Question

Product formed in the reaction: $C{H_3} - CH = C{H_2}\xrightarrow[{(ii){H_2}{O_2}/O{H^ - }}]{{(i){B_2}{H_6}}}$

Answer 1

Hint: The alkenes undergo addition reactions. This reaction is a hydroboration oxidation reaction. This two step reaction is used to convert an alkene into alcohol. The anti-Markovnikov’s rule is followed for addition in the first step.

Complete Solution :
We know that alkenes undergo addition reactions. This reaction is called hydroboration oxidation reaction where the reagent given reacts region-specifically to substrate alkene to give a product where $B{H_3}$is attached at terminal position and hydrogen at another carbon. The reaction of first step is given as -
\[C{H_3} - CH = C{H_2}\xrightarrow{{{B_2}{H_6}}}C{H_3} - C{H_2} - C{H_2}B{H_2}\]

The second step is where the $B{H_3}$ group is substituted by the hydroxyl group to give product alcohol. The reaction of second step is given as -
\[C{H_3} - C{H_2} - C{H_2}B{H_2}\xrightarrow{{{H_2}{O_2}/O{H^ - }}}C{H_3} - C{H_2} - C{H_2}OH\]
Thus, the product formed is propanol.

Additional Information:
- In the second step, if any other oxidant is used other than hydrogen peroxide, then the products can be carbonyl compounds. If dichromates are used, then with the terminal alkenes like in question, the product can be carboxylic acid but with a non-terminal alkene, the product is always a ketone.
- If we see the reactants and the products only, we can say that it is the only addition of water. Thus, it is also called a hydration reaction.

Note: This reaction is a syn addition reaction where hydrogen atom and the hydroxyl group, both the groups are attached on the same side of the double bond. It is a hydration reaction. The anti-Markovnikov’s rule says that the hydrogen should be substituted at the more substituted carbon atom.