Mission 1: How can C₂H₆O be both a liquid and a gas at room temperature?

Five kinds of formula

• Empirical: simplest whole-number ratio of atoms. Hexane: C₃H₇.
• Molecular: actual number of each atom. Hexane: C₆H₁₄.
• Full structural: every atom and bond. Condensed: CH₃CH₂CH₂CH₂CH₂CH₃, or CH₃(CH₂)₄CH₃.
• Skeletal: a zig-zag of bonds; carbons and their hydrogens are implied.
• Stereochemical: wedges and dashes show the 3D shape. Draw one only when asked.
• Build it as a 3D model (real or virtual): each carbon is tetrahedral, so chains zig-zag.

Reading a skeletal formula

• Every line end and corner is a carbon, unless a line ends in a label (OH, Cl): that end is the labelled atom.
• A group drawn at a corner, or on a short line off it, is attached to that carbon.
• H on carbon is not drawn: give each carbon enough H for four bonds. Other atoms, and H on them, are always drawn.
• Lines are bonds: count the points, not the lines.
• Worked example: five zig-zag lines, the last ending in OH. That is 5 C and 1 O.
• H: 3 + 2 + 2 + 2 + 2 + 1 (on O) = 12. So C₅H₁₂O, CH₃CH₂CH₂CH₂CH₂OH.

Functional groups

• A functional group gives a compound its typical reactions and decides its class.

Telling look-alikes apart

• Carboxyl: the C=O carbon also carries –OH. It counts as one group.
• Ester: the single-bonded O is bonded to a second carbon, not to H.
• Amido: N bonded to the C=O carbon. Amino: N on a carbon with no C=O.
• Alkoxy: C–O–C with no C=O.
• Saturated: only single C–C bonds. Unsaturated: at least one C=C or C≡C.
• Cyclohexane, C₆H₁₂, is saturated: its ring, not a double bond, lowers the H count.

Mission 2: Petrol and candle wax are both mostly alkanes. Why is one a runny liquid and the other a solid?

Homologous series

• A homologous series is a family with the same functional group.
• Each member differs from the next by CH₂.
• One general formula fits every member.
• Members have similar chemical properties; physical properties change gradually.

The series, part 1

The series, part 2

• Shared general formulae: aldehydes and ketones; acids and esters; alcohols and ethers. The functional group decides the series.

Why boiling point rises along a series

• Boiling separates whole molecules. No covalent bonds break.
• Each extra CH₂ adds electrons and contact area: stronger London (dispersion) forces.
• More energy is needed to separate the molecules, so the boiling point rises.
• In the alcohols, each molecule still has one –OH: its hydrogen bonding stays about the same.
• Melting points rise too, but less regularly: how molecules pack in the solid also matters.

Mission 3: Could you draw a molecule from its name alone?

The parts of a name

• Stem = carbons in the parent chain: meth 1, eth 2, prop 3, but 4, pent 5, hex 6.
• -an- only single C–C bonds; -en- one C=C; -yn- one C≡C.
• Suffix for the main group: -ol (hydroxyl), -al (aldehyde), -one (ketone), -oic acid (carboxyl). Alkanes end in -e.
• Prefixes for side groups: methyl, ethyl; fluoro, chloro, bromo, iodo.
• Locants are carbon numbers: 2-methylpentan-3-ol.

Four steps to a name

• 1. Find the longest chain that contains the functional-group carbon. It may bend through a branch.
• 2. Number from the end that gives the suffix group the lowest locant.
• 3. Then the lowest locant to the C=C or C≡C, then to the prefixes.
• 4. Prefixes in alphabetical order. Repeats: di-, tri- (not used for alphabetical order). Commas between numbers; hyphens between numbers and letters.

Worked examples: prefixes only

• CH₃CH₂CH(CH₃)CH(CH₃)CH₃: longest chain 5 → pentane.
• From the right, methyls on C2 and C3; from the left, C3 and C4. Lower set wins.
• Name: 2,3-dimethylpentane.
• CH₃CH(CH₃)CHBrCH₂CH₃: from the left {2, 3}; from the right {3, 4}.
• Name: 3-bromo-2-methylpentane. Prefixes go in alphabetical order (b before m), not number order.

Groups that take a suffix

• -ol: CH₃CH₂CH₂CH(OH)CH₃ is pentan-2-ol, not pentan-4-ol.
• -one: CH₃COCH₂CH₂CH₃ is pentan-2-one. Propanone needs no number.
• -al and -oic acid: that carbon is always C1, so no number: CH₃CH₂CH₂COOH is butanoic acid.
• With a branch: (CH₃)₂CHCOOH is 2-methylpropanoic acid.
• -ene, -yne: number the multiple bond by its lower carbon: CH₂=CHCH₂CH₂CH₃ is pent-1-ene; CH₃C≡CCH₃ is but-2-yne.

A C=C plus a suffix group

• The suffix group gets the lowest locant first; the C=C locant goes before -en-.
• CH₂=CHCH₂CH(OH)CH₃: from the CH₃ end, OH is on C2 and the C=C starts at C4: pent-4-en-2-ol.
• Not pent-1-en-4-ol: the suffix group outranks the double bond.
• Branch plus suffix: CH₃CH(CH₃)CH₂CH₂OH is 3-methylbutan-1-ol.

Mission 4: How many different molecules can you build from C₅H₁₂?

Structural isomers

• Structural isomers: same molecular formula, atoms connected differently.
• Chain (branched-chain) isomers: different skeleton. Hexane and 2-methylpentane, both C₆H₁₄.
• Position isomers: same group, different place. Propan-1-ol and propan-2-ol.
• Functional group isomers: different series. Propanal and propanone, both C₃H₆O.
• Rotating, flipping or bending a drawing does not make a new isomer.

Primary, secondary, tertiary

• Alcohols and halogenoalkanes: count the carbons bonded to the carbon carrying –OH or –X.
• One → primary; two → secondary; three → tertiary.
• Propan-1-ol is primary, propan-2-ol secondary, 2-methylpropan-2-ol tertiary. (CH₃)₃CBr is a tertiary halogenoalkane.
• Amines: count the carbons bonded to N. CH₃CH₂NH₂ 1°, (CH₃)₂NH 2°, (CH₃)₃N 3°.
• The class is not the locant: hexan-3-ol is secondary.

Mission 5: Spearmint and caraway get their smells from mirror-image molecules. How can your nose tell them apart?

Cis–trans isomers

• Stereoisomers: same atoms, same connections, different arrangement in space.
• Cis–trans needs restricted rotation: a C=C (its π bond stops rotation) or a ring.
• It also needs two different groups on each of the two carbons.
• Cis: matching groups on the same side. Trans: on opposite sides.
• 1,2-Dichloroethene, ClCH=CHCl, has both. Propene, CH₂=CHCH₃, does not: one carbon carries two H.
• Rings: 1,2-dichlorocyclopropane has cis and trans forms. E–Z names are not needed.

Chiral carbons and enantiomers

• A chiral carbon is bonded to four different groups.
• Its molecule exists as two non-superimposable mirror images: enantiomers.
• 2-Hydroxypropanoic acid (lactic acid), CH₃CH(OH)COOH: C2 carries H, OH, CH₃ and COOH.
• Wedge-dash drawing: two bonds in the plane, a tapered wedge towards you, a dash away. The shape is a tetrahedron (109.5°).
• Draw the other enantiomer as the mirror image. In a 3D model, no rotation makes the two match.

Spotting a chiral carbon

• Only a carbon with four single bonds can be chiral: CH₃, CH₂ and C=C carbons never are.
• Compare whole groups, not just the atom bonded to the carbon.
• 1-Chloro-2-methylbutane, CH₂ClCH(CH₃)CH₂CH₃: C2 carries H, CH₃, C₂H₅ and CH₂Cl: chiral.
• Pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃: C3 carries H, OH and two C₂H₅: not chiral.

Optical activity

• Each enantiomer rotates plane-polarized light: by equal angles, in opposite directions. This is optical activity.
• A racemic mixture contains equal amounts of both enantiomers: the rotations cancel.
• Enantiomers share melting point, boiling point and density.
• They react in the same way, except in a chiral environment such as an enzyme or a receptor.

Mission 6: Why does carbon dioxide trap heat, while nitrogen, most of the air, does not?

Molecular ion and fragments

• A molecule loses one electron to form the molecular ion, M⁺. Its m/z = M_r.
• M⁺ is usually the peak at highest m/z (ignore tiny isotope peaks just above it). It can be small.
• M⁺ can break into a fragment ion (detected) and a neutral piece (not detected).
• The tallest peak, the base peak, is often a fragment ion.
• Mass lost = M_r − fragment m/z. Data booklet (section 22): 15 CH₃, 17 OH, 18 H₂O, 28 C₂H₄ or CO, 29 C₂H₅ or CHO, 31 OCH₃, 45 COOH.

Worked example: ethanol

• Ethanol, CH₃CH₂OH: M_r = 46, so M⁺ at m/z 46.
• Base peak at m/z 31: mass lost 46 − 31 = 15, a CH₃ (data booklet).
• So the ion is CH₂OH⁺: the C–C bond broke.
• m/z 29: mass lost 46 − 29 = 17, an OH, leaving C₂H₅⁺.

Reading an IR spectrum

• Bonds stretch and bend. A bond absorbs infrared at wavenumbers that match its vibrations.
• Wavenumber (cm⁻¹) = 1 ÷ wavelength in cm. Higher wavenumber, higher energy.
• The spectrum plots % transmittance against wavenumber, from 4000 cm⁻¹ on the left. Absorptions are dips.
• Read the functional group region, above about 1500 cm⁻¹. Below it, the fingerprint region is too complex to assign band by band. Strong C–O bands (1050–1410) can still be picked out.

Bands to look for

Worked example: alcohol or ether?

• Propan-1-ol and methoxyethane are both C₃H₈O. Spectra X and Y are theirs.
• X: a strong, broad dip centred near 3350 cm⁻¹: O–H in an alcohol (3200–3600). X is propan-1-ol.
• Y: nothing at 3200–3600. C–H at 2850–3000 and C–O near 1100. No O–H: Y is the ether, methoxyethane.
• C–O (1050–1410) can't decide: alcohols, ethers and esters all have it.

Greenhouse gases

• The Earth's surface emits infrared. Greenhouse gases absorb it, then re-emit it in all directions.
• A molecule absorbs IR only if a vibration changes its dipole moment.
• CO₂ is non-polar, but its asymmetric stretch and bend change the dipole: it absorbs IR. Its symmetric stretch does not.
• H₂O and CH₄ absorb IR too. N₂ and O₂ do not: stretching them never makes them polar.

Mission 7: An MRI scanner images your body by detecting hydrogen nuclei. What can the same effect tell a chemist about a molecule?

What ¹H NMR shows

• In a strong magnetic field, ¹H nuclei absorb radio waves.
• H atoms in different chemical environments absorb at different frequencies: one signal per environment.
• Position: chemical shift, δ in ppm, measured from TMS at δ = 0.
• δ depends on what is nearby: an O atom or a C=O moves a signal to higher δ.
• The area under a signal (the integration trace) is proportional to the number of H in that environment.

Worked example: ethanol, low resolution

• CH₃CH₂OH has three H environments: three signals, areas 3 : 2 : 1.
• δ 3.7, 2H: the CH₂ on the carbon bonded to O (data booklet R–O–CH₂, 3.3–3.7).
• δ 2.6, 1H: the O–H (R–OH, 1.0–6.0; its position varies with conditions).
• δ 1.2, 3H: the CH₃ (R–CH₃, 0.9–1.0; the O two carbons away moves it slightly higher).

Counting environments

• H atoms are equivalent if symmetry or free rotation makes them interchangeable. Equivalent H share one signal.
• All three H on one CH₃ group are equivalent.
• 1,2-Dichloroethane, ClCH₂CH₂Cl: one signal (4H).
• 2-Methylpropane, (CH₃)₃CH: two signals, areas 9 : 1.

Splitting: the n + 1 rule

• In a high-resolution spectrum, a signal is split by H atoms on adjacent carbons.
• n neighbouring H → n + 1 peaks.
• 0 → singlet, 1 → doublet, 2 → triplet, 3 → quartet.
• Equivalent H do not split each other. An O–H usually gives a singlet.
• Splitting shows the neighbours' H; the group's own H count comes from the area.

Worked example: ethanol, high resolution

• δ 1.2, 3H, triplet: CH₃ next to CH₂ (2 neighbours → 3 peaks).
• δ 3.7, 2H, quartet: CH₂ next to CH₃ (3 neighbours → 4 peaks).
• δ 2.6, 1H, singlet: O–H.
• A 3H triplet with a 2H quartet is the sign of an ethyl group, CH₃CH₂–.

Mission 8: Propanal and propanone share the formula C₃H₆O. Which spectrum tells the two bottles apart at a glance?

A strategy

• 1. Molecular formula: from composition; M_r from M⁺.
• 2. IR: which bonds are present (O–H? C=O? both? neither?).
• 3. MS fragments: mass lost → pieces such as CH₃, C₂H₅, OH.
• 4. ¹H NMR: signals = environments; areas = H counts; splitting = neighbours; shift = what is nearby.
• 5. Propose a structure, then check it against every piece of data.

Worked example: compound W, C₂H₄O (part 1)

• MS: M⁺ at m/z 44 = M_r of C₂H₄O.
• MS: a peak at m/z 29: mass lost 44 − 29 = 15, a CH₃.
• IR: strong band at 1730 cm⁻¹: C=O (1700–1750). Nothing at 3200–3600: no O–H.
• So far: a C=O and a CH₃, with only two carbons.

Worked example: compound W (part 2)

• ¹H NMR: δ 9.8, 1H, quartet: an aldehyde H, R–CHO (9.4–10.0), next to a CH₃.
• δ 2.2, 3H, doublet: CH₃ next to C=O (2.2–2.7), next to one H.
• W is ethanal, CH₃CHO. Check: M_r 44 ✓; C=O, no O–H ✓; CHO⁺ at 29 ✓; 1H quartet, 3H doublet ✓.

The big ideas

• Formulas show more and more: empirical → molecular → structural → skeletal → stereochemical.
• The functional group sets the class; a homologous series adds CH₂ each time.
• Boiling points rise along a series: stronger London forces, same functional-group force.
• IUPAC: longest chain with the group; lowest locant to the suffix; prefixes in alphabetical order.
• Structural isomers: same formula, different connections. 1°, 2°, 3°: count carbons on that C (on N for amines).
• AHLCis–trans needs restricted rotation and two different groups on each carbon; a chiral carbon gives enantiomers.
• AHLMS gives M_r and mass lost; IR gives bonds; ¹H NMR gives environments, ratios and neighbours. A structure must fit them all.