Step 1: Understanding the Question:
We need to evaluate four different alkyl bromides to locate the single molecule that displays molecular chirality.
Step 2: Key Formula or Approach:
A molecule is considered chiral if it contains an asymmetric carbon atom (also known as a stereocenter or chiral center). A carbon atom is a stereocenter if it is $\text{sp}^3$ hybridized and bonded to four completely distinct structural groups ($A \neq B \neq C \neq D$).
Step 3: Detailed Explanation:
Let us analyze the chemical structures of each option group-by-group:
(A) 2-Bromo-2-methylbutane: The C2 atom is bonded to a bromine ($-\text{Br}$), a methyl group ($-\text{CH}_3$), another identical methyl group ($-\text{CH}_3$), and an ethyl group ($-\text{CH}_2\text{CH}_3$). Since it has two identical methyl groups, it is achiral.
(B) 2-Bromo-3-methylbutane: Let's trace the groups attached to the C2 carbon atom ($\text{CH}_3\text{-C}^*\text{H(Br)-CH(CH}_3)_2$). The C2 carbon is bonded to:
1. A Hydrogen atom ($-\text{H}$)
2. A Bromine atom ($-\text{Br}$)
3. A Methyl group ($-\text{CH}_3$)
4. An Isopropyl group ($-\text{CH(CH}_3)_2$)
Since all four groups attached to C2 are completely unique, C2 is a chiral center. Hence, this molecule is chiral.
(C) 3-Bromopentane: The C3 carbon is bonded to a hydrogen ($-\text{H}$), a bromine ($-\text{Br}$), and two symmetric ethyl groups ($-\text{CH}_2\text{CH}_3$). The duplicate ethyl groups make it achiral.
(D) 2-Bromopropane: The C2 carbon is bonded to a hydrogen ($-\text{H}$), a bromine ($-\text{Br}$), and two identical methyl groups ($-\text{CH}_3$). Thus, it is achiral.
Step 4: Final Answer:
2-Bromo-3-methylbutane contains a chiral carbon atom and is the chiral molecule, matching option (B).