Chemistry 9701/43 — October/November 2016

Background Concept

When a complex ion is placed in a solution containing a different ligand (or a high concentration of a ligand that can replace the existing ones), the new ligands can replace the original ligands around the central metal ion. This is called ligand exchange, ligand substitution, or ligand displacement.

For example, adding concentrated hydrochloric acid to a blue aqueous copper(II) solution replaces water ligands with chloride ligands, forming a yellow-green tetrachlorocuprate(II) complex. The reaction is an equilibrium:
$[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}$

Understanding the Question

The question provides an equilibrium between the hexaaquacopper(II) ion and the tetrachlorocuprate(II) ion, and asks for the name of the type of reaction occurring.

Approach

Recognise that water ligands are being replaced by chloride ligands. The standard IUPAC and CIE term for this process is 'ligand exchange' or 'ligand substitution'.

Step-by-Step Reasoning

• The reactants contain $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ and $\text{Cl}^-$.
• The products contain $[\text{CuCl}_4]^{2-}$ and $\text{H}_2\text{O}$.
• Six water molecules are replaced by four chloride ions around the copper centre.
• This is a classic ligand exchange (or substitution) reaction.

Key Takeaways

• Replacing ligands in a complex ion is called ligand exchange or ligand substitution.
• These reactions are often reversible and can be driven to one side or the other by changing the concentration of the ligands (Le Chatelier's principle).

Common Mistakes

• Calling it a 'precipitation' or 'redox' reaction. No precipitate forms here (both complexes are soluble), and the oxidation state of copper remains +2.
• Using vague terms like 'mixing' or 'reacting'. Specific chemical terminology is required.

Things to Be Careful About

• 'Ligand exchange' and 'ligand substitution' are both fully acceptable. 'Displacement' is also accepted in some mark schemes, though 'substitution' is more precise for coordination chemistry.

Background Concept

The colour of transition metal complexes arises from d-d electronic transitions. When ligands approach a metal ion, the five degenerate d orbitals split into two energy levels (e.g., $t_{2g}$ and $e_g$ in an octahedral field). Electrons can be excited from the lower energy d orbitals to the higher energy ones by absorbing photons of visible light. The colour observed is the complementary colour to the light absorbed.

Different ligands cause different amounts of d-orbital splitting ($\Delta E$). Water is a weaker-field ligand than chloride in this context (though chloride is actually weak-field, the geometry change from octahedral to tetrahedral also drastically changes the splitting pattern and magnitude). The change in ligand and geometry alters the energy gap, so a different wavelength of light is absorbed, resulting in a different observed colour.

Understanding the Question

The question asks for the colours of the hexaaquacopper(II) ion and the tetrachlorocuprate(II) ion involved in the equilibrium.

Approach

Recall the standard colours taught for these common copper(II) complexes.

Step-by-Step Reasoning

• $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ is the typical aqueous copper(II) ion, which is blue.
• $[\text{CuCl}_4]^{2-}$ is formed in concentrated chloride solutions (e.g., concentrated HCl). It is yellow (often described as yellow-green or green-yellow in practice, but 'yellow' is the standard mark-scheme answer).
• When mixed, the solution can appear green (a mixture of blue and yellow).

Key Takeaways

• Aqueous $\text{Cu}^{2+}$ ($[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$) is blue.
• $[\text{CuCl}_4]^{2-}$ is yellow.
• Colour changes in ligand exchange reactions are a direct consequence of changes in d-orbital splitting energy.

Common Mistakes

• Saying the solution is 'green' for both. The individual ions have distinct colours; green is only seen when both are present in comparable amounts.
• Confusing the colours with other transition metal complexes (e.g., $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$ is pale green, $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ is yellow/brown).

Things to Be Careful About

• Mark schemes often accept 'yellow/green' or 'green/yellow' for $[\text{CuCl}_4]^{2-}$ because the actual observed colour in many lab conditions is a yellow-green. However, 'yellow' is the primary expected answer.

Background Concept

The shape of a complex ion is determined by its coordination number (the number of coordinate bonds to the central metal ion) and the nature of the ligands.

• Coordination number 6 typically gives an octahedral shape (e.g., $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$, $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$).
• Coordination number 4 can give either a tetrahedral or a square planar shape.
  • Square planar is typical for $\text{d}^8$ metal ions like $\text{Pt}^{2+}$, $\text{Pd}^{2+}$, and sometimes $\text{Ni}^{2+}$ with strong-field ligands (e.g., $[\text{Cu}(\text{NH}_3)_4]^{2+}$ is often described as square planar or distorted octahedral depending on context, but $[\text{Pt}(\text{NH}_3)_4]^{2+}$ is strictly square planar).
  • Tetrahedral is typical for $\text{d}^9$ ions like $\text{Cu}^{2+}$ when bulky or weak-field ligands are present, such as in $[\text{CuCl}_4]^{2-}$.

Understanding the Question

The question asks for the shape of the $[\text{CuCl}_4]^{2-}$ ion. The formula shows four chloride ligands bonded to copper, so the coordination number is 4.

Approach

Identify the coordination number (4) and recall that $[\text{CuCl}_4]^{2-}$ adopts a tetrahedral geometry.

Step-by-Step Reasoning

• The central ion is $\text{Cu}^{2+}$.
• There are 4 $\text{Cl}^-$ ligands, each forming one coordinate bond.
• Coordination number = 4.
• For $\text{Cu}^{2+}$ ($\text{d}^9$) with chloride ligands, the geometry is tetrahedral.

Key Takeaways

• Coordination number 4 complexes can be tetrahedral or square planar.
• $[\text{CuCl}_4]^{2-}$ is tetrahedral.
• $[\text{Cu}(\text{NH}_3)_4]^{2+}$ is often taught as square planar (or flattened octahedral with 2 weak axial water molecules), highlighting that different ligands can lead to different shapes for the same metal ion.

Common Mistakes

• Assuming all 4-coordinate copper complexes are square planar. While $[\text{Cu}(\text{NH}_3)_4]^{2+}$ is square planar, $[\text{CuCl}_4]^{2-}$ is tetrahedral.
• Confusing coordination number with the number of atoms in the ligand (chloride is monodentate, so 4 chlorides = coordination number 4).

Things to Be Careful About

• Simply writing '4-coordinate' is not enough; the question asks for the 'shape', which is 'tetrahedral'.

Background Concept

The stability constant, $K_{\text{stab}}$, is the equilibrium constant for the formation of a complex ion from the central metal ion and its ligands. It measures the stability of the complex in solution; a larger $K_{\text{stab}}$ indicates a more stable complex.

For a general reaction:
$\text{M} + n\text{L} \rightleftharpoons [\text{ML}_n]$
$K_{\text{stab}} = \frac{[[\text{ML}_n]]}{[\text{M}][\text{L}]^n}$

Crucial rule for aqueous equilibria: Water ($\text{H}_2\text{O}$) is the solvent and is in large excess. Its concentration is effectively constant and is omitted from the equilibrium expression. This applies whether water is a ligand being replaced or a product of the reaction.

Understanding the Question

Write the $K_{\text{stab}}$ expression for:
$[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}$

Approach

1. Identify the products and reactants.
2. Place product concentrations in the numerator and reactant concentrations in the denominator.
3. Raise each concentration to the power of its stoichiometric coefficient.
4. Omit $\text{H}_2\text{O}$ because it is the solvent.

Step-by-Step Reasoning

• Numerator: $[\text{CuCl}_4]^{2-}$ (coefficient is 1).
• Denominator: $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ (coefficient is 1) multiplied by $[\text{Cl}^-]^4$ (coefficient is 4).
• Omit: $[\text{H}_2\text{O}]^6$ because water is the solvent.
• Result: $K_{\text{stab}} = \frac{[\text{CuCl}_4^{2-}]}{[\text{Cu}(\text{H}_2\text{O})_6^{2+}][\text{Cl}^-]^4}$

Key Takeaways

• Always exclude water from aqueous equilibrium expressions.
• The power of the ligand concentration in the denominator must match its stoichiometric coefficient in the balanced equation.

Common Mistakes

• Including $[\text{H}_2\text{O}]^6$ in the denominator or numerator. This is the most common error.
• Forgetting to raise $[\text{Cl}^-]$ to the power of 4.
• Writing the expression for the reverse reaction (inversing the fraction).

Things to Be Careful About

• Charges on the complex ions do not appear in the $K_{\text{stab}}$ expression, only the concentrations.
• Ensure the brackets denote concentration in mol dm$^{-3}$.

Background Concept

Ligands are ions or molecules that donate a pair of electrons to a central metal ion to form a coordinate (dative covalent) bond.

• Monodentate ligand: Donates one lone pair (e.g., $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$). 'Dentate' comes from 'tooth'; one tooth.
• Bidentate ligand: Donates two lone pairs, forming two coordinate bonds. 'Bi' = two. Example: ethane-1,2-diamine (en), ethanedioate (oxalate).
• Multidentate (polydentate) ligands: Donate three or more pairs (e.g., EDTA$^{4-}$ is hexadentate).

Understanding the Question

Define the term 'bidentate ligand'. The question is worth 2 marks, indicating two distinct points are required.

Approach

State the two key features: (1) the presence of two lone pairs of electrons, and (2) the ability of each to form a coordinate bond with the metal centre.

Step-by-Step Reasoning

• Point 1: The ligand must have two lone pairs of electrons available for donation.
• Point 2: Each of these lone pairs is donated to the metal ion to form a coordinate (dative) bond.
• Together, this gives a coordination number contribution of 2 from a single ligand molecule.

Key Takeaways

• 'Bidentate' literally means 'two-toothed'.
• The definition must mention both the lone pairs and the coordinate bonds. Just saying 'binds twice' is not precise enough for full marks.

Common Mistakes

• Saying 'has two bonds'. This is circular and doesn't explain the electronic basis (lone pairs).
• Forgetting to specify 'coordinate' or 'dative' bond. Regular covalent bonds are not sufficient for the definition in this context.
• Confusing bidentate with having two atoms (e.g., 'has two nitrogen atoms'). While en has two N atoms, the definition is about lone pairs and bonds.

Things to Be Careful About

• Use the exact terminology: 'lone pair', 'coordinate bond' or 'dative bond'.

Background Concept

The stability constant $K_{\text{stab}}$ is an equilibrium constant. A larger $K_{\text{stab}}$ value means the numerator (products) is larger relative to the denominator (reactants) at equilibrium. This indicates that the equilibrium lies further to the right (favouring the complex formation) and the complex is more stable.

This question also introduces the chelate effect: bidentate ligands like 'en' form more stable complexes than monodentate ligands like $\text{NH}_3$ with the same number of donor atoms, due to an increase in entropy ($\Delta S > 0$) when multiple monodentate ligands are replaced by fewer multidentate ligands.

Understanding the Question

Given $K_{\text{stab}}$ values:

• $[\text{Cu}(\text{NH}_3)_2(\text{H}_2\text{O})_4]^{2+}$: $7.94 \times 10^7$
• $[\text{Cu}(\text{en})(\text{H}_2\text{O})_4]^{2+}$: $3.98 \times 10^{10}$

What do these tell us about the relative positions of equilibria 1 and 2?

Approach

Compare the magnitudes of the two $K_{\text{stab}}$ values. The larger value corresponds to the equilibrium that lies further to the right.

Step-by-Step Reasoning

• $K_{\text{stab}}$ for equilibrium 2 ($3.98 \times 10^{10}$) is much larger than for equilibrium 1 ($7.94 \times 10^7$).
• A larger $K$ means the forward reaction is more favoured.
• Therefore, equilibrium 2 lies more to the right (RHS) than equilibrium 1.

Key Takeaways

• Larger $K_{\text{stab}}$ = more stable complex = equilibrium lies further to the right.
• The chelate effect explains why the bidentate 'en' complex is so much more stable than the monodentate $\text{NH}_3$ complex.

Common Mistakes

• Saying 'equilibrium 1 is more stable'. This is the opposite of what the numbers show.
• Not mentioning 'lies to the right' or 'favouring the forward reaction'. Just saying 'equilibrium 2 is bigger' is vague.

Things to Be Careful About

• Ensure you link the $K_{\text{stab}}$ value directly to the position of the equilibrium. $K > 1$ means products are favoured; a larger $K$ means more products are favoured.

Background Concept

Stereoisomerism in coordination complexes includes geometric (cis-trans) and optical isomerism.

• Geometric isomerism: Occurs when ligands can be arranged differently around the metal (e.g., cis/trans in square planar or octahedral complexes with two different types of ligand).
• Optical isomerism: Occurs when a complex is non-superimposable on its mirror image. This requires the complex to lack a plane of symmetry.

Complexes of the type $[\text{M}(\text{bidentate})_3]$ (like $[\text{Ni}(\text{en})_3]^{2+}$, $[\text{Co}(\text{en})_3]^{3+}$) are classic examples of optical isomerism. The three bidentate ligands wrap around the octahedral metal centre in a propeller-like fashion. There are two non-superimposable mirror images, designated as $\Delta$ (delta, right-handed) and $\Lambda$ (lambda, left-handed).

Understanding the Question

Nickel forms $[\text{Ni}(\text{en})_3]^{2+}$, an octahedral complex with three bidentate ethane-1,2-diamine ligands. Name the type of stereoisomerism.

Approach

Recognise the $[\text{M}(\text{en})_3]^{2+}$ pattern and recall that it exhibits optical isomerism.

Step-by-Step Reasoning

• The complex is octahedral with three identical bidentate ligands.
• It has no plane of symmetry.
• It exists as two non-superimposable mirror images (enantiomers).
• This is optical isomerism.

Key Takeaways

• $[\text{M}(\text{en})_3]^{n+}$ complexes are optically active.
• They are used to demonstrate optical isomerism in inorganic chemistry, analogous to chiral carbon atoms in organic chemistry.

Common Mistakes

• Calling it 'geometric' or 'cis-trans'. There is no cis/trans distinction here because all three ligands are identical and bidentate.
• Forgetting to specify 'optical' and just writing 'stereoisomerism'. The question asks for the type.

Things to Be Careful About

• Ensure you write 'optical' and not 'optically active' (though the latter describes the isomers, 'optical isomerism' is the name of the phenomenon/type).

Background Concept

Drawing optical isomers of $[\text{M}(\text{en})_3]^{2+}$ requires showing the octahedral geometry and the three bidentate ligands wrapping around it.

• Use a central metal atom (Ni).
• Draw three ethane-1,2-diamine (en) ligands. Each 'en' ligand is $\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2$. The two nitrogen atoms bond to the metal.
• To show 3D, use solid wedges (coming out of the page) and dashed wedges (going into the page), or use a perspective octahedron with bold and hashed bonds.

The two isomers are:

1. $\Delta$ (delta) isomer: If you look down a $C_3$ axis, the 'chelate rings' (the loops formed by M-N-C-C-N) appear to twist clockwise.
2. $\Lambda$ (lambda) isomer: The chelate rings twist anticlockwise.

These are mirror images and cannot be superimposed.

Understanding the Question

Draw 3D diagrams of the two stereoisomers of $[\text{Ni}(\text{en})_3]^{2+}$. The question is worth 3 marks: 1 for correct octahedral 3D, 1 for first correct structure with 3D, 1 for second correct with 3D.

Approach

1. Draw an octahedral framework.
2. Attach three 'en' ligands, ensuring each forms a 5-membered chelate ring (Ni-N-C-C-N).
3. Draw one isomer with a clockwise twist ($\Delta$).
4. Draw the mirror image with an anticlockwise twist ($\Lambda$).
5. Ensure 3D representation (wedges/dashes) is clear.

Step-by-Step Reasoning

• Mark 1: Octahedral shape with 3D bonds (e.g., one bond forward, one back, or using a 3D perspective).
• Mark 2: First enantiomer. Show three 'en' ligands. Each is a loop: N-Ni-N-C-C. The loops must be arranged so they don't have a plane of symmetry. For $\Delta$, the top-right loop, bottom loop, and top-left loop twist clockwise.
• Mark 3: Second enantiomer. The mirror image. For $\Lambda$, the loops twist anticlockwise.

Note: In CIE mark schemes, simplified 3D drawings are accepted as long as the mirror image relationship and octahedral geometry are clear. Using bold/hashed bonds for the axial/equatorial positions helps.

Key Takeaways

• $[\text{M}(\text{en})_3]^{2+}$ has $\Delta$ and $\Lambda$ optical isomers.
• The key is showing the propeller-like twist of the chelate rings.
• 3D representation is mandatory to score the marks for stereoisomerism.

Common Mistakes

• Drawing flat 2D structures without 3D bonds. This fails to show the non-superimposable nature.
• Drawing geometric isomers instead. You cannot have cis/trans here; it must be optical.
• Forgetting that 'en' is bidentate and drawing monodentate ligands.
• Not making the two structures mirror images of each other.

Things to Be Careful About

• The 'en' ligand must be drawn with the correct connectivity: N-Ni-N-CH2-CH2. Do not draw the C-C bond breaking.
• Ensure the mirror plane is implied between the two drawings.

Background Concept

A Brønsted-Lowry base is defined as a proton ($\text{H}^+$) acceptor. For a molecule to accept a proton, it must have a lone pair of electrons to form a new coordinate bond with the $\text{H}^+$.

Amines (organic derivatives of ammonia, $\text{NH}_3$) have a nitrogen atom with a lone pair. This makes them basic. Aliphatic amines (like ethane-1,2-diamine) are generally stronger bases than ammonia because the alkyl groups are electron-donating, increasing the electron density on the nitrogen and making the lone pair more available for protonation.

Understanding the Question

Explain how the amino groups in ethane-1,2-diamine ($\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$) allow it to act as a Brønsted-Lowry base. Worth 2 marks.

Approach

1. Identify the feature on the amino group that allows proton acceptance (the lone pair on nitrogen).
2. State the action: accepting/receiving a proton ($\text{H}^+$).

Step-by-Step Reasoning

• The amino group ($-\text{NH}_2$) contains a nitrogen atom.
• Nitrogen has a lone pair of electrons.
• This lone pair can accept a proton ($\text{H}^+$) from an acid to form a coordinate bond.
• This satisfies the Brønsted-Lowry definition of a base.

Key Takeaways

• Brønsted-Lowry base = proton acceptor.
• Amines are basic because of the lone pair on nitrogen.

Common Mistakes

• Saying 'donates electrons to H+'. While technically true (coordinate bond formation), the Brønsted-Lowry definition specifically requires 'accepts a proton'. 'Donates electrons' is the Lewis definition.
• Forgetting to mention the lone pair. Just saying 'it accepts a proton' doesn't explain how.

Things to Be Careful About

• Use 'lone pair' and 'accept/receive a proton' or 'H+'.

Background Concept

Amines react with acids to form ammonium salts. Since ethane-1,2-diamine has two amino groups ($-\text{NH}_2$), and the question specifies an excess of hydrochloric acid, both amino groups will be protonated.

Each $-\text{NH}_2$ group accepts one $\text{H}^+$ to become $-\text{NH}_3^+$.

Understanding the Question

Write an equation for the reaction of ethane-1,2-diamine with excess hydrochloric acid.

Approach

1. Write the organic reactant: $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$.
2. Add 2 molecules of HCl (or 2 $\text{H}^+$ ions) because there are two basic sites and acid is in excess.
3. Show both nitrogens becoming protonated ($-\text{NH}_3^+$).
4. Balance the equation.

Step-by-Step Reasoning

• Reactant: $\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2$
• Reagent: $2\text{HCl}$ (excess)
• Product: Both nitrogens gain an H and a positive charge. The chlorides are counter-ions.
• Molecular equation: $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 + 2\text{HCl} \rightarrow \text{Cl}^-\text{H}_3\text{N}^+-\text{CH}_2-\text{CH}_2-\text{NH}_3^+\text{Cl}^-$ (often written as $\text{ClH}_3\text{NCH}_2\text{CH}_2\text{NH}_3\text{Cl}$ or $[\text{H}_3\text{NCH}_2\text{CH}_2\text{NH}_3]^{2+} 2\text{Cl}^-$).
• Ionic equation: $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 + 2\text{H}^+ \rightarrow \text{H}_3\text{N}^+\text{CH}_2\text{CH}_2\text{NH}_3^+$

Key Takeaways

• Polyamines can be fully protonated if acid is in excess.
• Always check the number of basic sites (amino groups) in the molecule.

Common Mistakes

• Only protonating one amino group. The question says 'excess' HCl, implying both react.
• Forgetting the charges on the product. The product is a dication (or a salt with two Cl-).
• Writing $\text{NH}_4^+$ instead of $\text{NH}_3^+$ attached to the carbon chain.

Things to Be Careful About

• The mark scheme accepts both the full molecular equation with Cl- and the ionic equation with H+. Ensure charges are balanced.

Background Concept

When a diamine (molecule with two $-\text{NH}_2$ groups) reacts with a dicarboxylic acid (molecule with two $-\text{COOH}$ groups), they undergo condensation polymerisation to form a polyamide.

• The $-\text{NH}_2$ group reacts with the $-\text{COOH}$ group.
• A molecule of water ($\text{H}_2\text{O}$) is eliminated (condensation).
• An amide bond (or peptide bond in proteins) is formed: $-\text{CO}-\text{NH}-$.

The repeat unit of the polymer contains the remnants of both monomers.

Understanding the Question

Ethane-1,2-diamine ($\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$) reacts with ethanedioic acid ($\text{HOOC}-\text{COOH}$) to form polymer Z. Draw the structure showing two repeat units.

Approach

1. Identify the functional groups: $-\text{NH}_2$ and $-\text{COOH}$.
2. Determine the linkage: amide bond ($-\text{CO}-\text{NH}-$).
3. Draw the repeat unit: $-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CO}-\text{CO}-$.
4. Draw two of these units connected, with continuation bonds at the ends.

Step-by-Step Reasoning

• Monomer 1: $\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2$ (loses 2 H from the NH2 groups).
• Monomer 2: $\text{HOOC}-\text{COOH}$ (loses 2 OH from the COOH groups).
• Linkage: $-\text{CO}-\text{NH}-$ (amide bond).
• Repeat unit: $-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CO}-\text{CO}-$
• Two repeat units:
  $\text{H}-[\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CO}-\text{CO}-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CO}-\text{CO}]-\text{OH}$
  (The ends will be H- from the amine and -OH from the acid, or vice versa depending on where you start drawing. Usually, start with H-N... and end with ...CO-OH).

Displayed formula should show:
$\text{H}-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{C}(=\text{O})-\text{C}(=\text{O})-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{C}(=\text{O})-\text{C}(=\text{O})-\text{OH}$
with continuation bonds at the far left (H-) and far right (-OH) or simply bonds extending out.

Key Takeaways

• Diamine + dicarboxylic acid = polyamide.
• The repeat unit contains the amide linkage $-\text{CONH}-$.
• Always show continuation bonds for polymers.

Common Mistakes

• Drawing an ester linkage ($-\text{COO}-$) instead of an amide linkage ($-\text{CONH}-$). This would happen if the student confused the diamine with a diol.
• Forgetting the continuation bonds at the ends of the two repeat units.
• Drawing the monomers side-by-side without showing the loss of water (i.e., keeping the H on N and OH on C). The bond must be $-\text{NH}-\text{CO}-$, not $-\text{NH}_2-\text{COOH}-$.
• Not drawing two full repeat units. The question specifically asks for two.

Things to Be Careful About

• The amide bond is $-\text{C}(=\text{O})-\text{NH}-$. The carbonyl (C=O) comes from the acid, the NH comes from the amine.
• Ethanedioic acid is $\text{HOOC}-\text{COOH}$, so the acid part of the repeat unit is just $-\text{CO}-\text{CO}-$ (two carbonyls directly bonded). Do not add extra carbons.

Background Concept

There are two main types of polymerisation:

1. Addition polymerisation: Monomers add together without the loss of any atoms. Typical for alkenes (polyethene, PVC, polystyrene). The repeat unit has the same atoms as the monomer.
2. Condensation polymerisation: Monomers join together with the loss of a small molecule (usually water, sometimes HCl or methanol). Typical for polyesters (diol + dicarboxylic acid) and polyamides (diamine + dicarboxylic acid).

CIE also accepts addition-elimination as a synonym for condensation in some contexts, though 'condensation' is the preferred term.

Understanding the Question

Name the type of reaction occurring during the formation of polymer Z (from a diamine and a dicarboxylic acid).

Approach

Recognise that water is lost when the amide bond forms. This is condensation.

Step-by-Step Reasoning

• The reaction forms an amide bond ($-\text{CONH}-$) from $-\text{NH}_2$ and $-\text{COOH}$.
• $\text{H}_2\text{O}$ is eliminated.
• This is condensation polymerisation.

Key Takeaways

• Diamine + dicarboxylic acid -> polyamide + water -> condensation polymerisation.
• Diol + dicarboxylic acid -> polyester + water -> condensation polymerisation.

Common Mistakes

• Saying 'addition'. No atoms are lost in addition polymerisation; here, water is lost.
• Saying 'dehydration'. While water is removed, the specific polymerisation term is 'condensation'. 'Dehydration synthesis' is used in biology, but 'condensation' is the chemistry term.

Things to Be Careful About

• 'Condensation' is the key word. 'Addition-elimination' is also accepted in the mark scheme.

Background Concept

Biodegradable polymers can be broken down by microorganisms (bacteria, fungi) into natural substances like water, carbon dioxide, and biomass. They often contain hydrolyzable bonds (like ester or amide bonds) that enzymes can attack. Examples: polylactic acid (PLA), polyhydroxyalkanoates (PHAs), and the polyamide from this question (if designed to be biodegradable, though typical nylons are not; the question states Z is an example, perhaps implying a specific biodegradable polyamide or just using it as a hypothetical).

Non-biodegradable polymers are resistant to biological breakdown. They typically have strong carbon-carbon backbones (from addition polymerisation of alkenes) or very stable linkages. Examples:

• Polyalkenes: Polyethene (polythene), polypropene, PVC, polystyrene, PTFE (Teflon).
• Thermosets: Bakelite.
• Aramid fibres: Kevlar (very strong, resistant to degradation).

Understanding the Question

Name a polymer that is non-biodegradable. Polymer Z is given as biodegradable.

Approach

Recall a common plastic or synthetic polymer that persists in the environment.

Step-by-Step Reasoning

• Polyethene (polythene) is the most common example. It is used in plastic bags and bottles and persists for centuries.
• PVC (polyvinyl chloride) is another.
• Polystyrene is used in packaging and is non-biodegradable.
• Bakelite and Kevlar are also acceptable (as per mark scheme).

Key Takeaways

• Addition polymers (polyalkenes) are generally non-biodegradable because the C-C backbone is chemically inert and resistant to enzymatic attack.
• Condensation polymers can be biodegradable if they have hydrolyzable bonds (esters, amides), but not all are (e.g., Kevlar is a polyamide but is very resistant).

Common Mistakes

• Naming the polymer from the question (polyamide Z) as non-biodegradable. The question states Z is biodegradable.
• Saying 'plastic'. 'Plastic' is a category, not a specific polymer name. Must name a specific polymer (e.g., polyethene).
• Naming a biodegradable polymer like PLA or PHA.

Things to Be Careful About

• Ensure the name is a specific polymer, not a generic term like 'plastic' or 'synthetic'.
• Polyethene, PVC, and polystyrene are the safest, most common answers.