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Chapter 27: Problem 44

Write the initiation, propagation, and termination steps involved in the monobromination of \(2,3-\) dimethylbutane to give 2 -bromo- 2,3 -dimethylbutane.

Short Answer

Expert verified
The bromination of 2,3-dimethylbutane involves three steps. Initiation: Br2 becomes 2Br* under light. Propagation: The bromine radical abstracts hydrogen from 2,3-dimethylbutane to form a primary radical, which reacts with another bromine molecule to form 2 -bromo- 2,3 -dimethylbutane. Termination: Radicals recombine to form neutral species, breaking the chain reaction.

Step by step solution

01

Initiation

This step involves homolytic fission. Here, the bond in the bromine molecule (Br2) breaks evenly to form two bromine radicals. Hence the equation is \[ Br2 \xrightarrow{h\nu} 2Br* \], where \(h\nu\) signifies light energy, and the asterisk (*) represents a radical.
02

Propagation

There are actually two propagation steps. For this exercise, the first step involved in propagation is the abstraction of a hydrogen atom from 2,3-dimethylbutane by a bromine radical to form a primary radical (2,3-dimethylbutyl radical), shown here: \[ (CH3)2CHCH(CH3)2 + Br* \rightarrow (CH3)2C*HCH(CH3)2 + HBr \] Also, the primary radical (2,3-dimethylbutyl radical) formed as a result will react with another molecule of bromine to form 2 -bromo- 2,3 -dimethylbutane. So \[ (CH3)2C*HCH(CH3)2 + Br2 \rightarrow (CH3)2CBrCH(CH3)2 + Br* \]
03

Termination

The termination step involves the recombination of radicals to form neutral species. Since the radicals available towards the end of the reaction are few, for the bromination of 2,3-dimethylbutane, the termination step could be one of the 2 possibilities: \[ Br* + Br* \rightarrow Br2 \] or \[ (CH3)2C*HCH(CH3)2 + Br* \rightarrow (CH3)2CBrCH(CH3)2 \] Both these reactions lead to disruptions of the chain reaction.

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