H2O18 - Na -baseA - B CH-3 - O-C - O - CH2 - CH3 - A - C - D alcoholProduct C of the reaction is-

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Base Catalysed Hydrolysis of Esters

H2O18 + Na →b...

Question

$H_{2} O^{18} + N a \to (A) (b a s e) + (B)$
$C H - 3 - O | | C - O - C H_{2} - C H_{3} + (A) \to (C) + (D)$ alcohol
Product $(C)$ of the reaction is:

C H_{3} - O | | C - O 18 - H

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C H_{3} - O^{18} | | C - O - H

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C H_{3} - O | | C - 18 O

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C H_{3} - O | | C - O^{⊖} N \oplus a

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C H_{3} - O | | C - O - C H_{2} - C H_{3} + H - O \to (O = O_{18})

C H_{3} - O | | C - C H_{3} (A), C H_{3} - O | | C - H (B), C H_{3} - O | | C - C H_{2} - C l (C), H - O | | C - H (D), C l - C H_{2} - O | | C - H (E)

C H_{2} = O H | C - C H_{2} - O | | C - C H_{3} ⇌ C H_{3} - O | | C - C H_{2} - O | | C - C H_{3} ⇌ C H_{3} - O H | C = C H - O | | C - C H_{3}

I

I I

I I I

C H_{2} = O H | C - C H_{2} (I) - O | | C - C H_{3} ⇌ C H_{3} - O | | C - C H_{2} (I I) - O | | C - C H_{3}

⇌ C H_{3} - O H | C = C H (I I I) - O | | C - C H_{3}

I . H N = C = O \leftrightarrow H - O - C \equiv N

I I . ¯ N = C = O \leftrightarrow N \equiv C - O^{-}

I I I . C H_{3} - O | | C - ¯ C H_{2} \leftrightarrow C H_{3} -^{-} C = C H_{2}

I V . C H_{3} - O | | C - C H_{3} \leftrightarrow C H_{3} - O H | C = C H_{2}

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