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What is the reducing agent that will be used in this experiment? Please draw two isomers formed in this reaction. What is the major product? Why? _________ Compare the relative reducing strengths of LiAIH_4 and NaBH_4. What functional groups are reduced by each? What is it important to use anhydrous conditions with LiAIH_3? How might you minimize the hydrogen evolution when using NaBH_4 in water? How does temperature affect NaBH_4 reactivity?
Expert Answer
1) NaBH4 is used as the reducing agent in this experiment.
2) Sodium borohydride reduces camphor and two isomers, borneol and isoborneol. As shown below.
The major product is isoborneol, because the sodium borohydride attacks the bottom side, the less sterically hindered side, of the camphor, isoborneol is formed.
Post lab
1)
LiAlH4 is more reactive than NaBH4.
LiAlH4 reduces aldehydes, ketones, carboxylic acids and its derivatives, and nitriles.
NaBH4 reduces aldehydes and ketones only
2) LiAlH4 reduces water too vigorously, because it very strong reducing agent. As a result, it is used in anhydrous conditions.
3) The reaction of NaBH4 with water is given below.
NaBH4 + 2H2O ——> NaBO2 + 4 H2 (g)
We could minimize the hydrogen evolution by reducing the water intake.
4) When temperature is increased, the hydrolysis reaction occurs evolving hydrogen gas decreasing the stability of sodium borohydride.
