Which of the following is the most stable car | vedclass

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(A) The stability of carbocations is determined by the extent of resonance and inductive effects. $(A)$ $Ph_3C^+$ (Triphenylmethyl cation) is the most

Vedclass · Accepted Answer

$Ph_3C^+$

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(A) The stability of carbocations is determined by the extent of resonance and inductive effects. $(A)$ $Ph_3C^+$ (Triphenylmethyl cation) is the most stable because the positive charge is delocalized over three benzene rings through resonance. $(B)$ $Ph_2CH^+$ has resonance with two benzene rings. $(C)$ $PhCH_2^+$ has resonance with one benzene ring. $(D)$ $CH_3^+$ has no resonance stabilization. Therefore,$Ph_3C^+$ is the most stable.

Column $-I$	Column $-II$
$A$. Free radical	$1$. Trigonal planar
$B$. Carbocation	$2$. Pyramidal
$C$. Carbanion	$3$. Linear

Which of the following is the most stable carbocation?

Similar Questions

The order of decreasing stability of the carbanions is:
$1.$ $(CH_3)_3C^-$
$2.$ $(CH_3)_2CH^-$
$3.$ $CH_3CH_2^-$
$4.$ $C_6H_5CH_2^-$

What is the order of stability of $1^{\circ}, 2^{\circ}, 3^{\circ},$ and $\overset{\oplus}{C}H_3$ carbocations?

Which of the following species $(A)$,$(B)$,$(C)$,and $(D)$ contain an $sp^2$ hybridized carbon atom? $(A) CH_3^+$,$(B) CH_3^-$,$(C) \dot{C}H_3$,$(D) :CH_2$

Which of the following represents the incorrect order of stability?

Match the intermediates given in Column $-I$ with their probable structure in Column $-II$.

Column $-I$ Column $-II$

$A$. Free radical $1$. Trigonal planar

$B$. Carbocation $2$. Pyramidal

$C$. Carbanion $3$. Linear

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