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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.

(a) 

Solution :

Analysis of Bond Cleavage: CH₃-Br → CH₃⁺ + Br⁻

The chemical reaction provided involves the breaking of the carbon-bromine covalent bond in bromomethane. To analyze this, we examine the electronegativity difference between the bonded atoms. Carbon has an electronegativity of approximately $2.55$, while Bromine has an electronegativity of approximately $2.96$. [Per the Pauling Electronegativity Scale].

Step 1: Identifying the Cleavage Mechanism

Since Bromine is more electronegative than Carbon, the shared pair of electrons in the $\text{C-Br}$ bond is displaced toward the bromine atom. When the bond breaks, both electrons from the shared pair are transferred to the bromine atom. This process, where the bond pair is unevenly distributed, is defined as heterolysis or heterolytic fission.

Step 2: Curved-Arrow Notation

The electron flow is represented by a full-headed curved arrow starting from the $\text{C-Br}$ bond and pointing toward the Bromine atom, indicating the transfer of both electrons to the more electronegative species.

H₃C — Br ⟶ H₃C⁺ + Br⁻

Step 3: Identification of Reactive Intermediates

Following the heterolytic cleavage:

  • CH₃⁺ (Methyl Cation): The carbon atom is left with a sextet of electrons and a formal positive charge. This is classified as a carbocation.
  • Br⁻ (Bromide Anion): The bromine atom retains the shared pair, resulting in a negative charge.
Process Type Reactive Intermediate
Heterolysis Carbocation

Final Solution: The bond cleavage is Heterolysis. The reactive intermediate produced is a Carbocation ($CH_3^+$).


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