$C H_{3} C O O H L i A i H_{4} ⟶ (A)$ ,
$(A) + C H_{3} C O O H H_{3} O^{+} ⟶ (B) + H_{2} O$
Choose the correct option(s):

A is ethanal

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A is ethanol

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B is ethyl acetate

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B is propyl formate

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Solution

The correct option is C B is ethyl acetate
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:
$C H_{3} C O O H L i A l H_{4} ⟶ C H_{3} C H_{2} O H (A)$

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Similar questions

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 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

A = {3, 5}

A \times B = B \times A

| \to a + \to b | = | \to a - \to b |

A, B, C, D a n d E

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