Isobutene,in the presence of $H_2SO_4$,forms a mixture of two isomeric alkenes $(C_8H_{16})$. The major alkene is

  • A
    $CH_3-C(CH_3)_2-CH=C(CH_3)_2$
  • B
    $CH_3-C(CH_3)_2-CH_2-C(CH_3)=CH_2$
  • C
    $CH_3-CH(CH_3)-CH=CH-CH(CH_3)-CH_3$
  • D
    $CH_2=C(CH_3)-CH_2-CH_2-CH(CH_3)-CH_3$

Explore More

Similar Questions

The carbon-carbon bond length in an ethylene molecule is:

The reaction is given as:
$(CH_3)_2C=C(CH_3)_2 + (CH_3)_3N^{+}-O^{-} + H_2O \xrightarrow{OsO_4 (10^{-4} \text{ mole})} A + (CH_3)_3N$
Product $(A)$ is:

$CH_3-CHCl-CHCl-CH_3$ $\xrightarrow[dry \ ether]{Na} X$ $\xrightarrow[(1) \ BH_3, THF; (2) \ H_2O_2, OH^-_{(aq)}] Y$; $Y$ is:

Difficult
View Solution

$1-$Propanol can be prepared from propene by

Difficult
View Solution

When vinyl and allyl radicals are joined together,we get:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo