This question involves the synthesis of 3-octyne using terminal alkyne alkylation, a fundamental reaction in organic chemistry. The process uses sodium amide (NaNH₂) to deprotonate a terminal alkyne, forming an acetylide ion, which then acts as a nucleophile in an S_N2 reaction with a primary alkyl halide to form a new carbon-carbon bond and a longer alkyne.

Step 1: Understanding the Target Molecule
3-Octyne has the structure CH₃CH₂C≡CCH₂CH₂CH₂CH₃. The triple bond is between carbons 3 and 4. This internal alkyne can be synthesized by coupling two smaller fragments: an alkyl halide and a terminal alkyne.

Step 2: The General Reaction Mechanism
The synthesis follows this general pattern: $\text{R-C}\equiv \text{C-H}+\text{NaNH}{2}_{}\to \text{R-C}\equiv \text{C}{-}^{}\text{Na}{+}^{}+\text{NH}{3}_{}$
$\text{R-C}\equiv \text{C}{-}^{}+\text{R'-X}\to \text{R-C}\equiv \text{C-R'}+\text{X}{-}^{}$
To form 3-octyne (R-C≡C-R'), the acetylide ion (R-C≡C^–) must come from a terminal alkyne that provides the left part of the molecule, and the alkyl halide (R'-X) must provide the right part. The carbon atoms directly involved in the new bond are the terminal alkyne carbon and the carbon from the alkyl halide.

Step 3: Retrosynthetic Analysis of 3-Octyne
We break the molecule at the triple bond. The carbon chain can be disconnected in two ways. The correct disconnection is to identify which part was the original terminal alkyne. The alkylation adds the alkyl group from the halide to the terminal carbon of the alkyne. Therefore, the terminal alkyne used will become one side of the product, and the alkyl halide will become the other.
For CH₃CH₂-C≡C-CH₂CH₂CH₂CH₃:
Option A: The alkyne portion could be CH₃CH₂C≡CH (providing 3 carbons) and the alkyl halide could be BrCH₂CH₂CH₂CH₃ (providing 4 carbons). 3 + 4 = 7 carbons, but 3-octyne has 8 carbons. This is incorrect.
Option B: The correct disconnection is to see that the left side of the triple bond (carbons 1-3) comes from the terminal alkyne: CH₃CH₂C≡CH (1-pentyne). The right side of the triple bond (carbons 4-8) must come from the alkyl halide. Carbon 4 is the carbon that was attacked; therefore, the alkyl halide must be BrCH₂CH₂CH₂CH₃ (1-bromobutane). This gives a total of 3 (from alkyne) + 5 (from halide, including the attacked carbon) = 8 carbons.

Step 4: Evaluating the Options
Let's analyze each pair:
1. BrCH₂CH₂CH₂CH₂CH₃ (1-bromopentane) and CH₃CH₂C≡CH (1-pentyne) would yield CH₃CH₂C≡CCH₂CH₂CH₂CH₂CH₃ (3-decyne).
2. BrCH₂CH₂CH₃ (1-bromopropane) and CH₃CH₂CH₂C≡CH (1-hexyne) would yield CH₃CH₂CH₂C≡CCH₂CH₂CH₃ (4-nonyne).
3. BrCH₂CH₂CH₂CH₂CH₃ (1-bromopentane) and CH₃C≡CH (propyne) would yield CH₃C≡CCH₂CH₂CH₂CH₂CH₃ (2-heptyne).
4. BrCH₂CH₂CH₂CH₃ (1-bromobutane) and CH₃CH₂C≡CH (1-pentyne) would yield CH₃CH₂C≡CCH₂CH₂CH₂CH₃ (3-octyne).

Final Answer: The correct bromoalkane and alkyne are BrCH₂CH₂CH₂CH₃ and CH₃CH₂C≡CH, respectively.

Related Topics & Formulae

Alkyne Alkylation: This is a key method for elongating carbon chains in organic synthesis. The reaction requires a strong base like NaNH₂ to generate the nucleophilic acetylide ion from a terminal alkyne. The alkyl halide must be primary to favor the S_N2 mechanism and avoid elimination.

General Formula: $\text{R-C}\equiv \text{C-H}+\text{R' - X}\underset{\text{1. NaNH}}{\overset{2_{}}{\to }}\text{R-C}\equiv \text{C-R'}$
Where R' - X is a primary alkyl halide.