Match List-$I$ with List-$II$ and then select the correct answer from the codes given below the lists:
List-$I$ List-$II$ $A$. Butane $\to$ Isobutane $a$. Cracking $B$. Butane $\to$ Lower hydrocarbons $b$. Isomerisation $C$. $n$-Heptane $\to$ Toluene $c$. Reed reaction $D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$ $d$. Aromatisation
| List-$I$ | List-$II$ |
|---|---|
| $A$. Butane $\to$ Isobutane | $a$. Cracking |
| $B$. Butane $\to$ Lower hydrocarbons | $b$. Isomerisation |
| $C$. $n$-Heptane $\to$ Toluene | $c$. Reed reaction |
| $D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$ | $d$. Aromatisation |
- A$A-b, B-a, C-d, D-c$
- B$A-b, B-d, C-a, D-c$
- C$A-b, B-c, C-d, D-a$
- D$A-b, B-d, C-c, D-a$
Explore More
Similar Questions
If the product of the roots of the equation $(a + 1)x^2 + (2a + 3)x + (3a + 4) = 0$ is $2$,then the sum of the roots is:
$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right) \Rightarrow y$ is equal to
$KO_2$ exhibits paramagnetic behaviour. This is due to the paramagnetic nature of
Find out the major product of the following reaction:
$2 \text{ mole } CH_2=CH-CH_2Br \xrightarrow{Mg, T.H.F}$
$2 \text{ mole } CH_2=CH-CH_2Br \xrightarrow{Mg, T.H.F}$
The period of oscillation of a simple pendulum of length $l$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by
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