Which reagent is used for the following conversion?

$R - C H_{2} - C l | C | C l - R ⟶ R - C \equiv C - R$ ,

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aq. KOH

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alcoholic KOH followed by NaNH

_{2}

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Solution

The correct option is C alcoholic KOH followed by NaNH $_{2}$
Vicinal dihalides on dehydrohalogenation gives alkynes. The reagents used for the conversion are alcoholic KOH followed by sodaamide.
$R - C H_{2} - C l | C | C l - R (a l c .) K O H - ----- \to R - C H = C l | C - R N a N H_{2} - ---- \to R - C \equiv C - R$ ,

So, the correct option is $C$

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R - C H_{2} - C C l_{2} - R ⟶ R - C H = C C l - R

R - C H_{2} - O H \to R - C H O :

R - C H_{2} - O H \to R - C H_{3} :

R - O | | C - C l H_{2} - ---- \to R e a g e n t R - O | | C - H

The most suitable reagent for the conversion of $R - C H_{2} - O H \to R - C H O$ is

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