Find missing reagents.

x = {L i A l H}_{4}, y = {N a B H}_{4}

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x = {L i A l H}_{4} ╱ {A l C l}_{3}, y = {L i A l H}_{4}

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x = {L i A l H}_{4}, y = {L i A l H}_{4} {╱ A l C l}_{3}

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x = H_{2} ╱ N i, y = H_{2} ╱ P t

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Solution

The correct option is B $x = {L i A l H}_{4}, y = {L i A l H}_{4} {╱ A l C l}_{3}$

(Refer to Image)
$2^{o}$ alcohol is formed in presence of $A l C l_{3}$ because $A l C l_{3}$ attaches to ester oxygen and prefers less steric hindrance side.

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B C l_{3} + L i A l H_{4} \to X + L i C l + A l C l_{3}

X + H_{2} O \to H_{2} + Y R e d h e a t \to Z

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B e {C l}_{2} + L i A l H_{4} ⟶ X + L i C l + A l {C l}_{3}

N a B H_{4} + I_{2} \to X + Y + Z

B F_{3} + N a H 450 K - -- \to X + P

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N a B H_{4} + I_{2} \to X + Y + Z

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