In which of the following compound C–Br bond heterolysis will take place at fastest rate.
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Heterolysis of C–Br bond involves breaking it to form carbocation and bromide ion. The rate depends on carbocation stability: more stable carbocation forms faster.
Carbocation stability order: tertiary > secondary > primary > methyl > phenyl (phenyl is unstable due to loss of aromaticity).
EtBr (primary) gives 1° carbocation. Me₃C–Br (tertiary) gives stable 3° carbocation. Me–Br (methyl) gives unstable CH₃⁺. Ph–Br gives unstable phenyl cation.
Thus, Me₃C–Br undergoes fastest heterolysis due to formation of most stable tertiary carbocation.
Final answer: Me3C–Br