Sample of 2, 3-dibromo-3-methylpentane is heated with zinc dust. The resulting product is isolated and heated with HI in the presence of phosphorus. Indicate which is the structure that represent the final organic product formed in the reaction

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Detailed Solution

Answer: 3-iodo-3-methylpentane.

Zn, heat removes the two vicinal bromines (reductive dehalogenation) to give the alkene:

2,3-dibromo-3-methylpentane → 3-methyl-2-pentene (CH₃–CH= C(CH₃)–CH₂–CH₃).

HI (with P) adds across the C=C by Markovnikov addition. Protonation of the double bond gives the more stable tertiary carbocation at C-3, and iodide (I⁻) then attacks that carbocation.

3-methyl-2-pentene + HI → 3-iodo-3-methylpentane (CH₃–CH₂–C(I)(CH₃)–CH₂–CH₃).

Reason: Zn converts the vicinal dihalide to the corresponding alkene; HI adds to the alkene so that iodine ends up on the more substituted (tertiary) carbon (carbocation stability → Markovnikov product).

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