Borane is an electron deficient compound. It has only six valence electrons, so the born atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH3 is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is thought to occur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdrows electrons from the pi bond and the carbon at the other end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly substituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with a hydroxyl group by using H2O2/
.
The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the
same stereochemical position. Thus, hydroboration of alkene is an example of stereospecific reaction, in which different stereoisomers of starting compounds react to give different stereoisomers of the product.
(i) What will be the product of following reaction
Product –
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
Sol. 
(ii)
Sol. 
(iii)
Sol. Optically inactive 1º-alcohol
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems