Which of the following reagent(s) result in syn-addition of hydrogen atoms to a triple bond?

Lindlar Catalyst

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N a / l i q . N H_{3}

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N i / P t / P d and H_{2} (g)

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Baeyer's reagent

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Solution

The correct option is C $N i / P t / P d and H_{2} (g)$
$Lindlar catalyst \to H_{2} / P d (C a C O_{3}, Q u i n o l i n e)$
Here, the alkyne will undergo partial reduction to form cis-alkene. Both the $H$ atom gets add to the alkyne in same direction to give cis- product.

Reduction of alkyne using $N a / l i q . N H_{3}$ is called Birch reduction. In this $H$ gets add to the triple bond in opposite direction to give trans product.

Addition of $H_{2}$ to unsaturated hydrocarbon in the presence of catalyst $(N i / P t / P d)$ is called catalytic hydrogenation. This reaction is also called Sabatier Senderens reaction. This reaction goes through a four membered cyclic transition state to give a syn addition of $H_{2}$ .

$Baeyer's reagent \to Alkaline cold d i l . K M n O_{4}$ . This reagent oxides the alkene reactant by syn addition of $O H$ .
It reacts with ethene to form Ethane-1,2-diol (glycol).

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