Lesson 6: Haloalkane Reactions | Portfolio Course

Lesson at a Glance | He Tirohanga Whakamua

Do NowClassify 1-chloropropane, 2-chloropropane, 2-chloro-2-methylpropane10 min

Reagent DistinctionKOH(aq) substitution vs KOH(alc) elimination15 min

Saytzeff's Rule"The poor get poorer" Major alkene product15 min

Portfolio EntryWrite Section 6: Haloalkanes Substitution vs Elimination10 min

Exit DrillPredict major product for 2-chlorobutane

+ K O H (a l c)

5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

How to classify haloalkanes as Primary ( $1^{\circ}$ ), Secondary ( $2^{\circ}$ ), or Tertiary ( $3^{\circ}$ ) based on carbon attachments.
How solvent conditions switch reaction pathways:
• $KOH (a q)$ aqueous $\to$ Substitution forming an Alcohol.
• $KOH (a l c)$ alcoholic $\to$ Elimination forming an Alkene.
• ${NH}_{3} (a l c)$ alcoholic $\to$ 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 $KOH$ dissolved in water, you get propan-1-ol. What happens if you dissolve the exact same $KOH$ in ethanol solvent instead of water?"

Unpack: Alcoholic solvent ( $KOH (a l c)$ ) acts as a strong base, favouring Elimination of $H C l$ to form propene ( $C = C$ ). Aqueous solvent ( $KOH (a q)$ ) acts as a nucleophile, favouring Substitution of $- C l$ for $- O H$ to form propan-1-ol!

Saytzeff's Elimination Rule (15 min)

Reaction: 2-chlorobutane ( ${CH}_{3} CH (Cl) {CH}_{2} {CH}_{3}$ ) + $KOH (a l c)$

🌟 Major Product: But-2-ene

${CH}_{3} CH = {CHCH}_{3}$
Hydrogen is removed from C3 (which has 2 H's instead of C1 which has 3 H's). "The poor get poorer!"

⚠️ Minor Product: But-1-ene Both alkenes form. The proportion shifts with temperature, base and solvent, so justify which is major from the stability of the double bond rather than quoting a split.

${CH}_{2} = {CHCH}_{2} {CH}_{3}$
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.

KOH(aq), warm: substitution

In water, hydroxide behaves as a nucleophile and replaces the halogen. The product is an alcohol. Written $K O H (a q)$ or "aqueous potassium hydroxide" — the state symbol is doing real work here.

KOH in ethanol, hot: elimination

In alcoholic solution, hydroxide behaves as a base and removes a hydrogen from the neighbouring carbon. A $C = C$ forms and the product is an alkene. Written $K O H (a l c)$ 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.

🛑 Before any of this is run as a practical

Potassium hydroxide is corrosive to skin and eyes in both the aqueous and the ethanolic solution. Heating KOH(alc) combines a corrosive base with a flammable solvent, which is why the elimination pathway is normally shown as a demonstration. Ethanolic ammonia is corrosive and its vapour is an irritant. The substitution-versus-elimination reasoning below needs no wet chemistry at all.

📁 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 ( $1^{\circ}, 2^{\circ}, 3^{\circ}$ ) and map reactions with $KOH (a q)$ , $KOH (a l c)$ , and ${NH}_{3} (a l c)$ .

2. Saytzeff Application: Draw Major and Minor alkene products for elimination of 2-bromo-2-methylbutane $+ K O H (a l c)$ .

3. Excellence Reagent Rationale: 1-paragraph explanation of why aqueous conditions produce substitution while alcoholic conditions produce elimination.

Answers & marking notes (kaiako) — screen only, does not print

Matrix: KOH(aq) → substitution → alcohol · KOH(alc) → elimination → alkene · NH₃(alc) → substitution → amine. Tertiary haloalkanes eliminate most readily; primary favour substitution.
Saytzeff: 2-bromo-2-methylbutane + KOH(alc) → major 2-methylbut-2-ene (more substituted C=C), minor 2-methylbut-1-ene.
Excellence: the SOLVENT is the decision: water supplies the substitution pathway, alcohol the elimination — same reagent formula, different product class, so 'KOH' alone is not an answer.

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 ( $1^{\circ}$ ), secondary ( $2^{\circ}$ ), tertiary ( $3^{\circ}$ ) haloalkane, substitution, elimination, aqueous KOH, alcoholic KOH, alcoholic $N H_{3}$ , Saytzeff's Rule, major product, minor product.

Other teaching approach — the video-and-practical unit: Lesson 7: Haloalkanes & Reaction Schemes →

Lesson 6: Haloalkanes: Substitution & Elimination Reactions