DSE Chemistry | Organic Synthesis Tutorial Plan (for tutors) | Functional group routes, isomer counting, multi-step yield and mass calculations

2026-11-02

Tutor lesson-prep guides · Secondary Chemistry

Students recognise ethanol, ethanal and ethanoic acid, yet asked how many steps and what overall yield, they answer only “add an oxidising agent”. This lesson turns functional groups, step count and yield into one route that can be counted box by box.

Objectives — Students can: (1) work backwards from a target molecule to the functional group needed and write the reagent and condition for each step; (2) count the 4 structural isomers of C4H9OH and explain the method; (3) multiply step yields to get an overall yield and convert moles to a final mass (0.1 mol ethanol with steps at 0.8 and 0.75 gives 3.6 g of ethanoic acid).

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Moles	Moles equal mass divided by molar mass	n = m / Mr
Functional groups	The difference between hydroxyl, carbonyl and carboxyl	Alcohol, aldehyde or ketone, carboxylic acid
Redox	Whether an oxidising agent gains or loses electrons	It gains electrons and is reduced

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-5 min	Opening question	Write ethanol, ethanal and ethanoic acid and ask which is the carboxylic acid	Answer aloud with whatever they recall
5-15 min	Prerequisite check	Hand out three written checks (mole formula, functional groups, oxidising agent)	Written answers, marked on the spot
15-37 min	Three scenes	Teach functional groups and reagents, isomers, multi-step yield	After each scene, one sentence on the next reagent
37-49 min	Worked questions	Model four questions, each as target, steps, numbers	Copy, then rewrite from memory
49-55 min	Independent practice	Circulate, only asking whether the question is about route or yield	Do three questions and self-mark
55-60 min	Exit check	Collect answers and count errors live	Written, no talking

03 Teaching script (read it out)

Scene 1: Scene 1: Functional groups, reagents and conditions

Synthesis questions are not reaction recall but backward reasoning. If the target has a carboxyl group, ask what can give one: a primary alcohol through an aldehyde, or an aldehyde oxidised directly. So ethanol to ethanoic acid takes two steps: acidified potassium dichromate with distillation gives ethanal, then acidified potassium dichromate under reflux gives ethanoic acid. Conditions must be explicit: distillation removes the lower-boiling aldehyde before it oxidises further, while reflux keeps reactants in the flask so the oxidation goes to completion. The molar masses to hold are ethanol 46.0, ethanal 44.0 and ethanoic acid 60.0. The second layer is the reducing direction: an aldehyde with lithium aluminium hydride gives a primary alcohol and a ketone gives a secondary alcohol, so the class of alcohol in the product tells you whether the starting material was an aldehyde or a ketone. The third layer is testing: a carboxylic acid gives bubbles with sodium hydrogencarbonate, an aldehyde gives a silver mirror with Tollens reagent, and an alcohol gives hydrogen with sodium, so each step can be confirmed.

Why distil when making ethanal from ethanol instead of using reflux?

Scene 2: Scene 2: Counting isomers and drawing them

C4H9OH has 4 structural isomers: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol. Counting needs an order, never freehand drawing. Draw the longest straight chain first, four carbons with the hydroxyl on carbon 1 or 2, which gives two. Then shorten the main chain to three carbons and place the remaining carbon as a methyl branch, on carbon 1 or 2, which gives two more, so 4 in total. Three warnings: positional and chain isomers are counted separately; a hydroxyl on carbon 3 of a four-carbon chain is the same molecule as carbon 2 counted from the other end; and ethers share the formula C4H10O but are not alcohol isomers unless the question asks for all isomers. Close with one line: fix the carbon skeleton, place the functional group, then check for duplicates.

Why can 2-methylpropan-2-ol only have the hydroxyl in one position?

Scene 3: Scene 3: Multi-step yield and mass

The overall yield of a multi-step synthesis is the product of each step, never the sum and never the average. Example: 0.1 mol of ethanol at a step yield of 0.8 gives 0.08000000000000002 mol of ethanal; the second step at 0.75 gives 0.06000000000000001 mol of ethanoic acid. The overall yield is 0.06000000000000001 divided by 0.1, that is 0.6000000000000001, and multiplying 0.8 by 0.75 confirms the same 0.6000000000000001. For mass, 0.06000000000000001 mol times the molar mass 60.0 gives 3.6 g; at 100 per cent the figure would be 0.1 times 60.0, that is 6.0 g, so the mass yield of 3.6 over 6.0 again gives 0.6000000000000001 and both routes must agree. Insist on units and significant figures at every step, and require a closing sentence that states what fraction of the theoretical yield was obtained.

If step one yields fifty per cent and step two ninety per cent, can the overall yield exceed ninety per cent?

04 Question types and worked steps

Type	Example	Steps	Answer
Route design	Synthesise ethanoic acid from ethanol, giving reagents and conditions.	1 identify the target group; 2 work back one step; 3 write reagent and condition (distil or reflux); 4 label the number of steps.	Example: two steps, acidified dichromate, distil then reflux.
Isomer count	How many structural isomers does C4H9OH have?	1 draw the longest chain; 2 shorten and branch; 3 check for duplicates; 4 state the total.	Example: 4.
Multi-step yield	0.1 mol ethanol with step yields 0.8 and 0.75. Find the moles of acid and the overall yield.	1 multiply step by step; 2 record moles at each step; 3 divide final by starting; 4 express as a percentage.	Example: 0.06000000000000001 mol, overall 0.6000000000000001.
Mass calculation	What mass of ethanoic acid is obtained above?	1 moles times molar mass; 2 compare with the theoretical value; 3 state grams; 4 one line on yield.	Example: 3.6 g against a theoretical 6.0 g.

Sanity check (teach this): Sanity checks: an overall yield of 0.6000000000000001 must be below every individual step yield, 0.8 and 0.75, so a figure above 0.6 means the steps were added. 0.06000000000000001 mol times 60.0 gives 3.6 g, and an answer of 6.0 g has forgotten the yield. The isomer count of 4 must match two straight-chain plus two branched structures, so three or five means something was missed or double counted. The molar mass of ethanol is 46.0, so 60 belongs to the acid. Distillation removes product while reflux retains reactants, and the answer must say which and why.

05 Independent practice (3 questions)

Question	Answer	What to watch
Give the reagent and condition for oxidising ethanol to ethanal.	Acidified potassium dichromate with heat and distillation, collecting the aldehyde at once.	Check whether reflux was written and the aldehyde never collected.
Which of the 4 C4H9OH isomers cannot be oxidised to an aldehyde?	2-methylpropan-2-ol, a tertiary alcohol, and secondary alcohols oxidise to ketones rather than aldehydes.	Check whether a tertiary alcohol was treated as primary.
Two steps at 0.8 and 0.75. What is the overall yield?	0.6000000000000001, that is 60.00000000000001 per cent.	Check for a sum of 1.55.

06 Common mistakes and what to say

Mistake	Student writes	Say this
Adding yields	Answering that the overall yield is 0.8 plus 0.75.	Each step takes a cut, so multiply: 0.8 times 0.75 is 0.6000000000000001, always below either step.
Forgetting molar mass	Treating 0.06000000000000001 mol as 0.06000000000000001 g.	Moles are not grams. Multiply by 60.0 to get 3.6 g.
Double counting isomers	Listing butan-3-ol as a separate isomer.	Counted from the other end it is butan-2-ol, the same molecule. Check symmetry after fixing the skeleton.
Confusing condition with purpose	Using reflux to collect ethanal from ethanol.	Reflux lets the aldehyde oxidise further to the acid. Distil and collect at once instead.

07 Homework (paper, section and question numbers)

Secondary Chemistry index: finish the three questions on this page, then compare addition and condensation in the polymers chapter — Practice 1-3 on this page

08 Exit check (last 5 minutes)

Question	Answer	Decision
What condition gives ethanal from ethanol?	Acidified dichromate with heat and distillation	Cannot answer = reteach scene 1
How many structural isomers does C4H9OH have?	4	Cannot answer = reteach scene 2
How are multi-step yields combined?	Multiply each step	Cannot answer = reteach scene 3

09 Tutor FAQ

Which year is this for?

Mainly the S6 organic chemistry unit; a useful extension after alkanes in S5.

Must every reagent be memorised?

The core is oxidation, reduction and esterification; the rest can be derived from the functional group.

Are yield calculations required?

DSE regularly examines moles and mass across multi-step syntheses, so both directions need practice.

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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.