Match the List-$I$ with List-$II$:
List-$I$ Name reaction List-$II$ Product obtainable $A.$ Swarts reaction $I.$ Ethyl benzene $B.$ Sandmeyer's reaction $II.$ Ethyl iodide $C.$ Wurtz Fittig reaction $III.$ Cyanobenzene $D.$ Finkelstein reaction $IV.$ Ethyl fluoride
Choose the correct answer from the options given below:
| List-$I$ Name reaction | List-$II$ Product obtainable |
| $A.$ Swarts reaction | $I.$ Ethyl benzene |
| $B.$ Sandmeyer's reaction | $II.$ Ethyl iodide |
| $C.$ Wurtz Fittig reaction | $III.$ Cyanobenzene |
| $D.$ Finkelstein reaction | $IV.$ Ethyl fluoride |
Choose the correct answer from the options given below:
- A$A-II, B-III, C-I, D-IV$
- B$A-IV, B-I, C-III, D-II$
- C$A-IV, B-III, C-I, D-II$
- D$A-II, B-I, C-III, D-IV$
Explore More
Similar Questions
Assertion : Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene.
Reason : Alkyl halides are less reactive than acyl halides.
Reason : Alkyl halides are less reactive than acyl halides.
In the given reaction sequence,the major product $C$ is ..... .
Match the reactants in Column-$I$ with the products in Column-$II$.
Column-$I$ (Reactants) Column-$II$ (Products) $(A) CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}}$ $(i) \text{Cyclohexylmagnesium bromide}$ $(B) 2RX + 2Na \xrightarrow{\text{dry ether}}$ $(ii) CH_3CH_2MgBr$ $(C) CH_3CH_2Br \xrightarrow{\text{alc. KOH}}$ $(iii) R-R + 2NaX$ $(D) \text{Bromocyclohexane} + Mg \xrightarrow{\text{dry ether}}$ $(iv) CH_2=CH_2 + H_2O + KBr$
| Column-$I$ (Reactants) | Column-$II$ (Products) |
|---|---|
| $(A) CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}}$ | $(i) \text{Cyclohexylmagnesium bromide}$ |
| $(B) 2RX + 2Na \xrightarrow{\text{dry ether}}$ | $(ii) CH_3CH_2MgBr$ |
| $(C) CH_3CH_2Br \xrightarrow{\text{alc. KOH}}$ | $(iii) R-R + 2NaX$ |
| $(D) \text{Bromocyclohexane} + Mg \xrightarrow{\text{dry ether}}$ | $(iv) CH_2=CH_2 + H_2O + KBr$ |
Medium
View SolutionThe reaction of the compound shown in the image with $alc. KOH$ yields:
Difficult
View SolutionGiven below are two statements:
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.
DifficultJEE MAIN 2026
View SolutionVedclass 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