The final product formed on addition of excess of hydrogen bromide to propyne in presence of benzoyl peroxide is .

2-Bromopropene

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2- Bromopropane

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2, 3-Dibromopropane

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1, 2-Dibromopropane

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Solution

The correct option is D 1, 2-Dibromopropane
Addition of hydrogen bromide to propyne in presence of benzoyl peroxide follows anti-Markovnikov's rule. Thus, the H atom of halogen acid bonds to the carbon atom with the least number of hydrogen atoms in the unsaturated hydrocarbon and the halogen atom bonds to the other carbon atom.
It takes place as follows:

$C H_{3} - C \equiv C H excess HBr - --------- \to Berzoyl peroxide C H_{3} - C H = C H B r$
Now, as both the carbon atoms have one H atom attached, there is equal probability of getting H atom added to each C atom.
$C H_{3} - C H = C H B r H B r - --------- \to B e n z o y l p e r o x i d e C H_{3} C H_{2} C H B r_{2} 1, 1 - D i b r o m o p r o p o n e + C H_{3} B r | C H - B r | C H 1, 2 - Dibromo propane$

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