In the reaction,H_3C-Cequiv C-H xrightarrow[2 | vedclass

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Question

(C) The reaction sequence is as follows:$1$. Propyne $(CH_3C\equiv CH)$ reacts with $NaNH_2$ to form the sodium salt of propyne $(CH_3C\equiv C^-Na^+)

Vedclass · Accepted Answer

$x= CH_3I$; $y= Na$ in liq. $NH_3$

Vedclass · Answer

(C) The reaction sequence is as follows:$1$. Propyne $(CH_3C\equiv CH)$ reacts with $NaNH_2$ to form the sodium salt of propyne $(CH_3C\equiv C^-Na^+)$.$2$. This salt reacts with methyl iodide $(CH_3I)$ as reagent $x$ to perform an $S_N2$ reaction,yielding but$-2-$yne $(CH_3C\equiv CCH_3)$.$3$. But$-2-$yne is then reduced to trans-but$-2-$ene using sodium in liquid ammonia $(Na/liq. NH_3)$ as reagent $y$ (Birch reduction).Thus,$x= CH_3I$ and $y= Na/liq. NH_3$.

In the reaction,$H_3C-C\equiv C-H$ $\xrightarrow[2. x]{1. NaNH_2, \Delta}$ $\xrightarrow{3. y} H_3C-CH=CH-CH_3$ (trans form),$x$ and $y$,respectively are

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