Chemistry › Hydrocarbons › Reactions of alkynes
The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne.…
The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$ $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$ $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $\mathrm{CH}_3 \mathrm{C} \equiv \equiv \mathrm{CH}$ $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$ Solution
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2$ $-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ There are two (marked) possibility of halides
Thus, in the given context
$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \longrightarrow \\
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \overline{\mathrm{CNa}^{+}}
\end{array}
$
$
\begin{aligned}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \overline{\mathrm{CNa}} \\
\longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \\
\text { 3-octyne } \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
\end{aligned}
$
Hydrocarbon
Conceptual, reaction understanding II
Asked in: JEE Advanced 2010 (Paper 1)
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