Which of the following reagents are not correctly matched with the reaction?

C H_{3} C H = C H C H O \to C H_{3} C H = C H C O O H

: Ammoniacal

A g N O_{3}

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C H_{3} C H = C H C H O \to C H_{3} C H = C H C H_{2} O H : H_{2} / P t

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R - C O O H \to R - C H_{2} O H : N a B H_{4}

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C H_{3} C H_{2} C O C l \to C H_{3} C H_{2} C H O : H_{2}, P d / B a S O_{4}

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Solution

The correct option is B $C H_{3} C H = C H C H O \to C H_{3} C H = C H C H_{2} O H : H_{2} / P t$
$C H_{3} C H = C H C H O H_{2} / P t - --- \to C H_{3} C H_{2} C H_{2} C H_{2} O H$

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Similar questions

C H_{3} C H = C H C H O \to C H_{3} C H = C H C O O H : Ammoniacal A g N O_{3}

C H_{3} C H = C H C H O \to C H_{3} C H = C H C H_{2} O H : K M n O_{4}

R - C H O \to R - C H_{2} O H : N a B H_{4}

C H_{3} C H_{2} C O C l \to C H_{3} C H_{2} C H O : H_{2}, P d / B a S O_{4}

{C H}_{3} C H = C H C H O

{C H}_{3} - C H = C H C O O H

P h C H = C H C H O + C H_{3} C H = C H C H O base - ---- \to E t O H, Δ (A) 87 %

C H_{3} C H = C H C H O

P h - C H = C H C H O + C H_{3} C H = C H C H O b a s e ⟶ E t O H, Δ (A)

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