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Valency and Oxidation State

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Question

Choose the best reagent to carry out the above transformation:

\begin{matrix} I & I I & I I I A c e t y l e n e & H_{2} / P d / B a S O_{4} & O s O_{4} / N a H S O_{3} N a N H_{2} / N H_{3} \end{matrix}

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\begin{matrix} I & I I & I I I A c e t y l e n e & H_{2} / L i n d l a r & B H_{3} / N a O H, H_{2} O_{2} N a N H_{2} / N H_{3} & c a t a l y s t \end{matrix}

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\begin{matrix} I & I I & I I I A c e t y l e n e & L i / N H_{3} & K M n O_{4} / N a O H N a N H_{2} / N H_{3} \end{matrix}

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\begin{matrix} I & I I & I I I A c e t y l e n e & H_{2} / N i & O s O_{4} / N a H S O_{3} N a N H_{2} / N H_{3} \end{matrix}

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Solution

The correct option is A $\begin{matrix} I & I I & I I I A c e t y l e n e & H_{2} / P d / B a S O_{4} & O s O_{4} / N a H S O_{3} N a N H_{2} / N H_{3} \end{matrix}$
The best reagents to carry out the given transformation are I. Acetylene, $N a N H_{2} / N H_{3}$ , II. $H_{2} / P d / B a S O_{4}$ and III. $O s O_{4} / N a H S O_{3}$
The acidic H atoms of acetylene are abstracted with soda amide to form a dianion. This dianion then attacks two moles of n propyl bromide to form two C-C single bonds with loss of two bromide ions.

In the next step, carbon carbon triple bond is reduced to cis C=C double bond by hydrogenation with Pd in presence of barium sulphate. In the last step, C=C double bond is converted to a vicinal diol.

Hence, the correct option is A.

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