Carbonyl CompoundMedium
Question
Identify the correct reagents that would bring about the following transformation.
Options
A.(i) ${BH}_{3}\ $(ii) $H_{2}O_{2}/\overset{\ominus}{O}H\ \ \ $(iii) alk. ${KMnO}_{4}$ (iv) $H_{3}O^{\oplus}$
B.(i) $H_{2}O/H^{+}$ (ii) PCC
C.(i) $H_{2}O/H^{+}$ (ii) ${CrO}_{3}$
D.(i) ${BH}_{3}$ (ii) $H_{2}O_{2}/\overset{\ominus}{O}H$ (iii) PCC
Solution
Sol. [Key Concept: Hydroboration followed by controlled oxidation]
Hydroboration–oxidation with (i) BH₃ and (ii) H₂O₂/OH⁻ places OH at the terminal carbon of C₆H₁₁–CH₂–CH=CH₂, giving C₆H₁₁–CH₂–CH₂–CH₂OH. PCC then oxidises this primary alcohol to C₆H₁₁–CH₂–CH₂–CHO without the usual further oxidation to an acid. The required sequence is option (4).
Balanced stages, with R = C₆H₁₁–CH₂:
Hydroboration: 3 R–CH=CH₂ + BH₃ → B(CH₂CH₂R)₃.
Oxidation: B(CH₂CH₂R)₃ + 3 H₂O₂ + OH⁻ → 3 R–CH₂CH₂OH + B(OH)₄⁻.
PCC then converts R–CH₂CH₂OH into R–CH₂CHO.
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