Which one among the following carbocations is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The stability of carbocations is determined by resonance,hyperconjugation,and inductive effects. $(a)$ Diphenylmethyl carbocation is stabilized by

Vedclass · Accepted Answer

Tricyclopropylmethyl cation

Vedclass · Answer

(D) The stability of carbocations is determined by resonance,hyperconjugation,and inductive effects. $(a)$ Diphenylmethyl carbocation is stabilized by resonance with two phenyl rings. $(b)$ Tropylium cation is an aromatic system ($6\pi$ electrons,Huckel's rule),making it exceptionally stable. $(c)$ tert-Butyl carbocation is stabilized by hyperconjugation with $9$ $\alpha$-hydrogens. $(d)$ Tricyclopropylmethyl cation is highly stabilized due to the extensive delocalization of the positive charge into the cyclopropyl rings (bent bonds/sigma-conjugation). Among these,the tricyclopropylmethyl cation is exceptionally stable due to the unique electronic structure of the cyclopropyl groups,which provide significant stabilization to the carbocation center.

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

Which one among the following carbocations is the most stable?

Similar Questions

The correct order of stability of the given carbocations is:

What is the increasing order of stability for the following free radicals?

Arrange the following carbanions in the order of their stability:
$i$. $RC \equiv C^{\ominus}$
$ii$. $C_6H_5^{\ominus}$
$iii$. $R_2C = CH^{\ominus}$
$iv$. $R_3C - CH_2^{\ominus}$

Which of the following is the most stable carbocation?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module