Isopropyl bromide was treated separately with sodium tert-butoxide and sodium ethoxide under two different condition. Reaction I : at 400C gave almost exclusively compound A, C3H6 Reaction II : at 300C gave compound A along with small amount of ether B, C5H12O. compound A was readily oxidized by a neutral solution of cold dilute potassium permanganate to give a brown precipitate. 5. Which type of reaction is used to explain the formation of A? 6. Draw the structure of the activated complex formed in reaction II that leads to compound A? 7. What happened when product A is treated with cold and alkaline KMnO4 . 8. Write the IUPAC name of compound B 9. Write the equation used to prepare isopropyl bromide
The above reaction mechanism can be explained as follows ;
In compound B their is a 20% formation takes place which indicates it is SN2 reaction as their is a clear evident in the spectrum for the formation of C-O bond which has a absorption band at 1100 cm-1 and the compound A undergoes elimination reaction i.e E2 elimination which
follows the second order kinetics.
13 - A option suits best
21 - C is the activated complex which yields 80% of the product.