Question

1. What is oxidation?  How do you know if a molecule/atom has undergone oxidation.

2. Describe the reactivity of alcohols (1˚, 2˚, and 3˚) to oxidation.  

3. What are two ways besides IR that we can determine if we have the right product? Please explain what molecular property is measured and how borneol and camphor will show up differently in the technique.

4. When we take the IR of borneol and camphor, how will the spectra look different?

5. Why is the Camphor reaction considered “green”?

Answer 1

Ans:

1. Oxidation - Oxidation is any chemical reaction that involves the moving of electrons. Specifically, it means the substance that gives away electrons is oxidation. Another way oxidation is gain of number of oxygen atom or increasing bond with oxygen.

If oxidation number increases or number of oxygen atom increases which will indicate atom/ molecules undergone oxidation.

2. Primary and secondary alcohols are reactive towards oxidation,  Primary alcohols (R-CH₂-OH) can be oxidized either to aldehydes (R-CHO) or to carboxylic acids (R-CO₂H), while the oxidation of secondary alcohols (R¹R²CH-OH) normally terminates at the ketone (R¹R²C=O) stage. Tertiary alcohols (R¹R²R³C-OH) are resistant to oxidation.

3. Borneol is having alcohol as a function group while camphor is having ketone is function group, so if IR will show broad spectrum arround 3300 cm^{​​​​​-1}​​​ which is indicate presence of borneol and if IR will show spectrum at around 1780cm^-1 which is indicating presence of camphor.

4. Borneol will show spectra at around 3300 cm^{​​​​​-1} and camphor will show spectra at around 1780cm^-1.

5. In this reaction there is no toxic by-product that's why this is green reaction.