Correct Answer - A
`{:(underset(But-2-en e)(CH_(3)CH=CHCH_(3)),,underset(Prpen e)(CH_(3)CH=CH_(2)),,),(underset(2-Methy1but-2-en e)((CH_(3))_(2)C=CHCH_(3)),,underset(2-Methy1pr open e)((CH_(3))_(2)C=CH_(2)),,):}`
Alkenes with two identical ligands on the same double-bonded `C` have no geometric isomers. only in but`-2-`ene each doubly-bonded carbon has two different ligands
