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Q8.16:

For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

(b) 

Solution :

Step 1: Analysis of Bond Cleavage

In the chemical process provided, we are examining the cleavage of a $C-Br$ bond where the bonding electrons are transferred to the more electronegative atom (Bromine). The reaction is represented as follows:

$CH_3-CH_2-Br \longrightarrow CH_3-CH_2^+ + :Br^-$

Step 2: Identification of Electron Flow

To represent the electron flow using curved-arrow notation, a full-headed arrow ($\rightarrow$) is drawn from the center of the covalent bond toward the atom receiving the electron pair. This indicates the heterolytic movement of a shared electron pair.

CH₃-CH₂ Br Substrate: Bromoethane

Step 3: Classification of Cleavage and Intermediate

  • Type of Cleavage: This process is classified as heterolysis. [Per IUPAC definition, heterolytic fission occurs when a covalent bond breaks such that the shared pair of electrons remains with one of the fragments, resulting in ions].
  • Reactive Intermediate: The resulting species $CH_3-CH_2^+$ possesses a carbon atom with a positive charge and a sextet of valence electrons. This is identified as a carbocation (specifically, an ethyl carbocation).

Step 4: Comparative Summary Table

Attribute Description
Cleavage Type Heterolysis
Electron Displacement Unequal (both electrons to $Br$)
Intermediate Formed Ethyl Carbocation ($CH_3CH_2^+$)

Final Solution: The bond cleavage is a heterolysis resulting in the formation of an ethyl carbocation ($CH_3CH_2^+$) and a bromide anion ($Br^-$).


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