Question

Isopentane can form four structurally isomeric mono bromo derivatives. How many of them are optically active?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

The condition of chirality is that a chiral molecule must possess four different groups attached to a single carbon atom. After the mono bromination of isopentane, four different products are formed and they are given in the diagram names a,b,c,d. Among them $2-Bromo3-methylbutane\left(b\right)$ is optically active. Star marked carbon has four different groups ($Br,H,C{H}_3,\left(C{H}_3{)}_{2}CH\right)$ attached to it. So it is optically active. So, the number of optically active products is $1$.

Option A is the correct answer.