Question: 51. Identify the correct reagents that would bring about the following transformation
(1) (i) \mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+} (ii) \mathrm{CrO}_{3}
(2) (i) \mathrm{BH}_{3} (ii) \mathrm{H}_{2} \mathrm{O}_{2} / \stackrel{\ominus}{\mathrm{O}} \mathrm{H} (iii) PCC
(3) (i) \mathrm{BH}_{3} (ii) \mathrm{H}_{2} \mathrm{O}_{2} / \stackrel{\ominus}{\mathrm{O}} \mathrm{H} (iii) Alk. \mathrm{KMnO}_{4} (iv) \mathrm{H}_{3} \mathrm{O}^{\oplus}
(4) (i) \mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+} (ii) PCC
Answer: Option (2)
Explanation:
The given alkene undergoes transformation to an aldehyde, indicating anti-Markovnikov addition followed by controlled oxidation.
In the first step, treatment with \mathrm{BH}_{3} followed by
\mathrm{H}_{2} \mathrm{O}_{2} / \stackrel{\ominus}{\mathrm{O}} \mathrm{H}leads to hydroboration–oxidation of the alkene.
This reaction adds \mathrm{OH} group at the terminal carbon following anti-Markovnikov rule, forming a primary alcohol.
In the next step, oxidation of the primary alcohol with PCC converts it selectively into an aldehyde without further oxidation to carboxylic acid.
Thus, the correct sequence of reagents is hydroboration–oxidation followed by PCC oxidation.
Therefore, the correct answer is Option (2).