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

Solution :

Step 1: Analysis of the Transformation

The goal is to convert a primary alkyl halide (1-bromopropane, $CH_3CH_2CH_2Br$) into a secondary alkyl halide (2-bromopropane, $CH_3CHBrCH_3$). This requires two sequential reactions: dehydrohalogenation to form an alkene, followed by hydrohalogenation (electrophilic addition) governed by Markovnikov’s rule.

Step 2: Elimination of Hydrogen Bromide (Dehydrohalogenation)

Treat 1-bromopropane with an alcoholic solution of potassium hydroxide ($KOH$). This promotes a $\beta$-elimination reaction (specifically an $E2$ mechanism) to produce propene.

$CH_3-CH_2-CH_2-Br + alc. KOH \rightarrow CH_3-CH=CH_2 + KBr + H_2O$

[Justification: Alcoholic $KOH$ acts as a strong base, abstracting the $\beta$-hydrogen to form the double bond.]

Step 3: Electrophilic Addition of Hydrogen Bromide

To obtain 2-bromopropane, react the propene intermediate with hydrogen bromide ($HBr$). According to Markovnikov’s Rule, the negative part of the addendum ($Br^-$) attaches to the carbon atom with fewer hydrogen atoms.

$CH_3-CH=CH_2 + HBr \rightarrow CH_3-CHBr-CH_3$

[Justification: The reaction proceeds through a more stable secondary carbocation intermediate ($CH_3-CH^+-CH_3$) compared to the primary carbocation, ensuring regioselectivity for the 2-bromopropane product.]

Visual Representation of the Reaction Pathway

CH₃-CH₂-CH₂Br alc. KOH CH₃-CH=CH₂ HBr CH₃-CH(Br)-CH₃ Step 1: Elimination Step 2: Markovnikov Addition

Summary of Reaction Conditions

Step Reagent Mechanism
1 $alc. KOH, \Delta$ $\beta$-Elimination ($E2$)
2 $HBr$ (gaseous or liquid) Electrophilic Addition

Final Solution: The conversion is achieved by first treating 1-bromopropane with alcoholic $KOH$ to yield propene, followed by the addition of $HBr$ to the alkene, which regioselectively produces 2-bromopropane via Markovnikov's rule.


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