A) \[C{{H}_{3}}C{{H}_{2}}Cl\]
B) \[C{{H}_{2}}=CHC{{H}_{2}}Cl\]
C) \[{{(C{{H}_{3}})}_{3}}C-Cl\]
D) \[C{{H}_{2}}=CHCl\]
Correct Answer: D
Solution :
The non-reactivity of the chlorine atom in vinyl chloride may be explained from the molecular orbital point of view as follows. If the chlorine atom has \[s{{p}^{2}}\] hybridisation the C?Cl bond will be a \[\sigma \] bond and the two lone pairs of electron would occupy the other two \[s{{p}^{2}}\] orbitals. This would leave a p-orbital containing a lone pair and this orbital could not conjugate with the \[\pi \] bond of the ethylenic link. Thus, two molecular orbitals will be required to accommodate these four \[\pi \] electrons. Furthermore since, chlorine is more electronegative than carbon, the electron will tend to be found in the chlorine atom, has now lost full control of the lone pair and so, is less negative than it would have been. Since two carbon atoms have acquired a share in the lone pair, each atom acquires a small negative charge. Hence, owing to delocalisation of bonds (through conjugation), the vinyl chloride molecule has an increased stability. Before the chlorine atom can be displaced by some other group the lone pair must be localised again on the chlorine atom. This requires energy and so, the chlorine is more firmly found.
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