Question

Post-lab questions

1. We decided to use 5.2 molar equivalents based on our past experience performing this type of reduction. What is the theoretical absolute minimum number of molar equivalents one could use in a sodium borohydride reduction of a ketone like camphor? (Think about the structure of sodium borohydride).

2. Consider the reducing agent LiAlH4 as an alternative reagent, which is typically used in THF, followed by careful aqueous workup. If LiAlH4 would be used, what would the consequence be of using an alcoholic solvent (like in this weeks experiment) instead of an inert solvent like THF? Draw a mechanism describing what might happen. You might want to review the LiAlH4 reagent in your organic book, or use resources on the internet. (Be sure to provide a citation in your report if you do so).

3. When menthone is reduced by the selective hydride reducing agent “L-selectride” , two products are formed (a major one and a minor one). Predict the structure of the two molecules. What is the stereochemical relationship between the two molecules? (enantiomers, diastereomers, etc.).

Btw: in this reaction, NaBH4 is not capable of discriminating the faces of the ketone.

Answer 1

1. Sodium borohydride has four hydride. So, the theoretical absolute minimum number of molar equivalents one could use in a sodium borohydride reduction of a ketone like camphor is 0.25.

2. If LiAlH₄ used in an alcoholic solvent, it will react with the hydrogen of OH and produce hydrogen gas, The LiAlH4 will quenched with all the alcoholic solvent.

One important thing you should note, In LiAlH₄ , its not a hydride rather its a bare hydrogen with a lone pair of electron. That is why LiAlH₄ is a reducing agent where as sodium hydride (NaH - here also hydride), which doesn't reduces but acts as a base.

3.