Question

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

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

Answer 1

Answer: (B)

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
$\underset{\text{Atomic carbon}}{1s^{2}2s^{2}2p^2}\stackrel{\text{ Promotion }}{\to }1s^{2}2s^{1}2p^3\stackrel{\text{ Hybridisation}}{\to }\underset{\text{Trigonal hybridisation}}{1s^{2}2(s{p}^2{)}^3+2p^0}$