A) \[s{{p}^{3}}\] hybridised
B) \[s{{p}^{2}}\]hybridized
C) sp hybridised
D) \[s{{p}^{3}}d\] hybridised
Correct Answer: B
Solution :
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 \[\underset{Atomic\text{ }carbon}{\mathop{1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}}}\,\xrightarrow{Promotion}1{{s}^{2}}2{{s}^{1}}2{{p}^{3}}\xrightarrow{Hybridisation}\] \[\underset{Trigonal\text{ }hybridisation}{\mathop{1{{s}^{2}}2{{(s{{p}^{2}})}^{3}}+2{{p}^{0}}}}\,\]
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