Identify Z in the following reaction:
$CH_3CH_2CH=CH_2 \xrightarrow{B_2H_6} \xrightarrow{H_2O, H_2O_2, OH^-} Z

Question

Identify Z in the following reaction:
$CH_3CH_2CH=CH_2 \xrightarrow{B_2H_6} \xrightarrow{H_2O, H_2O_2, OH^-} Z$

Accepted Answer

The reaction sequence represents **Hydroboration-Oxidation** of an alkene.
1.  **Reaction:** But-1-ene ($CH_3CH_2CH=CH_2$) reacts with diborane ($B_2H_6$) followed by oxidation with alkaline hydrogen peroxide ($H_2O_2/OH^-$).
2.  **Mechanism:** 
    *   **Hydroboration:** Boron adds to the double bond. To minimize steric hindrance, the boron atom attaches to the less hindered carbon atom (the terminal carbon). This step effectively adds 'H' and 'B' across the double bond in an anti-Markovnikov fashion.
    *   **Oxidation:** The organoborane intermediate is oxidized by $H_2O_2$ in the presence of a base ($OH^-$). The C-B bond is replaced by a C-OH bond with retention of configuration.
3.  **Net Result:** The overall reaction corresponds to the addition of water ($H_2O$) to the double bond contrary to Markovnikov's rule (Anti-Markovnikov hydration). Therefore, the -OH group attaches to the primary carbon.
4.  **Product:** Butan-1-ol ($CH_3CH_2CH_2CH_2OH$).

(Contrast: Acid-catalyzed hydration would yield Butan-2-ol via Markovnikov addition).

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