DSE Chemistry | Organic Functional Groups Lesson-Prep Guide (Tutor Edition)

DSE Chemistry | Organic Functional Groups Lesson-Prep Guide (Tutor Edition) | Isomerism, Esterification and Yield

2026-11-01

Tutor lesson-prep guides · Secondary Chemistry

Students panic at a long carbon chain because they read every letter instead of hunting the functional group first. This lesson teaches one move: circle the group, count the carbons, then name.

Objectives — Students can (1) circle the functional group in a structure and give the IUPAC name; (2) turn percentage composition into an empirical and then a molecular formula; (3) write the esterification equation and use yield to get an actual mass.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Moles	How many moles in 18 g of water?	1 mol
Percentages	What fraction of 100 is 40?	0.40
Relative mass	What is the Mr of CH2O?	30

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Spot the group	Alcohol, aldehyde, ketone, acid, ester, amine, halide	Circle and name
10-22 min	Naming	Longest chain, number towards the group	Name three structures
22-34 min	Empirical formula	Percentage / Ar, then divide by the smallest	CH2O
34-46 min	Molecular formula	Mr / empirical mass	C6H12O6
46-56 min	Ester and yield	Acid + alcohol, then multiply by yield	11.44 g
56-60 min	Exit check	Collect and mark three questions	Work independently

03 Teaching script (read it out)

Scene 1: Circle the group before counting carbons

Alcohol -OH, aldehyde -CHO, ketone -C=O inside the chain, carboxylic acid -COOH, ester -COO-, amine -NH2, halide -X. A molecule may carry several groups; the suffix follows the highest priority: acid, then aldehyde, then ketone, then alcohol, then amine. So CH3CH2OH is ethanol, CH3COOH is ethanoic acid and CH3COOCH2CH3 is ethyl ethanoate.

Ask them to do: circle the groups in CH3CH(OH)COOH (hydroxyl plus carboxyl, a hydroxy acid).

Scene 2: Empirical then molecular: two steps, never one

Take C 40.00%, H 6.67%, O 53.33%. Divide each by its relative atomic mass: C 3.33, H 6.67, O 3.33. Divide by the smallest, 3.33: C 1, H 2.00, O 1.00, so the empirical formula is CH2O with mass 30. Then Mr 180 / 30 = 6, giving C6H12O6, glucose.

Ask them to compute: if Mr is 60, what is the molecular formula? (C2H4O2, ethanoic acid.)

Scene 3: Esterification: keep theory and actual apart

Carboxylic acid + alcohol gives ester + water, with concentrated sulfuric acid as catalyst and heat. 0.20 mol of acid with excess ethanol, ester Mr 88, yield 65%: actual moles 0.13, so mass = 0.13 x 88 = 11.44 g. Isomers: C4H10 has 2 and C5H12 has 3.

Remind them: yield multiplies the moles, never the molar mass.

04 Question types and worked steps

Type	Example	Steps	Answer
Identification	Circle the group and name it	Group first, chain second	Alcohol / acid / ester
Calculation	Percentages to empirical formula	Divide by Ar, then by the smallest	CH2O
Calculation	Empirical to molecular formula	Mr / empirical mass	C6H12O6
Calculation	Ester yield	mol x yield x Mr	11.44 g

Sanity check (teach this): Empirical subscripts must be the smallest whole numbers; a 1.5 means multiply everything by 2. The molecular multiple must be a whole number, so a result near 6.0 usually means a wrong percentage or Ar. Yield cannot exceed 100%; a bigger number means the limiting reagent was missed.

05 Independent practice (3 questions)

Question	Answer	What to watch
Empirical formula from C 40%, H 6.67%, O 53.33%	CH2O	divide by Ar then by the smallest
Molecular formula from CH2O and Mr 180	C6H12O6	Mr / 30
Mass of ester from 0.20 mol acid at 65% yield	11.44 g	mol x yield x Mr

06 Common mistakes and what to say

Mistake	Student writes	Say this
Naming from the carbon count alone	Calling CH3COOH methane	Circle the -COOH first; it is a carboxylic acid.
Leaving 1.5 in the formula	Writing C1.5H3O1.5	Subscripts must be whole numbers, so double them.
Multiplying the wrong quantity by yield	0.20 mol x 65% x 60 = 7.8 g	The ester Mr is 88, not 60.
Writing a one-way esterification	Using a single arrow	Esterification is reversible, so use the equilibrium arrow.

07 Homework (paper, section and question numbers)

DSE Chemistry practice paper (organic chemistry) — Section B groups and esterification
Secondary Chemistry chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
How many isomers of C4H10?	2	Three correct means move on to polymers
Molecular formula from CH2O and Mr 180	C6H12O6	One wrong means re-teach empirical formula
Ester mass from 0.20 mol acid at 65%	11.44 g	Two wrong means restart scene two

09 Tutor FAQ

Do they need every functional group name?

Seven cover most papers: alcohol, aldehyde, ketone, acid, ester, amine, halide.

Why concentrated sulfuric acid?

It catalyses the reaction and removes water, shifting the equilibrium right.

Must isomers be drawn?

Yes. DSE awards marks for the full structural formula, not for a name alone.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.