Arrange the following alcohols in decreasing | vedclass

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(C) The ease of ionization of alcohols under acidic conditions depends on the stability of the carbocation formed after the loss of a water molecule.

Vedclass · Accepted Answer

$II > III > I$

Vedclass · Answer

(C) The ease of ionization of alcohols under acidic conditions depends on the stability of the carbocation formed after the loss of a water molecule. $1$. In compound $(II)$,the carbocation formed is stabilized by the resonance effect of the oxygen atom,making it aromatic and highly stable. $2$. In compound $(III)$,the carbocation is non-aromatic but is relatively stable due to resonance with the two benzene rings. $3$. In compound $(I)$,the carbocation formed is anti-aromatic,which makes it highly unstable. Therefore,the stability order of the carbocations is $(II > III > I)$. Thus,the decreasing order of the ease of ionization is $(II > III > I)$.

Arrange the following alcohols in decreasing order of the ease of ionization under acidic conditions.

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