Question

The carbocation formed in $S_{N}1$ reaction of alkyl halide in the slow step is

• A. $sp^{3}$ hybridised
• B. $sp^{2}$ hybridized
• C. $sp$ hybridised
• D. $sp^{3}d$ hybridised

Answer 1

Answer: (B)

$sp^{2}$ hybridized

Answer 2

The carbocation formed in the $S_{N} 1$ reaction has a plane of symmetry, which means that it is a chiral. The structure of the carbocation results because the carbon in the alkyl halide is changing from $s p^{3}$ hybridisation to $s p^{2}$ hybridisation as the halide anion departs. Instead of one single s and three p orbitals of the excited state of carbon hybridising to form four $s p^{3}$ hybrid orbitals (tetrahedral hybridisation), as in the starting alkyl halide, the single $s$ and two p orbitals hybridise to form three $s p^{2}$ hybrid orbitals (trigonal
$\underset{\text{Atomic carbon}}{1s^{2} 2s^{2} 2p^{2}} \xrightarrow{\text { Promotion }}1s^{2} 2s^{1} 2p^{3} \xrightarrow{\text { Hybridisation}} \underset{\text{Trigonal hybridisation}}{1s^2 2(sp^2)^3 +2p^0}$