In the given compound,Cl will be liberated ea | vedclass

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(B) The given compound is a derivative of cyclopropene. Upon the loss of the chloride ion $(Cl^{-})$,the molecule forms a cyclopropenyl cation. This r

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$Cl^{-}$

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(B) The given compound is a derivative of cyclopropene. Upon the loss of the chloride ion $(Cl^{-})$,the molecule forms a cyclopropenyl cation. This resulting cation is highly stable because it follows $H$ückel's rule for aromaticity ($4n+2$ $\pi$ electrons,where $n=0$,giving $2$ $\pi$ electrons) and is planar with a fully conjugated system. Due to the formation of this stable aromatic cation,the $Cl$ atom is easily liberated as a chloride ion $(Cl^{-})$.

In the given compound,$Cl$ will be liberated easily in the form of:

Similar Questions

Consider the following compound $(X)$:
$H-C \equiv C-CH_2-CH(CH_3)-CH_3$
The most stable and least stable carbon radicals,respectively,produced by homolytic cleavage of the corresponding $C-H$ bond are:

The correct order of stability of the following carbocations is:

The correct stability order for the following carbocations is:

What is the minimum energy required for the homolytic bond fission of the $C-H$ bond in the following molecule?

The most stable carbocation among the following is:

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