Complete Study Notes for NZQA Level 2 Chemistry (91165) - Organic Compounds

1. NOMENCLATURE (IUPAC Naming & Drawing)

Key Functional Groups You Must Know

• Alkanes: Single C-C bonds only (suffix -ane)

• Alkenes: Contains C=C double bond (suffix -ene)

• Alcohols: Contains -OH group (suffix -ol)

• Haloalkanes: Contains F, Cl, Br, or I (prefix fluoro-, chloro-, bromo-, iodo-)

• Amines: Contains -NH₂ group (suffix -amine)

• Carboxylic Acids: Contains -COOH group (suffix -oic acid)

Rules for Naming

1. Find the longest continuous carbon chain (parent chain)

2. Number the chain from the end nearest the highest priority functional group

3. For alkenes, the double bond gets the lowest possible number

4. For alcohols, the -OH carbon gets the lowest possible number

5. For carboxylic acids, the -COOH carbon is always carbon 1

6. List substituents alphabetically with their position numbers

7. Use prefixes: di-, tri- for multiple identical groups

Example Names & Structures

• Ethanamine: CH₃-CH₂-NH₂

• 3,3-dimethylbutanoic acid: HOOC-CH₂-C(CH₃)₂-CH₃

• 1-pentyne: CH≡C-CH₂-CH₂-CH₃

• Propan-1-ol: CH₃-CH₂-CH₂-OH

• 2-chlorobutane: CH₃-CH(Cl)-CH₂-CH₃

2. CLASSIFICATION (Primary, Secondary, Tertiary)

For Alcohols (where -OH is attached)

• Primary (1°): The carbon bearing -OH is attached to ONE other carbon

  • Example: CH₃-CH₂-OH (ethanol)

• Secondary (2°): The carbon bearing -OH is attached to TWO other carbons

  • Example: CH₃-CH(OH)-CH₃ (propan-2-ol)

• Tertiary (3°): The carbon bearing -OH is attached to THREE other carbons

  • Example: (CH₃)₃-C-OH (methylpropan-2-ol)

For Haloalkanes (where halogen is attached)

• Primary (1°): The carbon bearing halogen is attached to ONE other carbon

• Secondary (2°): The carbon bearing halogen is attached to TWO other carbons

• Tertiary (3°): The carbon bearing halogen is attached to THREE other carbons

3. KEY REACTIONS (This is the most important section)

Reaction 1: Halogenoalkane + KOH (aqueous) → Alcohol

• Type: Nucleophilic substitution

• Reagent: KOH(aq) or NaOH(aq)

• Conditions: Heat

• Product: Alcohol (substitutes halogen for -OH)

• Example: CH₃-CH₂-Br + KOH(aq) → CH₃-CH₂-OH + KBr

Reaction 2: Halogenoalkane + KOH (alcoholic) → Alkene

• Type: Elimination

• Reagent: KOH(alc) or NaOH(alc) - dissolved in ethanol

• Conditions: Heat

• Product: Alkene (removes H and halogen to form C=C)

• Example: CH₃-CH₂-Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O

• Major vs Minor: Follows Zaitsev's rule (more substituted alkene is major)

Reaction 3: Alkene + H₂O / H⁺ → Alcohol

• Type: Hydration (addition reaction)

• Reagent: Water with acid catalyst (H₂SO₄ or H₃PO₄)

• Conditions: Heat

• Product: Alcohol (adds H and OH across double bond)

• Major vs Minor: Follows Markovnikov's rule (H adds to carbon with more H's, OH adds to carbon with fewer H's)

Reaction 4: Alkene + H₂ (catalyst) → Alkane

• Type: Hydrogenation (addition reaction)

• Reagent: H₂ gas with metal catalyst (Ni, Pd, or Pt)

• Conditions: Room temperature or heat

• Product: Alkane (adds H₂ across double bond)

Reaction 5: Alkene + Halogen (Cl₂ or Br₂) → Dihaloalkane

• Type: Halogenation (addition reaction)

• Reagent: Cl₂ or Br₂ (pure or in solution)

• Conditions: Room temperature, no heat needed (at room temp)

• Product: Dihaloalkane (adds halogen atoms across double bond)

• Observation: Orange/brown bromine water decolourises

Reaction 6: Alkene + HX (HCl or HBr) → Haloalkane

• Type: Hydrohalogenation (addition reaction)

• Reagent: HCl or HBr gas or solution

• Conditions: Room temperature (no heat needed)

• Product: Haloalkane (adds H and halogen across double bond)

• Major vs Minor: Follows Markovnikov's rule (H adds to carbon with more H's, halogen adds to carbon with fewer H's)

Reaction 7: Primary Alcohol → Aldehyde → Carboxylic Acid

• Step 1: Primary alcohol + Cr₂O₇²⁻/H⁺ → Aldehyde

• Reagent: Acidified potassium dichromate (Cr₂O₇²⁻/H⁺)

• Conditions: Distillation (to remove aldehyde before it oxidises further)

• Observation: Orange Cr₂O₇²⁻ turns green (Cr³⁺)

• Step 2: Aldehyde + Cr₂O₇²⁻/H⁺ → Carboxylic acid

• Conditions: Reflux (heat with condenser)

• Complete Oxidation: Primary alcohol + excess Cr₂O₇²⁻/H⁺ under reflux → Carboxylic acid

Reaction 8: Secondary Alcohol → Ketone

• Type: Oxidation

• Reagent: Acidified potassium dichromate (Cr₂O₇²⁻/H⁺) OR acidified KMnO₄

• Conditions: Reflux

• Product: Ketone

• Observation: Orange Cr₂O₇²⁻ turns green (Cr³⁺)

• Note: Secondary alcohols can only oxidise to ketones (cannot go further)

Reaction 9: Tertiary Alcohol

• Type: No reaction (resists oxidation)

• Reagent: Acidified potassium dichromate or acidified KMnO₄

• Observation: No colour change (solution remains orange/purple)

Reaction 10: Alcohol → Alkene (Elimination)

• Type: Dehydration (elimination)

• Reagent: Concentrated H₂SO₄

• Conditions: Heat

• Product: Alkene (removes H and OH to form C=C)

• Major vs Minor: Follows Zaitsev's rule (more substituted alkene is major)

4. GEOMETRIC (CIS-TRANS) ISOMERISM

Requirements for Geometric Isomerism

1. Must have a C=C double bond (prevents rotation)

2. Each carbon in the double bond must have TWO DIFFERENT groups attached

Checking if a compound can form geometric isomers

• Compound A: CH₂=CH-CH₂-CH₃ (but-1-ene)

  • Carbon 1 has two H atoms (same) ❌ Cannot form geometric isomers

• Compound B: CH₃-CH=CH-CH₂-CH₃ (pent-2-ene)

  • Carbon 2 has CH₃ and H (different) ✅

  • Carbon 3 has CH₂CH₃ and H (different) ✅

  • Can form geometric isomers

Drawing Cis and Trans Isomers

• Cis: Similar groups on the SAME side of the double bond

• Trans: Similar groups on OPPOSITE sides of the double bond

Example: pent-2-ene (CH₃-CH=CH-CH₂-CH₃)

• Cis: Both CH₃ groups on same side of double bond

• Trans: CH₃ groups on opposite sides of double bond

5. ADDITION POLYMERISATION

What is Addition Polymerisation?

• Many alkene monomers join together to form a long chain polymer

• The C=C double bond opens up to form single bonds

• No other product is formed (unlike condensation polymerisation)

Monomer Requirements

• Must contain a C=C double bond

• The pi bond breaks, allowing the molecule to join to others

Monomer vs Polymer Reactivity

• Monomer is MORE reactive because:

  • Contains reactive C=C double bond (pi bond is weaker and more reactive)

  • Pi bond electrons are exposed and available for reaction

• Polymer is LESS reactive (inert) because:

  • Only contains strong C-C and C-H single bonds (sigma bonds)

  • No reactive functional groups remain

  • This makes polymers suitable for uses where stability is needed (e.g., face masks, non-stick coatings)

Drawing Polymer Chains

• Break the double bond in the monomer

• Draw repeat units in brackets with "n" subscript

• Show three repeating units side by side

• Use brackets to show repeating structure

Example: Styrene → Polystyrene

• Monomer: CH₂=CH-C₆H₅ (styrene)

• Polymer: [-CH₂-CH(C₆H₅)-]n

Example: Propene → Polypropene

• Monomer: CH₂=CH-CH₃ (prop-1-ene)

• Polymer: [-CH₂-CH(CH₃)-]n

Example: Tetrafluoroethene → Teflon

• Monomer: CF₂=CF₂

• Polymer: [-CF₂-CF₂-]n

6. IDENTIFICATION TESTS (Chemical Properties)

Test 1: Acidified Potassium Dichromate (Cr₂O₇²⁻/H⁺)

• Reagent colour: Orange

• Positive test: Colour change to GREEN (Cr³⁺)

• What reacts:

  • Primary alcohols → Aldehyde → Carboxylic acid

  • Secondary alcohols → Ketone

  • Alkenes (also decolourises but via different mechanism)

• What doesn't react:

  • Tertiary alcohols (no colour change)

  • Alkanes (no colour change)

  • Haloalkanes (no colour change)

Test 2: Acidified Potassium Permanganate (MnO₄⁻/H⁺)

• Reagent colour: Purple

• Positive test: Colour change to COLOURLESS (Mn²⁺)

• What reacts:

  • Alkenes (decolourises quickly)

  • Primary alcohols (decolourises)

  • Secondary alcohols (decolourises)

• What doesn't react:

  • Tertiary alcohols (no colour change)

  • Alkanes (no colour change)

Test 3: Bromine Water (Br₂(aq))

• Reagent colour: Orange/brown

• Positive test: Colour change to COLOURLESS

• What reacts:

  • Alkenes only (addition reaction)

• What doesn't react:

  • Alkanes (no colour change)

  • Alcohols (no colour change)

  • Haloalkanes (no colour change)

Test 4: Litmus Paper

• Blue litmus:

  • Turns RED with carboxylic acids (acidic)

  • Stays BLUE with neutral compounds (alkenes, alcohols, haloalkanes)

• Red litmus:

  • Turns BLUE with amines (basic)

  • Stays RED with neutral compounds

Test 5: Sodium Carbonate (Na₂CO₃)

• What reacts: Carboxylic acids only

• Observation: Effervescence (bubbling) - CO₂ gas produced

• Equation: R-COOH + Na₂CO₃ → R-COO⁻Na⁺ + H₂O + CO₂

7. MAJOR vs MINOR PRODUCTS

Markovnikov's Rule (for addition to alkenes)

• When adding H-X (or H-OH) to an alkene:

  • H atom adds to the carbon with MORE hydrogen atoms

  • X atom (or OH) adds to the carbon with FEWER hydrogen atoms

• This forms the more stable carbocation intermediate

Example: Propene + HBr

• CH₂=CH-CH₃ + HBr

• Major product (2-bromopropane): H adds to C₁ (has 2 H's), Br adds to C₂ (has 1 H)

  • CH₃-CH(Br)-CH₃

• Minor product (1-bromopropane): H adds to C₂, Br adds to C₁

  • CH₂(Br)-CH₂-CH₃

Zaitsev's Rule (for elimination to form alkenes)

• When forming an alkene from an alcohol or haloalkane:

  • Major product is the more substituted alkene

  • More substituted = more alkyl groups attached to the C=C carbons

  • More substituted alkenes are more stable

Example: 2-chlorobutane + KOH(alc) → butenes

• CH₃-CH(Cl)-CH₂-CH₃

• Major product (but-2-ene): Elimination from carbon 2 gives CH₃-CH=CH-CH₃ (more substituted)

• Minor product (but-1-ene): Elimination from carbon 1 gives CH₂=CH-CH₂-CH₃ (less substituted)

8. PHYSICAL PROPERTIES (Boiling Point & Solubility)

Boiling Point Trends

• Hydrogen bonding increases boiling point:

  • Alcohols have hydrogen bonding → highest boiling points

  • Amines have hydrogen bonding (but weaker than alcohols) → moderate

  • Alkanes, haloalkanes, alkenes have only dispersion forces → lowest boiling points

• Increases with molecular size: Larger molecules = higher boiling points

• Branching lowers boiling point: More branched = lower boiling point (less surface area for intermolecular forces)

Solubility in Water

• Hydrogen bonding with water increases solubility:

  • Alcohols and amines can hydrogen bond with water

  • Small alcohols (≤ 4 carbons) are soluble

  • Longer chains (> 4 carbons) become less soluble (non-polar hydrocarbon part dominates)

• Alkanes, haloalkanes, alkenes:

  • Are non-polar (or weakly polar)

  • Do NOT dissolve in water

Distinguishing Liquids by Physical Properties

Example: Distinguish methanol (CH₃OH), ethanol (C₂H₅OH), and hexane (C₆H₁₄)

1. Add to water:

   • Methanol and ethanol: Mix with water (soluble) → homogeneous solution

   • Hexane: Forms two layers (insoluble) → separate layer

2. Distinguish methanol and ethanol by boiling point:

   • Ethanol has higher boiling point (larger molecule with stronger dispersion forces)

   • Methanol has lower boiling point

9. ISOMERISM TYPES

Constitutional (Structural) Isomers

• Same molecular formula, DIFFERENT connectivity

• Different arrangement of atoms

• Different names, different structures

Example: C₄H₁₀O isomers

1. Butan-1-ol: CH₃-CH₂-CH₂-CH₂-OH

2. Butan-2-ol: CH₃-CH(OH)-CH₂-CH₃

3. 2-methylpropan-1-ol: HO-CH₂-CH(CH₃)-CH₃

4. 2-methylpropan-2-ol: (CH₃)₃-C-OH

5. Ethoxyethane: CH₃-CH₂-O-CH₂-CH₃ (ether - but usually focus on alcohols for NCEA)

Drawing All Isomers for a Given Formula

Example: C₅H₁₁Cl (chloropentane isomers)

1. 1-chloropentane: CH₂(Cl)-CH₂-CH₂-CH₂-CH₃

2. 2-chloropentane: CH₃-CH(Cl)-CH₂-CH₂-CH₃

3. 3-chloropentane: CH₃-CH₂-CH(Cl)-CH₂-CH₃

4. 1-chloro-2-methylbutane: CH₂(Cl)-CH(CH₃)-CH₂-CH₃

5. 2-chloro-2-methylbutane: CH₃-C(Cl)(CH₃)-CH₂-CH₃

6. 2-chloro-3-methylbutane: CH₃-CH(Cl)-CH(CH₃)-CH₃

10. REACTION SCHEMES (Multi-step Conversions)

Common Conversion Sequences

Sequence 1: Alkane → Haloalkane → Alkene → Alcohol → ...

• Step 1: Alkane + Cl₂ (UV light) → Haloalkane (substitution - NOT tested much)

• Step 2: Haloalkane + KOH(alc) → Alkene (elimination)

• Step 3: Alkene + H₂O/H⁺ → Alcohol (hydration)

Sequence 2: Alkene → Alcohol → Aldehyde → Carboxylic Acid

• Step 1: Alkene + H₂O/H⁺ → Alcohol (hydration)

• Step 2: Alcohol + Cr₂O₇²⁻/H⁺ (distillation) → Aldehyde (oxidation)

• Step 3: Aldehyde + Cr₂O₇²⁻/H⁺ (reflux) → Carboxylic acid (oxidation)

Sequence 3: Alkene → Alcohol → Alkene (different position)

• Step 1: Alkene + H₂O/H⁺ → Alcohol (hydration)

• Step 2: Alcohol + conc. H₂SO₄/heat → Alkene (elimination)

• Note: The double bond may shift to form more substituted alkene (major product)

Sequence 4: Haloalkane → Amine

• Haloalkane + NH₃(alc) → Amine (substitution)

Example Pathway: 2-bromobutane → butan-2-one

1. 2-bromobutane + KOH(aq) → butan-2-ol (substitution)

2. Butan-2-ol + Cr₂O₇²⁻/H⁺ → butan-2-one (oxidation of secondary alcohol)

11. TIPS FOR EXCELLENCE QUESTIONS

To get Excellence, you need to:

1. Explain WHY not just describe WHAT

2. Justify your choice of major/minor products using rules (Markovnikov or Zaitsev)

3. Compare and contrast reactions (e.g., why one alkene forms geometric isomers but another doesn't)

4. Link structure to reactivity (e.g., why monomer is more reactive than polymer)

5. Link properties to uses (e.g., why Teflon is used for non-stick cookware)

Common "Why" Questions:

• Why does compound X form geometric isomers but compound Y doesn't?

  • Answer: Need C=C with different groups on each carbon. Check each alkene carbon.

• Why does compound X form two products but compound Y forms one?

  • Answer: Elimination reactions can form different alkenes depending on which H is removed. If the alkene formed is symmetrical, only one product. If not, two products possible.

• Why is the monomer more reactive than the polymer?

  • Answer: Monomer has reactive C=C pi bond; polymer only has strong C-C sigma bonds.

• Why is one product major and another minor?

  • Answer: Markovnikov's rule (more stable carbocation) OR Zaitsev's rule (more substituted alkene is more stable).

12. SAMPLE EXAM-STYLE QUESTIONS

Question: Distinguish between these compounds using Cr₂O₇²⁻/H⁺

• CH₃-CH₂-CH₂OH (propan-1-ol, primary alcohol)

  • Reaction occurs: Orange → Green

  • Product: Propanoic acid (if refluxed) or propanal (if distilled)

• (CH₃)₃-COH (methylpropan-2-ol, tertiary alcohol)

  • No reaction: Remains orange

Question: Geometric isomers for CH₃-CH=CH-CH₂-CH₃

• Yes, can form geometric isomers

• Cis: Both CH₃ groups on same side of double bond

• Trans: CH₃ groups on opposite sides

• Justification: Each carbon in C=C has two different groups attached

Question: Products of 2-chlorobutane + KOH(alc)

• Major: But-2-ene (CH₃-CH=CH-CH₃) - more substituted

• Minor: But-1-ene (CH₂=CH-CH₂-CH₃) - less substituted

• Justification: Zaitsev's rule - more substituted alkene is more stable and forms in greater amount

Question: Why doesn't 1-chlorobutane + KOH(alc) form two products?

• 1-chlorobutane only has one type of β-hydrogen (H on carbon 2)

• Elimination can only occur in one direction

• Only but-1-ene is formed

13. QUICK REFERENCE - REACTION SUMMARY TABLE

14. FINAL EXAM REMINDERS

1. Check your structures: Count all carbons, hydrogens, and functional groups

2. Know your reagent conditions: KOH(aq) vs KOH(alc) makes a BIG difference

3. Draw clearly: NZQA markers need to see your structures clearly

4. Use correct terminology: Primary/secondary/tertiary, addition/elimination/substitution/oxidation

5. Justify your answers: For Excellence, always explain WHY

6. Check for geometric isomers: Always check if each carbon in C=C has two different groups

7. Major vs Minor: Remember Markovnikov (addition) and Zaitsev (elimination) rules

8. Physical properties: Link to intermolecular forces (hydrogen bonding, dispersion forces)

9. Observations: Always state colour changes for tests (orange→green, purple→colourless, orange/brown→colourless)

10. Practice reaction schemes: These combine multiple concepts and are worth many marks