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Chapter 22: 8P (page 868)

Draw the products formed (including stereoisomers) when each compound is reduced with NaBH4in CH3OH.

Short Answer

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Answer

The products formed when each compound is treated with NaBH4 in CH3OH. are shown below:

Step by step solution

01

Step-by-Step Solution Step 1: Function of  NaBH4

NaBH4 is a strong reducing agent that reduces ketone into secondary alcohol and aldehyde into primary alcohol.

Since chiral alcohol is formed after reduction, hence two kinds of alcohols are formed. They are collectively called enantiomeric alcohols.

02

Mechanism of the reducing action of  NaBH4

In NaBH4 , B-H is weakly polar where the hydride group acts as a nucleophile and can readily attack the carbonyl carbon center to yield alcohol.

Hence, based on the explanations, the products formed when each compound is treated with NaBH4in CH3OH are shown below:

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Most popular questions from this chapter

What organocuprate reagent is needed to convert CH3CH2COCIto each ketone?

a.

b.

c.

Question: Cinnamoylcocaine, a natural product that occurs in coca leaves, can be converted tococaine by the following reaction sequence. Identify the structure of cinnamoylcocaine, aswell as intermediates X and Y.

Draw the products formed when A or B is treated with each reagent. In some cases, no reaction occurs.

  1. NaBH4, CH3OH
  2. [1] LiAlH4 ; [2] H2O
  3. [1] CH3MgBr ; [2] H2O
  4. [1]C6H5Li ; [2] H2O
  5. Na2Cr2O7, H2SO4,H2O

Draw the products formed from LiAlH4 reduction of each compound.

What product is formed when each compound is treated with either LiAlH4 (followed by H2O ), or NaBH4 in CH3OH ?
See all solutions

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