we will determine the ration of borneol to isoborneol formed by the reduction of

Question

we will determine the ration of borneol to isoborneol formed by the reduction of camphor by using 1H NMR. In borneol the exo hydrogesn 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 0.264g of product, and integration of the key hydrogen signals yields these values:

H at 4.0ppm: 1.00

H at 3.6ppm: 3.14

what percentage of the mixture is borneol? What percentage is isoborneol? please show calculations

Explanation / Answer

First of all, the molar mass of both borneol and isoborneol is 154.25 g/mol.

So:

0.264 g / 154.25 g/mol = 0.00171 moles of the mixture.

Now, because each signal in the NMR represents a single proton, and because the two molecules have the same molar mass, the molar ratio of borneol : isoborneol is simply the ratio of the integrals of the two peaks.

Thus, we have 3.14 : 1 isoborneol : borneol.

So, we have:

100% * 1/4.14 = 24.2 mol % borneol
100% * 3.14/4.14 = 75.8 mol % isoborneol.

In moles then, we have:

.242 * 0.00171 = 4.14 * 10^-4 moles borneol
.758 * 0.00171 = 1.30 * 10^-3 moles isoborneol.

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