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

Vedclass · Accepted Answer

$Cl^{-}$

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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

The incorrect statement about carbene $(CH_2)$ is:

Rank the following radicals in order of increasing stability (least $ < < < $ most).

The decreasing order of stability of the following carbocations is:
$A$. $m-CH_3O-C_6H_4-CH_2^+$
$B$. $p-CH_3O-C_6H_4-CH_2^+$
$C$. $C_6H_5-CH_2^+$
$D$. $p-NO_2-C_6H_4-CH_2^+$

The most stable carbonium ion is:

Assertion : Carbanions like ammonia have pyramidal shape.
Reason : The carbon atom carrying negative charge has an octet of electrons.

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