We will determine the ration of borneol to isoborneol formed by the reduction of camphor using 1H NMR. In borneol the exo hydrogen on the ring carbon bearing the hydroxyl group appears at 4.0 ppm, whereas the analogous hydrogen in isoborneol appears at 3.6 ppm. Suppose you isolate 0.264 g of product, and integration of the key hydrogen signals yields these values: H at 4.0 ppm: 1.00 H at 3.6 ppm: 3.14 A. How many moles of borneol/isoborneol mixture were formed? B. What percentage of the mixture is borneol? What percentage is isoborneol? C. How many moles of borneol were formed? Of isoborneol?
The Correct Answer and Explanation is :
To determine the composition of borneol and isoborneol in your product mixture, we can use the integration values from the 1H NMR spectrum. Here’s a step-by-step breakdown:
A. Moles of Borneol/Isoborneol Mixture Formed
The total mass of the product is 0.264 g. Given that the molar mass of borneol and isoborneol is approximately 154.24 g/mol, we can calculate the total moles of the mixture:
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B. Percentage Composition of Borneol and Isoborneol
In the 1H NMR spectrum, the signal at 4.0 ppm corresponds to the exo hydrogen in borneol, and the signal at 3.6 ppm corresponds to the analogous hydrogen in isoborneol. The integration values are:
- H at 4.0 ppm: 1.00
- H at 3.6 ppm: 3.14
The total integration value is 1.00 + 3.14 = 4.14. The percentage of borneol is calculated as:
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Therefore, the mixture contains approximately 24.2% borneol and 75.8% isoborneol.
C. Moles of Borneol and Isoborneol Formed
Using the total moles calculated in part A and the percentages from part B, we can determine the moles of each compound:
- Moles of Borneol: 0.00171 mol × 24.2% ≈ 0.000413 mol
- Moles of Isoborneol: 0.00171 mol × 75.8% ≈ 0.001297 mol
In summary, approximately 0.000413 moles of borneol and 0.001297 moles of isoborneol were formed in the reduction of camphor.