Which is the most suitable reagent for the following conversion?
$C H_{3} - C H = C H - C H_{2} - O | | C - C H_{3} \to C H_{3} - C H = C H - C H_{2} - O | | C - O H$

Tollen's reagent

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Benzoyl peroxide

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I_{2}

and

N a O H

solution

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S n

and

N a O H

solution

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Solution

The correct option is C $I_{2}$ and $N a O H$ solution

This reaction is called as lodoform reaction.

In this reaction the $- C O C H_{3}$ group gets oxidised to $- C O O N a$ without affecting the double bond and $C H I_{3}$ (yellow crystals) are formed. $- C O O N a$ gets converted into -COOH in the solution.

Option C is correct.

Suggest Corrections

Similar questions

C H_{3} - C H = C H - C H_{2} - O ∥ C - C H_{3} \to C H_{3} - C H = C H - C H_{2} - O ∥ C - O H

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

C H_{3} - O H | C H - C H_{2} - C H_{2} - O | | C - C H_{3} \to C H_{3} - O H | C H - (C H_{2})_{3} - C H_{3}

C H_{3} - C \equiv C - C H_{3}

C H_{3} - C H = C H - C H_{3}

C H_{3} - C H_{2} C \equiv C H

C H_{3} - C H = C H_{2}

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