.

E6. Board · Case study But-3-en-2-one, $\mathrm{CH_{2}{=}CH{-}CO{-}CH_{3}}$, is an industrial building block for plastics and vitamins. (i) Give the hybridisation of each carbon (from the left) and the shape around the third carbon. (2 marks) (ii) Count the σ and π bonds. (1 mark) (iii) Name its two functional groups and give its molecular formula. (1 mark)

(i) $\mathrm{C1}$ $sp^{2}$, $\mathrm{C2}$ $sp^{2}$, $\mathrm{C3}$ $sp^{2}$ (the carbonyl carbon), $\mathrm{C4}$ $sp^{3}$. Round $\mathrm{C3}$: trigonal planar, about $120^{\circ}$.
(ii) $\mathrm{C_{4}H_{6}O}$ has 11 atoms, no ring → 10σ; $\mathrm{C{=}C}$ and $\mathrm{C{=}O}$ → 2π.
(iii) Alkene ($\mathrm{C{=}C}$) and ketone ($\mathrm{C{=}O}$); $\mathrm{C_{4}H_{6}O}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Pharmaceutical companies screen millions of candidate molecules every year. Before spending money on a compound, chemists check simple numbers that can be read straight off its structural formula. Which numbers, and do familiar medicines pass the test?

Why it matters. In 1997 Christopher Lipinski at Pfizer noticed that most medicines taken by mouth obey a "rule of five": molar mass at most $500\ \mathrm{g\,mol^{-1}}$, at most 5 hydrogen-bond donors ($\mathrm{O{-}H}$ and $\mathrm{N{-}H}$ groups), at most 10 hydrogen-bond acceptors (counted simply as N and O atoms) and a limited preference for oil over water. Drug-discovery teams in Europe, the USA, India and China still use such filters to decide which molecules to make first — and they are applied to structural formulae just like the ones in this section.

What to do.

  1. Look up the condensed or bond-line formulae of paracetamol, aspirin, ibuprofen and caffeine (any reliable chemistry database). Classify each by skeleton and list its functional groups.
  2. Work out the molecular formula and molar mass of each ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{N}=14$, $\mathrm{O}=16$).
  3. Count the hydrogen-bond donors and acceptors of each.
  4. Decide whether each passes Lipinski's limits on mass, donors and acceptors. Which functional groups contribute most to the donor and acceptor counts?

Tools: Definitions 8.2–8.4, the functional-group table, Interactive 1 (for checking your atom counts on smaller fragments) and a chemistry database.

Hand in: A table (compound, formula, $M$, functional groups, donors, acceptors, pass/fail) and a paragraph on how one structural change — say, replacing $\mathrm{{-}COOH}$ by $\mathrm{{-}COOCH_{3}}$ — would change the counts.

Going further. Find a medicine that breaks the rule of five but still works (many antibiotics do). How does it reach its target?

Instructor note: paracetamol $\mathrm{C_{8}H_{9}NO_{2}}$, $151$; donors 2 (OH, NH), acceptors 3. Aspirin $\mathrm{C_{9}H_{8}O_{4}}$, $180$; donors 1, acceptors 4. Ibuprofen $\mathrm{C_{13}H_{18}O_{2}}$, $206$; donors 1, acceptors 2. Caffeine $\mathrm{C_{8}H_{10}N_{4}O_{2}}$, $194$; donors 0, acceptors 6. All four pass. Methylating $\mathrm{{-}COOH}$ removes one donor and adds $14\ \mathrm{g\,mol^{-1}}$. Rule-breakers taken by mouth include erythromycin and ciclosporin.

🧪 Self-check quiz

0/5 correct

Q1. The hybridisation of the carbonyl carbon in propanone, $\mathrm{CH_{3}COCH_{3}}$, is

Q2. Ethyne, $\mathrm{HC{\equiv}CH}$, contains

Q3. Which is a heterocyclic aromatic compound?

Q4. Successive members of a homologous series differ in molar mass by

Q5. In a bond-line formula, a vertex where three single lines meet stands for

← Chapter 8 — Organic Chemistry – Some Basic Principles and Techniques8.2 IUPAC Nomenclature and Isomerism →