Question

The number of all the possible stereoisomers formed by the given compound is :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

The $C{H}_3$ present at $s{p}^2$ carbons are planar and have no possibility to go above or below the plane.
But the $s{p}^3$ carbon attached to methyl groups are chiral. There are two chiral carbon atoms and when methyl group is present above and below the plane at two chiral centres, it shows centre of symmetry. Therefore, the possible isomers are shown below:


Hence, (b) is correct.