Flashcards · H432 · Topic 6

Organic chemistry and analysis flashcards

61 cards on Topic 6, subtopics 6.1.1 to 6.3.2. Pick a subtopic or study them all.

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6.1.1 Aromatic compounds

Three pieces of evidence against the Kekulé structure of benzene: All six C–C bond lengths are equal, between C–C and C=C; the enthalpy of hydrogenation is less exothermic than three times that of cyclohexene; benzene does not undergo electrophilic addition with bromine at room temperature.
Halogenation of benzene: reagents and electrophile: Br₂ with a halogen carrier (e.g. FeBr₃ or AlBr₃), room temperature. Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻. Products bromobenzene + HBr; FeBr₃ is regenerated.
Nitration of benzene: reagents and conditions: Concentrated HNO₃ and concentrated H₂SO₄, about 50 °C. HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. Product nitrobenzene.
Why does phenol react with bromine water without a halogen carrier?: A lone pair from the oxygen p-orbital is partly delocalised into the ring, increasing its electron density, so the ring polarises Br₂ and is more easily attacked. White precipitate of 2,4,6-tribromophenol.
How do NaOH(aq) and Na₂CO₃(aq) distinguish phenol from a carboxylic acid?: Both react with NaOH. Only the carboxylic acid reacts with Na₂CO₃ (effervescence); phenol does not.
Directing effects of –OH, –NH₂ and –NO₂: –OH and –NH₂ (electron-donating): direct to positions 2 and 4. –NO₂ (electron-withdrawing): directs to position 3.
Phenol with dilute nitric acid: Room temperature, no H₂SO₄ needed; gives a mixture of 2-nitrophenol and 4-nitrophenol, plus water.

6.1.2 Carbonyl compounds

Reduction of carbonyl compounds with NaBH₄: Aldehyde → primary alcohol; ketone → secondary alcohol. Nucleophilic addition of H⁻; equations use 2[H].
HCN with carbonyl compounds: reagent and product from ethanal: NaCN(aq)/H⁺(aq). Ethanal gives 2-hydroxypropanenitrile, CH₃CH(OH)CN. It adds a carbon to the chain.
Why is C=O attacked by nucleophiles but C=C by electrophiles?: C=O is polar (C δ+, O δ−), so nucleophiles attack the carbon. C=C is non-polar with an electron-rich π-bond, so it attracts electrophiles.
2,4-DNP test: An orange (yellow-orange) precipitate shows an aldehyde or ketone. Carboxylic acids and esters give no precipitate.
How does a 2,4-DNP derivative identify a carbonyl compound?: Filter, recrystallise and dry the precipitate, measure its melting point, and compare with a data table.
Tollens’ test: result and explanation: Warm in a water bath: an aldehyde gives a silver mirror; a ketone gives no change. The aldehyde is oxidised to a carboxylic acid; Ag⁺ is reduced to Ag (Ag⁺ + e⁻ → Ag).
Mass of silver from 0.290 g of propanal (Mr 58.0) with Tollens’ reagent (Ag 107.9): n(propanal) = 0.00500 mol; n(Ag) = 2 × 0.00500 = 0.0100 mol; mass = 1.08 g

6.1.3 Carboxylic acids and esters

Why are short-chain carboxylic acids soluble in water, and why does solubility fall with chain length?: The –COOH group hydrogen bonds with water. A longer non-polar chain cannot hydrogen bond, so solubility decreases.
Equation: ethanoic acid with potassium carbonate: 2CH₃COOH + K₂CO₃ → 2CH₃COOK + H₂O + CO₂
Esterification with a carboxylic acid vs with an acid anhydride: Acid + alcohol: heat with a concentrated H₂SO₄ catalyst; reversible; water also forms. Anhydride + alcohol: warm; not reversible; a carboxylic acid also forms; higher yield.
Acid vs alkaline hydrolysis of an ester: Hot dilute acid (reflux): carboxylic acid + alcohol; reversible. Hot aqueous alkali (reflux): carboxylate salt + alcohol; goes to completion.
Making an acyl chloride from a carboxylic acid: RCOOH + SOCl₂ → RCOCl + SO₂ + HCl, in dry conditions. Both by-products are gases, so the acyl chloride is easy to separate.
Products of an acyl chloride with water, an alcohol, ammonia and a primary amine: Water: carboxylic acid. Alcohol (or phenol): ester. Ammonia: primary amide. Primary amine: secondary amide. HCl forms each time (as an ammonium salt with excess base).
How is a phenyl ester made?: Phenol is not readily esterified by a carboxylic acid, so react the phenol with an acyl chloride.

6.2.1 Amines

Why are amines bases?: The lone pair on the nitrogen accepts a proton (H⁺), forming a dative covalent bond.
Salt formed from phenylamine and HCl: C₆H₅NH₃⁺Cl⁻, phenylammonium chloride. Warming with NaOH(aq) gives back the free amine.
Primary amine from a haloalkane: reagent and conditions: Excess ammonia in ethanol, heated in a sealed tube. RX + 2NH₃ → RNH₂ + NH₄X
Why ethanol, a sealed tube and excess ammonia?: Ethanol, not water, so OH⁻ does not form an alcohol. Sealed tube so volatile ammonia does not escape. Excess NH₃ so the primary amine is not substituted further.
Aromatic amine from a nitroarene: reagents and equation: Tin and concentrated HCl, heat under reflux, then excess NaOH(aq). ArNO₂ + 6[H] → ArNH₂ + 2H₂O
Why is NaOH added after reducing a nitroarene with Sn/HCl?: In concentrated HCl the amine forms as ArNH₃⁺; OH⁻ removes the extra proton to give the free amine.

6.2.2 Amino acids, amides and chirality

General formula of an α-amino acid: RCH(NH₂)COOH: the NH₂ and COOH groups on the same carbon.
Which group of an amino acid reacts with alkali, with acid, and with an alcohol?: Alkali: COOH → carboxylate salt. Acid: NH₂ → ammonium salt. Alcohol with conc. H₂SO₄ catalyst: COOH → ester.
Primary vs secondary amide: Primary: N bonded to one carbon (the C=O carbon), RCONH₂. Secondary: N bonded to two carbons, RCONHR′.
Optical isomers: A pair of non-superimposable mirror images about a chiral centre (a carbon atom carrying four different atoms or groups).
How many chiral centres in 2,4-dichloropentane, CH₃CHClCH₂CHClCH₃?: Two: C2 and C4 each carry H, Cl, CH₃ and the rest of the chain.
Why does glycine have no optical isomers?: R = H, so its α-carbon carries two H atoms and is not a chiral centre.

6.2.3 Polyesters and polyamides

Monomers that give a polyester and a polyamide: Polyester: dicarboxylic acid (or diacyl chloride) + diol, or one hydroxycarboxylic acid. Polyamide: dicarboxylic acid (or diacyl chloride) + diamine, or one aminocarboxylic acid.
Small molecule lost: diacid monomers vs diacyl chloride monomers: H₂O from –COOH; HCl from –COCl.
Products of hydrolysing a polyamide in hot aqueous acid and in hot aqueous alkali: Acid (e.g. HCl): –COOH and –NH₃⁺ Cl⁻. Alkali (e.g. NaOH): –COO⁻ Na⁺ and –NH₂.
Repeat unit from ethanedioic acid and pentane-1,5-diamine: –[CO–CO–NH–(CH₂)₅–NH]–, with water lost at each link.
How do you tell addition from condensation polymers from the chain?: An all-carbon backbone means addition. –COO– or –CONH– links in the backbone mean condensation (side-group esters or amides do not count).

6.2.4 Carbon–carbon bond formation

Nitrile from a haloalkane: reagent, conditions and product from 1-chlorobutane: NaCN or KCN in ethanol, reflux; nucleophilic substitution. 1-chlorobutane gives pentanenitrile, CH₃CH₂CH₂CH₂CN.
Reduction vs acid hydrolysis of a nitrile: Reduction: H₂ with Ni catalyst → primary amine, RCH₂NH₂. Hydrolysis: dilute acid, reflux → carboxylic acid RCOOH + NH₄⁺. Both keep the nitrile carbon.
Friedel–Crafts alkylation of benzene: reagents and electrophile: A haloalkane with a halogen carrier such as AlCl₃, dry. R–Cl + AlCl₃ → R⁺ + AlCl₄⁻. C₆H₆ + RCl → C₆H₅R + HCl.
Route: pentanal to 2-hydroxyhexanoic acid: NaCN(aq)/H⁺(aq) → 2-hydroxyhexanenitrile; then hydrolyse with dilute acid under reflux.
In a cyanide mechanism, where does the curly arrow start?: At the lone pair on the carbon of :CN⁻, never at the nitrogen.

6.2.5 Organic synthesis

Why does recrystallisation remove soluble impurities?: The product saturates the cold solution and crystallises; soluble impurities are too dilute to saturate it, so they stay dissolved.
Why filter under reduced pressure rather than by gravity?: It is faster and leaves a drier solid.
NaBH₄ vs H₂/Ni as reducing agents: NaBH₄ reduces only C=O in aldehydes and ketones. H₂/Ni reduces C=C and C≡N.
Route: benzene to 1-phenylbutan-1-ol: Butanoyl chloride with AlCl₃ (Friedel–Crafts acylation) → C₆H₅COCH₂CH₂CH₃; then NaBH₄ → C₆H₅CH(OH)CH₂CH₂CH₃.
Overall yield of a two-stage route with stages of 80% and 65%: 0.80 × 0.65 × 100 = 52.0%
Methyl 3-hydroxybutanoate with Na₂CO₃ and with hot NaOH(aq): Na₂CO₃: no effervescence (no COOH). Hot NaOH(aq): the ester is hydrolysed to sodium 3-hydroxybutanoate and methanol.

6.3.1 Chromatography and qualitative analysis

Stationary and mobile phases in TLC and GC: TLC: stationary solid (silica or alumina) on a plate; mobile solvent. GC: stationary high-boiling liquid on a solid support in a column; mobile unreactive carrier gas such as helium or nitrogen.
A polar compound on a silica TLC plate: high or low Rf?: Low Rf: it is adsorbed strongly onto the silica.
Solvent front 5.60 cm, spot centre 1.96 cm from the baseline. Rf?: 1.96 ÷ 5.60 = 0.35
How is the proportion of a component found from a gas chromatogram?: Its peak area ÷ total peak area × 100 (area, not peak height). For an exact concentration, use an external calibration curve of peak area against concentration.
Test for a haloalkane: Warm with AgNO₃(aq) in ethanol in a water bath: white AgCl, cream AgBr or yellow AgI precipitate.
How do you tell a primary from a secondary alcohol?: Both turn acidified dichromate green. Distil off the product and test it: a silver mirror with Tollens’ (aldehyde) means primary; a ketone means secondary.

6.3.2 Spectroscopy

Why is TMS used as the NMR standard?: Defined as δ = 0; its 12 equivalent H (and 4 equivalent C) give one sharp signal away from almost all sample signals; it is inert and volatile, so easily removed.
Why use a deuterated solvent such as CDCl₃?: It contains no ¹H, so the solvent gives no signal in the proton spectrum to swamp the sample.
What does a D₂O shake show?: O–H and N–H protons exchange with deuterium (R–OH + D₂O ⇌ R–OD + HDO), so their signal disappears on rerunning; C–H signals stay.
Splitting patterns: (CH₃)₂CH– and a CH₂ between CH₃ and CH₂: (CH₃)₂CH–: 6H doublet + 1H septet. CH₂ with 3 + 2 = 5 neighbouring H: a sextet.
Carbon-13 C=O at δ 160–185 vs 190–220: 160–185: acid, ester or amide. 190–220: aldehyde or ketone.
How many carbon environments in benzene-1,4-dicarboxylic acid?: Three: the two C=O carbons, the two ring carbons bonded to COOH, and the four ring C–H carbons.