Synthesis of 3 Octyne Using Alkynide Ion

To synthesize an internal alkyne via the alkylation of a terminal alkyne, the terminal alkyne is deprotonated by a strong base like sodium amide ($\text{NaNH}_2$) to form an alkynide ion (acetylide ion), which then undergoes an $\text{S}_\text{N}2$ nucleophilic substitution with a primary bromoalkane:

$\text{R}'-\text{C}\equiv\text{C}-\text{H} \xrightarrow{\text{NaNH}_2} \text{R}'-\text{C}\equiv\text{C}^- \text{Na}^+ + \text{NH}_3$
$\text{R}'-\text{C}\equiv\text{C}^- \text{Na}^+ + \text{R}-\text{Br} \xrightarrow{\text{S}_\text{N}2} \text{R}'-\text{C}\equiv\text{C}-\text{R} + \text{NaBr}$

The target molecule is 3-octyne, which has the following structural formula:
$\overset{1}{\text{C}}\text{H}_3-\overset{2}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\equiv\overset{4}{\text{C}}-\overset{5}{\text{C}}\text{H}_2-\overset{6}{\text{C}}\text{H}_2-\overset{7}{\text{C}}\text{H}_2-\overset{8}{\text{C}}\text{H}_3$

The carbon fragments attached to the carbon-carbon triple bond are:

• An ethyl group ($\text{CH}_3\text{CH}_2-$) containing $2$ carbons.
• A butyl group ($-\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$) containing $4$ carbons.

Thus, 3-octyne can be synthesized via one of two combinations:

1. 1-Butyne ($\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$) and 1-bromobutane ($\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_3$):
   $\text{CH}_3\text{CH}_2\text{C}\equiv\text{C}^- + \text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \longrightarrow \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}-\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \quad (\text{3-octyne})$

2. 1-Hexyne ($\text{CH}_3(\text{CH}_2)_3\text{C}\equiv\text{CH}$) and bromoethane ($\text{BrCH}_2\text{CH}_3$).

Evaluating the given options:

• Option A: $\text{Br(CH}_2)_4\text{CH}_3$ ($5$ carbons) and $\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$ ($4$ carbons) yields 3-nonyne ($9$ carbons).
• Option B: $\text{Br(CH}_2)_2\text{CH}_3$ ($3$ carbons) and $\text{CH}_3(\text{CH}_2)_2\text{C}\equiv\text{CH}$ ($5$ carbons) yields 4-octyne.
• Option C: $\text{Br(CH}_2)_4\text{CH}_3$ ($5$ carbons) and $\text{CH}_3\text{C}\equiv\text{CH}$ ($3$ carbons) yields 2-octyne.
• Option D: $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ ($4$ carbons) and $\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$ ($4$ carbons) correctly yields 3-octyne.

Therefore, the correct choice is D.