1. Principle used

A terminal alkyne reacts with sodium amide, $NaNH_2$, to form an acetylide ion:

$R-C \equiv C-H \xrightarrow{NaNH_2} R-C \equiv C^- Na^+$

This acetylide ion then undergoes $S_N2$ substitution with a primary bromoalkane:

$R-C \equiv C^- + R'-Br \rightarrow R-C \equiv C-R'$

So, to synthesize 3-octyne, we must choose:

• a terminal alkyne, and
• a primary bromoalkane,

such that the final product is:

$CH_3CH_2-C \equiv C-CH_2CH_2CH_2CH_3$

This is oct-3-yne (3-octyne).

2. Analyze the target molecule

3-octyne has 8 carbons total and the triple bond between C-3 and C-4:

$CH_3CH_2-C \equiv C-CH_2CH_2CH_2CH_3$

This can be formed by joining:

• an ethyl-substituted terminal alkyne: $CH_3CH_2C \equiv CH$ (1-butyne), and
• a butyl halide: $BrCH_2CH_2CH_2CH_3$ (1-bromobutane).

Reaction:

$CH_3CH_2C \equiv CH \xrightarrow{NaNH_2} CH_3CH_2C \equiv C^-$

Then:

$CH_3CH_2C \equiv C^- + BrCH_2CH_2CH_2CH_3 \rightarrow CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3$

This is exactly 3-octyne.

3. Check each option

Option A

Bromoalkane: $BrCH_2CH_2CH_2CH_2CH_3$ (1-bromopentane)

Alkyne: $CH_2CH_2C \equiv CH$

This alkyne as written is intended to be a 4-carbon terminal alkyne only if interpreted properly, but the notation is unclear and not the standard correct condensed formula. Even if interpreted as a terminal alkyne chain, combining pentyl + corresponding acetylide would not match the needed structure for 3-octyne.

So, A is not correct.

Option B

Bromoalkane: $BrCH_2CH_2CH_3$ (1-bromopropane)

Alkyne: $CH_3CH_2CH_2C \equiv CH$ (1-pentyne)

Product formed:

$CH_3CH_2CH_2C \equiv CCH_2CH_2CH_3$

This is an 8-carbon alkyne, but numbering from either end gives the triple bond at carbon 4:

$\text{oct-4-yne}$

So, B is incorrect.

Option C

Bromoalkane: $BrCH_2CH_2CH_2CH_2CH_3$ (1-bromopentane)

Alkyne: $CH_3C \equiv CH$ (propyne)

Product formed:

$CH_3C \equiv CCH_2CH_2CH_2CH_2CH_3$

This is 8 carbons total. Numbering to give the triple bond lowest locant gives triple bond at carbon 2:

$\text{oct-2-yne}$

So, C is incorrect.

Option D

Bromoalkane: $BrCH_2CH_2CH_2CH_3$ (1-bromobutane)

Alkyne: $CH_3CH_2C \equiv CH$ (1-butyne)

Product formed:

$CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3$

Numbering from the nearer end gives the triple bond at carbon 3:

$\text{3-octyne}$

So, D is correct.

4. Final answer

The correct combination is:

$BrCH_2CH_2CH_2CH_3 \text{ and } CH_3CH_2C \equiv CH$

Hence, the correct option is D.