Which of the following represent free radical substitution reaction $(FSR)$,electrophilic addition reaction $(EAR)$ and nucleophilic addition reaction $(NAR)$?
$I. CH_3-CH=CH_2 \xrightarrow{Cl_2/CCl_4} CH_3-CHCl-CH_2Cl$
$II. CH_3-CO-CH_3 \xrightarrow{HCN/OH^-} CH_3-C(OH)(CN)-CH_3$
$III. CH_3-CH_2-CH_3 \xrightarrow{Cl_2/h\nu} CH_3-CHCl-CH_3$
$I. CH_3-CH=CH_2 \xrightarrow{Cl_2/CCl_4} CH_3-CHCl-CH_2Cl$
$II. CH_3-CO-CH_3 \xrightarrow{HCN/OH^-} CH_3-C(OH)(CN)-CH_3$
$III. CH_3-CH_2-CH_3 \xrightarrow{Cl_2/h\nu} CH_3-CHCl-CH_3$
- A$I-EAR, II-NAR, III-NAR$
- B$I-EAR, II-NAR, III-FSR$
- C$I-FSR, II-EAR, III-EAR$
- D$I-NAR, II-FSR, III-EAR$
Explore More
Similar Questions
The heterolytic cleavage of a carbon-chlorine bond gives:
Medium
View SolutionMatch Column-$I$ and Column-$II$ with the correct relation:
Column-$I$ Column-$II$ $(i)$ Substitution reaction $(p)$ $CH_2=CH_2 + Br_2 \to CH_2Br-CH_2Br$ $(ii)$ Addition reaction $(q)$ $C_6H_6 + NO_2^+ \to C_6H_5NO_2 + H^+$ $(iii)$ Electrophilic addition reaction $(r)$ $CH_3Cl + NaOH \to CH_3OH + NaCl$ $(iv)$ Electrophilic substitution reaction $(s)$ $CH_3CH_2OH \xrightarrow[\Delta]{Al_2O_3} CH_2=CH_2$ $(v)$ Elimination reaction
| Column-$I$ | Column-$II$ |
|---|---|
| $(i)$ Substitution reaction | $(p)$ $CH_2=CH_2 + Br_2 \to CH_2Br-CH_2Br$ |
| $(ii)$ Addition reaction | $(q)$ $C_6H_6 + NO_2^+ \to C_6H_5NO_2 + H^+$ |
| $(iii)$ Electrophilic addition reaction | $(r)$ $CH_3Cl + NaOH \to CH_3OH + NaCl$ |
| $(iv)$ Electrophilic substitution reaction | $(s)$ $CH_3CH_2OH \xrightarrow[\Delta]{Al_2O_3} CH_2=CH_2$ |
| $(v)$ Elimination reaction |
Medium
View SolutionIn the given compound,how many sites are available for the attack of $CH_3O^{-}$?
Medium
View SolutionThe most likely protonation site in the following molecule is:
In the above reaction,the electrophile is substituted at the meta position only,due to:
$I$. Electron density is more at ortho $\&$ para positions
$II$. Electron density is relatively less at ortho $\&$ para positions
$III$. Electron density is less at meta position
$IV$. Electron density is relatively more at meta position
The correct answer is:
$I$. Electron density is more at ortho $\&$ para positions
$II$. Electron density is relatively less at ortho $\&$ para positions
$III$. Electron density is less at meta position
$IV$. Electron density is relatively more at meta position
The correct answer is:
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