Heterocyclic Compounds and Biomolecules

Two sub-headings of Section 3 of the GATE Chemistry (CY) paper. Heterocyclic Compounds: the nomenclature of mono- and bicyclic, mono- and di-heteroatomic compounds, and the structure, preparation, properties and reactions of furan, pyrrole, thiophene, pyridine, indole, quinoline and isoquinoline. Biomolecules: the structure, properties and reactions of mono- and disaccharides; the physicochemical properties of amino acids; the chemical synthesis of peptides; the chemical determination of the structure of peptides and proteins; the structural features of proteins and nucleic acids; and lipids, steroids, terpenoids, carotenoids and alkaloids. Natural products are treated at the level of structure and classification the syllabus names.

1. Naming heterocycles

The Hantzsch–Widman system builds a name from a prefix for each heteroatom — oxa (O), thia (S), aza (N), in that order of priority — and a stem giving ring size and saturation: three-membered -irene/-irane (N: -irine/-iridine), four -ete/-etane (-etidine for N), five -ole (unsaturated) and -olane (saturated; -olidine for N), six -ine (for N) and -inane or -ane (for O and S). So oxirane is ethylene oxide, thiirane the sulfur analogue, azetidine the saturated four-membered amine, 1,3-oxazole and 1,3-thiazole the O,N and S,N five-membered rings, and 1,3-diazole is imidazole. Numbering starts at the highest-priority heteroatom (O > S > N) and gives the other heteroatoms the lowest locants. Many rings keep trivial names that are also accepted: furan, thiophene, pyrrole, pyridine, pyrimidine (1,3-diazine), pyrazine (1,4-diazine), pyridazine (1,2-diazine) and pyrazole (1,2-diazole).

Fused bicyclic systems are named by prefixing the attached ring to the base component and lettering the fusion bond: indole is benzo[b]pyrrole, benzofuran benzo[b]furan, benzothiophene benzo[b]thiophene; quinoline is benzo[b]pyridine with N at position 1 and isoquinoline benzo[c]pyridine with N at 2; purine fuses pyrimidine with imidazole. Fused rings keep conventional numbering: in indole N is 1 and the reactive carbon is C3; in quinoline C4 is para to N and C5/C8 are the peri positions of the benzene ring.

2. Furan, pyrrole and thiophene

In furan, pyrrole and thiophene each ring atom contributes one p orbital and the heteroatom supplies a lone pair, giving six π electrons over five atoms: aromatic, planar and π-excessive. Aromatic stabilisation falls as the heteroatom becomes more electronegative and holds its lone pair more tightly: benzene > thiophene > pyrrole > furan; furan is the least aromatic and behaves as a diene in Diels–Alder reactions with maleic anhydride. Because the ring carbons carry extra π density, electrophilic substitution is far faster than for benzene — pyrrole > furan > thiophene > benzene — and occurs at C2, where the σ-complex has three resonance forms including one with the positive charge on the heteroatom (C3 attack gives two). Pyrrole is so reactive that it polymerises in strong acid, so mild reagents are used: Vilsmeier–Haack formylation (DMF/POCl₃), acetylation with Ac₂O, nitration with acetyl nitrate. Pyrrole’s N–H is weakly acidic (pKa ≈ 17.5), and the anion alkylates on N or C.

Preparation. The Paal–Knorr synthesis cyclises a 1,4-dicarbonyl compound: with acid (or P₂O₅) to a furan, with ammonia or a primary amine to a pyrrole, and with P₄S₁₀ or Lawesson’s reagent to a thiophene. The Knorr pyrrole synthesis condenses an α-amino ketone with a β-keto ester, and the Hantzsch pyrrole synthesis combines an α-halo ketone, a β-keto ester and ammonia. Industrially furfural (from pentose-rich biomass by acid dehydration) is the gateway to furan.

3. Pyridine, indole, quinoline and isoquinoline

Pyridine is isoelectronic with benzene; the nitrogen lone pair lies in an sp² orbital in the ring plane, not in the π system, so pyridine is aromatic and basic (pKaH 5.2) and is a good ligand and nucleophilic catalyst. The electronegative nitrogen makes the ring π-deficient: electrophilic substitution is very sluggish (made worse by protonation of N in acidic media) and goes to C3 under forcing conditions; the N-oxide, activated at C4, is the practical detour. Conversely, nucleophiles attack C2 and C4, where the negative charge of the intermediate can sit on nitrogen: the Chichibabin reaction (NaNH₂) gives 2-aminopyridine with loss of H⁻, and 2- and 4-halopyridines undergo S_NAr easily. The Hantzsch pyridine synthesis condenses two equivalents of a β-keto ester, an aldehyde and ammonia to a 1,4-dihydropyridine, oxidised to the pyridine.

Indole reacts with electrophiles at C3, not C2: attack at C3 gives a σ-complex in which the positive charge is stabilised by nitrogen without disrupting the benzene ring. Indoles are made by the Fischer indole synthesis, in which an arylhydrazone of a ketone or aldehyde rearranges in acid through its ene-hydrazine by a [3,3]-sigmatropic shift, then cyclises and loses NH₃. Quinoline and isoquinoline combine a benzene and a pyridine ring: electrophiles attack the benzene ring (C5 and C8), nucleophiles the pyridine ring (C2 and C4 of quinoline, C1 of isoquinoline). Quinolines come from the Skraup synthesis (aniline, glycerol, conc. H₂SO₄ and a mild oxidant: glycerol dehydrates to acrolein, which undergoes conjugate addition, cyclisation and oxidation), the Doebner–Miller and the Friedländer synthesis (2-aminobenzaldehyde with a ketone). Isoquinolines come from the Bischler–Napieralski reaction (a β-arylethylamide with POCl₃ gives a 3,4-dihydroisoquinoline) and the Pictet–Spengler reaction (a β-arylethylamine with an aldehyde gives a tetrahydroisoquinoline), the latter mirroring alkaloid biosynthesis.

4. Monosaccharides and disaccharides

Monosaccharides are polyhydroxy aldoses or ketoses. The D/L label refers to the configuration of the stereocentre farthest from the carbonyl, compared with D-glyceraldehyde; it is unrelated to the sign of rotation. An open-chain aldohexose has four stereocentres and 2⁴ = 16 stereoisomers, eight D and eight L; D-glucose, D-mannose (its C2 epimer) and D-galactose (its C4 epimer) are the common ones. In solution glucose exists almost entirely as the six-membered pyranose hemiacetal, formed by attack of O5 on C1, which creates a new stereocentre, the anomeric carbon: the α-anomer (C1–OH axial, trans to CH₂OH in the Haworth projection) has [α]_D +112° and the β-anomer (C1–OH equatorial) +18.7°. Either dissolves and changes rotation to the equilibrium value +52.7° — mutarotation — with about 36% α and 64% β, the β favoured because all its substituents are equatorial despite the anomeric effect.

Reactions. Sugars with a free hemiacetal are reducing sugars: they reduce Tollens’ reagent and Fehling’s solution. Br₂ water oxidises an aldose to the aldonic acid; dilute HNO₃ oxidises both ends to the aldaric acid; NaBH₄ gives the alditol. Three equivalents of phenylhydrazine form an osazone at C1 and C2, so glucose, mannose and fructose, which differ only at C1–C2, give the same osazone. The Kiliani–Fischer synthesis lengthens the chain by one carbon (cyanohydrin, hydrolysis, reduction; two C2 epimers result), and the Ruff degradation shortens it. Alcohols under acid give glycosides (acetals), which do not mutarotate. Disaccharides: maltose is α-D-Glc-(1→4)-D-Glc and cellobiose β-D-Glc-(1→4)-D-Glc, both reducing; lactose is β-D-Gal-(1→4)-D-Glc, reducing; sucrose is α-D-Glc-(1→2)-β-D-Fru, joined through both anomeric carbons and so non-reducing; its hydrolysis changes the rotation from +66.5° to negative (invert sugar), because fructose is strongly laevorotatory.

5. Amino acids, peptide synthesis and the structure of peptides, proteins and nucleic acids

The twenty coded α-amino acids (all L, i.e. S except cysteine, which is R) exist in water as zwitterions H₃N⁺–CHR–COO⁻: they are high-melting, water-soluble and poorly soluble in organic solvents. Each has pKa₁ (COOH, about 2) and pKa₂ (NH₃⁺, about 9–10), and side-chain ionisations for acidic and basic residues. The isoelectric point pI, where the net charge is zero and the amino acid does not move in electrophoresis, is the average of the two pKa values that bracket the neutral zwitterion: glycine (2.34, 9.60) has pI 5.97; aspartic acid (1.88, 3.65 side chain, 9.60) has pI (1.88 + 3.65)/2 = 2.77; lysine (2.18, 8.95, 10.53 side chain) has pI (8.95 + 10.53)/2 = 9.74.

Chemical peptide synthesis couples an N-protected amino acid to a C-protected one. The amine is protected as Boc (removed by TFA) or Fmoc (removed by piperidine); the carboxyl is activated with a carbodiimide such as DCC (often with HOBt to suppress racemisation through the oxazolone) or as an active ester; and side chains carry orthogonal protecting groups. In Merrifield solid-phase synthesis the C-terminal residue is anchored to a polystyrene resin, the chain is built from C to N by repeated deprotection and coupling, excess reagents are washed away, and the peptide is finally cleaved from the resin (HF for Boc chemistry, TFA for Fmoc chemistry).

Structure determination. Acid hydrolysis and an amino acid analyser give the composition. The N-terminus is identified by Sanger’s reagent (1-fluoro-2,4-dinitrobenzene) or, repeatedly, by the Edman degradation: phenyl isothiocyanate adds to the N-terminal amine and, in acid, cleaves that residue alone as a phenylthiohydantoin, leaving the shortened peptide for the next cycle. The C-terminus is found with carboxypeptidase. Longer chains are cut into overlapping fragments: trypsin cleaves after Lys and Arg, chymotrypsin after Phe, Tyr and Trp, and cyanogen bromide after Met. Disulfide bridges are located after cleaving them. Structural features: the amide bond is planar (partial C=N character) and usually trans; secondary structure is the α-helix (3.6 residues and 0.54 nm per turn, N–H···O=C bonds from residue i to i + 4) and the β-sheet; tertiary and quaternary structure follow. Nucleic acids: a nucleoside is a base on ribose or 2′-deoxyribose (β-N-glycoside), a nucleotide its phosphate ester, and the backbone links 3′ to 5′ through phosphodiesters. In double-stranded DNA the antiparallel strands pair A with T (two hydrogen bonds) and G with C (three), so A = T and G = C (Chargaff), in a right-handed B-form helix of about 10.5 base pairs per turn.

6. Lipids, steroids, terpenoids, carotenoids and alkaloids

Lipids are the water-insoluble biomolecules: triglycerides (glycerol triesters of fatty acids, saponified by NaOH to glycerol and soaps), phospholipids (a phosphate head group on a diacylglycerol, which self-assemble into bilayers) and waxes. Natural unsaturated fatty acids have cis double bonds, which kink the chain and lower the melting point (oleic acid is liquid, stearic solid). Steroids share the tetracyclic cyclopenta[a]phenanthrene (perhydro-1,2-cyclopentenophenanthrene) nucleus of 17 carbons in rings A–D, usually all-trans fused; cholesterol, the bile acids, the sex hormones and the corticosteroids differ in oxidation pattern and side chain.

Terpenoids are built from C₅ isoprene units, joined mostly head-to-tail (the isoprene rule), biosynthetically from isopentenyl and dimethylallyl diphosphates. Classes by size: monoterpenes C₁₀ (limonene, menthol, citral), sesquiterpenes C₁₅ (farnesol), diterpenes C₂₀ (phytol, retinol), triterpenes C₃₀ (squalene, formed tail-to-tail from two farnesyl units and cyclised to lanosterol and then the steroids), and tetraterpenes C₄₀, the carotenoids. β-Carotene has eleven conjugated C=C bonds, which move its absorption into the visible and make it orange; central cleavage gives two molecules of retinal (vitamin A aldehyde). Alkaloids are basic, nitrogen-containing natural products, mostly from plants and mostly derived from amino acids, classified by their heterocyclic skeleton — pyridine and piperidine (nicotine, coniine), tropane (atropine, cocaine), quinoline (quinine), isoquinoline (papaverine, morphine), indole (reserpine, strychnine) and purine (caffeine) — and characterised by their basicity, which allows extraction into aqueous acid and precipitation with base.

Key takeaways

  • Hantzsch–Widman: oxa > thia > aza, -ole/-olane for five-membered rings; indole is benzo[b]pyrrole, quinoline benzo[b]pyridine (N1), isoquinoline benzo[c]pyridine (N2).
  • Furan, pyrrole, thiophene: π-excessive, substitute at C2, reactivity pyrrole > furan > thiophene > benzene, aromaticity thiophene > pyrrole > furan; Paal–Knorr from 1,4-dicarbonyls.
  • Pyridine: basic, π-deficient, electrophiles at C3 only under force, nucleophiles at C2/C4 (Chichibabin); indole substitutes at C3; Fischer indole, Skraup, Friedländer, Bischler–Napieralski and Pictet–Spengler build the fused rings.
  • Glucose: 16 open-chain aldohexose stereoisomers; α +112°, β +18.7°, equilibrium +52.7° (≈ 64% β); glucose, mannose and fructose give one osazone; sucrose is non-reducing, maltose and lactose reducing.
  • pI averages the pKa values around the zwitterion (Gly 5.97, Asp 2.77, Lys 9.74); Boc/Fmoc, DCC and Merrifield resin; Edman sequencing, trypsin after Lys/Arg; A–T two and G–C three H-bonds; terpenes are C₅ multiples, carotenoids C₄₀.

Practice questions (21)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. What is the Hantzsch–Widman name of the saturated five-membered ring containing one oxygen atom (tetrahydrofuran)?

    1. Oxolane
    2. Oxirane
    3. Oxetane
    4. Oxane
    Show answer

    Answer: A — Oxolane

    "Oxa" plus the saturated five-membered stem "-olane" gives oxolane. Oxirane is the three-membered epoxide ring, oxetane the four-membered one, and oxane the six-membered tetrahydropyran.
  2. Isoquinoline is best described as

    1. benzo[c]pyridine, with nitrogen at position 2
    2. benzo[b]pyridine, with nitrogen at position 1
    3. benzo[b]pyrrole
    4. 1,3-diazanaphthalene
    Show answer

    Answer: A — benzo[c]pyridine, with nitrogen at position 2

    In isoquinoline the benzene ring is fused to the pyridine c-bond, so N sits at position 2, one atom away from the ring fusion. Quinoline (benzo[b]pyridine) has N at 1, next to the fusion; benzo[b]pyrrole is indole.
  3. Why does pyrrole undergo electrophilic substitution mainly at C2 rather than C3?

    1. The σ-complex from C2 attack is stabilised by three resonance forms, including one with the charge on nitrogen; C3 attack gives only two
    2. C2 is less hindered than C3
    3. The nitrogen lone pair is not part of the π system
    4. C3 carries a partial positive charge in pyrrole
    Show answer

    Answer: A — The σ-complex from C2 attack is stabilised by three resonance forms, including one with the charge on nitrogen; C3 attack gives only two

    Attack at C2 leaves an allylic cation over C3–C4–C5 plus an iminium form with N⁺, more delocalisation than the two forms from C3 attack. Steric effects are negligible in pyrrole, and the lone pair is part of the aromatic sextet — which is exactly why pyrrole is so reactive.
  4. Which statements about furan, pyrrole and thiophene are correct?

    1. Thiophene is the most aromatic of the three
    2. Furan can act as the diene in a Diels–Alder reaction
    3. All three can be made from 1,4-dicarbonyl compounds by the Paal–Knorr synthesis
    4. Pyrrole is more basic than pyridine
    Show answer

    Answer: A — Thiophene is the most aromatic of the three; B — Furan can act as the diene in a Diels–Alder reaction; C — All three can be made from 1,4-dicarbonyl compounds by the Paal–Knorr synthesis

    Sulfur holds its lone pair least tightly and gives the most aromatic ring; furan, the least aromatic, adds maleic anhydride as a diene; acid, amines or P₄S₁₀ close a 1,4-diketone to furan, pyrrole or thiophene. Pyrrole’s lone pair is in the sextet (pKaH ≈ −3.8), far less basic than pyridine (5.2).
  5. Pyridine is heated with sodium amide in an inert solvent (the Chichibabin reaction). The product is

    1. 2-aminopyridine
    2. 3-aminopyridine
    3. piperidine
    4. pyridine N-oxide
    Show answer

    Answer: A — 2-aminopyridine

    NH₂⁻ adds at C2, where the anionic σ-adduct can place the negative charge on nitrogen; loss of hydride (as H₂ with the NH protons) restores aromaticity. C3 attack gives no N-stabilised intermediate, which is why electrophiles, not nucleophiles, go to C3.
  6. Indole reacts with the Vilsmeier reagent (DMF/POCl₃). Formylation occurs at

    1. C3
    2. C2
    3. N1
    4. C5 of the benzene ring
    Show answer

    Answer: A — C3

    Attack at C3 gives a cation stabilised by the nitrogen lone pair (an iminium) while the benzene ring stays intact; attack at C2 would need to disrupt the benzene ring to put charge on nitrogen. So indole’s electrophilic chemistry is at C3, unlike pyrrole’s C2.
  7. Match each synthesis to the ring it builds. Which pairings are correct?

    1. Fischer synthesis — indole
    2. Skraup synthesis — quinoline
    3. Bischler–Napieralski reaction — dihydroisoquinoline
    4. Hantzsch synthesis — thiophene
    Show answer

    Answer: A — Fischer synthesis — indole; B — Skraup synthesis — quinoline; C — Bischler–Napieralski reaction — dihydroisoquinoline

    An arylhydrazone gives an indole via a [3,3] shift; aniline, glycerol and H₂SO₄ give quinoline; a β-arylethylamide and POCl₃ close a 3,4-dihydroisoquinoline. The Hantzsch syntheses make pyridines (via dihydropyridines) and pyrroles, not thiophenes.
  8. Why is pyridine much less reactive than benzene towards electrophilic substitution?

    1. The electronegative ring nitrogen withdraws π density and is protonated in acidic media
    2. Pyridine is not aromatic
    3. Its lone pair is part of the π sextet
    4. Pyridine has fewer π electrons than benzene
    Show answer

    Answer: A — The electronegative ring nitrogen withdraws π density and is protonated in acidic media

    Pyridine is aromatic with six π electrons like benzene, but the nitrogen makes the ring electron-poor, and under nitrating or sulfonating conditions it becomes the pyridinium ion, more deactivated still. Its lone pair lies in the ring plane, outside the sextet, which is why it is basic.
  9. How many stereoisomers are possible for an open-chain aldohexose, HOCH₂(CHOH)₄CHO?

    Numerical answer — type the value.

    Show answer

    Answer: 16

    C2, C3, C4 and C5 are stereocentres and no meso form is possible because the two ends differ (CHO and CH₂OH), so 2⁴ = 16: eight D-sugars (glucose, mannose, galactose …) and their eight L-enantiomers. Counting the cyclic anomers would add another centre.
  10. α-D-Glucopyranose has [α]_D = +112.2° and β-D-glucopyranose +18.7°. After mutarotation the equilibrium solution shows +52.7°. Assuming only these two forms are present, what percentage is the β-anomer, to the nearest whole number?

    Numerical answer — type the value.

    Show answer

    Answer: 64

    With x the fraction β: 112.2(1 − x) + 18.7x = 52.7, so x = (112.2 − 52.7)/(112.2 − 18.7) = 59.5/93.5 = 0.636, i.e. 64% β and 36% α. The β-anomer, with every substituent equatorial, is favoured. Answering 36 gives the α fraction.
  11. Which disaccharide is non-reducing?

    1. Sucrose
    2. Maltose
    3. Lactose
    4. Cellobiose
    Show answer

    Answer: A — Sucrose

    In sucrose the glucose C1 and the fructose C2 — both anomeric carbons — form the glycosidic bond, so neither ring can open to a free carbonyl. Maltose, lactose and cellobiose are (1→4)-linked and keep a free hemiacetal on the second residue, so they reduce Fehling’s solution and mutarotate.
  12. D-Glucose, D-mannose and D-fructose all give the same osazone with excess phenylhydrazine because

    1. they differ only at C1 and C2, which both become C=N–NHPh in the osazone
    2. they have the same specific rotation
    3. they are all ketoses
    4. phenylhydrazine epimerises every stereocentre
    Show answer

    Answer: A — they differ only at C1 and C2, which both become C=N–NHPh in the osazone

    Osazone formation converts both C1 and C2 into phenylhydrazones, erasing the stereocentre at C2 and the aldose/ketose difference; C3–C6 are identical in all three. Glucose and mannose are aldoses (C2 epimers) and fructose a ketose, with very different rotations.
  13. Glycine has pKa₁ = 2.34 and pKa₂ = 9.60. What is its isoelectric point, to two decimal places?

    Numerical answer — type the value.

    Show answer

    Answer: 5.97

    For a neutral amino acid the zwitterion lies between the two ionisations, so pI = (2.34 + 9.60)/2 = 5.97. At lower pH glycine is mostly cationic and moves to the cathode in electrophoresis; at higher pH it is anionic.
  14. Aspartic acid has pKa values 1.88 (α-COOH), 3.65 (side-chain COOH) and 9.60 (α-NH₃⁺). What is its isoelectric point, to two decimal places?

    Numerical answer — type the value.

    Show answer

    Answer: 2.77

    The neutral species (net charge 0) exists between the first and second ionisations, both carboxyl groups: pI = (1.88 + 3.65)/2 = 2.77. Averaging all three values (5.04) or the first and last (5.74) ignores which two pKa values bracket the zwitterion.
  15. Lysine has pKa values 2.18 (α-COOH), 8.95 (α-NH₃⁺) and 10.53 (side-chain NH₃⁺). What is its isoelectric point, to two decimal places?

    Numerical answer — type the value.

    Show answer

    Answer: 9.74

    At low pH lysine carries +2; losing the carboxyl proton gives +1, losing the α-NH₃⁺ proton gives the neutral form, and losing the side-chain proton gives −1. The neutral form lies between 8.95 and 10.53: pI = (8.95 + 10.53)/2 = 9.74. Using 2.18 and 8.95 would treat lysine like a neutral amino acid.
  16. A peptide Ala–Lys–Gly–Arg–Phe–Leu is digested with trypsin. How many peptide fragments are formed?

    Numerical answer — type the value.

    Show answer

    Answer: 3

    Trypsin cleaves on the C-terminal side of Lys and Arg: Ala–Lys | Gly–Arg | Phe–Leu, three fragments. Chymotrypsin would instead cut after Phe, giving Ala–Lys–Gly–Arg–Phe and Leu.
  17. In the Edman degradation, phenyl isothiocyanate is used to

    1. remove and identify the N-terminal residue while leaving the rest of the peptide intact for the next cycle
    2. cleave the peptide after every arginine
    3. identify the C-terminal residue
    4. activate the carboxyl group for coupling
    Show answer

    Answer: A — remove and identify the N-terminal residue while leaving the rest of the peptide intact for the next cycle

    PhNCS adds to the free N-terminal amine; mild acid then cyclises only that residue off as a thiazolinone, which rearranges to the phenylthiohydantoin identified by chromatography. Repeating the cycle reads the sequence from the N-end. Arginine cleavage is trypsin’s job and C-terminal analysis uses carboxypeptidase.
  18. A 10-base-pair segment of double-stranded DNA contains 4 G·C pairs and 6 A·T pairs. How many interstrand hydrogen bonds hold it together?

    Numerical answer — type the value.

    Show answer

    Answer: 24

    G·C pairs have three hydrogen bonds and A·T pairs two: 4 × 3 + 6 × 2 = 12 + 12 = 24. That is why GC-rich DNA melts at a higher temperature; assuming two per pair gives 20.
  19. β-Carotene is a C₄₀ tetraterpene. How many isoprene (C₅) units is it built from?

    Numerical answer — type the value.

    Show answer

    Answer: 8

    40/5 = 8 isoprene units — two C₂₀ halves joined tail-to-tail at the centre, as squalene (C₃₀, six units) is. Its eleven conjugated double bonds make it absorb near 450 nm and appear orange.
  20. Which statement about steroids is correct?

    1. They share a tetracyclic cyclopenta[a]phenanthrene nucleus of 17 carbons
    2. They are C₄₀ tetraterpenes
    3. They are basic nitrogen compounds
    4. They are made by the Merrifield method in cells
    Show answer

    Answer: A — They share a tetracyclic cyclopenta[a]phenanthrene nucleus of 17 carbons

    Rings A, B and C (six-membered) and D (five-membered) make the steroid nucleus; cholesterol, biosynthesised from the triterpene squalene via lanosterol, is the parent. Tetraterpenes are carotenoids and basic nitrogen compounds are alkaloids.
  21. Why can alkaloids usually be separated from neutral plant constituents by extraction with dilute aqueous acid?

    1. Their basic nitrogen is protonated to water-soluble ammonium salts, which base then converts back to the free bases
    2. They are all carboxylic acids
    3. They are insoluble in organic solvents
    4. They hydrolyse in acid to sugars
    Show answer

    Answer: A — Their basic nitrogen is protonated to water-soluble ammonium salts, which base then converts back to the free bases

    Alkaloids are defined by a basic nitrogen, usually in a heterocycle; acid converts them into salts that move into the aqueous layer, leaving fats and terpenes behind, and basifying releases the free alkaloid for re-extraction. Acidity is the property of carboxylic acids, not alkaloids.