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

Which of the following carbocations is least stable?

The stability of carbanions follows the order:
$I: CH_3CH_2CH_2\bar{C}H_2$
$II: CH_3\bar{C}HCH_2CH_3$
$III: (CH_3)_3\bar{C}$
$IV: CH_3\bar{C}(Ph)CH_2CH_3$

The correct order of stability of the given carbanions is:

The correct order of stability for the following carbanions is:
$CH_2=CH^{-}, CH_3-CH_2^{-}, CH\equiv C^{-}$

In the potential energy diagram shown,the point $X$ represents:

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