## CHM 001
### Question 1
**(a)**
(i) State three Bohr's postulates of hydrogen atom. *(1½ marks)*
(ii) Calculate the wavelength (nm) of the spectral line of hydrogen atom for which n₁ = 3 and n₂ = 6. (R = 1.09678 × 10⁷ m⁻¹) *(2 marks)*
**(b)** State the type of chemical bonds in each of the following:
(i) Na₂O (ii) BeCl₂ (iii) F₃BNH₃ (iv) CO₂ (v) NH₄⁺ (vi) NaCl *(3 marks)*
**(c)** The isotopic masses of two isotopes are 34.969 and 36.9689 amu respectively. Estimate the relative atomic mass of the element if the percentage abundance of the heavier isotope is 24.47%. *(2 marks)*
**(d)**
(i) Write the electronic configuration of strontium ₃₈Sr.
(ii) What block does it belong to? *(1½ marks)*
---
### Question 2
**(a)**
(i) Define the term ionization energy. *(1 mark)*
(ii) State two factors that affect ionization energy of an atom. *(1 mark)*
**(b)** 500 mg of iron (II) complex ferrous bisglycinate hydrochloride was dissolved in dilute H₂SO₄ and titrated with 0.0200 mol·dm⁻³ KMnO₄. 18.10 cm³ of KMnO₄ solution were required to reach the end point.
The equation for the titration reaction is:
**5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O**
Calculate:
i. Number of moles of Fe²⁺ in the capsule
ii. Mass of iron in the capsule
iii. Molar mass of the iron (II) complex, assuming 1 mole of the complex contains 1 mole of iron. (Fe = 55.9 g mol⁻¹) *(3½ marks)*
**(c)** Identify each of the following reactions as precipitation, neutralization, decomposition, or combination:
(i) Ba(OH)₂(aq) + 2HI(aq) → BaI₂(aq) + 2H₂O
(ii) 2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
(iii) Pd(NO₃)₂(aq) + H₂S(g) → PdS(s) + 2HNO₃(aq)
(iv) Cu(NO₃)₂(aq) → CuO(s) + NO₂(g) + ½O₂(g)
(v) FeCl₂(aq) + 2NaOH(aq) → Fe(OH)₂(s) + 2NaCl(aq) *(2½ marks)*
**(d)** The following results were obtained from a replicate analysis of blood sample for its lead content: 0.752, 0.756, 0.752, 0.769 ppm lead. Explain the precision of the results. *(2 marks)*
---
## CHM 002
### Question 3
**(a)**
(i) State the two factors that affect the solubility of a solid in a liquid. *(1 mark)*
(ii) A saturated solution of AgCl was found to have a concentration of 1.3 × 10⁻⁵ mol/dm³. What is the solubility product of AgCl? *(1½ marks)*
**(b)** Define the terms: (i) entropy (ii) enthalpy. *(2 marks)*
**(c)** The equation for the reaction between sulphur trioxide and CuO is:
**CuO(s) + SO₃(g) → CuSO₄(s)**
Given:
| Substance | ΔH° (kJ/mol) | S° (J/K·mol) |
|-----------|--------------|---------------|
| SO₃ | −157 | 42.63 |
| CuO | −395.2 | 256.2 |
| CuSO₄ | −771.36 | 109 |
(i) Calculate the standard free energy for this reaction at 37°C.
(ii) Comment on the spontaneity of the reaction based on the value in (i). *(3 marks)*
**(d)** If the same volume of NH₃ and an unknown gas X effuse at a rate of 2.25 cm³s⁻¹ and 1.40 cm³s⁻¹ respectively under the same experimental conditions, what is the molecular mass (Mᵣ) for X? Suggest a possible identity for X. [N=14, H=1.00] *(2½ marks)*
### Question 4
**(a)** Calculate ΔH for the reaction: **4HI + O₂ → 2I₂ + 2H₂O**, given:
(i) H₂ + I₂ → 2HI, ΔH = +52 kJ
(ii) 2H₂ + O₂ → 2H₂O, ΔH = −480 kJ *(3 marks)*
**(b)** Determine the equilibrium constant for the following reaction at 45°C:
**Sn²⁺(aq) + Cu²⁺(aq) ⇌ Sn⁴⁺(aq) + Cu(s)**
Given:
- E° Cu²⁺/Cu = +0.337 V
- E° Sn⁴⁺/Sn²⁺ = +0.5 V
- R = 8.314 J mol⁻¹ K⁻¹ *(3½ marks)*
**(c)** State three factors that affect the rate of a chemical reaction. *(1½ marks)*
**(d)** Consider the reaction: **CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)**
Explain the effect on the position of equilibrium of addition of:
(i) CH₃COOC₂H₅(l)
(ii) C₂H₅OH(l) *(2 marks)*
---
## CHM 003
### Question 5
**(a)**
(i) Define the term allotropy. *(1 mark)*
(ii) Name two allotropes each of carbon and tin. *(2 marks)*
(iii) What is the difference between the type of allotropy exhibited by carbon and tin? *(1 mark)*
**(b)** Give reason(s) for the following observations: *(3 marks)*
(i) Fluorine exhibits only −1 oxidation state while other members of the group exhibit −1 as well as other oxidation states.
(ii) Hydrogen chloride is a stronger acid than hydrogen fluoride.
(iii) Beryllium does not react with water, even on heating.
**(c)** Using suitable reaction equations, outline three methods of laboratory synthesis of H₂ from suitable metals. *(1½ marks)*
**(d)** State three properties of transition elements. *(1½ marks)*
### Question 6
**(a)** The molecular formulae of coordination compounds A and B are:
- A = [Cr(NH₃)₄Cl₂]Br
- B = [Cr(NH₃)₄(Br)(Cl)]Cl
(i) Give the IUPAC name of compounds A and B. *(2 marks)*
(ii) State the oxidation number of the metal ion in both compounds. *(1 mark)*
(iii) Which of these compounds would give a white precipitate with silver nitrate solution? *(½ mark)*
**(b)** Using a statement and appropriate equation(s) where necessary, distinguish between synthesis gas and water gas. *(2 marks)*
**(c)** Using reaction equations only, show how Li, Na and K react when heated in excess oxygen. *(1½ marks)*
**(d)** Describe, using chemical equations where applicable, how aluminium is extracted from its ore. *(3 marks)*
---
## CHM 004
### Question 7
**(a)**
(i) Define the term hybridization. *(1 mark)*
(ii) Indicate the hybridization of the carbon atom labelled (a)–(e) in the given structure (benzene ring connected to CH₂–C≡CH with labels). *(2½ marks)*
**(b)** The scheme represents various reactions of propanone. Give the formula of each product represented by letters A to E. *(2½ marks)*
(Reagents shown: ZnConc.HCl → A, CH₃NH₂ → E, LiAlH₄ → C, CH₃MgBr → D, CH₃COCH₃/NaOH → B)
**(c)** Draw the structure of the following compounds: *(3 marks)*
(i) 2-bromo-3-chloro-4,4-dimethylpentanal
(ii) Butane-1,2,3-triol
(iii) 2,4,6-tribromophenol
---
### Question 8
**(a)**
(i) Define biotechnology. *(1 mark)*
(ii) List two advantages in the use of biotechnology. *(1 mark)*
**(b)** Explain the following terms:
(i) Inductive effect
(ii) Electromeric effect
(iii) Homolytic cleavage *(1½ marks)*
**(c)** Classify each of the following carbohydrates as Monosaccharides, Disaccharides, or Polysaccharides:
(I) Cellulose (II) Fructose (III) Glucose (IV) Maltose (V) Sucrose *(2½ marks)*
**(d)**
(i) Write the structures of all the isomeric alcohols with molecular formula C₄H₁₀O. *(1½ marks)*
(ii) Classify them as primary, secondary and tertiary alcohols. *(1½ marks)*
(iii) Arrange the alcohols according to their increasing boiling points. *(1 mark)*
**(d)** [from CHM 001 section] A compound K, [C₇H₁₆O], is an alcohol which is oxidized by Chromic acid to yield compound L [C₇H₁₄O]. L forms crystalline 2,4-Dinitrophenylhydrazine compound and gives iodoform when treated with iodine-alkali but does not form a red precipitate with Fehling's reagent. With the given data, write possible structures for compounds K and L. *(1 mark)*
---
# ANSWERS
---
## CHM 001
### Q1(a)(i) — Bohr's Postulates
1. Electrons revolve around the nucleus in fixed circular orbits called stationary states without radiating energy.
2. Only orbits where the angular momentum of the electron is an integral multiple of h/2π are allowed (mvr = nh/2π).
3. Energy is emitted or absorbed only when an electron transitions between orbits: ΔE = hf.
### Q1(a)(ii) — Wavelength of spectral line (n₁=3, n₂=6)
Using the Rydberg formula:
**1/λ = R(1/n₁² − 1/n₂²)**
1/λ = 1.09678 × 10⁷ × (1/9 − 1/36)
= 1.09678 × 10⁷ × (4/36 − 1/36)
= 1.09678 × 10⁷ × 3/36
= 1.09678 × 10⁷ × 0.08333
= 9.139 × 10⁵ m⁻¹
**λ = 1/9.139 × 10⁵ = 1.094 × 10⁻⁶ m = 1094 nm** (infrared, Paschen series)
### Q1(b) — Types of Chemical Bonds
| Compound | Bond Type |
|----------|-----------|
| Na₂O | Ionic |
| BeCl₂ | Covalent (polar) |
| F₃BNH₃ | Dative/coordinate covalent (N→B) |
| CO₂ | Covalent (double bonds) |
| NH₄⁺ | Covalent + dative covalent |
| NaCl | Ionic |
### Q1(c) — Relative Atomic Mass
- Lighter isotope mass = 34.969; abundance = 100 − 24.47 = 75.53%
- Heavier isotope mass = 36.9689; abundance = 24.47%
RAM = (34.969 × 75.53/100) + (36.9689 × 24.47/100)
= 26.413 + 9.048
= **35.46 amu** (Chlorine)
### Q1(d) — Strontium
(i) ₃₈Sr: **1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s²**
(ii) It belongs to the **s-block**
---
### Q2(a)
(i) **Ionization energy** is the minimum energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state.
(ii) Factors:
- Nuclear charge (atomic number)
- Atomic radius / shielding effect
### Q2(b) — Iron Titration Calculation
**i. Moles of KMnO₄ used:**
= 0.0200 × 18.10/1000 = 3.62 × 10⁻⁴ mol
From equation: 5 mol Fe²⁺ : 1 mol MnO₄⁻
Moles of Fe²⁺ = 5 × 3.62 × 10⁻⁴ = **1.81 × 10⁻³ mol**
**ii. Mass of iron:**
= 1.81 × 10⁻³ × 55.9 = **0.1012 g = 101.2 mg**
**iii. Molar mass of complex:**
= mass/moles = 0.500 g / 1.81 × 10⁻³ mol = **276.2 g/mol**
### Q2(c) — Reaction Types
| Reaction | Type |
|----------|------|
| (i) Ba(OH)₂ + 2HI | Neutralization |
| (ii) 2Al + 3Cl₂ | Combination |
| (iii) Pd(NO₃)₂ + H₂S | Precipitation |
| (iv) Cu(NO₃)₂ → CuO + NO₂ + ½O₂ | Decomposition |
| (v) FeCl₂ + 2NaOH | Precipitation |
### Q2(d) — Precision
Mean = (0.752 + 0.756 + 0.752 + 0.769)/4 = **0.757 ppm**
Range = 0.769 − 0.752 = 0.017 ppm
The results show **moderate precision**. Three of the four values (0.752, 0.752, 0.756) are close together, but the fourth (0.769) is a slight outlier. The small range indicates reasonable reproducibility, though not excellent precision.
---
## CHM 002
### Q3(a)
(i) Factors affecting solubility: **Temperature** and **Nature of solvent and solute** (like dissolves like).
(ii) AgCl → Ag⁺ + Cl⁻
[Ag⁺] = [Cl⁻] = 1.3 × 10⁻⁵ mol/dm³
**Ksp = [Ag⁺][Cl⁻] = (1.3 × 10⁻⁵)² = 1.69 × 10⁻¹⁰ mol²dm⁻⁶**
### Q3(b)
- **Entropy (S):** A measure of the degree of disorder or randomness in a system.
- **Enthalpy (H):** The total heat content of a system at constant pressure; ΔH = heat absorbed or released in a reaction.
### Q3(c) — Gibbs Free Energy
**Step 1: Calculate ΔH°rxn**
ΔH°rxn = ΔHf°(CuSO₄) − [ΔHf°(CuO) + ΔHf°(SO₃)]
= −771.36 − [−395.2 + (−157)]
= −771.36 − (−552.2)
= **−219.16 kJ/mol**
**Step 2: Calculate ΔS°rxn**
ΔS° = S°(CuSO₄) − [S°(CuO) + S°(SO₃)]
= 109 − [256.2 + 42.63]
= 109 − 298.83
= **−189.83 J/mol·K = −0.18983 kJ/mol·K**
**Step 3: ΔG° at T = 310 K (37°C)**
ΔG° = ΔH° − TΔS°
= −219.16 − (310 × −0.18983)
= −219.16 + 58.85
= **−160.31 kJ/mol**
(ii) Since ΔG° is **negative (−160.31 kJ/mol)**, the reaction is **spontaneous** at 37°C.
### Q3(d) — Graham's Law / Molar Mass of X
Graham's Law: r₁/r₂ = √(M₂/M₁)
2.25/1.40 = √(Mₓ/17)
1.607 = √(Mₓ/17)
(1.607)² = Mₓ/17
2.582 = Mₓ/17
**Mₓ = 43.9 ≈ 44 g/mol**
Possible identity: **CO₂** (M = 44 g/mol) or **Propane (C₃H₈)**
---
### Q4(a) — Hess's Law
Target: 4HI + O₂ → 2I₂ + 2H₂O
Given:
1. H₂ + I₂ → 2HI, ΔH = +52 kJ → Reverse × 2: 4HI → 2H₂ + 2I₂, ΔH = −104 kJ
2. 2H₂ + O₂ → 2H₂O, ΔH = −480 kJ
Adding:
4HI + O₂ → 2I₂ + 2H₂O
**ΔH = −104 + (−480) = −584 kJ**
### Q4(b) — Equilibrium Constant at 45°C
**Cell reaction:** Cu²⁺ + Sn²⁺ → Cu + Sn⁴⁺
E°cell = E°cathode − E°anode = 0.337 − 0.5 = **−0.163 V**
Wait — since the reaction as written has Sn²⁺ being oxidized:
- Cathode: Cu²⁺ + 2e⁻ → Cu, E° = +0.337 V
- Anode: Sn²⁺ → Sn⁴⁺ + 2e⁻, E° = −0.5 V (reversed)
E°cell = 0.337 − 0.5 = **−0.163 V** (non-spontaneous as written)
Using: ΔG° = −nFE° = −RT ln K
ln K = nFE°/RT = (2 × 96485 × −0.163)/(8.314 × 318)
= −31,454.1/2643.9
= −11.897
**K = e⁻¹¹·⁹⁰ = 6.8 × 10⁻⁶**
### Q4(c) — Factors Affecting Rate of Reaction
1. **Temperature** — higher temperature increases rate
2. **Concentration** (or pressure for gases) — higher concentration increases rate
3. **Catalyst** — lowers activation energy, increases rate
*(Also valid: surface area, nature of reactants, light)*
### Q4(d) — Le Chatelier's Principle
Equilibrium: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
(i) Adding **CH₃COOC₂H₅** (product): equilibrium **shifts left** (backward), producing more reactants.
(ii) Adding **C₂H₅OH** (reactant): equilibrium **shifts right** (forward), producing more ester and water.
---
## CHM 003
### Q5(a)
(i) **Allotropy** is the existence of an element in two or more different physical forms in the same physical state, with different physical and chemical properties.
(ii) Allotropes:
- **Carbon:** Diamond and Graphite (also Fullerene/Buckminsterfullerene)
- **Tin:** White tin (β-tin) and Grey tin (α-tin)
(iii) Carbon exhibits **enantiotropy** (stable forms at different temperature ranges) while — actually, **carbon's allotropes** (diamond/graphite) show **monotropy** (graphite is the stable form at all temperatures; diamond is metastable), whereas **tin** exhibits **enantiotropy** (grey tin ⇌ white tin are interconvertible at 13.2°C).
### Q5(b) — Reasons for Observations
(i) **Fluorine only shows −1 oxidation state:** Fluorine is the most electronegative element and has no d-orbitals available in its valence shell. It cannot expand its octet, so it can only accept electrons (−1 state) and never donate or share in higher oxidation states as other halogens do.
(ii) **HCl is a stronger acid than HF:** Although the H–F bond is more polar, it is much stronger (higher bond energy) than H–Cl. HF dissociates poorly in water and also forms hydrogen bonds, reducing [H⁺]. HCl dissociates completely, making it a stronger acid.
(iii) **Beryllium does not react with water even on heating:** Beryllium forms a thin, impermeable oxide layer (BeO) on its surface that protects it from further reaction with water. It also has a very high ionization energy and small ionic radius, making it kinetically inert toward water.
### Q5(c) — Laboratory Synthesis of H₂ from Metals
1. **Zinc + dilute H₂SO₄:**
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
2. **Magnesium + dilute HCl:**
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
3. **Sodium + water (cold):**
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
### Q5(d) — Properties of Transition Elements
1. They have **variable oxidation states** due to incompletely filled d-orbitals.
2. They form **coloured ions** in solution.
3. They act as **catalysts** (e.g., Fe in Haber process, V₂O₅ in Contact process).
---
### Q6(a) — IUPAC Names of Coordination Compounds
**A = [Cr(NH₃)₄Cl₂]Br**
- Cr oxidation state: x + 0(4) + (−1)(2) = +1 (charge of complex cation); Br⁻ is counter ion
- x − 2 = +1, so x = **+3**
- Name: **Tetraamminedichloridochromium(III) bromide**
**B = [Cr(NH₃)₄(Br)(Cl)]Cl**
- Complex cation charge = +1; Cl⁻ counter ion
- x + 0(4) − 1 − 1 = +1 → x = **+3**
- Name: **Tetraamminebromidochloridochromium(III) chloride**
(ii) Oxidation number of Cr in **both A and B = +3**
(iii) **Compound B** would give a white precipitate with AgNO₃ solution, because it has Cl⁻ as the free (outer sphere) counter ion. Compound A has Br⁻ as the counter ion (giving a pale yellow precipitate). Actually — both have halide counter ions: A has Br⁻ (pale yellow ppt with AgNO₃) and B has Cl⁻ (**white precipitate**). So **B** gives the white precipitate (AgCl).
### Q6(b) — Synthesis Gas vs Water Gas
- **Water gas:** A mixture of CO and H₂ produced by passing steam over red-hot coke:
C(s) + H₂O(g) → CO(g) + H₂(g)
- **Synthesis gas (syngas):** Also a mixture of CO and H₂ but produced by steam reforming of natural gas (methane):
CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)
The key difference is the **source** and **ratio of H₂:CO** — water gas has a 1:1 ratio, syngas has a 3:1 ratio.
### Q6(c) — Reactions of Li, Na, K with Excess Oxygen
- **Li** forms the **oxide:**
4Li(s) + O₂(g) → 2Li₂O(s)
- **Na** forms the **peroxide:**
2Na(s) + O₂(g) → Na₂O₂(s)
- **K** forms the **superoxide:**
K(s) + O₂(g) → KO₂(s)
### Q6(d) — Extraction of Aluminium (Hall-Héroult Process)
1. **Ore:** Bauxite (Al₂O₃·2H₂O) is purified by the Bayer process to give pure alumina (Al₂O₃).
2. **Electrolyte:** Al₂O₃ is dissolved in molten cryolite (Na₃AlF₆) to lower the melting point from ~2000°C to ~950°C.
3. **Electrolysis:**
- At cathode (reduction): Al³⁺ + 3e⁻ → Al(l)
- At anode (oxidation): 2O²⁻ → O₂(g) + 4e⁻
The carbon anodes are gradually consumed as oxygen reacts with them:
C(s) + O₂(g) → CO₂(g)
Molten aluminium sinks to the bottom and is tapped off.
---
## CHM 004
### Q7(a)
(i) **Hybridization** is the mixing of atomic orbitals of similar energy to form new hybrid orbitals of equal energy and shape for bonding.
(ii) For the structure (phenyl–CH₂–C≡CH):
- **a** (benzene ring carbons): **sp²**
- **b** (benzene ring carbons): **sp²**
- **c** (–CH₂– carbon): **sp³**
- **d** (alkyne carbon ≡): **sp**
- **e** (terminal alkyne carbon): **sp**
### Q7(b) — Reactions of Propanone (CH₃COCH₃)
| Letter | Reagent | Product | Formula |
|--------|---------|---------|---------|
| A | ZnConc.HCl (Clemmensen) | Propane | CH₃CH₂CH₃ |
| B | CH₃COCH₃/NaOH (Aldol) | Diacetone alcohol | CH₃COCH₂C(CH₃)₂OH |
| C | LiAlH₄ | Propan-2-ol | CH₃CH(OH)CH₃ |
| D | CH₃MgBr (Grignard) | 2-methylbutan-2-ol | CH₃C(OH)(CH₃)CH₂CH₃ |
| E | CH₃NH₂ | N-methylpropan-2-imine (Schiff base) | CH₃N=C(CH₃)₂ |
### Q7(c) — Structures of Compounds
**(i) 2-bromo-3-chloro-4,4-dimethylpentanal:**
```
O
‖
H — C — CH(Br) — CH(Cl) — C(CH₃)₂ — CH₃
1 2 3 4 5
```
Full: OHC–CHBr–CHCl–C(CH₃)₂–CH₃
**(ii) Butane-1,2,3-triol:**
```
HO–CH₂–CH(OH)–CH(OH)–CH₃
1 2 3 4
```
**(iii) 2,4,6-tribromophenol:**
```
OH
|
Br–C–Br
/ \
C C
| |
Br–C C
\ /
C
```
Phenol with Br at positions 2, 4, and 6 relative to OH.
---
## CHM 004 / Question 8
### Q8(a)
(i) **Biotechnology** is the use of biological systems, living organisms, or their derivatives to develop or create products and processes for specific use.
(ii) Advantages:
1. Production of medicines and vaccines (e.g., insulin, antibiotics)
2. Improvement of crop yields through genetic modification
### Q8(b) — Electronic Effects
(i) **Inductive effect:** The permanent displacement of electrons along a chain of atoms through sigma (σ) bonds due to differences in electronegativity. Electron-withdrawing groups show −I effect; electron-donating groups show +I effect.
(ii) **Electromeric effect:** A temporary, reversible displacement of π electrons of a multiple bond towards one atom under the influence of an attacking reagent. It operates only in the presence of a reagent.
(iii) **Homolytic cleavage (homolysis):** The breaking of a covalent bond in which each atom retains one electron from the bonding pair, forming free radicals. e.g., A:B → A• + B•
### Q8(c) — Classification of Carbohydrates
| Carbohydrate | Classification |
|--------------|---------------|
| Cellulose | Polysaccharide |
| Fructose | Monosaccharide |
| Glucose | Monosaccharide |
| Maltose | Disaccharide |
| Sucrose | Disaccharide |
### Q8(d) — Isomeric Alcohols with Formula C₄H₁₀O
**(i) Structures (4 alcohol isomers):**
1. **Butan-1-ol:** CH₃CH₂CH₂CH₂OH
2. **Butan-2-ol:** CH₃CH₂CH(OH)CH₃
3. **2-Methylpropan-1-ol:** (CH₃)₂CHCH₂OH
4. **2-Methylpropan-2-ol:** (CH₃)₃COH
**(ii) Classification:**
| Compound | Type |
|----------|------|
| Butan-1-ol | Primary (1°) |
| Butan-2-ol | Secondary (2°) |
| 2-Methylpropan-1-ol | Primary (1°) |
| 2-Methylpropan-2-ol | Tertiary (3°) |
**(iii) Increasing Boiling Points:**
Tertiary < Secondary < Primary (branched) < Primary (straight-chain)
**2-Methylpropan-2-ol (82°C) < Butan-2-ol (99°C) < 2-Methylpropan-1-ol (108°C) < Butan-1-ol (118°C)**
---
### Compound K & L (CHM 001 Section d)
**Clues:**
- K = C₇H₁₆O (alcohol), oxidized by CrO₃ to give L = C₇H₁₄O (ketone, since it gives 2,4-DNP but NOT Fehling's red ppt → not an aldehyde)
- L gives **iodoform test** → L contains CH₃CO– group (methyl ketone)
Therefore **L = methyl ketone**: CH₃–CO–C₅H₁₁
**L = heptan-2-one:** CH₃COCH₂CH₂CH₂CH₂CH₃ (or a branched isomer)
**K = heptan-2-ol:** CH₃CH(OH)CH₂CH₂CH₂CH₂CH₃
This fits: secondary alcohol K → oxidized → methyl ketone L → gives iodoform ✓, gives 2,4-DNP ✓, no Fehling's reaction ✓.