$C H_{3} C O O H L i A i H_{4} ⟶ (A)$ ,
$(A) + C H_{3} C O O H H_{3} O^{+} ⟶ (B) + H_{2} O$
In the given reaction 'A' and 'B' respectively are ?

C H_{3} C O O C_{2} H_{5}, C_{2} H_{5} O H

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C H_{3} C H O, C_{2} H_{5} O H

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C_{2} H_{5} O H, C H_{3} C H O

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C_{2} H_{5} O H, C H_{3} C O O C_{2} H_{5}

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Solution

The correct option is D $C_{2} H_{5} O H, C H_{3} C O O C_{2} H_{5}$
Since $L i A l H_{4}$ is a reducing agent it will convert $C H_{3} C O O H$ into alcohol $(C H_{3} C H_{2} O H)$ and we know that reaction of alcohol and acid leads to the formation of esters hence (B) is $C H_{3} C O O C_{2} H_{5}$ . The reaction is given below:
$C H_{3} C O O H L i A l H_{4} ⟶ C H_{3} C H_{2} O H (A)$
$(B)$
Hence, the correct option is (d).

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Similar questions

C H_{3} C O O H L i A L H_{3} - ---- \to (A)

(A) + C H_{3} C O O H H_{3} O^{+} - -- \to (B) + H_{2} O

C H_{3} C O O H L i A i H_{4} ⟶ (A)

(A) + C H_{3} C O O H H_{3} O^{+} ⟶ (B) + H_{2} O

C H_{3} C H_{2} C O N H_{2} A \to C H_{3} - C H_{2} - N H_{2} B \to C H_{3} - C H_{2} - O H

A

B

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