MHT CET Chemistry Question (2026) — Solution

Step-by-step solution

The optical activity of a compound depends on the presence of a chiral carbon atom, which is bonded to four different groups. In 3-bromohexane, the C3 carbon is bonded to a hydrogen atom, a bromine atom, an ethyl group, and a propyl group. Since all four groups are different, it is chiral and optically active. In 2-bromo-2-methylbutane, the C2 carbon is bonded to a bromine atom, an ethyl group, and two identical methyl groups. Since two groups are the same, it does not have a chiral center and is optically inactive. In 2-bromopentane, the C2 carbon is bonded to a hydrogen atom, a bromine atom, a methyl group, and a propyl group. It is chiral and optically active. In 2-bromo-3-methylbutane, the C2 carbon is bonded to a hydrogen atom, a bromine atom, a methyl group, and an isopropyl group. It is chiral and optically active. Thus, 2-bromo-2-methylbutane is not optically active. Answer: 2 - Bromo - 2 - methylbutane