A sample of isoborneol prepared by reduction of camphor was analyzed by IR and showed a band at 1750 cm-1. This result was unexpected. Why?
Doesn't the C=O group absorb about 1750 cm^-1 or so. And doesn't camphor have a C=O group? What surprises me is that it was unexpected. Surely we don't expect 100% yield.
Camphor has a C=O group but isoborneol does not. Just an OH group. I was thinking that it was the C=O absorption that was a result of some camphor not reacting and being left with the product.
Camphor has a carbonyl group which would produce an IR band at 1750 cm^-1. During the reduction, the C=O group would change to C-OH which absorbs IR at a much higher spatial frequency. The reduction might have been incomplete, with enough camphor present to produce the 1750 cm^-1 band.
The unexpected presence of a band at 1750 cm-1 in the IR spectrum of isoborneol prepared by the reduction of camphor suggests an imperfection in the reduction process. To understand why this result is unexpected, let's break it down:
Isoborneol is a secondary alcohol, while camphor is a ketone. When camphor undergoes reduction, the carbonyl group (C=O) should be converted to an alcohol group (C-OH). In the IR spectrum, the C=O stretch typically occurs at a higher frequency, around 1700-1750 cm-1, while the C-OH stretch occurs at a lower frequency, around 3400-3600 cm-1.
Therefore, the presence of a band at 1750 cm-1 suggests that the carbonyl group of the camphor was not fully converted to an alcohol group. This could be due to incomplete reduction or the formation of some side products during the reaction.
To confirm this hypothesis and gain a better understanding, one could further investigate the reaction conditions and analyze other spectroscopic data (such as proton NMR or mass spectrometry) to identify any impurities or byproducts that might have been formed during the reduction of camphor. Additional purification techniques may be necessary to obtain a pure sample of isoborneol without the observed band at 1750 cm-1.