The shape of the lesson
By the end of the lesson, learners can
- Kinds of formula: Build a molecule in the simulator and switch between its formula types.
- Skeletal: corners and ends are C; H on C omitted.
- Structural: must be unambiguous.
- Homologous series: same general formula, differ by CH2.
- Functional group decides the reactions.
One picture every idea returns to
A molecular formula gives the actual number of each atom (C4H10O); a structural formula shows the arrangement unambiguously (CH3CH2CH2CH2OH); a displayed formula shows every bond; a skeletal formula shows only the carbon skeleton and functional groups, with carbon atoms at corners and line ends and hydrogens on carbon left out. A homologous series shares a general formula and functional group, with each member differing by CH2. The functional group is the atom or group responsible for the characteristic reactions.
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, One molecule, five drawings.. Organic chemists draw the same molecule in several ways, each useful for a different job. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 30 min | Teach and check: Kinds of formula | Build a molecule in the simulator and switch between its formula types. Project tab 1 of the lab (Organic chemistry lab, formulae) and let a student drive it while the class predicts. | Mix-Pair-Share |
| 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, 14 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 Kinds of formula, 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.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
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 |
|---|---|
| In a skeletal formula hydrogens on carbon are not drawn, but they are still there: work out how many each carbon needs to make four bonds. | Hydrogens on O or N are shown (–OH, –NH 2). A structural formula must be unambiguous: C 3 H 7 OH is not acceptable because it could be propan-1-ol or propan-2-ol. |
Organic formulae and functional groups: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
For but-1-ene give (a) the molecular formula, (b) the structural formula and (c) the empirical formula.
Draw the displayed formula of ethanoic acid and name its functional group.
A skeletal formula shows a four-carbon chain with a C=O at the end carbon. Name the functional group and give the molecular formula.
State two features of a homologous series.
Classify each alcohol as primary, secondary or tertiary: (a) butan-1-ol, (b) butan-2-ol, (c) 2-methylpropan-2-ol.
Identify all the functional groups in CH2=CHCH2OH and in HOOCCH2NH2.
Answer key and marking notes
- (a) C4H8.
- (b) CH2=CHCH2CH3.
- (c) CH2.
- H3C–C(=O)–O–H with all C–H bonds shown.
- Carboxylic acid group, –COOH.
- Aldehyde (–CHO): butanal.
- C4H8O.
- Same general formula and the same functional group.
- Successive members differ by CH2; similar chemical properties and a trend in physical properties.
- (a) primary.
- (b) secondary.
- (c) tertiary.
- CH2=CHCH2OH: alkene (C=C) and primary alcohol (–OH).
- HOOCCH2NH2: carboxylic acid (–COOH) and amine (–NH2).
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.
The skeletal formula of an alcohol is shown. What is its molecular formula?
- AC6H14O
- BC6H12O
- CC5H12O
- DC6H13O
Which compound belongs to the same homologous series as ethene, C2H4?
- AC5H10
- BC5H12
- CC5H8
- DC5H11OH
What is the empirical formula of butane, C4H10?
- AC2H5
- BC4H10
- CCH3
- DCH2.5
Which functional group is present in propanone, CH3COCH3?
- Aketone, C=O bonded to two carbon atoms
- Baldehyde, C=O at the end of the chain bonded to H
- Ccarboxylic acid, COOH
- Dalcohol, C-OH
How is 2-methylpropan-2-ol, (CH3)3COH, classified?
- Atertiary alcohol
- Bprimary alcohol
- Csecondary alcohol
- Dit is not an alcohol, because the OH is on a branched carbon
Which functional group is present in methyl propanoate, CH3CH2COOCH3?
- Aester, -COO- linking two carbon groups
- Bcarboxylic acid, -COOH
- Cketone and alcohol
- Daldehyde
Which statement about propanenitrile, CH3CH2CN, is correct?
- Ait contains the nitrile group, C≡N, and has three carbon atoms
- Bit is an amine, because it contains nitrogen
- Cit is an ionic compound containing the CN− ion
- Dit contains the nitrile group and has two carbon atoms, since the CN carbon is not counted
An alcohol in the homologous series CnH2n+1OH has Mr = 88.0. What is its molecular formula? (Ar H 1.0, C 12.0, O 16.0)
- AC5H11OH
- BC4H9OH
- CC6H13OH
- DC5H9OH
Quiz key
There are six carbons: five along the zigzag chain plus one branch carbon. The two chain-end carbons and the branch carbon are CH3 (3 × 3 = 9 H), the carbon at the bottom middle is CH2 (2 H), the carbon carrying OH has 1 H and the branch-point carbon has 1 H: 9 + 2 + 1 + 1 = 13 H on carbon, plus the H of OH gives 14. So the formula is C6H14O (4-methylpentan-2-ol), which fits CnH2n+1OH. C6H12O comes from forgetting that line ends carry three H atoms, the classic skeletal counting error. C5H12O misses the branch carbon at the top. C6H13O forgets the hydrogen in the OH group.
Alkenes have the general formula CnH2n; with n = 5 this gives C5H10 (for example pent-1-ene). Members of a homologous series share a general formula and a functional group and differ by CH2. C5H12 fits CnH2n+2, so it is an alkane with no C=C. C5H8 has two fewer hydrogens than an alkene, so it would need two C=C bonds or a ring and a C=C. C5H11OH fits CnH2n+1OH, the alcohols.
C : H = 4 : 10, which simplifies by dividing by 2 to 2 : 5, so the empirical formula is C2H5. C4H10 is the molecular formula, the actual number of atoms in one molecule, not the simplest ratio. CH3 is the empirical formula of ethane (C2H6), not butane. CH2.5 divides by 4 but leaves a fraction; an empirical formula must use whole numbers.
In CH3COCH3 the C=O carbon is bonded to two CH3 groups, so it is a ketone. An aldehyde has the carbonyl carbon bonded to at least one H, written CHO at the end of a chain (as in propanal, CH3CH2CHO), which is a functional group isomer of propanone. A carboxylic acid needs both C=O and OH on the same carbon (COOH). Writing CO in a condensed formula shows C=O, not C-OH, so it is not an alcohol.
The carbon bearing the OH is bonded to three other carbon atoms (three CH3 groups), so the alcohol is tertiary. A primary alcohol has the C-OH carbon bonded to one other carbon (for example propan-1-ol); counting all four carbons in the molecule instead of the neighbours is a common slip. A secondary alcohol has two carbons on the C-OH carbon (propan-2-ol). Branching does not stop a compound being an alcohol; any OH on an sp3 carbon of a chain is an alcohol group.
COO with carbon on both sides is the ester linkage, R-COO-R', here formed from propanoic acid and methanol. A carboxylic acid ends in COOH; here the H has been replaced by CH3, so there is no acidic OH. There is a C=O, but it is bonded to an O, so it is not a ketone, and there is no C-OH, so there is no alcohol group. An aldehyde needs a CHO group with H on the carbonyl carbon, which is absent.
Nitriles contain C≡N, and the nitrile carbon is counted in the chain, so CH3CH2CN has three carbons and is named propanenitrile. An amine contains N bonded to H and/or carbon by single bonds (for example CH3CH2NH2); containing nitrogen alone does not make a compound an amine. The C≡N is covalently bonded to the chain; the molecule is not ionic, unlike KCN. Leaving out the CN carbon gives two carbons, which is why the name is propanenitrile, not ethanenitrile.
Mr = 12.0n + 1.0(2n + 1) + 16.0 + 1.0 = 14n + 18. Setting 14n + 18 = 88 gives n = 5, so the alcohol is C5H11OH (a pentanol). C4H9OH has Mr 74.0, one CH2 short. C6H13OH has Mr 102.0, one CH2 too many. C5H9OH has Mr 86.0 and does not fit CnH2n+1OH, since it is two H short (it would contain a C=C or a ring).
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.