## CHM 001: GENERAL CHEMISTRY
**1. (a)** 2.21g of calcium is reacted with dilute hydrochloric acid, to give 6.15g of anhydrous metal chloride. Find the empirical formula of the metal chloride. **[3 Marks]**
**(b)** What are the oxidation numbers of nitrogen in the following compounds?
- (i) Dinitrogen oxide
- (ii) Sodium nitrite
- (iii) Dinitrogen trioxide (N₂O₃) **[3 Marks]**
**(c)** Use the balanced equation of reaction of chlorine gas with water to:
- (i) explain disproportionation of chlorine. **[2 Marks]**
- (ii) find the oxidation number of oxygen in sodium peroxide (Na₂O₂) and caesium superoxide (CsO₂). **[2 Marks]**
**[Total = 10 Marks]**
---
**2. (a) (i)** State the *Periodic law*. **[1 Mark]**
**(ii)** List any **TWO** atomic properties and state their trends in the periodic table. **[2 Marks]**
**(b) (i)** Define the term *Standard solution*. Mention **ONE** application of standard solution. **[1 Mark]**
**(ii)** Differentiate between empirical formula and molecular formula. Name **ONE** other chemical formula that describes atoms in a molecule of a compound. **[2 Marks]**
**(iii)** An organic compound J is known to contain carbon, hydrogen and oxygen only. When burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of J gives 0.88 g carbon dioxide and 0.54 g water. Find the empirical formula of compound J. (C = 12.0, O = 16.0, H = 1.0) **[4 Marks]**
**(c)** Given that 0.153g of dichloromethane displaced 45 cm³ of air at 20°C and 100.7 kNm⁻² pressure. If at 20°C the saturated vapour pressure of water is 2.319 kNm⁻², calculate the relative molecular mass of the dichloromethane.
(Avogadro's constant = 6.02 × 10²³ mol⁻¹, R = 8.314J /K/mol) **[4½ Marks]**
**[Total = 10 Marks]**
---
## CHM 002: PHYSICAL CHEMISTRY
**3. (a) (i)** State Hess' law of enthalpy summation. **[1 Mark]**
**(ii)** Hydrogen sulphide, H₂S, is a poisonous gas with the odour of rotten eggs. The reaction for the formation of H₂S from the elements is:
H₂(g) + ⅛S(rhombic) → H₂S(g)
Use Hess' law to obtain the enthalpy change for this reaction from the following enthalpy changes:
- H₂S(g) + 3/2 O₂(g) → H₂O(g) + SO₂(g) ΔH = −518 kJ mol⁻¹
- H₂(g) + ½ O₂(g) → H₂O(g) ΔH = −242 kJ mol⁻¹
- S(rhombic) + O₂(g) → SO₂(g) ΔH = −297 kJ mol⁻¹ **[2½ Marks]**
**(b)** Complete the following reactions:
- (i) ²³Na + ⁴He → ²⁶Mg + ?
- (ii) ⁶³Cu → ⁶³e + ?
- (iii) ¹⁰B + ⁴He → ¹³N + ?
- (iv) ¹⁰⁸Pd + ? → ¹⁰⁷Ag + ¹p
- (v) Give **THREE** general applications of radioactive elements. **[2 Marks] [1½ Marks]**
**(c)** Uranium has an atomic weight of 236 and half-life of 4.5 × 10⁹ years. Calculate the number of disintegrations produced per year from 1g of uranium. **[3 Marks]**
**[Total = 10 Marks]**
---
**4. (a)** What do you understand by the term "Standard enthalpy of combustion"? **[1 Mark]**
**(b)** A sample of butane gas, measured at 25°C and 98 kPa having a volume of 200 cm³, was completely burnt in air. The heat produced raised the temperature of 250g of water by 14.4°C.
- (i) Assuming no heat losses occurred during this experiment, calculate the mass of butane used.
- (ii) Determine the amount of heat released if the specific heat capacity of water is 4.2 Jkg⁻¹K⁻¹.
- (iii) Calculate the standard enthalpy of combustion of butane.
[R = 8.31J/K/mol, C = 12, H = 1] **[4½ Marks]**
---
## CHM 003: INORGANIC CHEMISTRY
**5. (a)** Define the following:
- (i) A ligand
- (ii) Co-ordination number **[2 Marks]**
**(b)** Given the complex ion [Cu(NH₃)₄]²⁺:
- (i) What is the coordination number of the complex ion?
- (ii) Give the IUPAC name of the complex ion.
- (iii) What is the oxidation state of the central metal ion?
- (iv) What is the shape of the complex ion? **[3 Marks]**
**(c)** Iodine, bromine and chlorine belong to **group 17** of the periodic table. Describe the reason(s) why iodine is a solid, bromine is a liquid and chlorine is a gas. **[2 Marks]**
**(d)** Explain why the alkaline earth metals are harder and have higher melting point than the alkali metals. **[3 Marks]**
**[Total = 10 Marks]**
---
**6. (a)**
- (i) Why do halogens form interhalogen compounds? **[½ Mark]**
- (ii) Give **FOUR** examples of interhalogen compounds. **[2 Marks]**
**(b)** Suggest reasons for the following:
- (i) Ionisation energy of Manganese (Mn) is greater than Iron (Fe) despite the trend ionisation energy increases across the period on the periodic table.
- (ii) Compounds of Sc³⁺ and Zn²⁺ are mostly white while those of Se²⁺ and Cu²⁺ are coloured.
- (iii) Ionization energy of gallium (Ga) is slightly higher than that of aluminium (Al).
---
## CHM 004: ORGANIC CHEMISTRY
**7. (a)** Draw the structures of **THREE** isomers of C₄H₈O. **[1½ Marks]**
**(b)** The chlorination of methane can be achieved through a free-radical reaction. Write a balanced equation for each stage. **[3 Marks]**
**(c)** Write balanced reaction equations using structural formulae, for the reaction of propene with:
- (i) bromine (ii) hydrogen bromide (iii) water (iv) Hydrogen (with catalyst) **[2½ Marks]**
**(d) (i)** Given the reaction scheme below:
**CH₂=CHCH₂CH₂COOH** →(H₂/Ni) **X** →(CH₃CH₂OH/H⁺) **Y**
Write the chemical formula and name of X and Y.
**(ii)** Give the name of the reactions producing X and Y respectively. **[3 Marks]**
**[Total = 10 Marks]**
---
**8. (a) (i)** Define the term *hybridisation*. **[1 Mark]**
**(ii)** Give the name of the following compound and the hybridisation pattern of first and last carbon atom: **CH₂=CHCH₂CH₂COOH** **[2 Marks]**
**(b)** Aldehydes and ketones are an important class of organic compounds.
- (i) Using butanal and 2-butanone as examples, what type of isomerism is exhibited by these two classes of organic compounds?
- (ii) Give the products formed when butanal and 2-butanone are reacted with:
- i. Na₂Cr₂O₇/H₂SO₄
- ii. NaBH₄, H₂O⁺ **[2½ Marks]**
**(c) (i)** Define the term *polymerisation*.
**(ii)** Give **TWO** differences between condensation polymerisation and addition polymerisation.
**(iii)** Using chemical equations **ONLY**, describe the polymerisation reaction for:
- a condensation polymer
- an addition polymer **[3½ Marks]**
**(d)** Using appropriate examples, describe a chemical test to distinguish between a reducing sugar and a non-reducing sugar. **[1 Mark]**
# COMPLETE ANSWERS: CHM 001–004
---
## CHM 001: GENERAL CHEMISTRY
### Question 1
**(a) Empirical Formula of Metal Chloride**
Given:
- Mass of calcium = 2.21 g
- Mass of metal chloride = 6.15 g
- Mass of chlorine = 6.15 − 2.21 = **3.94 g**
| Element | Mass (g) | Molar Mass | Moles | Ratio |
|---------|----------|------------|-------|-------|
| Ca | 2.21 | 40 | 0.055 | 1 |
| Cl | 3.94 | 35.5 | 0.111 | 2 |
**Empirical Formula = CaCl₂**
---
**(b) Oxidation Numbers of Nitrogen**
**(i) Dinitrogen oxide (N₂O):**
Let N = x
2x + (−2) = 0 → x = **+1**
**(ii) Sodium nitrite (NaNO₂):**
(+1) + x + 2(−2) = 0 → x = **+3**
**(iii) Dinitrogen trioxide (N₂O₃):**
2x + 3(−2) = 0 → x = **+3**
---
**(c) Chlorine with Water**
Balanced equation:
**Cl₂ + H₂O ⇌ HCl + HOCl**
**(i) Disproportionation of Chlorine:**
In this reaction, chlorine is simultaneously oxidised and reduced. In Cl₂, chlorine has oxidation state 0. In HCl, it is reduced to −1, while in HOCl (hypochlorous acid), it is oxidised to +1. Since the same element undergoes both oxidation and reduction, this is disproportionation.
**(ii) Oxidation number of oxygen in:**
- **Sodium peroxide (Na₂O₂):**
2(+1) + 2x = 0 → x = **−1**
- **Caesium superoxide (CsO₂):**
(+1) + 2x = 0 → x = **−½**
---
### Question 2
**(a)(i) Periodic Law:**
The physical and chemical properties of elements are a periodic function of their atomic numbers.
**(a)(ii) Two Atomic Properties and their Trends:**
1. **Atomic radius** – Decreases across a period (left to right) due to increasing nuclear charge; increases down a group due to addition of new electron shells.
2. **Ionisation energy** – Increases across a period due to stronger nuclear attraction; decreases down a group as electrons are farther from the nucleus.
---
**(b)(i) Standard Solution:**
A standard solution is a solution of accurately known concentration.
**Application:** Used in titrations (volumetric analysis) to determine the unknown concentration of another solution.
**(b)(ii) Empirical vs Molecular Formula:**
- **Empirical formula** shows the simplest whole-number ratio of atoms of each element in a compound (e.g., CH₂O).
- **Molecular formula** shows the actual number of atoms of each element in one molecule of a compound (e.g., C₆H₁₂O₆).
**Another chemical formula:** Structural formula — shows how atoms are bonded and arranged within a molecule.
---
**(b)(iii) Empirical Formula of Compound J**
Given:
- 0.46 g of J → 0.88 g CO₂ and 0.54 g H₂O
**Mass of Carbon:**
Moles of CO₂ = 0.88/44 = 0.02 mol → Mass of C = 0.02 × 12 = **0.24 g**
**Mass of Hydrogen:**
Moles of H₂O = 0.54/18 = 0.03 mol → Mass of H = 0.03 × 2 = **0.06 g**
**Mass of Oxygen:**
= 0.46 − 0.24 − 0.06 = **0.16 g**
| Element | Mass | Molar Mass | Moles | Ratio |
|---------|------|------------|-------|-------|
| C | 0.24 | 12 | 0.02 | 1 |
| H | 0.06 | 1 | 0.06 | 3 |
| O | 0.16 | 16 | 0.01 | 0.5 |
Multiply all by 2:
**Empirical Formula = C₂H₆O (Ethanol)**
---
**(c) Relative Molecular Mass of Dichloromethane**
Using the formula derived from Dalton's Law of Partial Pressures and Ideal Gas Law:
- Mass of DCM = 0.153 g
- Volume of air displaced = 45 cm³ = 45 × 10⁻⁶ m³
- Total pressure = 100.7 kNm⁻² = 100700 Pa
- Saturated vapour pressure of water = 2.319 kNm⁻² = 2319 Pa
- Partial pressure of DCM vapour = 100700 − 2319 = **98381 Pa**
- T = 20°C = 293 K
- R = 8.314 J/K/mol
Using PV = nRT:
n = PV/RT = (98381 × 45 × 10⁻⁶) / (8.314 × 293)
n = 4.427 / 2436 = **1.817 × 10⁻³ mol**
Molar mass = mass/moles = 0.153 / 1.817 × 10⁻³ = **84.2 g/mol**
**Relative molecular mass of dichloromethane ≈ 84** (theoretical = 85, CH₂Cl₂ ✓)
---
## CHM 002: PHYSICAL CHEMISTRY
### Question 3
**(a)(i) Hess' Law:**
The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.
**(a)(ii) Enthalpy of Formation of H₂S**
Target: H₂(g) + ⅛S(rhombic) → H₂S(g) ΔH = ?
Given:
- (1) H₂S(g) + 3/2 O₂(g) → H₂O(g) + SO₂(g) ΔH₁ = −518 kJ/mol
- (2) H₂(g) + ½O₂(g) → H₂O(g) ΔH₂ = −242 kJ/mol
- (3) S(rhombic) + O₂(g) → SO₂(g) ΔH₃ = −297 kJ/mol
Reverse equation (1):
H₂O(g) + SO₂(g) → H₂S(g) + 3/2O₂(g) ΔH = **+518**
Add equation (2): +( −242)
Add equation (3): +(−297)
ΔHf = +518 − 242 − 297 = **−21 kJ/mol**
**ΔHf(H₂S) = −21 kJ mol⁻¹**
---
**(b) Nuclear Reactions**
(i) ²³₁₁Na + ⁴₂He → ²⁶₁₂Mg + **¹₁H (proton)**
(ii) ⁶³₂₉Cu → ⁶³₂₈Ni + **⁰₊₁e (positron)** *(β⁺ decay)*
(iii) ¹⁰₅B + ⁴₂He → ¹³₇N + **¹₀n (neutron)**
(iv) ¹⁰⁸₄₆Pd + **²₁H (deuteron)** → ¹⁰⁷₄₇Ag + ¹₁p
**(v) Three Applications of Radioactive Elements:**
1. **Medical** – used in cancer treatment (radiotherapy) and diagnosis (e.g., PET scans)
2. **Carbon dating** – used in archaeology to determine the age of ancient materials
3. **Nuclear power** – used as fuel in nuclear reactors to generate electricity
---
**(c) Disintegrations from 1g of Uranium**
Given:
- Atomic weight = 236 g/mol
- Half-life t½ = 4.5 × 10⁹ years
- Mass = 1 g
Number of atoms (N):
N = (1/236) × 6.02 × 10²³ = **2.55 × 10²¹ atoms**
Decay constant:
λ = 0.693 / t½ = 0.693 / (4.5 × 10⁹) = **1.54 × 10⁻¹⁰ yr⁻¹**
Activity (disintegrations per year):
A = λN = 1.54 × 10⁻¹⁰ × 2.55 × 10²¹
**A = 3.93 × 10¹¹ disintegrations per year**
---
### Question 4
**(a) Standard Enthalpy of Combustion:**
The enthalpy change when **one mole** of a substance is completely burnt in excess oxygen under standard conditions (298 K, 100 kPa), with all reactants and products in their standard states.
---
**(b) Butane Combustion**
**Given:**
- T = 25°C = 298 K, P = 98 kPa = 98000 Pa
- V = 200 cm³ = 200 × 10⁻⁶ m³
- Mass of water = 250 g, ΔT = 14.4°C
- Specific heat capacity = 4.2 J g⁻¹ K⁻¹
- R = 8.31 J/K/mol, M(butane C₄H₁₀) = 58 g/mol
**(i) Mass of Butane Used:**
n = PV/RT = (98000 × 200 × 10⁻⁶) / (8.31 × 298)
n = 19.6 / 2476.4 = **7.92 × 10⁻³ mol**
Mass = n × M = 7.92 × 10⁻³ × 58 = **0.459 g**
**(ii) Heat Released:**
Q = mcΔT = 250 × 4.2 × 14.4 = **15,120 J = 15.12 kJ**
**(iii) Standard Enthalpy of Combustion:**
Moles of butane = 0.459/58 = 7.91 × 10⁻³ mol
ΔHc = −Q/n = −15120 / 7.91 × 10⁻³ = **−1,911,504 J/mol ≈ −1911.5 kJ/mol**
---
## CHM 003: INORGANIC CHEMISTRY
### Question 5
**(a) Definitions:**
**(i) A Ligand:**
A ligand is an ion or molecule that donates a lone pair of electrons to a central metal ion to form a coordinate (dative) bond in a complex ion. Examples: NH₃, H₂O, Cl⁻, CN⁻.
**(ii) Co-ordination Number:**
The coordination number is the total number of ligand donor atoms directly bonded to the central metal ion in a complex.
---
**(b) Complex Ion [Cu(NH₃)₄]²⁺**
**(i) Coordination number:** **4** (four NH₃ ligands)
**(ii) IUPAC name:** **Tetraamminecopper(II) ion**
**(iii) Oxidation state of Cu:**
x + 4(0) = +2 → x = **+2**
**(iv) Shape:** **Square planar**
---
**(c) Iodine (solid), Bromine (liquid), Chlorine (gas):**
All three are non-polar molecules held together by **van der Waals (London dispersion) forces**. The strength of these forces increases with increasing molecular size and number of electrons:
- Cl₂ (smallest, fewest electrons) → weakest forces → **gas** at room temperature
- Br₂ (intermediate size) → moderate forces → **liquid** at room temperature
- I₂ (largest, most electrons) → strongest forces → **solid** at room temperature
---
**(d) Alkaline Earth Metals vs Alkali Metals (Hardness & Melting Point):**
Alkaline earth metals (Group 2) are harder and have higher melting points than alkali metals (Group 1) because:
1. They have **two valence electrons** compared to one in alkali metals, contributing to a **stronger metallic bond**.
2. The greater number of delocalised electrons per atom creates a **more rigid electron sea**, making the metallic lattice stronger.
3. Their **smaller atomic radii** mean the nucleus attracts the delocalised electrons more strongly, resulting in a denser, harder structure.
---
### Question 6
**(a)(i) Why Halogens Form Interhalogen Compounds:**
Halogens form interhalogen compounds because different halogen atoms have similar but slightly different electronegativities and comparable atomic sizes, allowing them to form covalent bonds with each other. The bonding between unlike halogens is actually stronger than between like halogens in some cases.
**(a)(ii) Four Examples of Interhalogen Compounds:**
1. ClF (chlorine monofluoride)
2. BrF₃ (bromine trifluoride)
3. ICl (iodine monochloride)
4. IF₅ (iodine pentafluoride)
---
**(b) Reasons:**
**(i) IE of Mn > Fe despite the trend:**
Manganese (Mn) has the electronic configuration [Ar] 3d⁵ 4s². The 3d subshell is exactly **half-filled**, which gives it extra stability due to exchange energy. Removing an electron from this stable configuration requires more energy. Iron (Fe) has [Ar] 3d⁶ 4s²; removing one 3d electron from Fe actually relieves the electron-electron repulsion in the paired 3d orbital, making ionisation easier. Hence IE(Mn) > IE(Fe).
**(ii) Sc³⁺ and Zn²⁺ compounds are white; Sc²⁺ and Cu²⁺ are coloured:**
Colour in transition metal compounds arises from **d-d electron transitions** when electrons absorb visible light and move between split d orbitals. Sc³⁺ has no d electrons (d⁰) and Zn²⁺ has a completely filled d subshell (d¹⁰); in both cases, no d-d transitions are possible, so they appear **white/colourless**. However, Cu²⁺ (d⁹) and Sc²⁺ (hypothetical, d¹) have partially filled d orbitals, allowing d-d transitions and thus producing **colour**.
**(iii) IE of Ga slightly higher than Al:**
Gallium (Ga) follows the transition metals in the periodic table. The 3d electrons in Ga are poor at shielding the nucleus compared to the 2p electrons in Al. As a result, the effective nuclear charge felt by the outer 4p electrons of Ga is greater than expected, making it slightly harder to remove an electron from Ga than from Al.
---
## CHM 004: ORGANIC CHEMISTRY
### Question 7
**(a) Three Isomers of C₄H₈O** (1½ Marks)
Possible isomers include:
**1. Butanal (aldehyde):**
CH₃CH₂CH₂CHO
**2. Butanone (ketone):**
CH₃COCH₂CH₃
**3. Cyclobutanol (cyclic alcohol):**
Cyclobutane ring with –OH group
*(Other valid answers: methyl propenyl ether, tetrahydrofuran derivatives, etc.)*
---
**(b) Free-Radical Chlorination of Methane**
**Stage 1 – Initiation:**
Cl₂ → 2Cl• (UV light)
**Stage 2 – Propagation:**
Cl• + CH₄ → CH₃• + HCl
CH₃• + Cl₂ → CH₃Cl + Cl•
**Stage 3 – Termination:**
Cl• + Cl• → Cl₂
CH₃• + Cl• → CH₃Cl
CH₃• + CH₃• → C₂H₆
---
**(c) Reactions of Propene (CH₂=CHCH₃)**
**(i) With Bromine (Br₂):**
CH₂=CHCH₃ + Br₂ → CH₂Br–CHBr–CH₃
(1,2-dibromopropane) — electrophilic addition
**(ii) With Hydrogen Bromide (HBr):**
CH₂=CHCH₃ + HBr → CH₃–CHBr–CH₃
(2-bromopropane) — Markovnikov addition
**(iii) With Water (H₂O, acid catalyst):**
CH₂=CHCH₃ + H₂O → CH₃–CHOH–CH₃
(propan-2-ol) — electrophilic addition
**(iv) With Hydrogen (H₂, catalyst):**
CH₂=CHCH₃ + H₂ → CH₃–CH₂–CH₃
(propane) — catalytic hydrogenation
---
**(d)(i) Chemical Formula and Name of X and Y**
CH₂=CHCH₂CH₂COOH →(H₂/Ni) **X** →(CH₃CH₂OH/H⁺) **Y**
- **X:** CH₃CH₂CH₂CH₂COOH — **Pentanoic acid** (hydrogenation of the double bond)
- **Y:** CH₃CH₂CH₂CH₂COOC₂H₅ — **Ethyl pentanoate** (esterification with ethanol)
**(d)(ii) Name of Reactions:**
- Reaction producing X: **Hydrogenation (catalytic reduction)**
- Reaction producing Y: **Esterification (condensation reaction)**
---
### Question 8
**(a)(i) Hybridisation:**
Hybridisation is the mixing of atomic orbitals of similar energies within the same atom to form new orbitals of equal energy and shape, called hybrid orbitals, which are used in bond formation.
**(a)(ii) CH₂=CHCH₂CH₂COOH**
**Name:** Pent-4-enoic acid
**Hybridisation:**
- **First carbon (CH₂=):** sp² hybridised (part of C=C double bond)
- **Last carbon (–COOH):** sp² hybridised (part of C=O in carboxyl group)
---
**(b) Aldehydes and Ketones**
**(i) Type of Isomerism:**
Butanal (CH₃CH₂CH₂CHO) and 2-butanone (CH₃COCH₂CH₃) have the same molecular formula C₄H₈O but belong to different functional group classes. This is **functional group isomerism**.
**(ii) Products with:**
**i. Na₂Cr₂O₇/H₂SO₄ (acidified dichromate — oxidising agent):**
- Butanal → **Butanoic acid** (CH₃CH₂CH₂COOH) — aldehydes are oxidised
- 2-butanone → **No reaction** — ketones are resistant to mild oxidation
**ii. NaBH₄, H₃O⁺ (reducing agent):**
- Butanal → **Butan-1-ol** (CH₃CH₂CH₂CH₂OH)
- 2-butanone → **Butan-2-ol** (CH₃CHOHCH₂CH₃)
---
**(c) Polymerisation**
**(i) Definition:**
Polymerisation is the chemical process by which small molecules (monomers) join together repeatedly to form a large molecule (polymer).
**(ii) Two Differences between Condensation and Addition Polymerisation:**
| | Condensation | Addition |
|--|--|--|
| 1 | A small molecule (e.g., H₂O, HCl) is eliminated for each bond formed | No small molecule is lost; all atoms of monomers are incorporated into the polymer |
| 2 | Monomers must have two different functional groups (e.g., –NH₂ and –COOH) | Monomers must contain a C=C double bond (unsaturated) |
**(iii) Polymerisation Reactions:**
**Condensation polymer (e.g., nylon-6,6):**
n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → [–NH(CH₂)₆NHCO(CH₂)₄CO–]ₙ + n H₂O
**Addition polymer (e.g., polyethene):**
n CH₂=CH₂ → [–CH₂–CH₂–]ₙ
---
**(d) Distinguishing Reducing Sugar from Non-Reducing Sugar**
**Test: Benedict's test (or Fehling's solution)**
- Add Benedict's solution to the sugar solution and heat.
- A **reducing sugar** (e.g., glucose, maltose) will produce a **brick-red/orange precipitate** (Cu₂O), as it reduces Cu²⁺ ions.
- A **non-reducing sugar** (e.g., sucrose) will show **no colour change** — solution remains blue.
**If no reaction with Benedict's:** Hydrolyse the sugar with dilute HCl, neutralise with NaHCO₃, then repeat the Benedict's test. A positive result (brick-red) confirms it is a non-reducing sugar that was hydrolysed to reducing sugars.