The shape of the lesson
By the end of the lesson, learners can
- Nucleophilic addition of HCN: Step through the mechanism in the simulator.
- The bench tests: Run each reagent against an aldehyde, a ketone and a methyl ketone in the simulator and record the observations.
- Identify the unknown: Pick a mystery carbonyl in the simulator, choose your tests in a sensible order and name the compound from the results.
- HCN/KCN: CN− attacks Cδ+, then O− takes H+.
- Planar C=O: attack from both faces, racemic product.
- 2,4-DNPH: orange ppt for any C=O.
- Fehling's / Tollens': aldehydes only; CHI3 for CH3CO–.
One picture every idea returns to
The C=O bond is polar, so the carbon is δ+ and is attacked by nucleophiles. With HCN and a KCN catalyst, the CN− ion attacks the carbon, the π electrons move onto oxygen, and the O− then takes H+ from HCN: a 2-hydroxynitrile forms. Because the C=O group is planar, attack from either face is equally likely, so a chiral product is a racemic mixture. 2,4-DNPH gives an orange-yellow precipitate with any aldehyde or ketone. Only aldehydes reduce Fehling's (blue to brick-red Cu2O) and Tollens' (silver mirror) and turn acidified dichromate green. Alkaline iodine gives yellow CHI3 with the CH3CO– group.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A nucleophile meets the δ+ carbon.. The polar C=O invites nucleophiles. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Nucleophilic addition of HCN | Step through the mechanism in the simulator. Project tab 3 of the lab (Carbonyl compounds lab, nucleophilic addition) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The bench tests | Run each reagent against an aldehyde, a ketone and a methyl ketone in the simulator and record the observations. Project tab 4 of the lab (Carbonyl compounds lab, the bench tests) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Identify the unknown | Pick a mystery carbonyl in the simulator, choose your tests in a sensible order and name the compound from the results. Project tab 5 of the lab (Carbonyl compounds lab, identify the unknown) and let a student drive it while the class predicts. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Identify the unknown, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| HCN adds by nucleophilic addition, not electrophilic: the attacking species is CN −, and it bonds through carbon. | The product is racemic because the C=O is planar, not because the reaction is SN1. 2,4-DNPH does not distinguish aldehydes from ketones; Fehling's and Tollens' do. Fehling's turns brick-red, Tollens' gives silver; do not swap them. Propanal gives no CHI 3: it has no CH 3 CO– group. |
Nucleophilic addition and carbonyl tests: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Describe the mechanism of the reaction between propanone and HCN in the presence of KCN. Name the product.
Butanone reacts with HCN to give a product that does not rotate plane-polarised light. Explain why, even though the product contains a chiral centre.
Describe a test that distinguishes propanal from propanone. Give the reagent, conditions and observations for each compound.
Compound R, C4H8O, gives an orange precipitate with 2,4-DNPH, does not react with Fehling's solution, and gives a yellow precipitate with alkaline iodine. Identify R, explaining each result.
Excess Tollens' reagent is warmed with 0.880 g of ethanal (Mr 44.0). Each mole of aldehyde reduces two moles of Ag+. Calculate the mass of silver deposited (Ar Ag = 107.9).
Which of ethanal, propanal, propanone and pentan-3-one give a positive tri-iodomethane test? Explain.
Answer key and marking notes
- The C=O is polar; the carbon is δ+.
- A lone pair on the carbon of CN− attacks the δ+ carbon; the π electrons of C=O move onto the oxygen, forming an alkoxide intermediate.
- The O− takes H+ from HCN, regenerating CN−.
- Product: 2-hydroxy-2-methylpropanenitrile, (CH3)2C(OH)CN.
- The product CH3C(OH)(CN)CH2CH3 has four different groups on one carbon, so it is chiral.
- The C=O carbon in butanone is trigonal planar, so CN− attacks from above or below with equal probability.
- Equal amounts of the two enantiomers form (a racemic mixture), and their rotations cancel.
- Warm each with Tollens' reagent (or Fehling's solution).
- Propanal: silver mirror (or blue to brick-red precipitate with Fehling's).
- Propanone: no change; ketones are not oxidised.
- 2,4-DNPH positive: R is an aldehyde or ketone.
- No reaction with Fehling's: R is a ketone.
- CHI3 forms: R has CH3CO–. R is butanone, CH3COCH2CH3.
- n(ethanal) = 0.880 ÷ 44.0 = 0.0200 mol, so n(Ag) = 0.0400 mol.
- Mass = 0.0400 × 107.9 = 4.32 g.
- Ethanal and propanone: both contain CH3CO–.
- Propanal (CH3CH2CO–) and pentan-3-one (two ethyl groups) have no CH3 on the C=O carbon, so are negative.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
In the reaction of ethanal with HCN in the presence of KCN, which species attacks first, and which atom does it attack?
- ACN−, which attacks the δ+ carbon of the C=O group
- BH+, which attacks the δ− oxygen of the C=O group
- CCN−, which attacks the δ− oxygen of the C=O group
- DHCN, acting as an electrophile, which attacks the C=O π bond
Propanone, CH3COCH3, reacts with HCN in the presence of KCN. What is the product?
- A2-hydroxy-2-methylpropanenitrile, (CH3)2C(OH)CN
- B2-hydroxypropanenitrile, CH3CH(OH)CN
- Cbutanenitrile, CH3CH2CH2CN
- D2-cyanopropan-2-one, CH3COCH2CN
Which observation shows that an unknown liquid is an aldehyde or a ketone, without telling you which?
- Aan orange-yellow precipitate with 2,4-dinitrophenylhydrazine (2,4-DNPH)
- Ba silver mirror when warmed with Tollens' reagent
- Ca brick-red precipitate when warmed with Fehling's solution
- Dsteamy fumes with phosphorus(V) chloride
Propanal and propanone are each warmed with Fehling's solution. What is observed?
- Apropanal: blue solution gives a brick-red precipitate; propanone: stays blue
- Bpropanal: silver mirror; propanone: stays blue
- Cboth give a brick-red precipitate, as both contain C=O
- Dpropanal: stays blue; propanone: brick-red precipitate
Which compound gives a pale yellow precipitate when warmed with alkaline aqueous iodine?
- Abutanone, CH3COCH2CH3
- Bpropanal, CH3CH2CHO
- Cpentan-3-one, CH3CH2COCH2CH3
- Dmethanal, HCHO
The diagram shows the carbonyl group of ethanal being attacked by cyanide ions. Why is the 2-hydroxypropanenitrile formed optically inactive?
- Athe C=O carbon is trigonal planar, so CN− attacks from either face with equal probability, giving a racemic mixture of the two enantiomers
- B2-hydroxypropanenitrile has no chiral carbon atom
- Cthe reaction is SN1, so a planar carbocation forms
- DCN− can attack only from one face, so a single enantiomer forms which does not rotate light
Samples A to D are pentanal, pentan-2-one, pentan-3-one and pentan-2-ol, in some order. The table shows the test results. Which compound is B?
- Apentan-2-one
- Bpentanal
- Cpentan-3-one
- Dpentan-2-ol
Excess Tollens' reagent is warmed with 2.32 g of propanal, CH3CH2CHO (Mr 58.0). Each mole of aldehyde reduces two moles of Ag+ to Ag. What mass of silver is deposited? (Ar Ag = 107.9)
- A8.63 g
- B4.32 g
- C17.3 g
- D2.32 g
Quiz key
Oxygen is more electronegative than carbon, so the C=O carbon is δ+. The cyanide ion is a nucleophile: a lone pair on its carbon forms a bond to the δ+ carbon while the π electrons move onto oxygen. The O− intermediate then takes H+ from HCN. Attack by H+ first describes an acid-catalysed or electrophilic process, which is not the accepted mechanism here. A nucleophile is repelled by the δ− oxygen, not drawn to it. HCN is a very weak acid and is not the electrophile; this is nucleophilic, not electrophilic, addition.
The CN group bonds to the carbonyl carbon and H bonds to the oxygen, so the C=O becomes C(OH)(CN): (CH3)2C(OH)CN. Named as a nitrile, the CN carbon is C1, so it is 2-hydroxy-2-methylpropanenitrile, with four carbons in all. CH3CH(OH)CN is the product from ethanal, not propanone. Butanenitrile has lost the oxygen, which addition does not do. CH3COCH2CN keeps the C=O, which would be substitution on a methyl group, not addition.
2,4-DNPH (Brady's reagent) undergoes a condensation reaction with the C=O group of any aldehyde or ketone, giving an orange or yellow precipitate; the melting point of the purified solid can identify the compound. Tollens' and Fehling's are positive only for aldehydes, because they rely on the aldehyde being oxidised, so they do distinguish the two classes. Steamy fumes with PCl5 are a test for an O-H group, not C=O.
Propanal is an aldehyde and is oxidised (to the propanoate ion in the alkaline solution), while the blue Cu2+ complex is reduced to brick-red Cu2O. Propanone is a ketone, has no H on the C=O carbon, and is not oxidised, so the solution stays blue. A silver mirror is the result with Tollens' reagent, not Fehling's. Having C=O is not enough; ketones do not react. The last option swaps the aldehyde and the ketone.
Butanone contains CH3CO-, so it gives yellow CHI3 (tri-iodomethane) and the propanoate ion. Propanal has CH3CH2CO-: the CH3 is one carbon too far away. Pentan-3-one has an ethyl group on each side of the C=O and no CH3CO-. Methanal has no carbon-carbon bond at all, so no CH3 group. Ethanal is the only aldehyde that gives a positive test.
The product CH3CH(OH)CN has four different groups (CH3, H, OH and CN) on one carbon, so it is chiral. Because the carbonyl carbon and its three groups lie in one plane, attack from above and below is equally likely; the two enantiomers form in equal amounts and their opposite rotations of plane-polarised light cancel: a racemic mixture. Claiming no chiral centre is the classic error. This is addition, not substitution, so SN1 does not apply (though the planar-intermediate reasoning is similar). A single enantiomer would rotate plane-polarised light.
B gives an orange precipitate with 2,4-DNPH (C=O present), no silver mirror (a ketone, not an aldehyde) and a yellow precipitate with alkaline iodine (CH3CO- present): pentan-2-one. A is pentanal (silver mirror, no CH3CO-). D is pentan-3-one (ketone without a methyl next to C=O). C is pentan-2-ol: no C=O so no reaction with 2,4-DNPH, but its CH3CH(OH)- group is oxidised by the alkaline iodine and gives CHI3.
n(propanal) = 2.32 / 58.0 = 0.0400 mol. Two Ag+ ions each gain one electron for every aldehyde oxidised, so n(Ag) = 0.0800 mol and mass = 0.0800 × 107.9 = 8.63 g. 4.32 g ignores the 1 : 2 ratio. 17.3 g uses a 1 : 4 ratio. 2.32 g simply repeats the mass of aldehyde, as though mass were conserved between the two substances.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.