Carbon,
with mechanism.

S_N2 · Substitution Nucleophilic Bimolecular

• Single concerted step — bond-making and bond-breaking happen simultaneously.
• Nucleophile attacks from the opposite side of the leaving group.
• Stereochemistry inverts ("Walden inversion") — like an umbrella turning inside out.
• Rate = k [RX][Nu⁻] — both reactants in the rate law.
• Favoured by primary halogenoalkanes (least steric hindrance to backside attack).

S_N1 · Substitution Nucleophilic Unimolecular

• Two steps. Step 1 (slow): leaving group departs, forming a planar carbocation. Step 2 (fast): nucleophile attacks the cation.
• Carbocation is planar → attack from either face → racemic product from a chiral starting material.
• Rate = k [RX] — only the substrate appears (the nucleophile attacks after the RDS).
• Favoured by tertiary halogenoalkanes (most stable carbocation) and polar protic solvents.

Predicting which mechanism dominates

The two-step mechanism (HBr + propene)

Step 1. The π electrons of C=C attack the H of H–Br. A new C–H bond forms; the Br⁻ leaves with the bonding electrons. A carbocation forms on the other carbon.

Step 2. Br⁻ attacks the carbocation, forming the C–Br bond.

CH₃-CH=CH₂ + H–Br → [CH₃-CH⁺-CH₃] + Br⁻ → CH₃-CHBr-CH₃

Why Markovnikov works

Carbocation stability: 3° > 2° > 1° > methyl. Reason: alkyl groups donate electron density to the cation (hyperconjugation and inductive effect), spreading the positive charge over a larger region.

So when there's a choice, the H adds first to whichever carbon gives the more stable carbocation. For propene + HBr, the secondary carbocation (positive on middle C) wins over the primary (positive on end C).

Symmetric alkenes

For symmetric alkenes (e.g. ethene, but-2-ene), Markovnikov doesn't apply — both carbons are equivalent, so only one product can form.

Benzene's six π electrons are delocalised in a ring above and below the plane — a "doughnut" of electron density. This delocalisation gives extra stability (~150 kJ mol⁻¹ over a hypothetical localised cyclohexatriene).

Addition would destroy this delocalisation (break the cycle). Substitution preserves it (the leaving group is H⁺). So benzene undergoes electrophilic substitution, not addition.

Nitration (most famous example)

Reagents: conc. HNO₃ + conc. H₂SO₄, 50 °C.

Step 1 (generate electrophile): HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. The nitronium ion NO₂⁺ is the actual electrophile.

Step 2 (electrophile attack): Two π electrons of benzene attack NO₂⁺, forming a positively-charged intermediate (the Wheland or arenium ion) with broken delocalisation.

Step 3 (lose H⁺): H⁺ is lost from the same carbon, restoring the aromatic ring.

C₆H₆ + HNO₃ →(H₂SO₄, 50°C)→ C₆H₅NO₂ + H₂O

Halogenation

Bromination of benzene requires a Lewis acid catalyst (FeBr₃ or AlBr₃) to polarise Br₂: FeBr₃ + Br₂ → FeBr₄⁻ + Br⁺. The Br⁺ then acts as the electrophile (same mechanism as nitration).

Stereoisomers have the same structural formula but a different spatial arrangement of atoms. Two main types: cis-trans (E/Z) isomers and optical isomers.

Cis-trans isomerism

Arises from restricted rotation around C=C (or around a ring). Both conditions must hold:

• A double bond (or ring) preventing free rotation.
• Each carbon of the double bond carries two different groups.

cis: identical or similar groups on the same side. trans: on opposite sides.

Modern E/Z notation

For each end of the C=C, assign priority using Cahn-Ingold-Prelog rules (higher atomic number = higher priority). If the two higher-priority groups are on the same side: Z (zusammen, German for "together"). If on opposite sides: E (entgegen, "opposite").

Properties differ

Cis isomers often have a net dipole (b.p. typically higher). Trans isomers usually pack more efficiently in the solid state (higher m.p.). Both biological activity and reactivity can differ — important in fats: cis fats kink, are liquid; trans fats are straight, solid.

A molecule is chiral if it is not superimposable on its mirror image — like left and right hands. The two non-superimposable mirror images are enantiomers (or "optical isomers").

The most common cause: a stereocentre — an sp³ carbon bonded to four different groups.

How to tell?

If a molecule has just one carbon with four different groups attached, it has two enantiomers. Two or more stereocentres can give up to 2ⁿ stereoisomers.

Properties

Enantiomers have identical: m.p., b.p., density, solubility, IR spectrum.
Enantiomers differ in: rotation of plane-polarised light (one rotates clockwise, +; the other counter-clockwise, −); biological activity (in chiral environments like enzymes, taste/smell receptors, etc.).

Racemic mixture

50:50 mixture of two enantiomers. Optically inactive overall (rotations cancel). Many synthetic drugs are produced as racemic mixtures and must be resolved before use.

Why it matters · thalidomide

Thalidomide has one chiral centre. (R)-thalidomide is a sedative; (S)-thalidomide is teratogenic (causes severe birth defects). Originally sold as a racemic mixture in the 1950s, before the chemistry community appreciated the importance of single-enantiomer drugs. Thousands of children born with limb deformities. Now a cautionary tale that drives modern enantioselective synthesis.

A mass spectrometer ionises a sample, accelerates the ions, and separates them by mass-to-charge ratio (m/z). Output: a stick plot of relative abundance vs m/z.

Key peaks

• Molecular ion M⁺•: the whole molecule with one electron knocked off. Gives M_r directly.
• Fragment peaks: smaller pieces formed when M⁺• breaks apart. Mass loss between peaks tells you what was lost.
• Base peak: the tallest peak; relative abundance set to 100% by convention.

Common mass losses

Isotope patterns

Chlorine: ³⁵Cl : ³⁷Cl = 3 : 1. M and M+2 peaks in 3:1 ratio.
Bromine: ⁷⁹Br : ⁸¹Br ≈ 1 : 1. M and M+2 peaks in ~1:1 ratio.

Infrared photons make bonds bend and stretch. Each bond has a characteristic vibration frequency, expressed in wavenumbers (cm⁻¹). When IR light at that frequency is absorbed, you see a dip in the spectrum.

Key absorption ranges (data booklet values)

The 1000–1400 cm⁻¹ "fingerprint region" contains many overlapping peaks unique to each compound — used for identification by matching against a database.

In ¹H NMR, hydrogen nuclei resonate at characteristic frequencies depending on their chemical environment. Three pieces of information:

1. Number of peaks = number of distinct H environments. Equivalent H's (by symmetry) give one peak.
2. Chemical shift (ppm): position of the peak relative to TMS (set to 0). Electronegative groups nearby pull electron density away, deshielding the H and shifting its peak downfield (higher ppm).
3. Integration: the area under each peak is proportional to the number of equivalent H's contributing.

Common chemical shifts

In high-resolution NMR, each peak from a low-resolution spectrum may split into multiple lines because of magnetic coupling with neighbouring H's.

The n+1 rule

If a H has n non-equivalent H's on directly bonded adjacent carbons, its peak splits into n + 1 lines.

Worked logic · ethanol CH₃CH₂OH

• –CH₃ has 2 H's next door (the CH₂) → splits into triplet.
• –CH₂– has 3 H's next door (the CH₃) → splits into quartet.
• –OH: doesn't usually couple in routine spectra (exchanges with traces of water). Appears as a singlet, often broad. Position is variable.

Integration: CH₃ : CH₂ : OH = 3 : 2 : 1.

The combined-spectra problem is the classic IB Paper 2 question. The standard workflow:

1. Mass spec → M_r from M⁺•. Maybe molecular formula too (combined with degrees of unsaturation, isotope patterns).
2. IR → which functional groups are present? C=O around 1720? O–H broad above 3200? N–H? C=C around 1650?
3. ¹H NMR → how many H environments? What chemical shifts? What integration ratios? What splitting patterns?
4. Combine: M_r sets the size; IR narrows the family; NMR settles the connectivity.
5. Sketch a candidate structure and check it predicts the observed spectra.