Basicity of Amines, Zwitterions in Amino Acids, Polyamides/Polyesters & Cisplatin
Organic nitrogen compounds and condensation polymers bridge synthetic organic chemistry and biochemistry. Amines act as Brønsted-Lowry bases because the lone pair of electrons on the nitrogen atom can accept a proton (H⁺) to form a dative covalent bond, with base strength governed by how strongly attached alkyl or aryl groups push or withdraw electron density at nitrogen. Amino acids are amphoteric molecules containing both a basic amine group (-NH₂) and an acidic carboxylic acid group (-COOH), existing as high-melting ionic zwitterions at their isoelectric point and changing charge at low or high pH. Similarly, dicarboxylic acids (or diacyl chlorides) condense with diols or diamines to form biodegradable polyesters (Terylene) and polyamides (Nylon-6,6, Kevlar, and polypeptides). Examiners regularly test amine base-strength comparisons, amino acid structures across pH, polymer repeat units vs hydrolysis products, and the mechanism of anticancer drug cisplatin on DNA.
- Basicity of Amines: Phenylamine < Ammonia < Primary Alkylamines
- Amines act as Brønsted-Lowry bases (proton acceptors) because the lone pair of electrons on the nitrogen atom forms a dative covalent bond with an H⁺ ion: RNH₂ + H⁺ → RNH₃⁺. Base strength depends on the availability (electron density) of the nitrogen lone pair: 1. Primary aliphatic amines (e.g. ethylamine, CH₃CH₂NH₂) are the strongest bases because the alkyl group has a positive inductive effect (electron-pushing), increasing electron density on the N lone pair so it attracts H⁺ more strongly than in NH₃. 2. Ammonia (NH₃) is intermediate (no inductive push or delocalisation). 3. Aromatic amines (e.g. phenylamine, C₆H₅NH₂) are the weakest bases because the lone pair of electrons on the nitrogen p-orbital overlaps and delocalises into the π-electron ring of benzene, making the lone pair much less available to accept an H⁺.
- Preparation of Aliphatic Amines (Nitrile Reduction vs Halogenoalkane)
- Method 1 (Halogenoalkane + excess ethanolic NH₃): gives an impure mixture of primary, secondary, and tertiary amines plus quaternary ammonium salts due to further nucleophilic substitution by the primary amine product. Method 2 (Reduction of a Nitrile: Pure Single Product): React halogenoalkane with KCN in aqueous ethanol to form a nitrile (R-C≡N), then reduce the nitrile to a pure primary amine (R-CH₂NH₂) using either H₂ gas with a Nickel catalyst at high temperature/pressure (catalytic hydrogenation: RCN + 2H₂ → RCH₂NH₂) OR LiAlH₄ in dry ether followed by dilute acid.
- Amino Acids, Zwitterions & Structures at Low/Isoelectric/High pH
- 2-aminocarboxylic acids (H₂N-CHR-COOH) have both a basic -NH₂ group and an acidic -COOH group (all except glycine, where R = H, have a chiral carbon centre and are optically active). 1. At the isoelectric point (neutral pH ~6): internal proton transfer forms a dipolar ion called a zwitterion, ⁺H₃N-CHR-COO⁻, which has no overall net charge. Because zwitterions form a giant ionic lattice held by strong electrostatic attractions between ⁺NH₃ and COO⁻ groups, solid amino acids are crystalline solids with high melting points (~200-300°C) that dissolve in water. 2. At strongly acidic pH (e.g. pH 1, excess H⁺): the -COO⁻ group is protonated to form the cationic species ⁺H₃N-CHR-COOH. 3. At strongly alkaline pH (e.g. pH 12, excess OH⁻): the -NH₃⁺ group is deprotonated to form the anionic species H₂N-CHR-COO⁻.
Curly-Arrow Organic Mechanisms (Nucleophilic Substitution & Addition)
3 Golden Rules for Curly Arrows:
1. Start at electrons: Every curly arrow MUST start at a lone pair of electrons or the middle of a covalent bond (never start an arrow on a δ+ or positive charge).
2. Show dipoles: Always label δ+ and δ− on polar bonds like C–Br or H–Br.
3. Balance charges: If a neutral molecule loses a leaving ion (:Br⁻), the intermediate carbocation must carry a formal + charge.
Key Concepts & Exam Definitions (6)
- Basicity of Amines: Phenylamine < Ammonia < Primary Alkylamines
- Amines act as Brønsted-Lowry bases (proton acceptors) because the lone pair of electrons on the nitrogen atom forms a dative covalent bond with an H⁺ ion: RNH₂ + H⁺ → RNH₃⁺. Base strength depends on the availability (electron density) of the nitrogen lone pair: 1. Primary aliphatic amines (e.g. ethylamine, CH₃CH₂NH₂) are the strongest bases because the alkyl group has a positive inductive effect (electron-pushing), increasing electron density on the N lone pair so it attracts H⁺ more strongly than in NH₃. 2. Ammonia (NH₃) is intermediate (no inductive push or delocalisation). 3. Aromatic amines (e.g. phenylamine, C₆H₅NH₂) are the weakest bases because the lone pair of electrons on the nitrogen p-orbital overlaps and delocalises into the π-electron ring of benzene, making the lone pair much less available to accept an H⁺.
- Preparation of Aliphatic Amines (Nitrile Reduction vs Halogenoalkane)
- Method 1 (Halogenoalkane + excess ethanolic NH₃): gives an impure mixture of primary, secondary, and tertiary amines plus quaternary ammonium salts due to further nucleophilic substitution by the primary amine product. Method 2 (Reduction of a Nitrile: Pure Single Product): React halogenoalkane with KCN in aqueous ethanol to form a nitrile (R-C≡N), then reduce the nitrile to a pure primary amine (R-CH₂NH₂) using either H₂ gas with a Nickel catalyst at high temperature/pressure (catalytic hydrogenation: RCN + 2H₂ → RCH₂NH₂) OR LiAlH₄ in dry ether followed by dilute acid.
- Amino Acids, Zwitterions & Structures at Low/Isoelectric/High pH
- 2-aminocarboxylic acids (H₂N-CHR-COOH) have both a basic -NH₂ group and an acidic -COOH group (all except glycine, where R = H, have a chiral carbon centre and are optically active). 1. At the isoelectric point (neutral pH ~6): internal proton transfer forms a dipolar ion called a zwitterion, ⁺H₃N-CHR-COO⁻, which has no overall net charge. Because zwitterions form a giant ionic lattice held by strong electrostatic attractions between ⁺NH₃ and COO⁻ groups, solid amino acids are crystalline solids with high melting points (~200-300°C) that dissolve in water. 2. At strongly acidic pH (e.g. pH 1, excess H⁺): the -COO⁻ group is protonated to form the cationic species ⁺H₃N-CHR-COOH. 3. At strongly alkaline pH (e.g. pH 12, excess OH⁻): the -NH₃⁺ group is deprotonated to form the anionic species H₂N-CHR-COO⁻.
- Condensation Polymers: Polyesters (Terylene) & Polyamides (Nylon-6,6, Kevlar)
- Unlike addition polymers (made from one alkene monomer with a C=C bond and a non-reactive C-C backbone), condensation polymers form from monomers with two functional groups on each end, eliminating a small molecule (H₂O or HCl) per linkage: 1. Terylene (PET polyester): benzene-1,4-dicarboxylic acid + ethane-1,2-diol → ester linkage (-COO-). 2. Nylon-6,6 (polyamide): hexanedioic acid (or hexanedioyl dichloride) + hexane-1,6-diamine → amide linkage (-CONH-). 3. Kevlar (polyamide): benzene-1,4-dicarboxylic acid (or diacyl chloride) + benzene-1,4-diamine → rigid flat planar chains that align parallel with strong intermolecular hydrogen bonds between C=O and N-H groups.
- Biodegradability & Acid vs Alkaline Hydrolysis of Condensation Polymers
- Addition polymers (poly(alkene)s) have saturated, non-polar C-C single-bond backbones that cannot be hydrolysed and are non-biodegradable (disposed of by landfill, incineration for energy, or feedstock recycling/cracking). Polyesters and polyamides contain polar C^δ+=O^δ- bonds in their ester (-COO-) or amide (-CONH-) linkages, which are susceptible to nucleophilic attack by H₂O/OH⁻ and are biodegradable: 1. Acid hydrolysis (hot aqueous HCl) of a polyamide gives the dicarboxylic acid (HOOC-R-COOH) + the diprotonated diammonium salt (⁺H₃N-R'-NH₃⁺). 2. Alkaline hydrolysis (hot aqueous NaOH) of a polyamide gives the sodium dicarboxylate salt (⁻OOC-R-COO⁻ 2Na⁺) + the free diamine (H₂N-R'-NH₂).
- DNA Base Pairing & Anticancer Mechanism of Cisplatin
- In DNA, a nucleotide forms from deoxyribose, phosphate, and a base via condensation reactions; complementary strands are held by hydrogen bonds between A=T (2 H-bonds) and G≡C (3 H-bonds). Cisplatin is the square planar complex cis-[Pt(NH₃)₂Cl₂] (90° Cl-Pt-Cl angle): inside rapidly dividing cancer cells, water replaces the chloride ligands, and the Nitrogen atom (with a lone pair) on two adjacent Guanine bases on a DNA strand undergoes ligand substitution with the two Cl⁻ ligands on Pt²⁺, forming coordinate bonds to the platinum ion. This kinks/distorts the DNA double helix so it cannot unwind and replicate, preventing cell division (mitosis); given in small doses because it also damages healthy fast-dividing cells (hair follicles, bone marrow, gut lining).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Arrange phenylamine, ammonia, and ethylamine in order of increasing base strength, and explain your answer.
Why does preparing propylamine by reducing propanenitrile (CH₃CH₂CN) with H₂/Ni give a much higher yield than reacting 1-bromopropane with ammonia?
Draw/write the structural formula of the amino acid alanine (H₂NCH(CH₃)COOH) at pH 7 (isoelectric point), at pH 1, and at pH 13.
Why is glycine (2-aminoethanoic acid) the only proteinogenic amino acid that is not optically active?
Why are polyesters (like Terylene) and polyamides (like Nylon-6,6) biodegradable, whereas poly(propene) is non-biodegradable?
Explain how cisplatin prevents cancer cells from dividing, and why transplatin is therapeutically inactive.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Compare and explain the base strengths of phenylamine (C₆H₅NH₂), ammonia (NH₃), and propylamine (CH₃CH₂CH₂NH₂). Write an equation to show propylamine acting as a Brønsted-Lowry base with water. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Equation for propylamine with water: CH₃CH₂CH₂NH₂ + H₂O ⇌ CH₃CH₂CH₂NH₃⁺ + OH⁻ [1 mark]. 2. Base strength depends on how readily the lone pair of electrons on the nitrogen atom can accept a proton (H⁺) to form a dative covalent bond [1 mark]. 3. Order of base strength: Phenylamine (weakest) < Ammonia < Propylamine (strongest) [1 mark]. 4. Propylamine is a stronger base than ammonia because the propyl alkyl group exerts a positive inductive (electron-releasing) effect, pushing electron density onto the nitrogen atom and making the lone pair more available to attract H⁺ [1 mark]. 5. Phenylamine is the weakest base because the p-orbital containing the lone pair on the nitrogen atom overlaps with the delocalised π-electron system of the benzene ring, delocalising the lone pair into the ring and making it much less available to accept H⁺ [1 mark].
- Q2
The amino acid serine has the structural formula H₂NCH(CH₂OH)COOH. Explain why solid serine has a melting point of 228°C whereas a carboxylic acid of similar relative molecular mass is a liquid or low-melting solid, and write the structures of the organic species formed when serine reacts with: (a) excess HCl(aq), (b) excess NaOH(aq), and (c) methanol in the presence of concentrated H₂SO₄. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. In the solid state, serine exists as a dipolar zwitterion, ⁺H₃NCH(CH₂OH)COO⁻, due to internal proton transfer from the -COOH group to the -NH₂ group [1 mark]. 2. Strong electrostatic (ionic) attractions exist between the oppositely charged -NH₃⁺ and -COO⁻ groups of neighbouring zwitterions in a giant ionic lattice, requiring a large amount of thermal energy to break (compared to weaker intermolecular hydrogen bonds in a simple molecular carboxylic acid) [1 mark]. 3. (a) With excess HCl(aq) (low pH): ⁺H₃NCH(CH₂OH)COOH Cl⁻ [1 mark]. 4. (b) With excess NaOH(aq) (high pH): H₂NCH(CH₂OH)COO⁻ Na⁺ [1 mark]. 5. (c) With methanol and conc. H₂SO₄ (esterification of -COOH and protonation of basic -NH₂ by the acid catalyst): ⁺H₃NCH(CH₂OH)COOCH₃ [1 mark].
- Q3
Kevlar is a condensation polymer used in bulletproof vests. Name the two monomers used to manufacture Kevlar, draw the structure of one repeat unit of Kevlar, explain why Kevlar fibres have exceptionally high tensile strength, and draw the two organic products formed when Kevlar is hydrolysed by heating under reflux with excess aqueous sodium hydroxide (NaOH). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Monomers of Kevlar: Benzene-1,4-dicarboxylic acid (or benzene-1,4-dioyl dichloride) and benzene-1,4-diamine [1 mark]. 2. Repeat unit of Kevlar: - OC - C₆H₄ - CO - NH - C₆H₄ - NH - (showing the trailing bonds at both ends and the central -CONH- amide linkage) [1 mark]. 3. High tensile strength explanation: The rigid, flat planar benzene rings and trans-amide linkages allow the polymer chains to pack closely together in parallel alignment, forming a dense network of strong intermolecular hydrogen bonds between the C=O (δ⁻) oxygen of one chain and the N-H (δ⁺) hydrogen of an adjacent chain [1 mark]. 4. Alkaline hydrolysis product 1 (from the dicarboxylic acid): Sodium benzene-1,4-dicarboxylate, ⁻OOC - C₆H₄ - COO⁻ 2Na⁺ (must be the carboxylate salt, not the free acid) [1 mark]. 5. Alkaline hydrolysis product 2 (from the diamine): Benzene-1,4-diamine, H₂N - C₆H₄ - NH₂ (un-protonated free amine in alkali) [1 mark].
- Q4
Describe how thin-layer chromatography (TLC) and ninhydrin can be used to identify the amino acids present in a hydrolysed tripeptide, and explain how two amino acids with identical Rf values in one solvent can be separated using two-dimensional (2D) TLC. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Spot the hydrolysed tripeptide mixture alongside known pure reference amino acids onto a pencil baseline near the bottom of a silica-coated TLC plate, place in a beaker with solvent below the baseline, cover with a lid, and allow the solvent front to rise [1 mark]. 2. Because amino acids are colourless, dry the plate and spray it with ninhydrin solution (and warm) or view under UV light to make the amino acid spots visible as purple spots [1 mark]. 3. Calculate the Rf value for each spot using Rf = (distance moved by amino acid spot from baseline) ÷ (distance moved by solvent front from baseline) and match against the reference amino acids [1 mark]. 4. In 2D TLC, a square TLC plate is run in Solvent 1, dried, rotated by 90°, and run a second time in a different Solvent 2 (with different polarity), separating the two amino acids because they have different relative solubilities in the second mobile phase [1 mark].
Common Exam Questions (4)
Why do secondary aliphatic amines (like diethylamine, (CH₃CH₂)₂NH) have an even stronger base strength in the gas phase/solution than primary amines, whereas tertiary amines in water are slightly weaker than secondary amines?
Secondary amines have two electron-pushing alkyl groups attached to the nitrogen atom, giving an even stronger positive inductive effect and higher electron density on the N lone pair than a primary amine. In aqueous solution, tertiary amines (R₃N) are slightly less basic than secondary amines because three bulky alkyl groups hinder hydration (hydrogen bonding with water molecules) of the R₃NH⁺ cation.
When drawing the products of acid or alkaline hydrolysis of an ester or amide in an exam, what is the #1 trap to watch out for?
Always check the pH of the hydrolysis reagent. In acid hydrolysis (HCl/H₂SO₄), any amine product is basic and gets protonated to -NH₃⁺, while carboxylic acids stay as -COOH. In alkaline hydrolysis (NaOH), any carboxylic acid product is acidic and gets deprotonated to the carboxylate salt -COO⁻Na⁺, while amines stay as neutral -NH₂!
How can a single monomer form a condensation polymer without needing two different monomers?
If a single monomer molecule contains both functional groups, such as a hydroxycarboxylic acid (e.g. lactic acid, HO-CH(CH₃)-COOH, which polymerises to biodegradable polylactic acid, PLA) or an aminocarboxylic acid (e.g. 6-aminohexanoic acid, H₂N-(CH₂)₅-COOH, which polymerises to Nylon-6) - it can condense with itself.
When drawing two complementary DNA nucleotides hydrogen-bonded together, what three rules do examiners check?
1. The two sugar-phosphate strands must be drawn ANTIPARALLEL (one deoxyribose ring pointing 'up', the opposite strand's deoxyribose ring inverted 'upside down'). 2. The exact correct number of hydrogen bonds must be dashed between the specified atoms shown in the Data Booklet (2 H-bonds for A=T; 3 H-bonds for G≡C). 3. Each H-bond must link a lone pair on N or O to a δ⁺ H attached to N.
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