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Q6.19(xvi):
How the following conversions can be carried out? (xvi) 2-Bromopropane to 1-bromopropane

Solution :

Step 1: Dehydrohalogenation of 2-Bromopropane

To convert 2-bromopropane ($CH_3-CHBr-CH_3$) to 1-bromopropane ($CH_3-CH_2-CH_2Br$), we must first remove the halogen atom to create an alkene, which serves as a reactive intermediate. 2-bromopropane is treated with an alcoholic solution of potassium hydroxide ($KOH$).

Reaction: $CH_3-CHBr-CH_3 + alc. KOH \xrightarrow{\Delta} CH_3-CH=CH_2 + KBr + H_2O$

[Justification: This is a $\beta$-elimination reaction (dehydrohalogenation), which follows Saytzeff's rule, though in this symmetrical secondary alkyl halide, it yields propene.]

Step 2: Anti-Markovnikov Addition of HBr (Peroxide Effect)

Once propene ($CH_3-CH=CH_2$) is obtained, we perform a radical addition of hydrogen bromide ($HBr$) in the presence of benzoyl peroxide ($C_6H_5CO)_2O_2$. This forces the bromine atom to attach to the terminal carbon atom (the less substituted carbon).

Reaction: $CH_3-CH=CH_2 + HBr \xrightarrow{Peroxide} CH_3-CH_2-CH_2Br$

[Justification: This follows the Kharasch effect or Peroxide Effect, where the addition of $HBr$ to an unsymmetrical alkene proceeds via a free-radical mechanism, resulting in anti-Markovnikov orientation.]

Visual Representation of Reaction Sequence

2-Bromopropane alc. KOH Propene HBr / Peroxide 1-Bromopropane

Summary of Reaction Conditions

Step Reagents/Conditions Mechanism Type
1 $alc. KOH, \Delta$ Dehydrohalogenation ($E2$ mechanism)
2 $HBr, (C_6H_5CO)_2O_2$ Free Radical Addition (Anti-Markovnikov)

Final Solution: The conversion of 2-bromopropane to 1-bromopropane is achieved in two steps: first by elimination to propene using $alc. KOH$, followed by the peroxide-catalyzed anti-Markovnikov addition of $HBr$.


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