Step 1: Understanding the Concept:
Diborane (\( \text{B}_{2}\text{H}_{6} \)) reacts with alkenes (olefins) in a process called hydroboration.
Step 2: Detailed Explanation:
Hydroboration involves the addition of B-H bonds across the carbon-carbon double bond of an olefin.
Diborane effectively behaves as two \( \text{BH}_{3} \) units. Each \( \text{BH}_{3} \) unit can react with three molecules of an alkene to form a trialkylborane.
If the olefin is ethylene (\( \text{CH}_{2}=\text{CH}_{2} \)), the reaction is:
\[ 6\text{CH}_{2}=\text{CH}_{2} + \text{B}_{2}\text{H}_{6} \longrightarrow 2(\text{CH}_{3}\text{CH}_{2})_{3}\text{B} \]
The product is triethyl borane. More generally, for any olefin, a trialkyl borane is formed.
Step 3: Final Answer:
Diborane reacts with olefins to form trialkyl boranes, like triethyl borane.