Naming, formulae and homologous series

A homologous series is a family with the same general formula and functional group, differing by CH₂ each step, with a gradual change in physical properties.

• Empirical — the simplest whole-number ratio of atoms (CH₂ for butene).
• Molecular — the actual number of each atom (C₄H₈).
• Structural — shows the arrangement, e.g. CH₃CH₂CH=CH₂.
• Skeletal — lines only; carbons at the vertices.

IUPAC rules: find the longest chain containing the functional group; number from the end giving the group the lowest locant; list substituents alphabetically.

Name it

Structural isomerism and E/Z

Structural isomers share a molecular formula but differ in how the atoms are connected. Three flavours:

• Chain — the carbon skeleton is rearranged (butane / 2-methylpropane).
• Position — the same group sits on a different carbon (propan-1-ol / propan-2-ol).
• Functional group — a different group altogether (ethanol / methoxymethane, both C₂H₆O).

E/Z (a stereoisomerism) arises because there is no rotation about a C=C double bond. It requires two different groups on each of the doubly-bonded carbons. Use the Cahn-Ingold-Prelog rules — highest atomic number wins priority. Same side = Z (zusammen); opposite sides = E (entgegen).

Which kind of structural isomerism?

Tap a pair of isomers, then the type of structural isomerism it shows.

Does it show E/Z?

Alkanes and free-radical substitution

Alkanes are saturated and fairly unreactive (strong, non-polar C–C and C–H bonds). With UV light, chlorine substitutes stepwise by a free-radical chain mechanism:

Why this is a poor synthesis: further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, and termination gives ethane (from 2 •CH₃). You get a mixture, so the yield of any one product is low.

Spot the propagation step

Empirical formulae and percentage composition

To get the molecular formula, divide the true M_r by the M_r of the empirical formula and multiply through.

Your turn — percentage by mass

Your turn — molecular formula

Alkenes and electrophilic addition

The C=C double bond is a σ bond plus a π bond — a region of high electron density above and below the plane. That π cloud attracts electrophiles (electron-pair acceptors).

With HBr, the π electrons attack the δ+ hydrogen; the H–Br bond breaks heterolytically, giving a carbocation which the Br⁻ then attacks.

Markovnikov: carbocation stability runs tertiary > secondary > primary, because alkyl groups are electron-releasing and spread the positive charge. The major product comes from the more stable intermediate.

The test for a C=C: bromine water is decolourised from orange to colourless. And repeated addition across C=C gives addition polymers such as poly(ethene).

Which product dominates?

Halogenoalkanes — nucleophilic substitution and elimination

The C–X bond is polar (Cδ+–Xδ−), so the carbon is open to attack by a nucleophile (an electron-pair donor with a lone pair): OH⁻, CN⁻, NH₃.

• NaOH(aq), reflux → alcohol (nucleophilic substitution / hydrolysis).
• KCN in ethanol, reflux → nitrile — and note this adds a carbon to the chain.
• Excess NH₃ in ethanol, heated in a sealed tube → amine.
• NaOH in ethanol, reflux → alkene (elimination: OH⁻ acts as a base, removing H⁺ from the adjacent carbon).

Rate of hydrolysis follows the C–X bond enthalpy, not electronegativity: C–I is the weakest bond so iodoalkanes hydrolyse fastest; C–F is very strong, so fluoroalkanes are extremely slow.

S_N2 vs S_N1: primary halogenoalkanes react by S_N2 (one step, backside attack, inversion). Tertiary ones react by S_N1 (via a stable tertiary carbocation).

Choose the reagent

Your turn — yield

Alcohols and carboxylic acids

Alcohols are classified by the number of carbons attached to the C–OH carbon: primary (1), secondary (2), tertiary (3).

Oxidation with acidified potassium dichromate(VI) (orange Cr₂O₇²⁻ → green Cr³⁺):

• Primary, distil as it forms → aldehyde. Primary, reflux with excess oxidant → carboxylic acid.
• Secondary, reflux → ketone (which resists further oxidation).
• Tertiary → no reaction — there is no H on the C–OH carbon to remove.

Dehydration with hot concentrated H₂SO₄ (or Al₂O₃) gives an alkene. Carboxylic acids are weak acids that still fizz with carbonates — releasing CO₂.

The apparatus is the answer: "distil" gives you the aldehyde because it escapes before it can be oxidised again; "reflux" holds it in the flask so it goes all the way to the acid.

Oxidise the alcohol

Reagents and conditions

Tap the reagent on the left, then its organic product on the right.

Mass spectrometry and infrared spectroscopy

Mass spectrometry: the peak at the highest m/z is the molecular ion, M⁺, and gives the relative molecular mass directly. Smaller peaks are fragments: m/z 15 is CH₃⁺, 29 is C₂H₅⁺ or CHO⁺, and 43 is CH₃CO⁺ or C₃H₇⁺.

Infrared: bonds absorb IR at characteristic wavenumbers.

• C=O — a strong, sharp peak at 1680–1750 cm⁻¹.
• O–H (alcohol) — broad, 3200–3600 cm⁻¹.
• O–H (carboxylic acid) — very broad, 2500–3300 cm⁻¹.
• N–H (amine) — 3300–3500 cm⁻¹.

The fingerprint region below 1500 cm⁻¹ is too complex to interpret bond by bond, but it is unique to each compound — a computer match against a database identifies the substance.

Your turn — the molecular ion

Read the IR

The big ideas to know

That is the whole of C3. Press Finish to see your score.