List the reactive intermediates produced by the fission of a covalent bond and describe their main characteristics.
(N/A) The reactive intermediates produced by the fission of a covalent bond are Carbocations,Carbanions,and Free radicals.
Characteristics:
$1$. These intermediates are highly unstable and reactive species formed during the course of a chemical reaction.
$2$. The stability of these intermediates is influenced by the inductive effect and hyperconjugation of the attached alkyl groups.
$3$. In these species,the central carbon atom is typically bonded to three other atoms or groups.
| Intermediate | $1^{\circ}$ (Primary) | $2^{\circ}$ (Secondary) | $3^{\circ}$ (Tertiary) |
|---|---|---|---|
| Carbocation | $CH_3^+, CH_3CH_2^+$ | $(CH_3)_2CH^+$ | $(CH_3)_3C^+$ |
| Carbanion | $:CH_3^-, CH_3CH_2^-$ | $(CH_3)_2CH^-$ | $(CH_3)_3C^-$ |
| Free radical | $CH_3\bullet, CH_3CH_2\bullet$ | $(CH_3)_2CH\bullet$ | $(CH_3)_3C\bullet$ |
Characteristics:
$1$. These intermediates are highly unstable and reactive species formed during the course of a chemical reaction.
$2$. The stability of these intermediates is influenced by the inductive effect and hyperconjugation of the attached alkyl groups.
$3$. In these species,the central carbon atom is typically bonded to three other atoms or groups.
Explore More
Similar Questions
Based upon an understanding of product stability,predict the product formed when the following dianion reacts with one equivalent of acid $(H^+)$.
Medium
View SolutionThe correct order of stability of the given carbanions is:
Which of the following compounds forms the most stable carbocation: $(CH_3)_3C-Br$,$(C_6H_5)_3CBr$,$(C_6H_5)_2CHBr$,and $C_6H_5CH_2Br$?
Medium
View SolutionArrange the following in decreasing order of stability.
$(i)$ $CH_3^{\bullet}$,$(CH_3)_3C^{\bullet}$,$(CH_3)_2CH^{\bullet}$ and $CH_3CH_2^{\bullet}$
$(ii)$ $CH_3CH_2^{+}$,$(CH_3)_3C^{+}$,$(CH_3)_2CH^{+}$ and $CH_3^{+}$
$(i)$ $CH_3^{\bullet}$,$(CH_3)_3C^{\bullet}$,$(CH_3)_2CH^{\bullet}$ and $CH_3CH_2^{\bullet}$
$(ii)$ $CH_3CH_2^{+}$,$(CH_3)_3C^{+}$,$(CH_3)_2CH^{+}$ and $CH_3^{+}$
Easy
View SolutionThe decreasing order of hydride affinity for the following carbocations is:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
Choose the correct answer from the options given below:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
Choose the correct answer from the options given below:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo