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Reductions Involving Aldehydes or Ketones

In the above ...

Question

In the above reaction, X and Y respectively are

B_{2} H_{6} / H_{2} O_{2} / O H^{-} and P C C / C H_{2} C l_{2}

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H_{2} O / H^{+}

and

C r O_{3}

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B_{2} H_{6}

and

P C C

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P C C / C H_{2} C l_{2} and B_{2} H_{6} / H_{2} O_{2} / O H^{-}

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Solution

The correct option is A $B_{2} H_{6} / H_{2} O_{2} / O H^{-} and P C C / C H_{2} C l_{2}$

The reagents $B H_{3}, T H F$ followed by $N a O H, H_{2} O_{2}$ are hydroxylating agents. This reaction is called hydroboration-oxidation reatction. Alkenes undergo oxidation and anti-markovnikov's alcohol is the end product.

PCC(Pyridinium chlorochromate) is a mild oxidizing agent. It converts $1^{o}$ alcohols to aldehydes and $2^{o}$ alcohols to ketones, but is not strong enough to convert primary alcohols to carboxylic acids.
The reaction is as follows:
Hence, option (a) i.e. $B_{2} H_{6} / H_{2} O_{2} / O H^{-} and P C C / C H_{2} C l_{2}$ is correct.

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