Why Multi-Step Reaction Pathways Have Low Overall Yields
Even when every reaction in an organic synthesis works well, the overall yield can be surprisingly low. Learn why individual yields multiply, how to compare alternative reaction pathways, and how to calculate the final amount of product using worked HSC Chemistry examples.
Start with 0.100 mol of an organic compound and put it through three reactions. The first gives a 90% yield, the second gives 85%, and the third gives 80%. All three reactions work reasonably well. Yet you finish with just 0.0612 mol of the desired product, an overall yield of 61.2%.
Why does so much starting material disappear from the final result?
The answer is that each reaction works with the material left by the previous reaction, not with the original starting amount. Small losses accumulate. By the time the pathway reaches its final product, even several successful reactions can leave you with a surprisingly small amount.
Understanding this changes how you approach organic synthesis questions. You need to consider both whether a reaction pathway is chemically possible and how efficiently it delivers the final product.
01Why individual reaction yields multiply
Imagine passing a batch of 100 components through a series of manufacturing stages. If the first stage retains 90 components, the second stage can only work with those 90. If it retains 90% of them, you now have 81.
Chemical synthesis follows the same mathematical pattern. Each reaction produces an amount of intermediate that becomes the starting material for the next reaction.
Suppose two consecutive reactions each have a 90% yield.
After the first reaction, 90% of the original material remains in the desired intermediate:
\[
100\times0.90=90
\]
The second reaction produces 90% of those 90 units:
\[
90\times0.90=81
\]
The overall yield is therefore 81%, not 90% or 180%.
This gives us the general relationship:
\[
Y_{\text{overall}}=y_1\times y_2\times\cdots\times y_n
\]
Here:
- \(Y_{\text{overall}}\) is the overall yield expressed as a decimal.
- \(y_1,y_2,\ldots,y_n\) are the individual reaction yields expressed as decimals.
- \(n\) is the number of reaction steps.
To convert a percentage yield into a decimal, divide it by 100. For example, 85% becomes 0.85.
The effect becomes clearer as the number of reactions increases.
| Number of reaction steps | Yield of each step | Overall yield |
|---|---|---|
| 1 | 90% | 90.0% |
| 2 | 90% | 81.0% |
| 3 | 90% | 72.9% |
| 4 | 90% | 65.6% |
| 5 | 90% | 59.0% |
Notice that even five reactions with individual yields of 90% produce an overall yield of only about 59%.
The important point is that a reaction pathway does not have one yield that can be judged by looking at its best-performing step. Its overall efficiency depends on the combined effect of every step.
This calculation assumes that each step converts one mole of the organic intermediate into one mole of the next product, as in the examples below. If the balanced equation has a different mole ratio, calculate the theoretical amount using the stoichiometric coefficients before applying that step’s percentage yield.
02Following the material through a real reaction pathway
Let’s make this more concrete by considering a synthesis of butanoic acid from 1-bromobutane.
The target molecule contains four carbon atoms and a carboxyl functional group, \(\ce{-COOH}\). Working backwards, we need a precursor that can produce that functional group without changing the four-carbon skeleton.
A suitable pathway is:
\[
\ce{CH3CH2CH2CH2Br}
\]
\[
\ce{->[NaOH(aq)] CH3CH2CH2CH2OH}
\]
\[
\ce{->[K2Cr2O7/H+, distil] CH3CH2CH2CHO}
\]
\[
\ce{->[K2Cr2O7/H+, reflux] CH3CH2CH2COOH}
\]
The compounds are:
| Stage | Compound | Functional group | Carbon atoms |
|---|---|---|---|
| Starting material | 1-bromobutane | Haloalkane | 4 |
| Intermediate 1 | Butan-1-ol | Primary alcohol | 4 |
| Intermediate 2 | Butanal | Aldehyde | 4 |
| Final product | Butanoic acid | Carboxylic acid | 4 |
There are three important decisions behind this pathway.
First, the carbon skeleton must be preserved. Each molecule contains four carbon atoms. Substitution and oxidation change the functional groups, but none of these steps needs to add or remove a carbon atom.
Second, the reagents must produce the correct intermediate. Aqueous sodium hydroxide promotes substitution of the bromine atom with an \(\ce{-OH}\) group, producing butan-1-ol. Using ethanolic hydroxide would favour elimination instead, producing an alkene. That would be a different pathway.
Third, the oxidation conditions must match the desired product. Butan-1-ol is a primary alcohol. Controlled oxidation followed by distillation produces butanal. Further oxidation under reflux produces butanoic acid.
A tempting mistake is to start with the word oxidation and assume that any alcohol can eventually produce the desired carboxylic acid. But oxidation of a secondary alcohol, such as butan-2-ol, produces a ketone, not a carboxylic acid under the usual HSC reaction conditions.
The functional group of the starting intermediate therefore matters just as much as the reagent.
For a fuller explanation of how to identify these transformations and choose the appropriate reagents, see How to Read Organic Reaction Pathways in HSC Chemistry.
03How chemists can measure the losses at each stage
A low final yield tells us that the complete pathway has not delivered much product. It does not automatically tell us which reaction caused the greatest loss.
To investigate that, we need to measure the amount of material after each reaction rather than weighing only the final product.
Consider an illustrative investigation using the butanoic acid pathway.
The starting amount is measured before the first reaction. The first reaction takes place in a heated reaction flask fitted with a vertical reflux condenser. The condenser cools escaping vapour and returns the liquid to the flask.
After the reaction, the butan-1-ol is separated and purified. Its amount is measured before it enters the next stage.
For the oxidation of butan-1-ol to butanal, the apparatus changes. The reaction flask is connected to a distillation head, a water-cooled condenser, and a receiving vessel. Butanal is volatile, so it can leave the reaction mixture as it forms, pass through the condenser, and collect in the receiver.
This matters because leaving the aldehyde in contact with the oxidising mixture could allow further oxidation to butanoic acid.
For the final oxidation, a reflux condenser is used again. Vapour condenses and flows back into the reaction flask, allowing the oxidation to continue rather than collecting the volatile material separately.
The distinction between these arrangements is important:
| Apparatus | What happens to the vapour? | Purpose |
|---|---|---|
| Distillation | Vapour passes through a condenser and collects in a separate vessel. | Remove and collect the aldehyde before further oxidation. |
| Reflux | Vapour condenses and returns to the reaction flask. | Allow oxidation to continue towards the carboxylic acid. |
An initial investigation that measures only the final mass cannot identify which stage is responsible for the losses.
A better design measures the amount of each isolated intermediate. We can then calculate the yield of every reaction separately and test whether the overall yield matches the multiplication model.
The following data are illustrative, not measurements from an actual experiment.
| Reaction | Starting amount | Isolated product | Individual yield |
|---|---|---|---|
| 1-bromobutane → butan-1-ol | 0.1000 mol | 0.0900 mol | 90% |
| Butan-1-ol → butanal | 0.0900 mol | 0.0765 mol | 85% |
| Butanal → butanoic acid | 0.0765 mol | 0.0612 mol | 80% |
The results support the cumulative-yield model. No individual reaction has failed completely, but the final amount is substantially below the original theoretical amount.
Worked example: Calculate the overall yield of a three-step synthesis
A chemist starts with 0.100 mol of 1-bromobutane and carries out the three reactions described above.
The individual yields are 90%, 85%, and 80%. Calculate the overall percentage yield and the mass of butanoic acid produced. Assume that each reaction has a 1:1 mole ratio between its organic starting material and desired product.
Step 1
\[
90\%=0.90,\quad
85\%=0.85,\quad
80\%=0.80
\]
Step 2
\[
\begin{aligned}
Y_{\text{overall}}
&=0.90\times0.85\times0.80\\
&=0.612
\end{aligned}
\]
Therefore:
\[
Y_{\text{overall}}=61.2\%
\]
Step 3
The starting amount is 0.100 mol, and the theoretical mole ratio from starting material to final product is 1:1.
\[
\begin{aligned}
n(\text{butanoic acid})
&=0.100\times0.612\\
&=0.0612\text{ mol}
\end{aligned}
\]
Step 4
Butanoic acid has the molecular formula \(\ce{C4H8O2}\).
Its molar mass is:
\[
M=4(12.01)+8(1.008)+2(16.00)
\]
\[
M=88.11\text{ g mol}^{-1}
\]
Using \(m=nM\):
\[
\begin{aligned}
m&=0.0612\times88.11\\
&=5.39\text{ g}
\end{aligned}
\]
The chemist obtains 5.39 g of butanoic acid, corresponding to an overall yield of 61.2%.
The key observation is that the second reaction acts on 0.0900 mol, not the original 0.100 mol. The third acts on an even smaller amount, 0.0765 mol. That is why the final yield is much lower than any individual reaction yield.
04Why working backwards can improve the overall yield
Once you understand how yields accumulate, the next decision is whether your proposed pathway contains unnecessary reactions.
Suppose your target is butanoic acid. You could deliberately stop the oxidation of butan-1-ol at butanal, isolate the aldehyde, and then oxidise it again.
But if the final product is already known to be a carboxylic acid, you can oxidise the primary alcohol directly under reflux.
That removes an intermediate isolation and one separate reaction step.
This gives two possible routes:
\[
\begin{aligned}
\text{Route A: }&
\ce{1-bromobutane -> butan-1-ol}\\
&\ce{-> butanoic acid}
\end{aligned}
\]
\[
\begin{aligned}
\text{Route B: }&
\ce{1-bromobutane -> butan-1-ol}\\
&\ce{-> butanal -> butanoic acid}
\end{aligned}
\]
Fewer reactions often improve the overall yield because there are fewer opportunities to lose material. However, fewer steps do not guarantee a better result. A longer route can outperform a shorter one if its individual reactions are more selective or give substantially higher yields.
The actual percentages must decide the comparison.
Worked example: Is a longer reaction pathway always less efficient?
A chemist wants to produce propanoic acid from 1-bromopropane. Both proposed pathways begin with substitution of 1-bromopropane to form propan-1-ol, with a yield of 90%.
Two oxidation options are available:
- Route A: Direct oxidation of propan-1-ol to propanoic acid under reflux, with a yield of 70%.
- Route B: Controlled oxidation and distillation to propanal, with a yield of 95%, followed by oxidation of propanal to propanoic acid under reflux, with a yield of 90%.
The chemist starts with 0.100 mol of 1-bromopropane. Determine which route produces more propanoic acid. Assume all reactions have a 1:1 mole ratio and that the stated yields include losses during isolation.
Step 1
Route A has two reaction steps:
\[
\begin{aligned}
Y_A&=0.90\times0.70\\
&=0.63
\end{aligned}
\]
The overall yield is 63%.
Step 2
\[
\begin{aligned}
n_A&=0.100\times0.63\\
&=0.0630\text{ mol}
\end{aligned}
\]
Step 3
Route B has three reaction steps:
\[
\begin{aligned}
Y_B&=0.90\times0.95\times0.90\\
&=0.7695
\end{aligned}
\]
The overall yield is 76.95%.
Step 4
\[
\begin{aligned}
n_B&=0.100\times0.7695\\
&=0.07695\text{ mol}
\end{aligned}
\]
Route B produces more propanoic acid, despite having an additional reaction step.
It gives 0.07695 mol compared with Route A’s 0.0630 mol.
This is an important qualification to the idea that fewer steps always mean higher yields. Route B performs better because its two oxidation steps have sufficiently high individual yields to outweigh the disadvantage of isolating an additional intermediate.
05A practical method for evaluating a reaction pathway
When an HSC question asks you to assess or propose a multi-step synthesis, use the following sequence.
| Decision | What to check | Why it matters |
|---|---|---|
| 1. Identify the target | Write its structural formula and functional group. | Establishes what the final reaction must produce. |
| 2. Work backwards | Identify a suitable immediate precursor. | Prevents you choosing a reaction that produces the wrong functional group. |
| 3. Check the carbon skeleton | Count the carbon atoms in every organic intermediate and reactant. | Prevents impossible changes in carbon count. |
| 4. Check the reagents | Match each reagent and condition to the required transformation. | A correct-looking arrow may produce a different product under different conditions. |
| 5. Calculate each yield | Apply the percentage yield to the material entering that particular step. | Accounts for the decreasing amount of intermediate. |
| 6. Calculate the overall yield | Multiply the individual decimal yields. | Determines the fraction of the original theoretical product actually obtained. |
| 7. Compare alternatives | Calculate the overall yield of every chemically valid route. | Identifies the more efficient pathway. |
There is a useful distinction between a pathway that is inefficient and one that is chemically impossible.
A pathway with three valid reactions, each giving 80%, has an overall yield of:
\[
0.80^3\times100\%=51.2\%
\]
That pathway is possible but inefficient.
A pathway that attempts to convert a primary alcohol into a ketone using ordinary oxidation conditions is chemically incorrect, even if someone assigns it a theoretical yield of 95%. You cannot rescue an invalid reaction by giving it a high percentage.
Similarly, a pathway that requires an additional carbon atom must identify a reactant that supplies it. Oxidising a four-carbon compound does not automatically produce a five-carbon compound.
For a more detailed explanation of atom conservation across organic synthesis, see How to Check Atom Conservation in Multi-Step Organic Synthesis.
One final distinction is worth keeping clear. The yield of an individual reaction can include losses during purification and isolation, while the overall yield measures the combined result across the entire pathway. A low isolated yield in a single esterification practical has additional possible explanations, including equilibrium and product recovery. See Why Esterification Can Give a Low Isolated Yield for that separate problem.
06Questions and solutions
Question 1
Propene is converted into propan-2-ol by hydration, followed by oxidation to propanone.
The hydration reaction has a yield of 80%, and the oxidation reaction has a yield of 75%.
A chemist begins with 0.250 mol of propene. Assume both reactions have a 1:1 mole ratio between the organic reactant and desired organic product.
Calculate:
a) The amount of propan-2-ol produced after the first reaction.
b) The amount of propanone produced.
c) The overall percentage yield of the pathway.
Solution 1
The pathway produces 0.150 mol of propanone, giving an overall yield of 60.0%.
Part a)
The first reaction has an 80% yield.
\[
\begin{aligned}
n(\text{propan-2-ol})
&=0.250\times0.80\\
&=0.200\text{ mol}
\end{aligned}
\]
Therefore, 0.200 mol of propan-2-ol is produced.
Part b)
The second reaction acts on the 0.200 mol of propan-2-ol, not the original 0.250 mol of propene.
\[
\begin{aligned}
n(\text{propanone})
&=0.200\times0.75\\
&=0.150\text{ mol}
\end{aligned}
\]
Therefore, 0.150 mol of propanone is produced.
Part c)
Multiply the two individual yields:
\[
\begin{aligned}
Y_{\text{overall}}
&=0.80\times0.75\\
&=0.60
\end{aligned}
\]
\[
Y_{\text{overall}}=60.0\%
\]
The pathway retains 60% of its original theoretical product amount. The remaining 40% is lost across the two stages.
The common mistake is to apply the second yield to the original amount of propene. Each yield must be applied to the amount entering its own reaction.
Question 2
A student needs to synthesise propanoic acid from 1-bromopropane.
They propose the following sequence:
\[
\ce{CH3CH2CH2Br ->[KOH(ethanol)] CH3CH=CH2}
\]
\[
\ce{CH3CH=CH2 ->[K2Cr2O7/H+] CH3CH2COOH}
\]
The student claims that the pathway should work because the first reaction produces an alkene and the second uses a familiar oxidising agent.
a) Explain why this proposed pathway is unsuitable.
b) Write a chemically appropriate pathway from 1-bromopropane to propanoic acid, including the reagents and conditions.
c) If the two reactions in your corrected pathway have yields of 88% and 85%, respectively, calculate the amount and mass of propanoic acid produced from 0.0800 mol of 1-bromopropane.
Assume 1:1 mole ratios for the organic transformations. The molar mass of propanoic acid is 74.08 g mol\(^{-1}\).
Solution 2
The proposed pathway is unsuitable because it uses elimination to form propene, rather than producing the primary alcohol needed for a straightforward oxidation to propanoic acid.
Part a)
Ethanolic potassium hydroxide favours elimination in a haloalkane reaction.
The bromine atom and a hydrogen atom are removed, forming a carbon-carbon double bond:
\[
\ce{CH3CH2CH2Br -> CH3CH=CH2}
\]
The product is propene, not propan-1-ol.
The proposed second step also does not represent the usual HSC oxidation pathway from an alkene to a carboxylic acid. Simply adding an oxidising reagent does not guarantee the required product.
Part b)
We need a primary alcohol because its oxidation can produce a carboxylic acid.
The correct sequence is:
\[
\ce{CH3CH2CH2Br}
\]
\[
\ce{->[NaOH(aq)] CH3CH2CH2OH}
\]
\[
\ce{->[K2Cr2O7/H+, reflux] CH3CH2COOH}
\]
Aqueous sodium hydroxide promotes substitution, replacing bromine with the hydroxyl group. The resulting propan-1-ol is then oxidised under reflux to propanoic acid.
The carbon count remains three throughout the pathway.
Part c)
Step 1: Convert the individual yields into decimals.
\[
88\%=0.88,\qquad85\%=0.85
\]
Step 2: Calculate the overall yield.
\[
\begin{aligned}
Y_{\text{overall}}
&=0.88\times0.85\\
&=0.748
\end{aligned}
\]
Therefore, the overall yield is 74.8%.
Step 3: Calculate the final amount in moles.
\[
\begin{aligned}
n(\text{propanoic acid})
&=0.0800\times0.748\\
&=0.05984\text{ mol}
\end{aligned}
\]
Step 4: Convert moles into mass.
Using \(m=nM\):
\[
\begin{aligned}
m&=0.05984\times74.08\\
&=4.43\text{ g}
\end{aligned}
\]
The final product is 0.05984 mol, or 4.43 g, of propanoic acid.
The calculation only becomes meaningful after the chemical pathway has been corrected. A high calculated yield cannot make an unsuitable reaction sequence valid.
Question 3
A synthesis converts 1-bromopropane into propanoic acid through three isolated stages:
\[
\ce{1-bromopropane -> propan-1-ol -> propanal -> propanoic\ acid}
\]
A chemist begins with 0.120 mol of 1-bromopropane. The following amounts are measured after each stage:
| Stage | Amount of isolated product |
|---|---|
| Propan-1-ol | 0.108 mol |
| Propanal | 0.0810 mol |
| Propanoic acid | 0.0729 mol |
Assume each reaction has a 1:1 mole ratio.
a) Calculate the yield of each reaction.
b) Calculate the overall yield of the pathway.
c) The chemist can improve exactly one reaction by five percentage points without affecting the others. Which reaction should be improved to produce the greatest increase in final propanoic acid? Justify your answer quantitatively.
Solution 3
The individual yields are 90%, 75%, and 90%. Improving the second reaction by five percentage points produces the greatest increase in final propanoic acid.
Part a)
Calculate each reaction yield using the amount entering that particular stage.
First reaction:
\[
\begin{aligned}
y_1&=\frac{0.108}{0.120}\\
&=0.90
\end{aligned}
\]
Therefore, the first reaction has a 90% yield.
Second reaction:
\[
\begin{aligned}
y_2&=\frac{0.0810}{0.108}\\
&=0.75
\end{aligned}
\]
Therefore, the second reaction has a 75% yield.
Third reaction:
\[
\begin{aligned}
y_3&=\frac{0.0729}{0.0810}\\
&=0.90
\end{aligned}
\]
Therefore, the third reaction has a 90% yield.
Part b)
Multiply the three individual yields:
\[
\begin{aligned}
Y_{\text{overall}}
&=0.90\times0.75\times0.90\\
&=0.6075
\end{aligned}
\]
\[
Y_{\text{overall}}=60.75\%
\]
The overall yield is 60.75%.
Part c)
We must compare the final amount produced when each reaction is improved separately.
The current final amount is 0.0729 mol.
If the first reaction improves from 90% to 95%:
\[
\begin{aligned}
n_1&=0.120\times0.95\times0.75\times0.90\\
&=0.07695\text{ mol}
\end{aligned}
\]
The increase is:
\[
0.07695-0.0729=0.00405\text{ mol}
\]
If the second reaction improves from 75% to 80%:
\[
\begin{aligned}
n_2&=0.120\times0.90\times0.80\times0.90\\
&=0.07776\text{ mol}
\end{aligned}
\]
The increase is:
\[
0.07776-0.0729=0.00486\text{ mol}
\]
If the third reaction improves from 90% to 95%:
\[
\begin{aligned}
n_3&=0.120\times0.90\times0.75\times0.95\\
&=0.07695\text{ mol}
\end{aligned}
\]
The increase is:
\[
0.07695-0.0729=0.00405\text{ mol}
\]
The second reaction gives the greatest improvement, increasing the final amount by 0.00486 mol.
The tempting approach is to assume that the first or last reaction must be the best place to improve because it is closest to the starting material or final product. Neither position determines the answer.
The increase depends on the amount of material reaching the improved step and the yields of all the other steps. Here, improving the weakest reaction has the greatest effect because it increases the fraction surviving the middle stage while preserving the other two yields.
Question 4
A chemist studies the oxidation of propan-1-ol to propanoic acid through propanal.
The following masses are recorded:
| Substance | Measured mass |
|---|---|
| Propan-1-ol entering the first reaction | 6.01 g |
| Propanal collected after the first reaction | 4.65 g |
| Propanoic acid collected after the second reaction | 4.44 g |
The relevant molar masses are:
- Propan-1-ol: 60.10 g mol\(^{-1}\)
- Propanal: 58.08 g mol\(^{-1}\)
- Propanoic acid: 74.08 g mol\(^{-1}\)
Assume each reaction has a 1:1 mole ratio.
a) Calculate the yield of each reaction.
b) Calculate the overall percentage yield.
c) A student calculates the overall yield by dividing 4.44 g by 6.01 g and multiplying by 100. Explain precisely why this method is incorrect.
d) Explain why the final product can have a greater mass than the original alcohol without violating conservation of mass.
Solution 4
The first reaction has a yield of approximately 80.0%, the second approximately 75.0%, and the overall yield is approximately 60.0%. Comparing the initial and final masses directly does not calculate the reaction yield.
Part a)
First convert the measured masses into moles.
For propan-1-ol:
\[
\begin{aligned}
n_1&=\frac{6.01}{60.10}\\
&=0.100\text{ mol}
\end{aligned}
\]
For propanal:
\[
\begin{aligned}
n_2&=\frac{4.65}{58.08}\\
&=0.0801\text{ mol}
\end{aligned}
\]
For propanoic acid:
\[
\begin{aligned}
n_3&=\frac{4.44}{74.08}\\
&=0.0599\text{ mol}
\end{aligned}
\]
The first reaction yield is:
\[
\begin{aligned}
y_1&=\frac{0.0801}{0.100}\times100\\
&\approx80.1\%
\end{aligned}
\]
The second reaction yield is:
\[
\begin{aligned}
y_2&=\frac{0.0599}{0.0801}\times100\\
&\approx74.8\%
\end{aligned}
\]
Part b)
The overall yield can be calculated by multiplying the individual decimal yields:
\[
\begin{aligned}
Y_{\text{overall}}
&=0.801\times0.748\\
&\approx0.599
\end{aligned}
\]
Therefore:
\[
Y_{\text{overall}}\approx59.9\%
\]
Small differences from 60.0% arise from rounding the recorded masses.
Part c)
The student’s calculation compares the mass of two chemically different compounds.
The molar mass of propan-1-ol is 60.10 g mol\(^{-1}\), whereas the molar mass of propanoic acid is 74.08 g mol\(^{-1}\).
A reaction yield compares the actual amount of product with the theoretical amount of that product. It does not compare the raw masses of the starting material and final product.
Because the molecular composition changes during oxidation, the mass ratio does not represent the fraction of starting molecules that successfully reached the final product.
The correct approach is to convert the masses into moles, account for the reaction stoichiometry, and then calculate the percentage yields.
Part d)
The final product can have a greater mass because oxidation incorporates oxygen into the organic molecule.
During the conversion of propan-1-ol to propanal and then propanoic acid, hydrogen atoms are removed and oxygen is incorporated.
The additional oxygen comes from the oxidising agent. It is not created during the reaction.
Conservation of mass applies to the complete reaction system, including the oxidising agent and all other reactants and products. It does not require the isolated organic product to have the same mass as the original organic reactant.
Question 5
A chemist receives a mixture containing 1.00 mol of two bromobutane isomers:
- 0.250 mol of 1-bromobutane
- 0.750 mol of 2-bromobutane
The chemist wants to produce butanone and considers processing the two isomers through separate reaction pathways.
Pathway A: Starting with 1-bromobutane
\[
\ce{1-bromobutane -> but-1-ene -> butan-2-ol -> butanone}
\]
The individual yields are:
- Elimination to but-1-ene: 90%
- Hydration to butan-2-ol: 80%
- Oxidation to butanone: 90%
Pathway B: Starting with 2-bromobutane
\[
\ce{2-bromobutane -> butan-2-ol -> butanone}
\]
The individual yields are:
- Substitution to butan-2-ol: 75%
- Oxidation to butanone: 90%
Assume that:
- The hydration of but-1-ene produces butan-2-ol as the desired major product.
- Aqueous sodium hydroxide is used for substitution of 2-bromobutane, and the stated yield already accounts for competing elimination.
- Every reaction has a 1:1 mole ratio between the organic reactant and desired organic product.
- Both pathways can be carried out independently, and their final butanone products can be combined without further loss.
- No additional material enters either pathway.
a) Calculate the amount of butanone produced through Pathway A.
b) Calculate the amount of butanone produced through Pathway B.
c) Determine the total amount of butanone produced and the overall percentage yield relative to the original 1.00 mol mixture.
d) A student multiplies all five reaction yields together to calculate the overall yield. Explain why this is incorrect, even though every reaction appears somewhere in the synthesis diagram.
Solution 5
The two pathways produce a combined total of 0.668 mol of butanone, giving an overall yield of approximately 66.8%.
The important point is that the two pathways are parallel routes, not five consecutive reactions experienced by every molecule.
Part a)
Pathway A begins with 0.250 mol of 1-bromobutane.
Multiply the three individual yields:
\[
\begin{aligned}
Y_A&=0.90\times0.80\times0.90\\
&=0.648
\end{aligned}
\]
The overall yield of Pathway A is 64.8%.
The amount of butanone produced is:
\[
\begin{aligned}
n_A&=0.250\times0.648\\
&=0.162\text{ mol}
\end{aligned}
\]
Therefore, Pathway A produces 0.162 mol of butanone.
Part b)
Pathway B begins with 0.750 mol of 2-bromobutane.
Multiply its two individual yields:
\[
\begin{aligned}
Y_B&=0.75\times0.90\\
&=0.675
\end{aligned}
\]
The overall yield of Pathway B is 67.5%.
The amount of butanone produced is:
\[
\begin{aligned}
n_B&=0.750\times0.675\\
&=0.50625\text{ mol}
\end{aligned}
\]
Therefore, Pathway B produces 0.50625 mol of butanone.
Part c)
Both pathways produce the same final compound, so their product amounts can be added:
\[
\begin{aligned}
n_{\text{total}}
&=0.162+0.50625\\
&=0.66825\text{ mol}
\end{aligned}
\]
The original mixture contains 1.00 mol of bromobutane isomers. Under the stated 1:1 stoichiometric assumptions, the theoretical maximum is therefore 1.00 mol of butanone.
The overall yield is:
\[
\begin{aligned}
Y_{\text{overall}}
&=\frac{0.66825}{1.00}\times100\\
&=66.825\%
\end{aligned}
\]
Therefore, the combined pathway produces 0.668 mol of butanone with an overall yield of approximately 66.8%.
Part d)
Multiplying all five yields is incorrect because no individual molecule passes through all five reactions.
The starting material is divided into two different groups. The 1-bromobutane molecules follow Pathway A, while the 2-bromobutane molecules follow Pathway B.
Each group experiences only the reactions in its own pathway.
The correct method is therefore:
- Calculate the product amount from Pathway A using its three yields.
- Calculate the product amount from Pathway B using its two yields.
- Add the two product amounts.
- Divide by the total theoretical product amount from the original mixture.
The tempting calculation treats the two branches as one continuous sequence. That would incorrectly assume that the 1-bromobutane molecules also undergo the substitution reaction assigned to Pathway B, and that the 2-bromobutane molecules also undergo the elimination and hydration reactions assigned to Pathway A.
This is the crucial difference between consecutive and parallel reaction pathways.
A reaction yield multiplies along an individual route. Product amounts add when separate routes converge on the same final compound.
07Where to go next
Once you can calculate overall yields, the next useful skill is deciding which of several chemically valid pathways is worth choosing.
That requires combining three ideas:
- The target functional group determines which precursor and reaction are suitable.
- Carbon conservation and reagent conditions determine whether the pathway is chemically possible.
- Individual reaction yields determine how much final product the pathway is likely to deliver.
In HSC Chemistry, a strong synthesis answer needs all three. A pathway with an impressive calculated yield is no use if it produces the wrong compound, and a chemically correct pathway may still be inefficient if too much material is lost at each stage.
Course note (November 2026): The examples use organic reaction concepts from the Chemistry Stage 6 Syllabus (2017), which remains the taught syllabus through 2027. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. Check the NESA syllabus implementation information for the requirements applying to your cohort.