Lesson at a Glance | He Tirohanga Whakamua
Ngā Whāinga Ako | Learning Intentions
Students will know
- How to classify haloalkanes as Primary (), Secondary (), or Tertiary () based on carbon attachments.
- How solvent conditions switch reaction pathways:
• aqueous Substitution forming an Alcohol.
• alcoholic Elimination forming an Alkene.
• alcoholic Substitution forming an Amine. - Saytzeff's Rule: In elimination reactions, Hydrogen is preferentially removed from the adjacent carbon atom with FEWER hydrogens attached ("the poor get poorer"), forming the Major Alkene Product.
Students will demonstrate
- By predicting major and minor alkene products for 2-chlorobutane elimination.
- By completing Section 6 of their Level 2 Organic Chemistry Mastery Portfolio.
Curriculum alignment
- NZC (2007) · Science · Level 7 · Material World: “Investigate and measure the chemical and physical properties of a range of groups of substances, for example, acids and bases, oxidants and reductants, and selected organic and inorganic compounds.”
Do Now | Tīmatanga Whakaaro (10 min)
Solvent Switch Prompt:
"If you react 1-chloropropane with dissolved in water, you get propan-1-ol. What happens if you dissolve the exact same in ethanol solvent instead of water?"
Unpack: Alcoholic solvent () acts as a strong base, favouring Elimination of to form propene (). Aqueous solvent () acts as a nucleophile, favouring Substitution of for to form propan-1-ol!
Saytzeff's Elimination Rule (15 min)
Reaction: 2-chlorobutane () +
Hydrogen is removed from C3 (which has 2 H's instead of C1 which has 3 H's). "The poor get poorer!"
Hydrogen is removed from C1. Forms a less substituted, less thermodynamically stable double bond.
Reagent Distinction: the Solvent Decides (15 min)
Where this usually goes wrong: students see "KOH" and stop reading. The same reagent gives two entirely different products depending on the solvent and temperature, and exam questions specify it deliberately.
In water, hydroxide behaves as a nucleophile and replaces the halogen. The product is an alcohol. Written or "aqueous potassium hydroxide" — the state symbol is doing real work here.
In alcoholic solution, hydroxide behaves as a base and removes a hydrogen from the neighbouring carbon. A forms and the product is an alkene. Written or "ethanolic". Where two alkenes are possible, Saytzeff's rule predicts the major one.
Do this now: predict BOTH products for 2-bromobutane, one for each set of conditions. Then underline the single word in each set that told you which pathway to use.
📁 Organic Chemistry Portfolio — Section 6: Haloalkane Reactions & Saytzeff
Students open their Level 2 Chemistry Portfolio and complete Section 6:
Section 6 Requirements:
1. Classification & Reaction Matrix: Classify 3 haloalkanes () and map reactions with , , and .
2. Saytzeff Application: Draw Major and Minor alkene products for elimination of 2-bromo-2-methylbutane .
3. Excellence Reagent Rationale: 1-paragraph explanation of why aqueous conditions produce substitution while alcoholic conditions produce elimination.
Exit Verification | Ka Mutu Hoki (5 min)
Exit Check:
"My Section 6 correctly identifies but-2-ene as the major elimination product of 2-chlorobutane using Saytzeff's Rule."
Teacher Planning & NCEA Alignment
NCEA Level 2 Chemistry Alignment (4 Credits External):
- Haloalkane Reactions: Predict reagents, conditions, and structural formulas for substitution and elimination reactions.
- Saytzeff's Rule: Justify major and minor alkene products in elimination reactions.
Vocabulary: Primary (), secondary (), tertiary () haloalkane, substitution, elimination, aqueous KOH, alcoholic KOH, alcoholic , Saytzeff's Rule, major product, minor product.
Other teaching approach: Guided Viewing & Practical Chemistry →