Question

Most stable carbocation is formed by heterolysis of
A. \[\;\;{\left( {{\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}} \right)_{\mathbf{3}}}{\mathbf{CBr}}\]
B. \[{\left( {{{\mathbf{C}}_{\mathbf{6}}}{{\mathbf{H}}_{\mathbf{5}}}}
\right)_{\mathbf{3}}}{\mathbf{CBr}}\]
C. \[\;\;{({{\mathbf{C}}_{\mathbf{6}}}{{\mathbf{H}}_{\mathbf{5}}})_{\mathbf{2}}}{\mathbf{C}}
H{\mathbf{Br}}\]
D.\[{{\mathbf{C}}_{\mathbf{6}}}{{\mathbf{H}}_{\mathbf{5}}}{\mathbf{C}}{{\mathbf{H}}_{\mathbf{2}}}{ \mathbf{Br}}\]

Answer 1

Hint: In the given question we have to determine the stable carbocation which is formed by heterolysis of the bonds. This breaking of covalent bonds takes place and thus form the most stable carbocation.

Complete step by step solution:
Heterolytic bond cleavage or we can say heterolysis is the breaking of a single bond with the two electrons in the bond distributed unequally between the two atoms bound by the bond. The carbocation generated can be stabilized by resonance due to the presence of lone pair the more electronegative the atom the more can provide electrons easily Hence might be expected to generate more stable carbocation. We know that the stability of the carbocation increases when moving from primary carbons to secondary carbons to tertiary carbons so the most stable carbocation which can be formed from heterolysis of \[{\left( {{{\mathbf{C}}_{\mathbf{6}}}{{\mathbf{H}}_{\mathbf{5}}}} \right)_{\mathbf{3}}}{\mathbf{CBr}}\] due to inductive effect along with resonance and hyperconjugation on the carbons.
Most stable carbocation is formed by heterolysis is \[{\left( {{{\mathbf{C}}_{\mathbf{6}}}{{\mathbf{H}}_{\mathbf{5}}}} \right)_{\mathbf{3}}}{\mathbf{CBr}}\].

Hence option b is correct.

Note: In heterolysis a chemical bond cleavage of a neutral molecule generating a cation and an anion occurs. Carbocations are the most important reactive intermediates in the field of organic chemistry and are common in the bond‐forming and bond‐breaking reactions by involving their participation at some stage of the reaction. Carbocations can also react by electron transfer to form free radicals. The rate of reaction for many reactions of unimolecular heterolysis depends on the rate of ionization of the covalent bond. The preventive reaction step is generally the formation of ion pairs.