Problem 15
Question
Syntheses have been carried out with one hundred or more sequential reactions. If the yield in each step is \(99 \%\), what would be the overall yield after one hundred steps? Repeat the calculation for a yield of \(99.9 \%\) in each step. What do you conclude from these calculations about the importance of yield in multistep syntheses? (It is interesting to contemplate how even simple organisms can synthesize molecules by what appear to be sequences of 10,000 or more separate steps with very few, if any, errors, and what means the organisms have to check the accuracy of the sequence after each incorporation of a new subunit.)
Step-by-Step Solution
Verified Answer
Overall yields for 99% and 99.9% per step are 36.6% and 90.5%, respectively. Small yield improvements per step significantly improve overall yield in multistep syntheses.
1Step 1: Understanding Percentage Yield
In chemical reactions, the percentage yield represents the efficiency of a reaction. It is defined as the actual yield divided by the theoretical yield, multiplied by 100. For this problem, each step has a yield of either 99% or 99.9%, meaning that in each step, a small percentage of the product is lost.
2Step 2: Calculating Overall Yield for 99%
When the yield for each step is 99%, the overall yield after 100 steps is calculated by raising the yield per step to the power of the number of steps: \((0.99)^{100}\).
3Step 3: Performing the Calculation for 99%
Compute \((0.99)^{100}\) using a calculator. This gives an overall yield of approximately 36.6%.
4Step 4: Calculating Overall Yield for 99.9%
Similarly, for a 99.9% yield per step, the overall yield after 100 steps is \((0.999)^{100}\).
5Step 5: Performing the Calculation for 99.9%
Compute \((0.999)^{100}\) using a calculator. This results in an overall yield of approximately 90.5%.
6Step 6: Analyzing the Implications
These calculations show that even a small difference in the efficiency of each individual reaction step can significantly impact the overall yield of a multistep synthesis. A small error compiles exponentially over many steps, reducing the efficiency dramatically.
Key Concepts
Understanding Percentage Yield in Chemical ReactionsChemical Reaction Efficiency in Multistep ProcessesCalculating Overall Yield in Multistep Syntheses
Understanding Percentage Yield in Chemical Reactions
The concept of percentage yield is pivotal in understanding the efficiency of chemical reactions. It measures how close the actual output of a reaction comes to the theoretical maximum. The formula for calculating percentage yield is: \[ \text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100 \] Here, the actual yield is the quantity of product you actually obtain from the reaction, while the theoretical yield is the amount predicted by stoichiometric calculations, assuming perfect conditions and no losses.
A high percentage yield indicates a highly efficient reaction, meaning that most of the reactants have been converted to products. In multistep syntheses, each step has its own percentage yield, and small inefficiencies can multiply across steps. Hence, understanding and optimizing percentage yield is vital to the success of complex reaction sequences.
A high percentage yield indicates a highly efficient reaction, meaning that most of the reactants have been converted to products. In multistep syntheses, each step has its own percentage yield, and small inefficiencies can multiply across steps. Hence, understanding and optimizing percentage yield is vital to the success of complex reaction sequences.
Chemical Reaction Efficiency in Multistep Processes
Efficiency in chemical reactions becomes crucial in multistep syntheses. Each reaction step is an opportunity for loss, which affects the overall yield. A reaction efficiency of 99% implies that in each step, there is a 1% loss. While this may seem negligible, over multiple steps, these losses accumulate significantly.
In our example, a sequence of 100 reactions with 99% efficiency per step results in an overall yield of \((0.99)^{100}\), leading to a substantial cumulative loss, reducing the yield to around 36.6%.
Optimizing reactions to achieve higher step efficiencies is thus critical in multicursory processes to prevent substantial losses over numerous steps.
In our example, a sequence of 100 reactions with 99% efficiency per step results in an overall yield of \((0.99)^{100}\), leading to a substantial cumulative loss, reducing the yield to around 36.6%.
- A fractional yield of 0.99 each step quickly accumulates into less tangible product.
- Efficiencies less than 100% compound exponentially across steps.
Optimizing reactions to achieve higher step efficiencies is thus critical in multicursory processes to prevent substantial losses over numerous steps.
Calculating Overall Yield in Multistep Syntheses
In multistep chemical syntheses, the overall yield is a critical measure of success. It describes the product yield from a series of reactions, systematically considering each individual step yield. The formula for the overall yield is: \[ \text{Overall Yield} = (\text{Yield per Step})^{\text{Number of Steps}} \] For example, if each of 100 reaction steps has a yield of 99%, you calculate the overall yield using:
\[ \text{Overall Yield} = (0.99)^{100} \]
The result, around 36.6%, demonstrates the dramatic effect of small inefficiencies over many sequential reactions. A similar calculation with each step yielding 99.9% results in a much higher overall yield of about 90.5%. This illustrates the exponential impact of yield loss accumulation in multi-stage syntheses.
\[ \text{Overall Yield} = (0.99)^{100} \]
The result, around 36.6%, demonstrates the dramatic effect of small inefficiencies over many sequential reactions. A similar calculation with each step yielding 99.9% results in a much higher overall yield of about 90.5%. This illustrates the exponential impact of yield loss accumulation in multi-stage syntheses.
- Small improvements in yield per step can vastly enhance the overall yield.
- Accurate calculations help predict product quantity effectively.
Other exercises in this chapter
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