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Q6.19(xix):
How the following conversions can be carried out? (xix) tert-Butyl bromide to isobutyl bromide

Solution :

Given Variables & Initial Setup

The objective is to convert tert-butyl bromide (2-bromo-2-methylpropane) to isobutyl bromide (1-bromo-2-methylpropane). The conversion involves shifting the functional group ($-Br$) from a tertiary carbon to a primary carbon via an alkene intermediate.

Reactant: $(CH_3)_3C-Br$
Product: $(CH_3)_2CH-CH_2-Br$

Step 1: Dehydrohalogenation to form an Alkene

To move the bromine atom, we must first eliminate it to create 2-methylpropene. This is achieved using an alcoholic potassium hydroxide ($KOH$) solution, which induces an $E2$ elimination mechanism [Per Saytzeff's Rule/Zaitsev's Rule].

$(CH_3)_3C-Br + alc. KOH \xrightarrow{\Delta} CH_2=C(CH_3)_2 + KBr + H_2O$

This reaction creates an alkene intermediate: 2-methylpropene.

Step 2: Anti-Markovnikov Hydrobromination

The addition of $HBr$ to an unsymmetrical alkene typically follows Markovnikov's rule. To place the bromine on the terminal carbon (primary position), we must utilize the Kharasch Effect (Peroxide effect). In the presence of benzoyl peroxide, the mechanism switches from electrophilic addition to a free-radical addition [Per Free Radical Addition Mechanism].

$CH_2=C(CH_3)_2 + HBr \xrightarrow{Peroxide} (CH_3)_2CH-CH_2-Br$

Analytical Summary of the Conversion Sequence

Step Reagents Used Transformation Mechanism/Principle
1 $alc. KOH, \Delta$ Elimination Dehydrohalogenation ($E2$)
2 $HBr, (C_6H_5CO)_2O_2$ Addition Anti-Markovnikov Radical Addition

Visual Representation of the Molecular Transition

tert-Butyl bromide (CH3)3C-Br 2-Methylpropene CH2=C(CH3)2 Isobutyl bromide (CH3)2CHCH2Br

Final Solution: The conversion is achieved in two steps: (1) Treatment with alcoholic $KOH$ to form 2-methylpropene via dehydrohalogenation, followed by (2) treatment with $HBr$ in the presence of an organic peroxide to yield 1-bromo-2-methylpropane (isobutyl bromide) via anti-Markovnikov addition.


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