In the following reaction sequence, X and Y, respectively, are:

H_{2} O_{2}; L i A l H_{4}

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C_{6} H_{5} C O O O H; L i A l H_{4}

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C_{6} H_{5} C O O O H; Z n / H g . H C l

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Alkaline

K M n O_{4}; L i A l H_{4}

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Solution

The correct option is A $C_{6} H_{5} C O O O H; L i A l H_{4}$
Peracids convert alkene to epoxide. It is an oxidation reaction.
Lithium aluminium hydride opens epoxide ring to form an alcohol. The -OH group goes to more substituted C atom.
In the given reaction sequence, X and Y, are $C_{6} H_{5} C O O O H L i A l H_{4}$ respectively.

Hence, option (B) is the correct answer.

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Similar questions

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

X, Y

Z

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

A + H_{2} O \to B + H_{2}

B \to C

P h C H_{2} C l N a C N ⟶ L i A l H_{4} ⟶ (C H_{3} C O)_{2} O ⟶

C H_{3} - B r K C N - -- \to A H_{3} O^{+} - -- \to B L i A l H_{4} - --- \to e t h e r C

C H_{3} C O O H + L i A l H_{4}

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