Grade 12 — Term 3
4 study notes · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 4 lessons (NIE Units 7, 8, 9, 10). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
Lessons in This Term
Basic Concepts of Organic Chemistry
What you must know
- IUPAC principal group priority hierarchy: -COOH > -COOR > -CONH₂ > -CHO > >C=O > -OH > -NH₂ > C=C > C≡C > halides, -NO₂, alkyls.
- What does an equimolar mixture of two enantiomers do to plane-polarized light?: Enantiomers rotate plane-polarized light in equal and opposite directions; an equimolar mixture is an optically inactive racemic mixture.
- What does homolytic fission produce, and what does heterolytic fission produce?: Homolytic fission generates free radicals; heterolytic fission generates ions (electrophiles and nucleophiles).
Try one question
Which of the following organic compounds is capable of displaying both geometrical (cis-trans) isomerism and optical isomerism (enantiomerism)?
- A. 3-chloropent-2-ene
- B. 1-chlorobut-2-ene
- C. 2-chlorobut-2-ene
- D. 2-chloropent-3-ene
Show the answer
Answer: D. 2-chloropent-3-ene. 2-chloropent-3-ene has the structure CH₃-CH(Cl)-CH=CH-CH₃. Around the C3=C4 double bond, each sp² carbon is bonded to two different groups (C3 is bonded to -H and -CH(Cl)CH₃; C4 is bonded to -H and -CH₃), satisfying the criteria for geometrical (cis-trans) isomerism. Furthermore, C2 is bonded to four distinct groups (-H, -Cl, -CH₃, and -CH=CH-CH₃), making it an asymmetric/chiral carbon center that exhibits optical activity.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Hydrocarbons and Halohydrocarbons
What you must know
- What intermediate does electrophilic substitution on benzene pass through?: Benzene electrophilic substitution mechanism proceeds via a non-aromatic arenium ion (sigma complex) intermediate before losing a proton to restore aromaticity.
- Reagent distinction: aqueous KOH promotes nucleophilic substitution to alcohols, whereas hot ethanolic KOH promotes elimination to alkenes.
- Unsaturation tests: alkenes decolorize bromine water (Br₂/H₂O or Br₂/CCl₄) without light and turn cold alkaline KMnO₄ brown (MnO₂ precipitate).
Try one question
In the electrophilic aromatic substitution reaction between methylbenzene (toluene) and a nitration mixture (conc. HNO₃ + conc. H₂SO₄) at 50 °C, what is the attacking electrophile and the directing effect of the methyl group?
- A. Electrophile is NO₂⁻; methyl group is deactivating and meta-directing
- B. Electrophile is NO₃⁻; methyl group is activating and meta-directing
- C. Electrophile is NO₂⁺; methyl group is activating and ortho/para-directing
- D. Electrophile is HNO₂; methyl group is deactivating and ortho/para-directing
Show the answer
Answer: C. Electrophile is NO₂⁺; methyl group is activating and ortho/para-directing. Concentrated H₂SO₄ protonates concentrated HNO₃ to generate the reactive nitronium ion electrophile: HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻. The methyl (-CH₃) substituent is electron-donating via inductive (+I) and hyperconjugation effects, which selectively stabilizes carbocation intermediates at the ortho and para positions, directing substitution to ortho-nitrotoluene and para-nitrotoluene.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Oxygen-Containing Organic Compounds
What you must know
- Lucas test distinguishes alcohols: 3° gives immediate turbidity, 2° gives turbidity within 5 minutes, 1° remains clear at room temperature.
- Which two reagents tell an aldehyde from a ketone, and what does each show?: Tollens' reagent (ammoniacal silver nitrate) and Fehling's solution are reduced by aldehydes (silver mirror and red Cu₂O precipitate) but not ketones.
- What does the iodoform test detect, and what does a positive result look like?: Iodoform test (I₂/NaOH) yields a pale yellow CHI₃ precipitate with compounds containing CH₃-CH(OH)- or CH₃-C=O groups.
Try one question
An unknown neutral organic compound X (C₃H₆O) forms an orange precipitate with 2,4-dinitrophenylhydrazine (2,4-DNP) and produces a shiny silver mirror with Tollens' reagent ([Ag(NH₃)₂]⁺). Identify compound X and the organic product formed when X is reduced by sodium borohydride (NaBH₄).
- A. Compound X is propanal; reduction product is propan-1-ol
- B. Compound X is prop-2-en-1-ol; reduction product is propan-1-ol
- C. Compound X is propanone; reduction product is propan-2-ol
- D. Compound X is methyl ethanoate; reduction product is methanol
Show the answer
Answer: A. Compound X is propanal; reduction product is propan-1-ol. Reaction with 2,4-DNP indicates a carbonyl compound (aldehyde or ketone). Reaction with Tollens' reagent confirms that X is an aldehyde (ketones like propanone cannot be oxidized by Tollens' reagent). For C₃H₆O, the only aldehyde is propanal (CH₃CH₂CHO). Reduction of propanal with NaBH₄ converts the -CHO group to a primary alcohol, yielding propan-1-ol (CH₃CH₂CH₂OH).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Nitrogen-Containing Organic Compounds
What you must know
- Nitrous acid diagnostic test: Primary aliphatic amines react at 0–5 °C to evolve N₂ gas bubbles vigorously; Primary aromatic amines form stable arenediazonium salts.
- What does Hofmann bromamide degradation do to a primary amide?: Hofmann bromamide degradation converts a primary amide into a primary amine with one less carbon atom: R-CONH₂ + Br₂ + 4KOH → R-NH₂ + K₂CO₃ + 2KBr + 2H₂O.
- Carbylamine test: Heating primary amines with CHCl₃ and alcoholic KOH produces foul-smelling isocyanides (carbylamines).
Try one question
Arrange the following nitrogen-containing bases in order of decreasing basicity in dilute aqueous solution: diethylamine ((C₂H₅)₂NH), ethylamine (C₂H₅NH₂), ammonia (NH₃), and phenylamine (aniline, C₆H₅NH₂).
- A. Ethylamine > Diethylamine > Ammonia > Phenylamine
- B. Phenylamine > Ammonia > Ethylamine > Diethylamine
- C. Diethylamine > Ammonia > Ethylamine > Phenylamine
- D. Diethylamine > Ethylamine > Ammonia > Phenylamine
Show the answer
Answer: D. Diethylamine > Ethylamine > Ammonia > Phenylamine. In aqueous solution, aliphatic secondary amines are more basic than primary amines because the +I inductive effect of two alkyl groups increases electron density on nitrogen, while steric hindrance and hydration energies remain favorable: (C₂H₅)₂NH > C₂H₅NH₂ > NH₃. In phenylamine (C₆H₅NH₂), the lone pair on nitrogen is delocalized into the aromatic benzene ring through resonance, making it drastically less available for protonation, so it is the weakest base.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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