Which of the following compounds forms the mo | vedclass

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

Question

(C) The stability of a carbocation is determined by the extent of resonance and inductive effects.$1$. $(CH_3)_3C^+$ is a tertiary alkyl carbocation s

Vedclass · Accepted Answer

$(C_6H_5)_3C^+$

Vedclass · Answer

(C) The stability of a carbocation is determined by the extent of resonance and inductive effects.$1$. $(CH_3)_3C^+$ is a tertiary alkyl carbocation stabilized by hyperconjugation and inductive effects.$2$. $C_6H_5CH_2^+$ is a benzyl carbocation stabilized by resonance with one phenyl ring.$3$. $(C_6H_5)_2CH^+$ is a benzhydryl carbocation stabilized by resonance with two phenyl rings.$4$. $(C_6H_5)_3C^+$ is a trityl carbocation stabilized by resonance with three phenyl rings.Since $(C_6H_5)_3C^+$ has the maximum number of phenyl rings,it exhibits the greatest degree of resonance stabilization,making it the most stable carbocation among the given options.

Which of the following compounds forms the most stable carbocation: $(CH_3)_3C-Br$,$(C_6H_5)_3CBr$,$(C_6H_5)_2CHBr$,and $C_6H_5CH_2Br$?

Similar Questions

Intermediates formed in the reactions $P$ and $Q$ are:
$P: CH_3-CH=CH_2 \xrightarrow{HCl}$
$Q: CH_3-CH=CH_2 \xrightarrow[\text{Peroxide}]{HCl}$

Arrange the following carbocations in decreasing order of their stability:
$\stackrel{+}{C}H_3, (CH_3)_3\stackrel{+}{C}, CH_3\stackrel{+}{C}H_2, (CH_3)_2\stackrel{+}{C}H$

For the following radicals,the correct order of their stability is

Based upon an understanding of product stability,predict the product formed when the following dianion reacts with one equivalent of acid $(H^+)$.

Which of the following orders regarding relative stability of free radicals is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module