Flashcards · H432 · Topic 4

Core organic chemistry flashcards

44 cards on Topic 4, subtopics 4.1.1 to 4.2.4. Pick a subtopic or study them all.

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4.1.1 Basic concepts of organic chemistry

Homologous series: A series of organic compounds having the same functional group, with each successive member differing by CH₂.
Aliphatic, alicyclic and aromatic: Aliphatic: C and H in straight chains, branched chains or non-aromatic rings. Alicyclic: aliphatic, in non-aromatic rings. Aromatic: contains a benzene ring.
Name CH₃CH(OH)CH₂CHClCH₃: 4-chloropentan-2-ol (numbered from the end nearer the OH).
Homolytic vs heterolytic fission: Homolytic: one electron goes to each atom, giving two radicals (Br–Br → Br• + Br•). Heterolytic: both electrons go to one atom, giving a cation and an anion (H–Cl → H⁺ + Cl⁻).
A haloalkane CₙH₂ₙ₊₁Br has Mr 108.9 (Br = 79.9). Find its formula.: 14n + 1 + 79.9 = 108.9, so 14n = 28 and n = 2: C₂H₅Br.
Where can a curly arrow start?: At a covalent bond, a lone pair or a negative charge. Never at an atom with no lone pair shown, or at a positive charge.

4.1.2 Alkanes

σ-bond, and why there is free rotation in alkanes: A σ-bond forms by direct overlap of orbitals between the two bonding atoms, on the line joining the nuclei. Rotation does not change that overlap, so rotation is free.
Why does a branched alkane boil lower than its straight-chain isomer?: Same number of electrons, but the branched molecule is more compact with fewer points of contact, so weaker London forces and less energy to overcome them.
High bond enthalpy: C–C and C–H σ-bonds are strong and need a lot of energy to break.
Very low polarity: C and H have very similar electronegativities, so the σ-bonds have little or no dipole and nothing attracts nucleophiles or electrophiles.
Radical substitution of propane to 2-chloropropane: the two propagation steps: CH₃CH₂CH₃ + Cl• → CH₃CH•CH₃ + HCl CH₃CH•CH₃ + Cl₂ → CH₃CHClCH₃ + Cl• (Conditions: UV radiation.)
Why is radical substitution of limited use in synthesis?: Further substitution gives di- and tri-substituted products, and substitution at different positions gives isomers, so a mixture forms.
Why is carbon monoxide dangerous?: It is toxic because it binds to haemoglobin and reduces oxygen transport in the blood; it is colourless and odourless, so it is not noticed without a detector.

4.1.3 Alkenes

π-bond: Sideways overlap of adjacent p-orbitals, with electron density above and below the bonding carbons. It restricts rotation about C=C.
Cis–trans isomerism vs E/Z isomerism: E/Z: stereoisomerism from restricted rotation about C=C, with two different groups on each C=C carbon. Cis–trans: a special case of E/Z where two of the substituent groups on each C=C carbon are the same.
Why are alkenes more reactive than alkanes?: The π-bond has a relatively low bond enthalpy, and its exposed electrons above and below the plane attract electrophiles.
Addition reactions of alkenes: reagents and conditions: H₂: Ni catalyst, heat → alkane. Cl₂ or Br₂: room temperature → dihaloalkane. HX: room temperature → haloalkane. Steam: acid catalyst (H₃PO₄) → alcohol.
Major product of 2-methylpent-2-ene with HI: 2-iodo-2-methylpentane, via the more stable tertiary carbocation (Markownikoff’s rule).
Electrophile: An electron pair acceptor.
Ways of dealing with waste polymers: Combustion for energy; use as organic feedstock for new plastics and chemicals; removing toxic products such as HCl from burning PVC; biodegradable and photodegradable polymers.

4.2.1 Alcohols

Why do alcohols have higher boiling points than alkanes of similar Mr?: Similar London forces, plus hydrogen bonds between –OH groups, so more energy is needed to separate the molecules.
Oxidation of a primary alcohol: aldehyde vs carboxylic acid: Aldehyde: gentle heat, distil it out as it forms. Carboxylic acid: excess K₂Cr₂O₇/H₂SO₄, heat under reflux. Orange → green in both.
Equation: butan-1-ol oxidised under reflux with excess oxidising agent: CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O (butanoic acid)
Why does a tertiary alcohol not oxidise with acidified dichromate?: There is no H atom on the C–OH carbon to remove; the solution stays orange.
Dehydration of an alcohol: reagent and products from 2-methylheptan-2-ol: Concentrated H₃PO₄ or H₂SO₄ catalyst, heat. Gives 2-methylhept-1-ene and 2-methylhept-2-ene.
Making a bromoalkane from an alcohol: Heat with NaBr and H₂SO₄. NaBr + H₂SO₄ → NaHSO₄ + HBr, then ROH + HBr → RBr + H₂O.

4.2.2 Haloalkanes

Nucleophile: An electron pair donor (e.g. :OH⁻ or H₂O). It must have a lone pair to donate; it need not be negative.
Hydrolysis of a haloalkane with aqueous alkali: reagent, conditions and mechanism: NaOH(aq), heat under reflux. Nucleophilic substitution: lone pair on O of OH⁻ attacks the δ+ carbon; the C–X bond breaks and X⁻ leaves. R–X + OH⁻ → R–OH + X⁻
Why do iodoalkanes hydrolyse fastest?: The C–X bond must break. Bond enthalpy decreases from C–F to C–I, so the C–I bond needs least energy to break. Rate: C–I > C–Br > C–Cl > C–F.
PAG7: comparing hydrolysis rates with aqueous silver nitrate in ethanol: Same temperature (water bath), volumes and chain length; time until a precipitate appears. Rate ∝ 1/time. Ethanol is a common solvent. AgI (yellow) forms fastest, AgCl (white) slowest.
How do chlorine radicals from CFCs break down ozone?: UV breaks C–Cl homolytically, releasing Cl•. Cl• + O₃ → ClO• + O₂ ClO• + O → Cl• + O₂ Overall O₃ + O → 2O₂; Cl• is regenerated, so it is a catalyst.
Source of NO radicals that also deplete ozone: High-temperature reactions of N₂ and O₂, such as in lightning and aircraft engines.

4.2.3 Organic synthesis

Heating under reflux vs distillation: Reflux: vertical condenser, open top; vapour condenses and returns, so the mixture can be heated for a long time without losing volatile substances. Distillation: sloping condenser to a receiver; thermometer at the side arm; removes a product.
Steps to purify an organic liquid: Separating funnel: wash with water, then NaHCO₃(aq) to remove acid (vent the CO₂); keep the organic layer; dry with anhydrous MgSO₄ or CaCl₂ until clear; decant; redistil and collect the fraction at the boiling point.
In a separating funnel, how do you know which layer is organic?: By density: the denser liquid is at the bottom. Use the density data, not a guess.
Washing vs drying an organic product: Washing removes water-soluble impurities such as acid or salts. Drying removes the traces of water left after washing.
Two-stage route: haloalkane to ketone: The haloalkane must be secondary. Hydrolyse with NaOH(aq), heat under reflux, to the secondary alcohol; then oxidise with K₂Cr₂O₇/H₂SO₄ under reflux to the ketone.
Mass of 6.00 cm³ of a liquid of density 0.785 g cm⁻³: 6.00 × 0.785 = 4.71 g

4.2.4 Analytical techniques

What happens when a bond absorbs infrared radiation?: It vibrates more (stretches and bends). It absorbs only at the frequencies matching its own vibrations, which depend on bond strength and the masses of the atoms.
IR: alcohol O–H vs carboxylic acid O–H (OCR typical values): Alcohol O–H: broad, 3200–3600 cm⁻¹. Acid O–H: very broad, 2500–3300 cm⁻¹, together with a strong C=O at about 1700 cm⁻¹ (range 1630–1820).
How is IR linked to global warming?: C=O, O–H and C–H bonds in CO₂, H₂O and CH₄ absorb IR emitted by the Earth’s surface and vibrate more; the energy is re-emitted in all directions, some back to the surface.
Two real-world uses of IR spectroscopy: Monitoring gases such as CO and NO from car emissions, and breathalysers measuring ethanol in breath (used as evidence in drink-driving cases).
Molecular ion peak vs M+1 peak: Molecular ion peak: the peak at highest m/z (ignoring M+1); gives Mr. M+1: a small peak one unit higher, from molecules containing a carbon-13 atom.
Common fragment ions at m/z 15, 29 and 43: 15: CH₃⁺. 29: C₂H₅⁺ or CHO⁺. 43: C₃H₇⁺ or CH₃CO⁺. Only the positive ion is detected; the radical lost gives no peak.